Advances in Matrices Finite and Infinite with Applications

Journal of Applied Mathematics

Guest Editors P N Shivakumar Panayiotis Psarrakos K C Sivakumar and Yang Zhang

Advances in Matrices Finite and Infinitewith Applications

Journal of Applied Mathematics

Advances in Matrices Finite and Infinitewith Applications

Guest Editors P N Shivakumar Panayiotis PsarrakosK C Sivakumar and Yang Zhang

Copyright copy 2013 Hindawi Publishing Corporation All rights reserved

This is a special issue published in ldquoJournal of Applied Mathematicsrdquo All articles are open access articles distributed under the CreativeCommons Attribution License which permits unrestricted use distribution and reproduction in any medium provided the originalwork is properly cited

Editorial Board

Saeid Abbasbandy IranMina B Abd-El-Malek EgyptMohamed A Abdou EgyptSubhas Abel IndiaMostafa Adimy FranceCarlos J S Alves PortugalMohamad Alwash USAIgor Andrianov GermanySabri Arik TurkeyFrancis TK Au Hong KongOlivier Bahn CanadaRoberto Barrio SpainAlfredo Bellen ItalyJafar Biazar IranHester Bijl The NetherlandsAnjan Biswas Saudi ArabiaStephane PA Bordas USAJames Robert Buchanan USAAlberto Cabada SpainXiao Chuan Cai USAJinde Cao ChinaAlexandre Carvalho BrazilSong Cen ChinaQianshun S Chang ChinaTai-Ping Chang TaiwanShih-sen Chang ChinaRushan Chen ChinaXinfu Chen USAKe Chen UKEric Cheng Hong KongFrancisco Chiclana UKJen-Tzung Chien TaiwanC S Chien TaiwanHan H Choi Republic of KoreaTin-Tai Chow ChinaM S H Chowdhury MalaysiaCarlos Conca ChileVitor Costa PortugalLivija Cveticanin SerbiaEric de Sturler USAOrazio Descalzi ChileKai Diethelm GermanyVit Dolejsi Czech RepublicBo-Qing Dong ChinaMagdy A Ezzat Egypt

Meng Fan ChinaYa Ping Fang ChinaAntonio J M Ferreira PortugalMichel Fliess FranceM A Fontelos SpainHuijun Gao ChinaBernard J Geurts The NetherlandsJamshid Ghaboussi USAPablo Gonzalez-Vera SpainLaurent Gosse ItalyK S Govinder South AfricaJose L Gracia SpainYuantong Gu AustraliaZhihong GUAN ChinaNicola Guglielmi ItalyFrederico G Guimaraes BrazilVijay Gupta IndiaBo Han ChinaMaoan Han ChinaPierre Hansen CanadaFerenc Hartung HungaryXiaoqiao He Hong KongLuis Javier Herrera SpainJ Hoenderkamp The NetherlandsYing Hu FranceNing Hu JapanZhilong L Huang ChinaKazufumi Ito USATakeshi Iwamoto JapanGeorge Jaiani GeorgiaZhongxiao Jia ChinaTarun Kant IndiaIdo Kanter IsraelAbdul Hamid Kara South AfricaHamid Reza Karimi NorwayJae-Wook Kim UKJong Hae Kim Republic of KoreaKazutake Komori JapanFanrong Kong USAVadim Krysko RussiaJin L Kuang SingaporeMiroslaw Lachowicz PolandHak-Keung Lam UKTak-Wah Lam Hong KongPGL Leach Cyprus

Yongkun Li ChinaWan-Tong Li ChinaJ Liang ChinaChing-Jong Liao TaiwanChong Lin ChinaMingzhu Liu ChinaChein-Shan Liu TaiwanKang Liu USAYansheng Liu ChinaFawang Liu AustraliaShutian Liu ChinaZhijun Liu ChinaJulian Lopez-Gomez SpainShiping Lu ChinaGert Lube GermanyNazim Idrisoglu Mahmudov TurkeyOluwole Daniel Makinde South AfricaFrancisco J Marcellan SpainGuiomar Martın-Herran SpainNicola Mastronardi ItalyMichael McAleer The NetherlandsStephane Metens FranceMichael Meylan AustraliaAlain Miranville FranceRam N Mohapatra USAJaime E Munoz Rivera BrazilJavier Murillo SpainRoberto Natalini ItalySrinivasan Natesan IndiaJiri Nedoma Czech RepublicJianlei Niu Hong KongRoger Ohayon FranceJavier Oliver SpainDonal OrsquoRegan IrelandMartin Ostoja-Starzewski USATurgut Ozis TurkeyClaudio Padra ArgentinaReinaldo Martinez Palhares BrazilFrancesco Pellicano ItalyJuan Manuel Pena SpainRicardo Perera SpainMalgorzata Peszynska USAJames F Peters CanadaMark A Petersen South AfricaMiodrag Petkovic Serbia

Vu Ngoc Phat VietnamAndrew Pickering SpainHector Pomares SpainMaurizio Porfiri USAMario Primicerio ItalyMorteza Rafei The NetherlandsRoberto Reno ItalyJacek Rokicki PolandDirk Roose BelgiumCarla Roque PortugalDebasish Roy IndiaSamir H Saker EgyptMarcelo A Savi BrazilWolfgang Schmidt GermanyEckart Schnack GermanyMehmet Sezer TurkeyNaseer Shahzad Saudi ArabiaFatemeh Shakeri IranJian Hua Shen ChinaHui-Shen Shen ChinaFernando Simoes PortugalTheodore E Simos Greece

Abdel-Maksoud A Soliman EgyptXinyu Song ChinaQiankun Song ChinaYuri N Sotskov BelarusPeter J C Spreij The NetherlandsNiclas Stromberg SwedenRay KL Su Hong KongJitao Sun ChinaWenyu Sun ChinaXianHua Tang ChinaAlexander Timokha NorwayMariano Torrisi ItalyJung-Fa Tsai TaiwanCh Tsitouras GreeceKuppalapalle Vajravelu USAAlvaro Valencia ChileErik Van Vleck USAEzio Venturino ItalyJesus Vigo-Aguiar SpainMichael N Vrahatis GreeceBaolin Wang ChinaMingxin Wang China

Qing-Wen Wang ChinaGuangchen Wang ChinaJunjie Wei ChinaLi Weili ChinaMartin Weiser GermanyFrank Werner GermanyShanhe Wu ChinaDongmei Xiao ChinaGongnan Xie ChinaYuesheng Xu USASuh-Yuh Yang TaiwanBo Yu ChinaJinyun Yuan BrazilAlejandro Zarzo SpainGuisheng Zhai JapanJianming Zhan ChinaZhihua Zhang ChinaJingxin Zhang AustraliaShan Zhao USAChongbin Zhao AustraliaRenat Zhdanov USAHongping Zhu China

Advances in Matrices Finite and Infinite with Applications P N Shivakumar Panayiotis PsarrakosK C Sivakumar and Yang ZhangVolume 2013 Article ID 156136 3 pages

On Nonnegative Moore-Penrose Inverses of Perturbed Matrices Shani Jose and K C SivakumarVolume 2013 Article ID 680975 7 pages

A New Construction of Multisender Authentication Codes from Polynomials over Finite FieldsXiuli WangVolume 2013 Article ID 320392 7 pages

Geometrical and Spectral Properties of the Orthogonal Projections of the Identity Luis GonzalezAntonio Suarez and Dolores GarcıaVolume 2013 Article ID 435730 8 pages

A Parameterized Splitting Preconditioner for Generalized Saddle Point ProblemsWei-Hua Luo and Ting-Zhu HuangVolume 2013 Article ID 489295 6 pages

Solving Optimization Problems on Hermitian Matrix Functions with ApplicationsXiang Zhang and Shu-Wen XiangVolume 2013 Article ID 593549 11 pages

Left and Right Inverse Eigenpairs Problem for κ-Hermitian Matrices Fan-Liang Li Xi-Yan Huand Lei ZhangVolume 2013 Article ID 230408 6 pages

Completing a 2times 2 Block Matrix of Real Quaternions with a Partial Specified InverseYong Lin and Qing-Wen WangVolume 2013 Article ID 271978 5 pages

Fuzzy Approximate Solution of Positive Fully Fuzzy Linear Matrix EquationsXiaobin Guo and Dequan ShangVolume 2013 Article ID 178209 7 pages

Ranks of a Constrained Hermitian Matrix Expression with Applications Shao-Wen YuVolume 2013 Article ID 514984 9 pages

An Efficient Algorithm for the Reflexive Solution of the Quaternion Matrix Equation AXB + CXHD = FNing Li Qing-Wen Wang and Jing JiangVolume 2013 Article ID 217540 14 pages

The Optimization on Ranks and Inertias of a Quadratic Hermitian Matrix Function and Its ApplicationsYirong YaoVolume 2013 Article ID 961568 6 pages

Constrained Solutions of a System of Matrix Equations Qing-Wen Wang and Juan YuVolume 2012 Article ID 471573 19 pages

Contents

On the Hermitian R-Conjugate Solution of a System of Matrix Equations Chang-Zhou DongQing-Wen Wang and Yu-Ping ZhangVolume 2012 Article ID 398085 14 pages

Perturbation Analysis for the Matrix Equation X minussummi=1 A

lowasti XAi +

sumnj=1 B

lowastj XBj = I

Xue-Feng Duan and Qing-Wen WangVolume 2012 Article ID 784620 13 pages

Refinements of Kantorovich Inequality for Hermitian Matrices Feixiang ChenVolume 2012 Article ID 483624 11 pages

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 156136 3 pageshttpdxdoiorg1011552013156136

EditorialAdvances in Matrices Finite and Infinite with Applications

P N Shivakumar1 Panayiotis Psarrakos2 K C Sivakumar3 and Yang Zhang1

1 Department of Mathematics University of Manitoba Winnipeg MB Canada R3T 2N22Department of Mathematics School of Applied Mathematical and Physical SciencesNational Technical Univeristy of Athens Zografou Campus 15780 Athens Greece

3 Department of Mathematics Indian Institute of Technology Madras Chennai 600 036 India

Correspondence should be addressed to P N Shivakumar shivakuccumanitobaca

Received 13 June 2013 Accepted 13 June 2013

Copyright copy 2013 P N Shivakumar et al This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

In the mathematical modeling of many problems includingbut not limited to physical sciences biological sciencesengineering image processing computer graphics medicalimaging and even social sciences lately solution methodsmost frequently involve matrix equations In conjunctionwith powerful computing machines complex numericalcalculations employing matrix methods could be performedsometimes even in real time It is well known that infinitematrices arise more naturally than finite matrices and havea colorful history in development from sequences seriesand quadratic forms Modern viewpoint considers infinitematrices more as operators defined between certain spe-cific infinite-dimensional normed spaces Banach spacesor Hilbert spaces Giant and rapid advancements in thetheory and applications of finite matrices have been madeover the past several decades These developments havebeen well documented in many books monographs andresearch journals The present topics of research includediagonally dominantmatrices theirmany extensions inverseof matrices their recursive computation singular matricesgeneralized inverses inverse positive matrices with specificemphasis on their applications to economics like the Leon-tief input-output models and use of finite difference andfinite element methods in the solution of partial differentialequations perturbationmethods and eigenvalue problems ofinterest and importance to numerical analysts statisticiansphysical scientists and engineers to name a few

The aim of this special issue is to announce certainrecent advancements in matrices finite and infinite andtheir applications For a review of infinite matrices andapplications see P N Shivakumar and K C Sivakumar

(Linear Algebra Appl 2009) Specifically an example of anapplication can be found in the classical problem ldquoshapeof a drumrdquo in P N Shivakumar W Yan and Y Zhang(WSES transactions in Mathematics 2011) The focal themesof this special issue are inverse positivematrices including119872-matrices applications of operator theorymatrix perturbationtheory and pseudospectra matrix functions and generalizedeigenvalue problems and inverse problems including scat-tering and matrices over quaternions In the following wepresent a brief overview of a few of the central topics of thisspecial issue

For119872-matrices let us start by mentioning the most citedbooks byR Berman andR J Plemmons (SIAM 1994) and byR A Horn and C R Johnson (Cambridge 1994) as excellentsources One of the most important properties of invertible119872-matrices is that their inverses are nonnegative There areseveral works that have considered generalizations of someof the important properties of 119872-matrices We only pointout a few of them here The article by D Mishra and K CSivakumar (Oper Matrices 2012) considers generalizationsof inverse nonnegativity to the Moore-Penrose inverse whilemore general classes of matrices are the objects of discussionin D Mishra and K C Sivakumar (Linear Algebra Appl2012) and A N Sushama K Premakumari and K CSivakumar (Electron J Linear Algebra 2012)

A brief description of the papers in the issue is as followsIn the work of Z Zhang the problem of solving cer-

tain optimization problems on Hermitian matrix functionsis studied Specifically the author considers the extremalinertias and ranks of a given function 119891(119883 119884) C119898times119899 rarr

C119898times119899 (which is defined in terms of matrices 119860 119861 119862 and 119863

2 Journal of Applied Mathematics

inC119898times119899) subject to119883 119884 satisfying certain matrix equationsAs applications the author presents necessary and sufficientconditions for 119891 to be positive definite positive semidefiniteand so forth As another application the author presents acharacterization for 119891(119883 119884) = 0 again subject to 119883 119884satisfying certain matrix equations

The task of determining parametrized splitting precon-ditioners for certain generalized saddle point problems isundertaken byW-H Luo and T-Z Huang in their workThewell-known and very widely applicable Sherman-Morrison-Woodbury formula for the sum of matrices is used indetermining a preconditioner for linear equations wherethe coefficient matrix may be singular and nonsymmetricIllustrations of the procedure are provided for Stokes andOseen problems

The work reported by A Gonzalez A Suarez and DGarcia deals with the problem of estimating the Frobeniuscondition of the matrix 119860119873 given a matrix 119860 isin R119899times119899 where119873 satisfies the condition that 119873 = argmin

119883isin119878119868 minus 119860119883

119865

Here 119878 is a certain subspace of R119899 and sdot 119865denotes the

Frobenius norm The authors derive certain spectral andgeometrical properties of the preconditioner119873 as above

Let 119861 119862 119863 and 119864 be (not necessarily square) matriceswith quaternion entries so that the matrix 119872 = (

119860 119861

119862 119863)

is square Y Lin and Q-W Wang present necessary andsufficient conditions so that the top left block 119860 of119872 existssatisfying the conditions that119872 is nonsingular and thematrix119864 is the top left block of119872minus1 A general expression for suchan 119860 when it exists is obtained A numerical illustration ispresented

For a complex hermitian 119860 of order 119899 with eigenvalues1205821le 1205822le sdot sdot sdot le 120582

119899 the celebrated Kantorovich inequality is

1 le 119909lowast

119860119909119909lowast

119860minus1

119909 le (120582119899+1205821)2

41205821120582119899 for all unit vectors 119909 isin

C119899 Equivalently 0 le 119909lowast119860119909119909lowast119860minus1119909minus1 le (120582119899minus1205821)2

41205821120582119899 for

all unit vectors 119909 isin C119899 F Chen in his work on refinements ofKantorovich inequality for hermitian matrices obtains somerefinements of the second inequality where the proofs useonly elementary ideas (httpwwwmathntuagrsimppsarr)

It is pertinent to point out that among other things aninteresting generalization of 119872-matrices is obtained by SJose and K C Sivakumar The authors consider the prob-lem of nonnegativity of the Moore-Penrose inverse of aperturbation of the form 119860minus119883119866119884

119879 given that 119860dagger ge 0 Usinga generalized version of the Sherman-Morrison-Woodburyformula conditions for (119860 minus 119883119866119884

119879

)dagger to be nonnegative

are derived Applications of the results are presented brieflyIterative versions of the results are also studied

It is well known that the concept of quaternions wasintroduced by Hamilton in 1843 and has played an importantrole in contemporary mathematics such as noncommuta-tive algebra analysis and topology Nowadays quaternionmatrices are widely and heavily used in computer sciencequantum physics altitude control robotics signal and colourimage processing and so on Many questions can be reducedto quaternion matrices and solving quaternion matrix equa-tionsThis special issue has provided an excellent opportunityfor researchers to report their recent results

Let H119899times119899 denote the space of all 119899 times 119899 matrices whoseentries are quaternions Let 119876 isin H119899times119899 be Hermitian and self-invertible 119883 isin H119899times119899 is called reflexive (relative to 119876) if 119883 =

119876119883119876 The authors F-L Li X-Y Hu and L Zhang presentan efficient algorithm for finding the reflexive solution of thequaternion matrix equation 119860119883119861 + 119862119883

lowast

119863 = 119865 They alsoshow that given an appropriate initialmatrix their procedureyields the least reflexive solutionwith respect to the Frobeniusnorm

In the work on the ranks of a constrained hermitianmatrix expression S W Yu gives formulas for the maximaland minimal ranks of the quaternion Hermitian matrixexpression 119862

4minus 1198604119883119860lowast

4 where 119883 is a Hermitian solution

to quaternion matrix equations 1198601119883 = 119862

1 1198831198611= 1198622 and

1198603119883119860lowast

3= 1198623 and also applies this result to some special

system of quaternion matrix equationsY Yao reports his findings on the optimization on ranks

and inertias of a quadratic hermitian matrix function Hepresents one solution for optimization problems on the ranksand inertias of the quadratic Hermitian matrix function 119876 minus119883119875119883lowast subject to a consistent system of matrix equations

119860119883 = 119862 and 119883119861 = 119863 As applications he derives necessaryand sufficient conditions for the solvability to the systems ofmatrix equations and matrix inequalities 119860119883 = 119862 119883119861 = 119863and119883119875119883lowast = (gt lt ge le)119876

Let 119877 be a nontrivial real symmetric involution matrixA complex matrix 119860 is called 119877-conjugate if 119860 = 119877119860119877In the work on the hermitian 119877-conjugate solution of asystem of matrix equations C-Z Dong Q-W Wang andY-P Zhang present necessary and sufficient conditions forthe existence of the hermitian 119877-conjugate solution to thesystem of complex matrix equations 119860119883 = 119862 and 119883119861 = 119863An expression of the Hermitian 119877-conjugate solution is alsopresented

A perturbation analysis for the matrix equation 119883 minus

sum119898

119894=1119860lowast

119894119883119860119894+ sum119899

119895=1119861lowast

119895119883119861119895= 119868 is undertaken by X-F Duan

and Q-W Wang They give a precise perturbation bound fora positive definite solution A numerical example is presentedto illustrate the sharpness of the perturbation bound

Linear matrix equations and their optimal approxima-tion problems have great applications in structural designbiology statistics control theory and linear optimal controland as a consequence they have attracted the attention ofseveral researchers for many decades In the work of Q-WWang and J Yu necessary and sufficient conditions of and theexpressions for orthogonal solutions symmetric orthogonalsolutions and skew-symmetric orthogonal solutions of thesystem of matrix equations 119860119883 = 119861 and 119883119862 = 119863 arederived When the solvability conditions are not satisfied theleast squares symmetric orthogonal solutions and the leastsquares skew-symmetric orthogonal solutions are obtainedA methodology is provided to compute the least squaressymmetric orthogonal solutions and an illustrative exampleis given to illustrate the proposed algorithm

Many real-world engineering systems are too complex tobe defined in precise terms and in many matrix equationssome or all of the system parameters are vague or impreciseThus solving a fuzzy matrix equation becomes important In

Journal of Applied Mathematics 3

the work reported by X Guo and D Shang the fuzzy matrixequation 119860 otimes 119883 otimes 119861 = 119862 in which 119860 119861 and 119862 are 119898 times 119898119899 times 119899 and 119898 times 119899 nonnegative LR fuzzy numbers matricesrespectively is investigated This fuzzy matrix system whichhas a wide use in control theory and control engineering isextended into three crisp systems of linear matrix equationsaccording to the arithmetic operations of LR fuzzy numbersBased on the pseudoinversematrix a computingmodel to thepositive fully fuzzy linear matrix equation is provided andthe fuzzy approximate solution of original fuzzy systems isobtained by solving the crisp linear matrix systems In addi-tion the existence condition of nonnegative fuzzy solutionis discussed and numerical examples are given to illustratethe proposed method Overall the technique of LR fuzzynumbers and their operations appears to be a strong tool ininvestigating fully fuzzy matrix equations

Left and right inverse eigenpairs problem arises in anatural way when studying spectral perturbations ofmatricesand relative recursive problems F-L Li X-Y Hu and LZhang consider in their work left and right inverse eigenpairsproblem for 119896-hermitian matrices and its optimal approx-imate problem The class of 119896-hermitian matrices containshermitian and perhermitian matrices and has numerousapplications in engineering and statistics Based on thespecial properties of 119896-hermitian matrices and their left andright eigenpairs an equivalent problem is obtained Thencombining a new inner product of matrices necessary andsufficient conditions for the solvability of the problem aregiven and its general solutions are derived Furthermorethe optimal approximate solution a calculation procedure tocompute the unique optimal approximation and numericalexperiments are provided

Finally let us point out a work on algebraic coding theoryRecall that a communication system in which a receivercan verify the authenticity of a message sent by a group ofsenders is referred to as a multisender authentication codeIn the work reported by X Wang the author constructsmulti-sender authentication codes using polynomials overfinite fields Here certain important parameters and theprobabilities of deceptions of these codes are also computed

Acknowledgments

We wish to thank all the authors for their contributions andthe reviewers for their valuable comments in bringing thespecial issue to fruition

P N ShivakumarPanayiotis Psarrakos

K C SivakumarYang Zhang

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 680975 7 pageshttpdxdoiorg1011552013680975

Research ArticleOn Nonnegative Moore-Penrose Inverses of Perturbed Matrices

Shani Jose and K C Sivakumar

Department of Mathematics Indian Institute of Technology Madras Chennai Tamil Nadu 600036 India

Correspondence should be addressed to K C Sivakumar kcskumariitmacin

Received 22 April 2013 Accepted 7 May 2013

Academic Editor Yang Zhang

Copyright copy 2013 S Jose and K C Sivakumar This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

Nonnegativity of theMoore-Penrose inverse of a perturbation of the form119860minus119883119866119884119879 is consideredwhen119860dagger ge 0 Using a generalizedversion of the Sherman-Morrison-Woodbury formula conditions for (119860 minus 119883119866119884119879)dagger to be nonnegative are derived Applications ofthe results are presented briefly Iterative versions of the results are also studied

1 Introduction

We consider the problem of characterizing nonnegativityof the Moore-Penrose inverse for matrix perturbations ofthe type 119860 minus 119883119866119884

119879 when the Moore-Penrose inverse of119860 is nonnegative Here we say that a matrix 119861 = (119887

119894119895) is

nonnegative and denote it by 119861 ge 0 if 119887119894119895ge 0 forall119894 119895 This

problemwasmotivated by the results in [1] where the authorsconsider an 119872-matrix 119860 and find sufficient conditions forthe perturbed matrix (119860 minus 119883119884119879) to be an119872-matrix Let usrecall that a matrix 119861 = (119887

119894119895) is said to 119885-matrix if 119887

119894119895le 0

for all 119894 119895 119894 = 119895 An119872-matrix is a nonsingular 119885-matrix withnonnegative inverse The authors in [1] use the well-knownSherman-Morrison-Woodbury (SMW) formula as one of theimportant tools to prove their main resultThe SMW formulagives an expression for the inverse of (119860minus119883119884119879) in terms of theinverse of119860 when it exists When119860 is nonsingular119860minus119883119884119879

is nonsingular if and only 119868 minus 119884119879119860minus1119883 is nonsingular In thatcase

(119860 minus 119883119884119879

)minus1

= 119860minus1

minus 119860minus1

119883(119868 minus 119884119879

119860minus1

119883)minus1

119884119879

119860minus1

(1)

Themain objective of the present work is to study certainstructured perturbations 119860 minus 119883119884119879 of matrices 119860 such thattheMoore-Penrose inverse of the perturbation is nonnegativewhenever the Moore-Penrose inverse of 119860 is nonnegativeClearly this class of matrices includes the class of matricesthat have nonnegative inverses especially119872-matrices In ourapproach extensions of SMW formula for singular matrices

play a crucial role Let us mention that this problem has beenstudied in the literature (See for instance [2] formatrices and[3] for operators over Hilbert spaces) We refer the reader tothe references in the latter for other recent extensions

In this paper first we present alternative proofs ofgeneralizations of the SMW formula for the cases of theMoore-Penrose inverse (Theorem 5) and the group inverse(Theorem 6) in Section 3 In Section 4 we characterize thenonnegativity of (119860 minus 119883119866119884119879)dagger This is done in Theorem 9and is one of the main results of the present work As aconsequence we present a result for119872-matrices which seemsnew We present a couple of applications of the main resultin Theorems 13 and 15 In the concluding section we studyiterative versions of the results of the second section Weprove two characterizations for (119860minus119883119884119879)dagger to be nonnegativein Theorems 18 and 21

Before concluding this introductory section let us givea motivation for the work that we have undertaken hereIt is a well-documented fact that 119872-matrices arise quiteoften in solving sparse systems of linear equations Anextensive theory of119872-matrices has been developed relativeto their role in numerical analysis involving the notion ofsplitting in iterativemethods and discretization of differentialequations in the mathematical modeling of an economyoptimization and Markov chains [4 5] Specifically theinspiration for the present study comes from the work of [1]where the authors consider a system of linear inequalitiesarising out of a problem in third generation wireless commu-nication systemsThematrix defining the inequalities there is

2 Journal of Applied Mathematics

an 119872-matrix In the likelihood that the matrix of thisproblem is singular (due to truncation or round-off errors)the earlier method becomes inapplicable Our endeavour isto extend the applicability of these results to more generalmatrices for instance matrices with nonnegative Moore-Penrose inverses Finally as mentioned earlier sincematriceswith nonnegative generalized inverses include in particular119872-matrices it is apparent that our results are expected toenlarge the applicability of themethods presently available for119872-matrices even in a very general framework including thespecific problem mentioned above

2 Preliminaries

Let R R119899 and R119898times119899 denote the set of all real numbers the119899-dimensional real Euclidean space and the set of all 119898 times 119899matrices over R For 119860 isin R119898times119899 let 119877(119860) 119873(119860) 119877(119860)perp and119860119879 denote the range space the null space the orthogonal

complement of the range space and the transpose of thematrix 119860 respectively For x = (119909

1 1199092 119909

119899)119879

isin R119899 wesay that x is nonnegative that is x ge 0 if and only if x

119894ge 0

for all 119894 = 1 2 119899 As mentioned earlier for a matrix 119861 weuse 119861 ge 0 to denote that all the entries of 119861 are nonnegativeAlso we write 119860 le 119861 if 119861 minus 119860 ge 0

Let 120588(119860) denote the spectral radius of the matrix 119860 If120588(119860) lt 1 for 119860 isin R119899times119899 then 119868 minus 119860 is invertible Thenext result gives a necessary and sufficient condition for thenonnegativity of (119868 minus 119860)minus1 This will be one of the results thatwill be used in proving the first main result

Lemma 1 (see [5 Lemma 21 Chapter 6]) Let 119860 isin R119899times119899 benonnegative Then 120588(119860) lt 1 if and only if (119868 minus 119860)minus1 exists and

(119868 minus 119860)minus1

=

infin

sum

119896=0

119860119896

ge 0 (2)

More generally matrices having nonnegative inversesare characterized using a property called monotonicity Thenotion of monotonicity was introduced by Collatz [6] A real119899 times 119899matrix 119860 is called monotone if 119860x ge 0 rArr x ge 0 It wasproved by Collatz [6] that 119860 is monotone if and only if 119860minus1exists and 119860minus1 ge 0

One of the frequently used tools in studying monotonematrices is the notion of a regular splitting We only refer thereader to the book [5] for more details on the relationshipbetween these concepts

The notion of monotonicity has been extended in a vari-ety of ways to singular matrices using generalized inversesFirst let us briefly review the notion of two importantgeneralized inverses

For 119860 isin R119898times119899 the Moore-Penrose inverse is the unique119885 isin R119899times119898 satisfying the Penrose equations 119860119885119860 = 119860119885119860119885 = 119885 (119860119885)119879 = 119860119885 and (119885119860)119879 = 119885119860 The uniqueMoore-Penrose inverse of119860 is denoted by119860dagger and it coincideswith 119860minus1 when 119860 is invertible

The following theorem by Desoer and Whalen whichis used in the sequel gives an equivalent definition for theMoore-Penrose inverse Let us mention that this result wasproved for operators between Hilbert spaces

Theorem 2 (see [7]) Let 119860 isin R119898times119899 Then 119860dagger isin R119899times119898 is theunique matrix119883 isin R119899times119898 satisfying

(i) 119885119860119909 = 119909 forall119909 isin 119877(119860119879)(ii) 119885119910 = 0 forall119910 isin 119873(119860119879)

Now for119860 isin R119899times119899 any119885 isin R119899times119899 satisfying the equations119860119885119860 = 119860 119885119860119885 = 119885 and 119860119885 = 119885119860 is called the groupinverse of 119860 The group inverse does not exist for everymatrix But whenever it exists it is unique A necessary andsufficient condition for the existence of the group inverse of119860 is that the index of119860 is 1 where the index of a matrix is thesmallest positive integer 119896 such that rank (119860119896+1) = rank (119860119896)

Some of the well-known properties of theMoore-Penroseinverse and the group inverse are given as follows 119877(119860119879) =119877(119860dagger

)119873(119860119879) = 119873(119860dagger) 119860dagger119860 = 119875119877(119860119879) and 119860119860dagger = 119875

119877(119860) In

particular x isin 119877(119860119879) if and only if x = 119860dagger119860x Also 119877(119860) =119877(119860

) 119873(119860) = 119873(119860

) and 119860119860 = 119860119860

= 119875119877(119860)119873(119860)

Herefor complementary subspaces 119871 and119872 of R119896 119875

119871119872denotes

the projection ofR119896 onto 119871 along119872 119875119871denotes 119875

119871119872if119872 =

119871perp For details we refer the reader to the book [8]Inmatrix analysis a decomposition (splitting) of amatrix

is considered in order to study the convergence of iterativeschemes that are used in the solution of linear systems of alge-braic equations As mentioned earlier regular splittings areuseful in characterizing matrices with nonnegative inverseswhereas proper splittings are used for studying singularsystems of linear equations Let us next recall this notionFor a matrix 119860 isin R119898times119899 a decomposition 119860 = 119880 minus 119881 iscalled a proper splitting [9] if 119877(119860) = 119877(119880) and 119873(119860) =119873(119880) It is rather well-known that a proper splitting existsfor every matrix and that it can be obtained using a full-rank factorization of the matrix For details we refer to [10]Certain properties of a proper splitting are collected in thenext result

Theorem 3 (see [9 Theorem 1]) Let 119860 = 119880 minus 119881 be a propersplitting of 119860 isin R119898times119899 Then

(a) 119860 = 119880(119868 minus 119880dagger119881)(b) 119868 minus 119880dagger119881 is nonsingular and(c) 119860dagger = (119868 minus 119880dagger119881)minus1119880dagger

The following result by Berman and Plemmons [9] givesa characterization for 119860dagger to be nonnegative when 119860 has aproper splitting This result will be used in proving our firstmain result

Theorem 4 (see [9 Corollary 4]) Let 119860 = 119880 minus 119881 be a propersplitting of 119860 isin R119898times119899 where 119880dagger ge 0 and 119880dagger119881 ge 0 Then119860dagger

ge 0 if and only if 120588(119880dagger119881) lt 1

3 Extensions of the SMW Formula forGeneralized Inverses

The primary objects of consideration in this paper are gen-eralized inverses of perturbations of certain types of a matrix

Journal of Applied Mathematics 3

119860 Naturally extensions of the SMW formula for generalizedinverses are relevant in the proofs Inwhat follows we presenttwo generalizations of the SMW formula for matrices Wewould like to emphasize that our proofs also carry overto infinite dimensional spaces (the proof of the first resultis verbatim and the proof of the second result with slightmodifications applicable to range spaces instead of ranks ofthe operators concerned) However we are confining ourattention to the case of matrices Let us also add that theseresults have been proved in [3] for operators over infinitedimensional spaces We have chosen to include these resultshere as our proofs are different from the ones in [3] and thatour intention is to provide a self-contained treatment

Theorem 5 (see [3 Theorem 21]) Let 119860 isin R119898times119899 119883 isin R119898times119896and 119884 isin R119899times119896 be such that

119877 (119883) sube 119877 (119860) 119877 (119884) sube 119877 (119860119879

) (3)

Let 119866 isin R119896times119896 and Ω = 119860 minus 119883119866119884119879 If

119877 (119883119879

) sube 119877 ((119866dagger

minus 119884119879

119860dagger

119883)119879

)

119877 (119884119879

) sube 119877 (119866dagger

minus 119884119879

119860dagger

119883) 119877 (119883119879

) sube 119877 (119866)

119877 (119884119879

) sube 119877 (119866119879

)

(4)

then

Ωdagger

= 119860dagger

+ 119860dagger

119883(119866dagger

minus 119884119879

119860dagger

119883)dagger

119884119879

119860dagger

(5)

Proof Set 119885 = 119860dagger + 119860dagger119883119867dagger119884119879119860dagger where119867 = 119866dagger

minus 119884119879

119860dagger

119883From the conditions (3) and (4) it follows that 119860119860dagger119883 = 119883119860dagger

119860119884 = 119884119867dagger119867119883119879 = 119883119879119867119867dagger119884119879 = 119884119879119866119866dagger119883119879 = 119883119879 and119866dagger

119866119884119879

= 119884119879

Now

119860dagger

119883119867dagger

119884119879

minus 119860dagger

119883119867dagger

119884119879

119860dagger

119883119866119884119879

= 119860dagger

119883119867dagger

(119866dagger

119866119884119879

minus 119884119879

119860dagger

119883119866119884119879

)

= 119860dagger

119883119867dagger

119867119866119884119879

= 119860dagger

119883119866119884119879

(6)

Thus 119885Ω = 119860dagger

119860 Since 119877(Ω119879) sube 119877(119860119879

) it follows that119885Ω(x) = 119909 forallx isin 119877(Ω119879)

Let y isin 119873(Ω119879

) Then 119860119879y minus 119884119866119879119883119879y = 0 so that119883119879

119860dagger119879

(119860119879

minus 119884119866119879

119883119879

)y = 0 Substituting 119883119879 = 119866119866dagger119883119879 andsimplifying it we get119867119879119866119879119883119879y = 0 Also119860119879y = 119884119866119879119883119879y =119884119867119867dagger

119866119879

119883119879y = 0 and so 119860daggery = 0 Thus 119885y = 0 for

y isin 119873(Ω119879) Hence by Theorem 2 119885 = Ωdagger

The result for the group inverse follows

Theorem 6 Let 119860 isin R119899times119899 be such that 119860 exists Let 119883119884 isin R119899times119896 and 119866 be nonsingular Assume that 119877(119883) sube 119877(119860)119877(119884) sube 119877(119860

119879

) and 119866minus1 minus 119884119879119860119883 is nonsingular Suppose that

rank (119860 minus 119883119866119884119879) = rank (119860) Then (119860 minus 119883119866119884119879) exists andthe following formula holds

(119860 minus 119883119866119884119879

)= 119860

+ 119860

119883(119866minus1

minus 119884119879

119860119883)minus1

119884119879

119860 (7)

Conversely if (119860minus119883119866119884119879) exists then the formula above holdsand we have rank (119860 minus 119883119866119884119879) = rank (119860)

Proof Since 119860 exists 119877(119860) and 119873(119860) are complementarysubspaces of R119899times119899

Suppose that rank (119860 minus 119883119866119884119879) = rank (119860) As 119877(119883) sube119877(119860) it follows that 119877(119860 minus 119883119866119884119879) sube 119877(119860) Thus 119877(119860 minus119883119866119884119879

) = 119877(119860) By the rank-nullity theorem the nullity ofboth 119860 minus 119883119866119884119879 and 119860 are the same Again since 119877(119884) sube119877(119860119879

) it follows that 119873(119860 minus 119883119866119884119879

) = 119873(119860) Thus119877(119860 minus 119883119866119884

119879

) and 119873(119860 minus 119883119866119884119879) are complementary sub-spaces This guarantees the existence of the group inverse of119860 minus 119883119866119884

119879Conversely suppose that (119860 minus 119883119866119884119879) exists It can be

verified by direct computation that 119885 = 119860+ 119860

119883(119866minus1

minus

119884119879

119860119883)minus1

119884119879

119860 is the group inverse of 119860 minus 119883119866119884119879 Also we

have (119860minus119883119866119884119879)(119860minus119883119866119884119879) = 119860119860 so that119877(119860minus119883119866119884119879) =119877(119860) and hence the rank (119860 minus 119883119866119884119879) = rank (119860)

We conclude this section with a fairly old result [2] as aconsequence of Theorem 5

Theorem 7 (see [2 Theorem 15]) Let 119860 isin R119898times119899 of rank 119903119883 isin R119898times119903 and 119884 isin R119899times119903 Let119866 be an 119903times119903 nonsingular matrixAssume that 119877(119883) sube 119877(119860) 119877(119884) sube 119877(119860119879) and 119866minus1 minus 119884119879119860dagger119883is nonsingular Let Ω = 119860 minus 119883119866119884119879 Then

(119860 minus 119883119866119884119879

)dagger

= 119860dagger

+ 119860dagger

119883(119866minus1

minus 119884119879

119860dagger

119883)minus1

119884119879

119860dagger

(8)

4 Nonnegativity of (119860minus119883119866119884119879)dagger

In this section we consider perturbations of the form 119860 minus

119883119866119884119879 and derive characterizations for (119860 minus 119883119866119884119879)dagger to be

nonnegative when 119860dagger ge 0 119883 ge 0 119884 ge 0 and 119866 ge 0 In orderto motivate the first main result of this paper let us recall thefollowing well known characterization of119872-matrices [5]

Theorem 8 Let 119860 be a 119885-matrix with the representation 119860 =119904119868 minus 119861 where 119861 ge 0 and 119904 ge 0 Then the following statementsare equivalent

(a) 119860minus1 exists and 119860minus1 ge 0

(b) There exists 119909 gt 0 such that 119860119909 gt 0

(c) 120588(119861) lt 119904

Let us prove the first result of this article This extendsTheorem 8 to singular matrices We will be interested inextensions of conditions (a) and (c) only

Theorem 9 Let 119860 = 119880 minus 119881 be a proper splitting of 119860 isin R119898times119899

with 119880dagger ge 0 119880dagger119881 ge 0 and 120588(119880dagger119881) lt 1 Let 119866 isin R119896times119896 benonsingular and nonnegative 119883 isin R119898times119896 and 119884 isin R119899times119896 be

4 Journal of Applied Mathematics

nonnegative such that 119877(119883) sube 119877(119860) 119877(119884) sube 119877(119860119879) and119866minus1 minus119884119879

119860dagger

119883 is nonsingular LetΩ = 119860minus119883119866119884119879Then the followingare equivalent

(a) Ωdagger ge 0(b) (119866minus1 minus 119884119879119860dagger119883)minus1 ge 0(c) 120588(119880dagger119882) lt 1 where119882 = 119881 + 119883119866119884

119879

Proof First we observe that since 119877(119883) sube 119877(119860) 119877(119884) sube119877(119860119879

) and 119866minus1 minus 119884119879119860dagger119883 is nonsingular by Theorem 7 wehave

Ωdagger

= 119860dagger

+ 119860dagger

119883(119866minus1

minus 119884119879

119860dagger

119883)minus1

119884119879

119860dagger

(9)

We thus haveΩΩdagger = 119860119860dagger andΩdaggerΩ = 119860dagger119860 Therefore

119877 (Ω) = 119877 (119860) 119873 (Ω) = 119873 (119860) (10)

Note that the first statement also implies that 119860dagger ge 0 byTheorem 4

(a) rArr (b) By taking 119864 = 119860 and 119865 = 119883119866119884119879 we get

Ω = 119864 minus 119865 as a proper splitting for Ω such that 119864dagger = 119860dagger ge 0and 119864dagger119865 = 119860dagger119883119866119884119879 ge 0 (since 119883119866119884119879 ge 0) Since Ωdagger ge 0by Theorem 4 we have 120588(119860dagger119883119866119884119879) = 120588(119864

dagger

119865) lt 1 Thisimplies that 120588(119884119879119860dagger119883119866) lt 1 We also have 119884119879119860dagger119883119866 ge 0Thus by Lemma 1 (119868minus119884119879119860dagger119883119866)minus1 exists and is nonnegativeBut we have 119868 minus 119884119879119860dagger119883119866 = (119866

minus1

minus 119884119879

119860dagger

119883)119866 Now 0 le(119868 minus 119884

119879

119860dagger

119883119866)minus1

= 119866minus1

(119866minus1

minus 119884119879

119860dagger

119883)minus1 This implies that

(119866minus1

minus 119884119879

119860dagger

119883)minus1

ge 0 since 119866 ge 0 This proves (b)(b) rArr (c) We have119880minus119882 = 119880minus119881minus119883119866119884

119879

= 119860minus119883119884119879

=

Ω Also 119877(Ω) = 119877(119860) = 119877(119880) and 119873(Ω) = 119873(119860) = 119873(119880)So Ω = 119880 minus 119882 is a proper splitting Also 119880dagger ge 0 and119880dagger

119882 = 119880dagger

(119881+119883119866119884119879

) ge 119880dagger

119881 ge 0 Since (119866minus1minus119884119879119860dagger119883)minus1 ge0 it follows from (9) that Ωdagger ge 0 (c) now follows fromTheorem 4

(c) rArr (a) Since119860 = 119880minus119881 we haveΩ = 119880minus119881minus119883119866119884119879 =119880minus119882 Also we have119880dagger ge 0119880dagger119881 ge 0 Thus119880dagger119882 ge 0 since119883119866119884119879

ge 0 Now by Theorem 4 we are done if the splittingΩ = 119880 minus119882 is a proper splitting Since 119860 = 119880 minus 119881 is a propersplitting we have 119877(119880) = 119877(119860) and 119873(119880) = 119873(119860) Nowfrom the conditions in (10) we get that 119877(119880) = 119877(Ω) and119873(119880) = 119873(Ω) Hence Ω = 119880 minus 119882 is a proper splitting andthis completes the proof

The following result is a special case of Theorem 9

Theorem 10 Let119860 = 119880minus119881 be a proper splitting of119860 isin R119898times119899

with 119880dagger ge 0 119880dagger119881 ge 0 and 120588(119880dagger119881) lt 1 Let 119883 isin R119898times119896 and119884 isin R119899times119896 be nonnegative such that 119877(119883) sube 119877(119860) 119877(119884) sube119877(119860119879

) and 119868minus119884119879119860dagger119883 is nonsingular LetΩ = 119860minus119883119884119879 Thenthe following are equivalent

(a) Ωdagger ge 0(b) (119868 minus 119884119879119860dagger119883)minus1 ge 0(c) 120588(119880dagger119882) lt 1 where119882 = 119881 + 119883119884

119879

The following consequence of Theorem 10 appears to benew This gives two characterizations for a perturbed 119872-matrix to be an119872-matrix

Corollary 11 Let119860 = 119904119868minus119861where119861 ge 0 and 120588(119861) lt 119904 (ie119860is an119872-matrix) Let 119883 isin R119898times119896 and 119884 isin R119899times119896 be nonnegativesuch that 119868 minus 119884119879119860dagger119883 is nonsingular Let Ω = 119860 minus 119883119884119879 Thenthe following are equivalent

(a) Ωminus1 ge 0(b) (119868 minus 119884119879119860dagger119883)minus1 ge 0(c) 120588(119861(119868 + 119883119884119879)) lt 119904

Proof From the proof ofTheorem 10 since 119860 is nonsingularit follows that ΩΩdagger = 119868 and ΩdaggerΩ = 119868 This shows that Ωis invertible The rest of the proof is omitted as it is an easyconsequence of the previous result

In the rest of this section we discuss two applicationsof Theorem 10 First we characterize the least element ina polyhedral set defined by a perturbed matrix Next weconsider the following Suppose that the ldquoendpointsrdquo of aninterval matrix satisfy a certain positivity property Then allmatrices of a particular subset of the interval also satisfythat positivity condition The problem now is if that we aregiven a specific structured perturbation of these endpointswhat conditions guarantee that the positivity property for thecorresponding subset remains valid

The first result is motivated by Theorem 12 below Let usrecall that with respect to the usual order an element xlowast isinX sube R119899 is called a least element ofX if it satisfies xlowast le x forall x isin X Note that a nonempty set may not have the leastelement and if it exists then it is unique In this connectionthe following result is known

Theorem 12 (see [11 Theorem 32]) For 119860 isin R119898times119899 and b isinR119898 let

Xb = x isin R119899

119860x + y ge b 119875119873(119860)

x = 0

119875119877(119860)

y = 0 119891119900119903 119904119900119898119890 y isin R119898 (11)

Then a vector xlowast is the least element ofXb if and only if xlowast =119860daggerb isin Xb with 119860dagger ge 0

Now we obtain the nonnegative least element of apolyhedral set defined by a perturbed matrix This is animmediate application of Theorem 10

Theorem 13 Let 119860 isin R119898times119899 be such that 119860dagger ge 0 Let 119883 isin

R119898times119896 and let 119884 isin R119899times119896 be nonnegative such that 119877(119883) sube119877(119860) 119877(119884) sube 119877(119860119879) and 119868 minus 119884119879119860dagger119883 is nonsingular Supposethat (119868 minus 119884119879119860dagger119883)minus1 ge 0 For b isin R119898 119887 ge 0 let

Sb = x isin R119899

Ωx + y ge b 119875119873(Ω)

x = 0

119875119877(Ω)

y = 0 119891119900119903 119904119900119898119890 y isin R119898 (12)

where Ω = 119860 minus 119883119884119879 Then xlowast = (119860 minus 119883119884119879)daggerb is the least

element of S119887

Proof From the assumptions using Theorem 10 it fol-lows that Ωdagger ge 0 The conclusion now follows fromTheorem 12

Journal of Applied Mathematics 5

To state and prove the result for interval matrices let usfirst recall the notion of interval matrices For119860 119861 isin R119898 times 119899an interval (matrix) 119869 = [119860 119861] is defined as 119869 = [119860 119861] = 119862 119860 le 119862 le 119861 The interval 119869 = [119860 119861] is said to be range-kernelregular if 119877(119860) = 119877(119861) and 119873(119860) = 119873(119861) The followingresult [12] provides necessary and sufficient conditions for119862dagger

ge 0 for 119862 isin 119870 where 119870 = 119862 isin 119869 119877(119862) = 119877(119860) =

119877(119861)119873(119862) = 119873(119860) = 119873(119861)

Theorem 14 Let 119869 = [119860 119861] be range-kernel regular Then thefollowing are equivalent

(a) 119862dagger ge 0 whenever 119862 isin 119870(b) 119860dagger ge 0 and 119861dagger ge 0

In such a case we have 119862dagger = suminfin119895=0(119861dagger

(119861 minus 119862))119895

119861dagger

Now we present a result for the perturbation

Theorem 15 Let 119869 = [119860 119861] be range-kernel regular with119860dagger ge0 and 119861dagger ge 0 Let 119883

1 1198832 1198841 and 119884

2be nonnegative matrices

such that 119877(1198831) sube 119877(119860) 119877(119884

1) sube 119877(119860

119879

) 119877(1198832) sube 119877(119861) and

119877(1198842) sube 119877(119861

119879

) Suppose that 119868minus1198841198791119860dagger

1198831and 119868minus119884119879

2119860dagger

1198832are

nonsingular with nonnegative inverses Suppose further that1198832119884119879

2le 1198831119884119879

1 Let 119869 = [119864 119865] where 119864 = 119860 minus 119883

1119884119879

1and

119865 = 119861 minus 1198832119884119879

2 Finally let = 119885 isin 119869 119877(119885) = 119877(119864) =

119877(119865) 119886119899119889 119873(119885) = 119873(119864) = 119873(119865) Then

(a) 119864dagger ge 0 and 119865dagger ge 0(b) 119885dagger ge 0 whenever 119885 isin

In that case 119885dagger = suminfin119895=0(119861dagger

119861 minus 119865dagger

119885)119895

119865dagger

Proof It follows fromTheorem 7 that

119864dagger

= 119860dagger

+ 119860dagger

1198831(119868 minus 119884

119879

1119860dagger

1198831)minus1

119884119879

1119860dagger

119865dagger

= 119861dagger

+ 119861dagger

1198832(119868 minus 119884

119879

2119861dagger

1198832)minus1

119884119879

2119861dagger

(13)

Also we have 119864119864dagger = 119860119860dagger 119864dagger119864 = 119860

dagger

119860 119865dagger119865 = 119861dagger

119861and 119865119865dagger = 119861119861

dagger Hence 119877(119864) = 119877(119860) 119873(119864) = 119873(119860)119877(119865) = 119877(119861) and 119873(119865) = 119873(119861) This implies that theinterval 119869 is range-kernel regular Now since 119864 and 119865 satisfythe conditions of Theorem 10 we have 119864dagger ge 0 and 119865dagger ge 0proving (a) Hence by Theorem 14 119885dagger ge 0 whenever 119885 isin Again by Theorem 14 we have 119885dagger = suminfin

119895=0(119865dagger

(119865 minus 119885))119895

119865dagger

=

suminfin

119895=0(119861dagger

119861 minus 119865dagger

119885)119895

119865dagger

5 Iterations That Preserve Nonnegativity ofthe Moore-Penrose Inverse

In this section we present results that typically provideconditions for iteratively defined matrices to have nonneg-ative Moore-Penrose inverses given that the matrices thatwe start with have this property We start with the followingresult about the rank-one perturbation case which is a directconsequence of Theorem 10

Theorem 16 Let 119860 isin R119899times119899 be such that 119860dagger ge 0 Let x isin R119898

and y isin R119899 be nonnegative vectors such that x isin 119877(119860) y isin119877(119860119879

) and 1 minus y119879119860daggerx = 0 Then (119860 minus xy119879)dagger ge 0 if and only ify119879119860daggerx lt 1

Example 17 Let us consider 119860 = (13) (1 minus1 1

minus1 4 minus1

1 minus1 1

) It can be

verified that 119860 can be written in the form 119860 = (1198682

119890119879

1

) (1198682+

e1119890119879

1)minus1

119862minus1

(1198682+ e1119890119879

1)minus1

(1198682e1) where e

1= (1 0)

119879 and 119862 isthe 2 times 2 circulant matrix generated by the row (1 12) Wehave 119860dagger = (12) ( 2 1 21 2 1

2 1 2

) ge 0 Also the decomposition 119860 =

119880minus119881 where119880 = (13) ( 1 minus1 2minus1 4 minus2

1 minus1 2

) and119881 = (13) ( 0 0 10 2 minus10 0 1

) isa proper splitting of 119860 with 119880dagger ge 0 119880dagger119881 ge 0 and 120588(119880dagger119881) lt1

Let x = y = (13 16 13)119879 Then x and let y are

nonnegative x isin 119877(119860) and y isin 119877(119860119879) We have 1 minus y119879119860daggerx =512 gt 0 and 120588(119880dagger119882) lt 1 for119882 = 119881 + xy119879 Also it can beseen that (119860 minus xy119879)dagger = (120) ( 47 28 4728 32 28

47 28 47

) ge 0 This illustratesTheorem 10

Let x1 x2 x

119896isin R119898 and y

1 y2 y

119896isin R119899 be nonneg-

ative Denote 119883119894= (x1 x2 x

119894) and 119884

119894= (y1 y2 y

119894) for

119894 = 1 2 119896 Then119860minus119883119896119884119879

119896= 119860minussum

119896

119894=1x119894y119879119894 The following

theorem is obtained by a recurring application of the rank-one result of Theorem 16

Theorem 18 Let119860 isin R119898times119899 and let119883119894 119884119894be as above Further

suppose that x119894isin 119877(119860 minus 119883

119894minus1119884119879

119894minus1) y119894isin 119877(119860 minus 119883

119894minus1119884119879

119894minus1)119879 and

1 minus y119879119894(119860 minus 119883

119894minus1119884119879

119894minus1)daggerx119894be nonzero Let (119860 minus 119883

119894minus1119884119879

119894minus1)dagger

ge 0for all 119894 = 1 2 119896 Then (119860 minus 119883

119896119884119879

119896)dagger

ge 0 if and only ify119879119894(119860 minus 119883

119894minus1119884119879

119894minus1)daggerx119894lt 1 where 119860 minus 119883

0119884119879

0is taken as 119860

Proof Set 119861119894= 119860 minus 119883

119894119884119879

119894for 119894 = 0 1 119896 where 119861

0is

identified as 119860 The conditions in the theorem can be writtenas

x119894isin 119877 (119861

119894minus1) y

119894isin 119877 (119861

119879

119894minus1)

1 minus y119879119894119861dagger

119894minus1x119894= 0

forall119894 = 1 2 119896

(14)

Also we have 119861dagger119894ge 0 for all 119894 = 0 1 119896 minus 1

Now assume that 119861dagger119896ge 0 Then by Theorem 16 and the

conditions in (14) for 119894 = 119896 we have y119896119861dagger

119896minus1x119896lt 1 Also

since by assumption 119861dagger119894ge 0 for all 119894 = 0 1 119896 minus 1 we have

y119879119894119861dagger

119894minus1x119894lt 1 for all 119894 = 1 2 119896

The converse part can be proved iteratively Then condi-tion y119879

11198610

daggerx1lt 1 and the conditions in (14) for 119894 = 1 imply

that 1198611

dagger

ge 0 Repeating the argument for 119894 = 2 to 119896 proves theresult

The following result is an extension of Lemma 23 in [1]which is in turn obtained as a corollaryThis corollary will beused in proving another characterization for (119860minus119883

119896119884119879

119896)dagger

ge 0

6 Journal of Applied Mathematics

Theorem 19 Let 119860 isin R119899times119899 bc isin R119899 be such that minusb ge 0minusc ge 0 and 120572(isin R) gt 0 Further suppose that b isin 119877(119860) andc isin 119877(119860119879) Let 119860 = ( 119860 b

c119879 120572 ) isin R(119899+1)times(119899+1) Then 119860dagger ge 0 if andonly if 119860dagger ge 0 and c119879119860daggerb lt 120572

Proof Let us first observe that 119860119860daggerb = b and c119879119860dagger119860 = c119879Set

119883 =(

119860dagger

+119860daggerbc119879119860dagger

120572 minus c119879119860daggerbminus119860daggerb

120572 minus c119879119860daggerbminus

c119879119860dagger

120572 minus c119879119860daggerb1

120572 minus c119879119860daggerb) (15)

It then follows that 119860119883 = ( 119860119860dagger 00119879 1 ) and 119883119860 = (119860dagger119860 0

0119879 1 ) Usingthese two equations it can easily be shown that119883 = 119860dagger

Suppose that 119860dagger ge 0 and 120573 = 120572 minus c119879119860daggerb gt 0 Then119860dagger

ge 0 Conversely suppose that 119860dagger ge 0 Then we musthave 120573 gt 0 Let e

1 e2 e

119899+1 denote the standard basis

of R119899+1 Then for 119894 = 1 2 119899 we have 0 le 119860dagger

(e119894) =

(119860daggere119894+(119860daggerbc119879119860daggere

119894120573) minus(c119879119860daggere

119894120573))119879 Since c le 0 it follows

that 119860daggere119894ge 0 for 119894 = 1 2 119899 Thus 119860dagger ge 0

Corollary 20 (see [1 Lemma 23]) Let 119860 isin R119899times119899 benonsingular Let b c isin R119899 be such that minusb ge 0 minusc ge 0

and 120572(isin R) gt 0 Let 119860 = (119860 bc119879 120572 ) Then 119860minus1 isin R(119899+1)times(119899+1)

is nonsingular with 119860minus1 ge 0 if and only if 119860minus1 ge 0 andc119879119860minus1b lt 120572

Now we obtain another necessary and sufficient condi-tion for (119860 minus 119883

119896119884119879

119896)dagger to be nonnegative

Theorem 21 Let 119860 isin R119898times119899 be such that 119860dagger ge 0 Let x119894 y119894

be nonnegative vectors inR119899 andR119898 respectively for every 119894 =1 119896 such that

x119894isin 119877 (119860 minus 119883

119894minus1119884119879

119894minus1) y

119894isin 119877(119860 minus 119883

119894minus1119884119879

119894minus1)119879

(16)

where 119883119894= (x1 x

119894) 119884119894= (y1 y

119894) and 119860 minus 119883

0119884119879

0is

taken as119860 If119867119894= 119868119894minus119884119879

119894119860dagger

119883119894is nonsingular and 1minus y119879

119894119860daggerx119894

is positive for all 119894 = 1 2 119896 then (119860 minus 119883119896119884119879

119896)dagger

ge 0 if andonly if

y119879119894119860daggerx119894lt 1 minus y119879

119894119860dagger

119883119894minus1119867minus1

119894minus1119884119879

119894minus1119860daggerx119894 forall119894 = 1 119896 (17)

Proof The range conditions in (16) imply that 119877(119883119896) sube 119877(119860)

and 119877(119884119896) sube 119877(119860

119879

) Also from the assumptions it followsthat 119867

119896is nonsingular By Theorem 10 (119860 minus 119883

119896119884119879

119896)dagger

ge 0 ifand only if119867minus1

119896ge 0 Now

119867119896= (

119867119896minus1

minus119884119879

119896minus1119860daggerx119896

minusy119879119896119860dagger

119883119896minus1

1 minus y119879119896119860daggerx119896

) (18)

Since 1minusy119879119896119860daggerx119896gt 0 usingCorollary 20 it follows that119867minus1

119896ge

0 if and only if y119879119896119860dagger

119883119896minus1119867minus1

119896minus1119884119879

119896minus1119860daggerx119896lt 1 minus y119879

119896119860daggerx119896and

119867minus1

119896minus1ge 0

Now applying the above argument to the matrix 119867119896minus1

we have that 119867minus1

119896minus1ge 0 holds if and only if 119867minus1

119896minus2ge 0

holds and y119879119896minus1119860dagger

119883119896minus2(119868119896minus2

minus 119884119879

119896minus2119860dagger

119883119896minus2)minus1

119884119879

119896minus2119860daggerx119896minus1 lt

1minus y119879119896minus1119860daggerx119896minus1 Continuing the above argument we get (119860minus

119883119896y119879119896)dagger

ge 0 if and only if y119879119894119860dagger

119883119894minus1119867minus1

119894minus1119884119879

119894minus1119860daggerx119894lt 1minusy119879

119894119860daggerx119894

forall119894 = 1 119896 This is condition (17)

We conclude the paper by considering an extension ofExample 17

Example 22 For a fixed 119899 let 119862 be the circulant matrixgenerated by the row vector (1 (12) (13) (1(119899 minus 1)))Consider

119860 = (119868

119890119879

1

) (119868 + e1119890119879

1)minus1

119862minus1

(119868 + e1119890119879

1)minus1

(119868 e1) (19)

where 119868 is the identity matrix of order 119899 minus 1 e1is (119899 minus 1) times 1

vector with 1 as the first entry and 0 elsewhere Then 119860 isin

R119899times119899 119860dagger is the 119899 times 119899 nonnegative Toeplitz matrix

119860dagger

=

((((((

(

11

119899 minus 1

1

119899 minus 2

1

21

1

21

1

119899 minus 1

1

3

1

2

1

119899 minus 1

1

119899 minus 2

1

119899 minus 3 1

1

119899 minus 1

11

119899 minus 1

1

119899 minus 2

1

21

))))))

)

(20)

Let 119909119894be the first row of 119861dagger

119894minus1written as a column vector

multiplied by 1119899119894 and let 119910119894be the first column of 119861dagger

119894minus1

multiplied by 1119899119894 where 119861119894= 119860 minus 119883

119894119884119879

119894 with 119861

0being

identified with 119860 119894 = 0 2 119896 We then have x119894ge 0

y119894ge 0 119909

1isin 119877(119860) and 119910

1isin 119877(119860

119879

) Now if 119861dagger119894minus1

ge 0 foreach iteration then the vectors 119909

119894and 119910

119894will be nonnegative

Hence it is enough to check that 1 minus 119910119879119894119861dagger

119894minus1119909119894gt 0 in order

to get 119861dagger119894ge 0 Experimentally it has been observed that for

119899 gt 3 the condition 1 minus 119910119879119894119861dagger

119894minus1119909119894gt 0 holds true for any

iteration However for 119899 = 3 it is observed that 119861dagger1ge 0 119861dagger

2ge

0 1 minus 1199101198791119860dagger

1199091gt 0 and 1 minus 119910dagger

2119861dagger

11199092gt 0 But 1 minus 119910119879

3119861dagger

21199093lt 0

and in that case we observe that 119861dagger3is not nonnegative

References

[1] J Ding W Pye and L Zhao ldquoSome results on structured119872-matrices with an application to wireless communicationsrdquoLinear Algebra and its Applications vol 416 no 2-3 pp 608ndash614 2006

[2] S K Mitra and P Bhimasankaram ldquoGeneralized inverses ofpartitionedmatrices and recalculation of least squares estimatesfor data ormodel changesrdquo Sankhya A vol 33 pp 395ndash410 1971

[3] C Y Deng ldquoA generalization of the Sherman-Morrison-Woodbury formulardquo Applied Mathematics Letters vol 24 no9 pp 1561ndash1564 2011

[4] A Berman M Neumann and R J SternNonnegative Matricesin Dynamical Systems JohnWiley amp Sons New York NY USA1989

Journal of Applied Mathematics 7

[5] A Berman and R J Plemmons Nonnegative Matrices inthe Mathematical Sciences Society for Industrial and AppliedMathematics (SIAM) 1994

[6] L Collatz Functional Analysis and Numerical MathematicsAcademic Press New York NY USA 1966

[7] C A Desoer and B H Whalen ldquoA note on pseudoinversesrdquoSociety for Industrial and Applied Mathematics vol 11 pp 442ndash447 1963

[8] A Ben-Israel and T N E Greville Generalized InversesTheoryand Applications Springer New York NY USA 2003

[9] A Berman and R J Plemmons ldquoCones and iterative methodsfor best least squares solutions of linear systemsrdquo SIAM Journalon Numerical Analysis vol 11 pp 145ndash154 1974

[10] D Mishra and K C Sivakumar ldquoOn splittings of matrices andnonnegative generalized inversesrdquo Operators and Matrices vol6 no 1 pp 85ndash95 2012

[11] D Mishra and K C Sivakumar ldquoNonnegative generalizedinverses and least elements of polyhedral setsrdquo Linear Algebraand its Applications vol 434 no 12 pp 2448ndash2455 2011

[12] M R Kannan and K C Sivakumar ldquoMoore-Penrose inversepositivity of interval matricesrdquo Linear Algebra and its Applica-tions vol 436 no 3 pp 571ndash578 2012

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 320392 7 pageshttpdxdoiorg1011552013320392

Research ArticleA New Construction of Multisender Authentication Codes fromPolynomials over Finite Fields

Xiuli Wang

College of Science Civil Aviation University of China Tianjin 300300 China

Correspondence should be addressed to Xiuli Wang xlwangcauceducn

Received 3 February 2013 Accepted 7 April 2013

Academic Editor Yang Zhang

Copyright copy 2013 Xiuli Wang This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Multisender authentication codes allow a group of senders to construct an authenticated message for a receiver such that thereceiver can verify the authenticity of the received message In this paper we construct one multisender authentication code frompolynomials over finite fields Some parameters and the probabilities of deceptions of this code are also computed

1 Introduction

Multisender authentication code was firstly constructed byGilbert et al [1] in 1974 Multisender authentication systemrefers towho a groupof senders cooperatively send amessageto a receiver then the receiver should be able to ascertainthat the message is authentic About this case many scholarsand researchers had made great contributions to multisenderauthentication codes such as [2ndash6]

In the actual computer network communications mul-tisender authentication codes include sequential model andsimultaneous model Sequential model is that each senderuses his own encoding rules to encode a source state orderlythe last sender sends the encoded message to the receiverand the receiver receives themessage and verifies whether themessage is legal or not Simultaneousmodel is that all sendersuse their own encoding rules to encode a source state andeach sender sends the encoded message to the synthesizerrespectively then the synthesizer forms an authenticatedmessage and verifies whether the message is legal or not Inthis paper we will adopt the second model

In a simultaneous model there are four participants agroup of senders 119880 = 119880

1 1198802 119880

119899 the key distribution

center he is responsible for the key distribution to sendersand receiver including solving the disputes between them areceiver 119877 and a synthesizer where he only runs the trustedsynthesis algorithm The code works as follows each senderand receiver has their own Cartesian authentication code

respectively Let (119878 119864119894 119879119894 119891119894) (119894 = 1 2 119899) be the sendersrsquo

Cartesian authentication code (119878 119864119877 119879 119892) be the receiverrsquos

Cartesian authentication code ℎ 1198791times 1198792times sdot sdot sdot times 119879

119899rarr

119879 be the synthesis algorithm and 120587119894

119864 rarr 119864119894be a

subkey generation algorithm where 119864 is the key set of thekey distribution center When authenticating a message thesenders and the receiver should comply with the protocolThe key distribution center randomly selects an encodingrule 119890 isin 119864 and sends 119890

119894= 120587119894(119890) to the 119894th sender 119880

119894(119894 =

1 2 119899) secretly then he calculates 119890119877by 119890 according to

an effective algorithm and secretly sends 119890119877to the receiver

119877 If the senders would like to send a source state 119904 to thereceiver 119877 119880

119894computes 119905

119894= 119891119894(119904 119890119894) (119894 = 1 2 119899) and

sends 119898119894= (119904 119905

119894) (119894 = 1 2 119899) to the synthesizer through

an open channel The synthesizer receives the message 119898119894=

(119904 119905119894) (119894 = 1 2 119899) and calculates 119905 = ℎ(119905

1 1199052 119905

119899) by the

synthesis algorithm ℎ and then sends message 119898 = (119904 119905) tothe receiver he checks the authenticity by verifying whether119905 = 119892(119904 119890

119877) or not If the equality holds the message is

authentic and is accepted Otherwise the message is rejectedWe assume that the key distribution center is credible and

though he know the sendersrsquo and receiverrsquos encoding rules hewill not participate in any communication activities Whentransmitters and receiver are disputing the key distributioncenter settles it At the same time we assume that the systemfollows the Kerckhoff principle in which except the actualused keys the other information of the whole system ispublic

2 Journal of Applied Mathematics

In a multisender authentication system we assume thatthe whole senders are cooperative to form a valid messagethat is all senders as a whole and receiver are reliable Butthere are some malicious senders who together cheat thereceiver the part of senders and receiver are not credible andthey can take impersonation attack and substitution attackIn the whole system we assume that 119880

1 1198802 119880

119899 are

senders 119877 is a receiver 119864119894is the encoding rules set of the

sender 119880119894 and 119864

119877is the decoding rules set of the receiver

119877 If the source state space 119878 and the key space 119864119877of receiver

119877 are according to a uniform distribution then the messagespace119872 and the tag space119879 are determined by the probabilitydistribution of 119878 and 119864

119877 119871 = 119894

1 1198942 119894

119897 sub 1 2 119899

119897 lt 119899 119880119871

= 1198801198941 1198801198942 119880

119894119897 119864119871

= 1198641198801198941

1198641198801198942

119864119880119894119897

Now consider that let us consider the attacks from maliciousgroups of senders Here there are two kinds of attack

The opponentrsquos impersonation attack to receiver119880119871 after

receiving their secret keys encode a message and send it tothe receiver 119880

119871are successful if the receiver accepts it as

legitimate message Denote by 119875119868the largest probability of

some opponentrsquos successful impersonation attack to receiverit can be expressed as

119875119868= max119898isin119872

1003816100381610038161003816119890119877 isin 119864119877| 119890119877sub 119898

10038161003816100381610038161003816100381610038161003816119864119877

1003816100381610038161003816

(1)

The opponentrsquos substitution attack to the receiver 119880119871

replace 119898 with another message 1198981015840 after they observe a

legitimatemessage119898119880119871are successful if the receiver accepts

it as legitimate message it can be expressed as

119875119878= max119898isin119872

max1198981015840=119898isin119872

10038161003816100381610038161003816119890119877isin 119864119877| 119890119877sub 119898119898

1015840

10038161003816100381610038161003816

1003816100381610038161003816119890119877 isin 119864119877| 119890119877sub 119898

1003816100381610038161003816

(2)

There might be 119897 malicious senders who together cheatthe receiver that is the part of senders and the receiverare not credible and they can take impersonation attackLet 119871 = 119894

1 1198942 119894

119897 sub 1 2 119899 119897 lt 119899 and 119864

119871=

1198641198801198941

1198641198801198942

119864119880119894119897

Assume that 119880119871

= 1198801198941 1198801198942 119880

119894119897

after receiving their secret keys send a message 119898 to thereceiver 119877119880

119871are successful if the receiver accepts it as legiti-

mate message Denote by 119875119880(119871) the maximum probability of

success of the impersonation attack to the receiver It can beexpressed as

119875119880(119871)

=max119890119871isin119864119871

max119890119871isin119890119880

max119898isin119872

1003816100381610038161003816119890119877isin119864119877 | 119890119877sub119898 119901 (119890119877 119890119875) =0

10038161003816100381610038161003816100381610038161003816119890119877 isin 119864

119877| 119901 (119890119877 119890119875) =0

1003816100381610038161003816

(3)

Notes119901(119890119877 119890119875) = 0 implies that any information 119904 encoded by

119890119879can be authenticated by 119890

119877

In [2] Desmedt et al gave two constructions for MRA-codes based on polynomials and finite geometries respec-tively To construct multisender or multireceiver authenti-cation by polynomials over finite fields many researchershave done much work for example [7ndash9] There are other

constructions of multisender authentication codes that aregiven in [3ndash6] The construction of authentication codesis combinational design in its nature We know that thepolynomial over finite fields can provide a better algebrastructure and is easy to count In this paper we constructone multisender authentication code from the polynomialover finite fields Some parameters and the probabilities ofdeceptions of this code are also computed We realize thegeneralization and the application of the similar idea andmethod of the paper [7ndash9]

2 Some Results about Finite Field

Let 119865119902be the finite field with 119902 elements where 119902 is a power

of a prime 119901 and 119865 is a field containing 119865119902 denote by 119865

lowast

119902be

the nonzero elements set of 119865119902 In this paper we will use the

following conclusions over finite fields

Conclusion 1 A generator 120572 of 119865lowast

119902is called a primitive

element of 119865119902

Conclusion 2 Let 120572 isin 119865119902 if some polynomials contain 120572 as

their root and their leading coefficient are 1 over 119865119902 then the

polynomial having least degree among all such polynomialsis called a minimal polynomial over 119865

119902

Conclusion 3 Let |119865| = 119902119899 then 119865 is an 119899-dimensional

vector space over 119865119902 Let 120572 be a primitive element of 119865

119902

and 119892(119909) the minimal polynomial about 120572 over 119865119902 then

dim119892(119909) = 119899 and 1 120572 1205722

120572119899minus1 is a basis of 119865 Further-

more 1 120572 1205722 120572119899minus1 is linear independent and it is equalto 120572 120572

2

120572119899minus1

120572119899 (120572 is a primitive element 120572 = 0) is also

linear independent moreover 120572119901 1205721199012

120572119901119899minus1

120572119901119899

is alsolinear independent

Conclusion 4 Consider (1199091+ 1199092+ sdot sdot sdot + 119909

119899)119898

= (1199091)119898

+

(1199092)119898

+ sdot sdot sdot + (119909119899)119898 where 119909

119894isin 119865119902 (1 le 119894 le 119899) and 119898 is a

nonnegative power of character 119901 of 119865119902

Conclusion 5 Let119898 le 119899 Then the number of119898times119899matricesof rank119898 over 119865

119902is 119902119898(119898minus1)2prod119899

119894=119899minus119898+1(119902119894

minus 1)

More results about finite fields can be found in [10ndash12]

3 Construction

Let the polynomial 119901119895(119909) = 119886

1198951119909119901119899

+ 1198861198952119909119901(119899minus1)

+ sdot sdot sdot +

119886119895119899119909119901

(1 le 119895 le 119896) where the coefficient 119886119894119897

isin 119865119902

(1 le 119897 le 119899) and these vectors by the composition of theircoefficient are linearly independent The set of source states119878 = 119865

119902 the set of 119894th transmitterrsquos encoding rules 119864

119880119894=

1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) 119909119894

isin 119865lowast

119902 (1 le 119894 le 119899) the set

of receiverrsquos encoding rules 119864119877

= 1199011(120572) 1199012(120572) 119901

119896(120572)

where 120572 is a primitive element of 119865119902 the set of 119894th transmit-

terrsquos tags 119879119894= 119905119894| 119905119894isin 119865119902 (1 le 119894 le 119899) the set of receiverrsquos

tags 119879 = 119905 | 119905 isin 119865119902

Journal of Applied Mathematics 3

Define the encoding map 119891119894 119878 times 119864

119880119894rarr 119879119894 119891119894(119904 119890119880119894) =

1199041199011(119909119894) + 1199042

1199012(119909119894) + sdot sdot sdot + 119904

119896

119901119896(119909119894) 1 le 119894 le 119899

The decoding map 119891 119878 times 119864119877

rarr 119879 119891(119904 119890119877) = 119904119901

1(120572) +

1199042

1199012(120572) + sdot sdot sdot + 119904

119896

119901119896(120572)

The synthesizing map ℎ 1198791times 1198792times sdot sdot sdot times 119879

119899rarr 119879

ℎ(1199051 1199052 119905

119899) = 1199051+ 1199052+ sdot sdot sdot + 119905

119899

The code works as followsAssume that 119902 is larger than or equal to the number of

the possible message and 119899 le 119902

31 Key Distribution The key distribution center randomlygenerates 119896 (119896 le 119899) polynomials 119901

1(119909) 1199012(119909) 119901

119896(119909)

where 119901119895(119909) = 119886

1198951119909119901119899

+ 1198861198952119909119901(119899minus1)

+ sdot sdot sdot + 119886119895119899119909119901

(1 le 119895 le 119896)and make these vectors by composed of their coefficient islinearly independent it is equivalent to the column vectors

of the matrix (

11988611 11988621 sdotsdotsdot 1198861198961

11988612 11988622 sdotsdotsdot 1198861198962

1198861119899 1198862119899 sdotsdotsdot 119886119896119899

) is linearly independent He

selects 119899 distinct nonzero elements 1199091 1199092 119909

119899isin 119865119902again

and makes 119909119894(1 le 119894 le 119899) secret then he sends privately

1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) to the sender 119880

119894(1 le 119894 le 119899) The

key distribution center also randomly chooses a primitiveelement 120572 of 119865

119902satisfying 119909

1+ 1199092+ sdot sdot sdot + 119909

119899= 120572 and sends

1199011(120572) 1199012(120572) 119901

119896(120572) to the receiver 119877

32 Broadcast If the senderswant to send a source state 119904 isin 119878

to the receiver 119877 the sender 119880119894calculates 119905

119894= 119891119894(119904 119890119880119894) =

119860119904(119909119894) = 119904119901

1(119909119894)+1199042

1199012(119909119894)+ sdot sdot sdot+119904

119896

119901119896(119909119894) 1 le 119894 le 119899 and then

sends 119860119904(119909119894) = 119905119894to the synthesizer

33 Synthesis After the synthesizer receives 1199051 1199052 119905

119899 he

calculates ℎ(1199051 1199052 119905

119899) = 1199051+ 1199052+ sdot sdot sdot + 119905

119899and then sends

119898 = (119904 119905) to the receiver 119877

34 Verification When the receiver 119877 receives 119898 = (119904 119905) hecalculates 1199051015840 = 119892(119904 119890

119877) = 119860

119904(120572) = 119904119901

1(120572) + 119904

2

1199012(120572) + sdot sdot sdot +

119904119896

119901119896(120572) If 119905 = 119905

1015840 he accepts 119905 otherwise he rejects itNext we will show that the above construction is a well

defined multisender authentication code with arbitration

Lemma 1 Let 119862119894= (119878 119864

119875119894 119879119894 119891119894) then the code is an A-code

1 le 119894 le 119899

Proof (1) For any 119890119880119894

isin 119864119880119894 119904 isin 119878 because 119864

119880119894= 1199011(119909119894)

1199012(119909119894) 119901

119896(119909119894) 119909119894isin 119865lowast

119902 so 119905

119894= 1199041199011(119909119894) + 1199042

1199012(119909119894) + sdot sdot sdot +

119904119896

119901119896(119909119894) isin 119879119894= 119865119902 Conversely for any 119905

119894isin 119879119894 choose 119890

119880119894=

1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) 119909119894isin 119865lowast

119902 where 119901

119895(119909) = 119886

1198951119909119901119899

+

1198861198952119909119901(119899minus1)

+ sdot sdot sdot + 119886119895119899119909119901

(1 le 119895 le 119896) and let 119905119894= 119891119894(119904 119890119880119894) =

1199041199011(119909119894) + 1199042

1199012(119909119894) + sdot sdot sdot + 119904

119896

119901119896(119909119894) it is equivalent to

(119909119901119899

119894 119909119901119899minus1

119894 119909

119901

119894)(

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)(

119904

1199042

119904119896

) = 119905119894

(4)

It follows that

(

(

119909119901119899

1119909119901119899minus1

1sdot sdot sdot 119909119901

1

119909119901n

2119909119901119899minus1

2sdot sdot sdot 119909119901

2

119909119901119899

119899119909119901119899minus1

119899sdot sdot sdot 119909119901

119899

)

)

times (

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)(

119904

1199042

119904119896

) = (

1199051

1199052

119905119899

)

(5)

Denote

119860 = (

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)

119883 = (

119909119901119899

1119909119901119899minus1

1sdot sdot sdot 119909119901

1

119909119901119899

2119909119901119899minus1

2sdot sdot sdot 119909119901

2

119909119901119899

119899119909119901119899minus1

119899sdot sdot sdot 119909119901

119899

)

119878 = (

119904

1199042

119904119896

) 119905 = (

1199051

1199052

119905119899

)

(6)

The above linear equation is equivalent to 119883119860119878 = 119905because the column vectors of 119860 are linearly independent119883 is equivalent to a Vandermonde matrix and 119883 is inversetherefore the above linear equation has a unique solution so119904 is only defined that is 119891

119894(1 le 119894 le 119899) is a surjection

(2) If 1199041015840 isin 119878 is another source state satisfying 1199041199011(119909119894) +

1199042

1199012(119909119894)+sdot sdot sdot+119904

119896

119901119896(119909119894) = 1199041015840

1199011(119909119894)+11990410158402

1199012(119909119894)+sdot sdot sdot+119904

1015840119896

119901119896(119909119894) =

119905119894 and it is equivalent to (119904 minus 119904

1015840

)1199011(119909119894) + (1199042

minus 11990410158402

)1199012(119909119894) + sdot sdot sdot +

(119904119896

minus 1199041015840119896

)119901119896(119909119894) = 0 then

(119909119901119899

119894 119909119901119899minus1

119894 119909

119901

119894)

times (

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)(

119904 minus 1199041015840

1199042

minus 11990410158402

119904119896

minus 1199041015840119896

) = 0

(7)

4 Journal of Applied Mathematics

Thus

(

(

119909119901119899

1119909119901119899minus1

1sdot sdot sdot 119909119901

1

119909119901n

2119909119901119899minus1

2sdot sdot sdot 119909119901

2

119909119901119899

119899119909119901119899minus1

119899sdot sdot sdot 119909119901

119899

)

)

times (

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)(

119904 minus 1199041015840

1199042

minus 11990410158402

119904119896

minus 1199041015840119896

) = (

0

0

0

)

(8)

Similar to (1) we know that the homogeneous linear equation119883119860119878 = 0 has a unique solution that is there is only zerosolution so 119904 = 119904

1015840 So 119904 is the unique source state determinedby 119890119880119894and 119905119894 thus 119862

119894(1 le 119894 le 119899) is an A-code

Lemma 2 Let 119862 = (119878 119864119877 119879 119892) then the code is an A-code

Proof (1) For any 119904 isin 119878 119890119877isin 119864119877 from the definition of 119890

119877

we assume that 119864119877

= 1199011(120572) 1199012(120572) 119901

119896(120572) where 120572 is a

primitive element of 119865119902 119892(119904 119890

119877) = 119904119901

1(120572) + 119904

2

1199012(120572) + sdot sdot sdot +

119904119896

119901119896(120572) isin 119879 = 119865

119902 on the other hand for any 119905 isin 119879 choose

119890119877= 1199011(120572) 1199012(120572) 119901

119896(120572) where 120572 is a primitive element

of 119865119902 119892(119904 119890

119877) = 119904119901

1(120572) + 119904

2

1199012(120572) + sdot sdot sdot + 119904

119896

119901119896(120572) = 119905 it is

equivalent to

(120572119901119899

120572119901119899minus1

sdot sdot sdot 120572119901

)

times (

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)(

119904

1199042

119904119896

) = 119905

119860 = (

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)

(9)

that is (120572119901119899

120572119901119899minus1

120572119901

)119860(

119904

1199042

119904119896

) = 119905 From Conclusion

3 we know that (120572119901119899

120572119901119899minus1

120572119901

) is linearly independentand the column vectors of 119860 are also linearly independenttherefore the above linear equation has unique solution so 119904

is only defined that is 119892 is a surjection

(2) If 1199041015840 is another source state satisfying 119905 = 119892(1199041015840

119890119877) then

(120572119901119899

120572119901119899minus1

120572119901

)119860(

1199041015840

11990410158402

1199041015840119896

)

= (120572119901119899

120572119901119899minus1

120572119901

)119860(

119904

1199042

119904119896

)

(10)

that is (120572119901119899

120572119901119899minus1

120572119901

)119860([[

[

1199041015840

11990410158402

1199041015840119896

]]

]

minus [

[

119904

1199042

119904119896

]

]

) = 0 Sim-

ilar to (1) we get that the homogeneous linear equation(120572119901119899

120572119901119899minus1

120572119901

)119860(1198781015840

minus 119878) = 0 has a unique solution thatis there is only zero solution so 119878 = 119878

1015840 that is 119904 = 1199041015840 So

119904 is the unique source state determined by 119890119877and 119905 thus

119862 = (119878 119864119877 119879 119892) is an A-code

At the same time for any valid119898 = (119904 119905) we have knownthat 120572 = 119909

1+ 1199092+ sdot sdot sdot + 119909

119899 and it follows that 1199051015840 = 119904119901

1(120572) +

1199042

1199012(120572)+sdot sdot sdot+119904

119896

119901119896(120572) = 119904119901

1(1199091+1199092+sdot sdot sdot+119909

119899)+1199042

1199012(1199091+1199092+

sdot sdot sdot + 119909119899) + sdot sdot sdot + 119904

119896

119901119896(1199091+ 1199092+ sdot sdot sdot + 119909

119899) We also have known

that 119901119895(119909) = 119886

1198951119909119901119899

+1198861198952119909119901(119899minus1)

+ sdot sdot sdot + 119886119895119899119909119901

(1 le 119895 le 119896) fromConclusion 4 (119909

1+ 1199092+ sdot sdot sdot + 119909

119899)119901119898

= (1199091)119901119898

+ (1199092)119901119898

+ sdot sdot sdot +

(119909119899)119901119898

where119898 is a nonnegative power of character 119901 of 119865119902

andwe get119901119895(1199091+1199092+sdot sdot sdot+119909

119899) = 119901119895(1199091)+119901119895(1199092)+sdot sdot sdot+119901

119895(119909119899)

therefore 1199051015840 = 1199041199011(120572) + 119904

2

1199012(120572) + sdot sdot sdot + 119904

119896

119901119896(120572) = (119904119901

1(1199091) +

1199042

1199012(1199091)+sdot sdot sdot+119904

119896

119901119896(1199091))+(119904119901

1(1199092)+1199042

1199012(1199092)+sdot sdot sdot+119904

119896

119901119896(1199092))+

sdot sdot sdot+(1199041199011(119909119899)+1199042

1199012(119909119899)+ sdot sdot sdot+119904

119896

119901119896(119909119899)) = 1199051+1199052+sdot sdot sdot+119905

119899= 119905

and the receiver 119877 accepts119898

From Lemmas 1 and 2 we know that such construction ofmultisender authentication codes is reasonable and there are119899 senders in this system Next we compute the parametersof this code and the maximum probability of success inimpersonation attack and substitution attack by the group ofsenders

Theorem 3 Some parameters of this construction are|119878| = 119902 |119864

119880119894| = [119902

119896(119896minus1)2

prod119899

119894=119899minus119896+1(119902119894

minus 1)] (119902minus1

1) =

[119902119896(119896minus1)2

prod119899

119894=119899minus119896+1(119902119894

minus 1)](119902 minus 1) (1 le 119894 le 119899) |119879119894| = 119902 (1 le

119894 le 119899) |119864119877| = [119902

119896(119896minus1)2

prod119899

119894=119899minus119896+1(119902119894

minus 1)]120593(119902 minus 1) |119879| = 119902Where 120593(119902 minus 1) is the 119864119906119897119890119903 119891119906119899119888119905119894119900119899 of 119902 minus 1 it representsthe number of primitive element of 119865

119902here

Proof For |119878| = 119902 |119879119894| = 119902 and |119879| = 119902 the results

are straightforward For119864119880119894 because119864

119880119894=1199011(119909119894) 1199012(119909119894)

119901119896(119909119894) 119909119894isin 119865lowast

119902 where 119901

119895(119909) = 119886

1198951119909119901119899

+ 1198861198952119909119901(119899minus1)

+ sdot sdot sdot +

119886119895119899119909119901

(1 le 119895 le 119896) and these vectors by the composition oftheir coefficient are linearly independent it is equivalent to

the columns of 119860 = (

11988611 11988621 sdotsdotsdot 1198861198961

11988612 11988622 sdotsdotsdot 1198861198962

1198861119899 1198862119899 sdotsdotsdot 119886119896119899

) is linear independent

Journal of Applied Mathematics 5

From Conclusion 5 we can conclude that the number of 119860satisfying the condition is 119902

119896(119896minus1)2

prod119899

119894=119899minus119896+1(119902119894

minus 1) On theother hand the number of distinct nonzero elements 119909

119894(1 le

119894 le 119899) in 119865119902is ( 119902minus11

) = 119902 minus 1 so |119864119880119894| = [119902

119896(119896minus1)2

prod119899

119894=119899minus119896+1(119902119894

minus

1)](119902 minus 1) For 119864119877 119864119877

= 1199011(120572) 1199012(120572) 119901

119896(120572) where 120572

is a primitive element of 119865119902 For 120572 from Conclusion 1 a

generator of119865lowast119902is called a primitive element of119865

119902 |119865lowast119902| = 119902minus1

by the theory of the group we know that the number ofgenerator of 119865lowast

119902is 120593(119902minus1) that is the number of 120572 is 120593(119902minus1)

For 1199011(119909) 1199012(119909) 119901

119896(119909) From above we have confirmed

that the number of these polynomials is 119902119896(119896minus1)2prod119899119894=119899minus119896+1

(119902119894

minus

1) therefore |119864119877| = [119902

119896(119896minus1)2

prod119899

119894=119899minus119896+1(119902119894

minus 1)]120593(119902 minus 1)

Lemma 4 For any 119898 isin 119872 the number of 119890119877contained 119898 is

120593(119902 minus 1)

Proof Let 119898 = (119904 119905) isin 119872 119890119877

= 1199011(120572) 1199012(120572) 119901

119896(120572)

where 120572 is a primitive element of 119865119902 isin 119864

119877 If 119890119877

sub 119898then 119904119901

1(120572) + 119904

2

1199012(120572) + sdot sdot sdot + 119904

119896

119901119896(120572) = 119905 hArr (120572

119901119899

120572119901119899minus1

120572119901

)119860(

119904

1199042

119904119896

) = 119905 For any 120572 suppose that there is

another 1198601015840 such that (120572

119901119899

120572119901119899minus1

120572119901

)1198601015840

(

119904

1199042

119904119896

) = 119905

then (120572119901119899

120572119901119899minus1

120572119901

)(119860 minus 1198601015840

)(

119904

1199042

119904119896

) = 0 because

120572119901119899

120572119901119899minus1

120572119901 is linearly independent so (119860minus119860

1015840

)(

119904

1199042

119904119896

) =

0 but (

119904

1199042

119904119896

) is arbitrarily therefore 119860 minus 1198601015840

= 0 that is

119860 = 1198601015840 and it follows that 119860 is only determined by 120572

Therefore as 120572 isin 119864119877 for any given 119904 and 119905 the number of

119890119877contained in119898 is 120593(119902 minus 1)

Lemma 5 For any119898 = (119904 119905) isin 119872 and1198981015840

= (1199041015840

1199051015840

) isin 119872 with119904 = 1199041015840 the number of 119890

119877contained119898 and119898

1015840 is 1

Proof Assume that 119890119877

= 1199011(120572) 1199012(120572) 119901

119896(120572) where

120572 is a primitive element of 119865119902 isin 119864

119877 If 119890119877

sub 119898 and119890119877

sub 1198981015840 then 119904119901

1(120572) + 119904

2

1199012(120572) + sdot sdot sdot + 119904

119896

119901119896(120572) = 119905 hArr

(120572119901119899

120572119901119899minus1

120572119901

)119860(

119904

1199042

119904119896

) = 119905 11990410158401199011(120572) + 119904

10158402

1199012(120572) + sdot sdot sdot +

1199041015840119896

119901119896(120572) = 119905 hArr (120572

119901119899

120572119901119899minus1

120572119901

)119860(

1199041015840

11990410158402

1199041015840119896

) = 119905 It is

equivalent to (120572119901119899

120572119901119899minus1

120572119901

)119860(

119904minus1199041015840

1199042minus11990410158402

119904119896minus1199041015840119896

) = 119905 minus 1199051015840 because

119904 = 1199041015840 so 119905 = 119905

1015840 otherwise we assume that 119905 = 1199051015840 and

since 120572119901119899

120572119901119899minus1

120572119901 and the column vectors of 119860 both

are linearly independent it forces that 119904 = 1199041015840 this is a

contradiction Therefore we get

(119905 minus 1199051015840

)minus1

[[[[

[

(120572119901119899

120572119901119899minus1

120572119901

)119860 (

119904 minus 1199041015840

1199042

minus 11990410158402

119904119896

minus 1199041015840119896

)

]]]]

]

= 1

(lowast)

since 119905 1199051015840 is given (119905 minus 119905

1015840

)minus1 is unique by equation (lowast) for

any given 119904 1199041015840 and 119905 119905

1015840 we obtain that (120572119901119899

120572119901119899minus1

120572119901

)119860 isonly determined thus the number of 119890

119877contained119898 and119898

1015840

is 1

Lemma6 For any fixed 119890119880= 1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) 119909119894isin

119865lowast

119902 (1 le 119894 le 119899) containing a given 119890

119871 then the number of 119890

119877

which is incidence with 119890119880is 120593(119902 minus 1)

Proof For any fixed 119890119880

= 1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) 119909119894isin

119865lowast

119902 (1 le 119894 le 119899) containing a given 119890

119871 we assume that

119901119895(119909119894) = 1198861198951119909119901119899

119894+1198861198952119909119901(119899minus1)

119894+sdot sdot sdot+119886

119895119899119909119901

119894(1 le 119895 le 119896 1 le 119894 le 119899)

119890119877= 1199011(120572) 1199012(120572) 119901

119896(120572) where 120572 is a primitive element

of 119865119902 From the definitions of 119890

119877and 119890119880and Conclusion 4

we can conclude that 119890119877is incidence with 119890

119880if and only if

1199091+ 1199092+ sdot sdot sdot + 119909

119899= 120572 For any 120572 since Rank(1 1 1) =

Rank(1 1 1 120572) = 1 lt 119899 so the equation1199091+1199092+sdot sdot sdot+119909

119899=

120572 always has a solution From the proof of Theorem 3 weknow the number of 119890

119877which is incident with 119890

119880(ie the

number of all 119864119877) is [119902119896(119896minus1)2prod119899

119894=119899minus119896+1(119902119894

minus 1)] 120593(119902 minus 1)

Lemma 7 For any fixed 119890119880= 1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) 119909119894isin

119865lowast

119902 (1 le 119894 le 119899) containing a given 119890

119871and119898 = (119904 119905) the num-

ber of 119890119877which is incidence with 119890

119880and contained in119898 is 1

Proof For any 119904 isin 119878 119890119877

isin 119864119877 we assume that 119890

119877=

1199011(120572) 1199012(120572) 119901

119896(120572) where 120572 is a primitive element

of 119865119902 Similar to Lemma 6 for any fixed 119890

119880=

1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) 119909119894isin 119865lowast

119902 (1 le 119894 le 119899) containing

a given 119890119871 we have known that 119890

119877is incident with 119890

119880if and

only if

1199091+ 1199092+ sdot sdot sdot + 119909

119899= 120572 (11)

Again with 119890119877sub 119898 we can get

1199041199011(120572) + 119904

2

1199012(120572) + sdot sdot sdot + 119904

119896

119901119896(120572) = 119905 (12)

By (11) and (12) and the property of 119901119895(119909) (1 le 119895 le 119896) we

have the following conclusion

1199041199011(

119899

sum

119894=1

119909119894) + 1199042

1199012(

119899

sum

119894=1

119909119894) + sdot sdot sdot + 119904

119896

119901119896(

119899

sum

119894=1

119909119894)

= 119905 lArrrArr (1199011(

119899

sum

119894=1

119909119894) 1199012(

119899

sum

119894=1

119909119894) 119901

119896(

119899

sum

119894=1

119909119894))

6 Journal of Applied Mathematics

times (

119904

1199042

119904119896

)

= 119905 lArrrArr (

119899

sum

119894=1

1199011(119909119894)

119899

sum

119894=1

1199012(119909119894)

119899

sum

119894=1

119901119896(119909119894))(

119904

s2119904119896

)

=119905lArrrArr( (

119899

sum

119894=1

119909119894)

119899

(

119899

sum

119894=1

119909119894)

119899minus1

(

119899

sum

119894=1

119909119894))119860(

119904

1199042

119904119896

)

= 119905 lArrrArr [(

119899

sum

119894=1

1199011(119909119894)

119899

sum

119894=1

1199012(119909119894)

119899

sum

119894=1

119901119896(119909119894))

minus((

119899

sum

119894=1

119909119894)

119899

(

119899

sum

119894=1

119909119894)

119899minus1

(

119899

sum

119894=1

119909119894))119860]

times (

119904

1199042

119904119896

) = 0

(13)

because 119904 is any given Similar to the proof of Lemma 4we can get (sum

119899

119894=11199011(119909119894) sum119899

119894=11199012(119909119894) sum

119899

119894=1119901119896(119909119894)) minus

((sum119899

119894=1119909119894)119899(sum119899119894=1

119909119894)119899minus1

(sum119899

119894=1119909119894))119860=0 that is ((sum119899

119894=1119909119894)119899

(sum119899

119894=1119909119894)119899minus1

(sum119899

119894=1119909119894))119860 = (sum

119899

119894=11199011(119909119894) sum119899

119894=11199012(119909119894)

sum119899

119894=1119901119896(119909119894)) but 119901

1(119909119894) 1199012(119909119894) 119901

119896(119909119894) and 119909

119894(1 le 119894 le 119899)

also are fixed thus 120572 and 119860are only determined so thenumber of 119890

119877which is incident with 119890

119880and contained in 119898

is 1

Theorem 8 In the constructed multisender authenticationcodes if the sendersrsquo encoding rules and the receiverrsquos decodingrules are chosen according to a uniform probability distribu-tion then the largest probabilities of success for different typesof deceptions respectively are

119875119868=

1

119902119896(119896minus1)2prod119899

119894=119899minus119896+1(119902119894 minus 1)

119875119878=

1

120593 (119902 minus 1)

119875119880(119871) =

1

[119902119896(119896minus1)2prod119899

119894=119899minus119896+1(119902119894 minus 1)] 120593 (119902 minus 1)

(14)

Proof By Theorem 3 and Lemma 4 we get

119875119868= max119898isin119872

1003816100381610038161003816119890119877 isin 119864119877| 119890119877sub 119898

10038161003816100381610038161003816100381610038161003816119864119877

1003816100381610038161003816

=120593 (119902 minus 1)

[119902119896(119896minus1)2prod119899

119894=119899minus119896+1(119902119894 minus 1)] 120593 (119902 minus 1)

=1

119902119896(119896minus1)2prod119899

119894=119899minus119896+1(119902119894 minus 1)

(15)

By Lemmas 4 and 5 we get

119875119878= max119898isin119872

max1198981015840=119898isin119872

10038161003816100381610038161003816119890119877isin 119864119877| 119890119877sub 119898119898

1015840

10038161003816100381610038161003816

1003816100381610038161003816119890119877 isin 119864119877| 119890119877sub 119898

1003816100381610038161003816

=1

120593 (119902 minus 1)

(16)

By Lemmas 6 and 7 we get

119875119880(119871)

=max119890119871isin119864119871

max119890119871isin119890119880

max119898isin119872

1003816100381610038161003816119890119877 isin 119864119877| 119890119877sub119898 119901 (119890

119877 119890119875) =0

10038161003816100381610038161003816100381610038161003816119890119877 isin 119864

119877| 119901 (119890119877 119890119875) =0

1003816100381610038161003816

=1

[119902119896(119896minus1)2prod119899

119894=119899minus119896+1(119902119894 minus 1)] 120593 (119902 minus 1)

(17)

Acknowledgments

This paper is supported by the NSF of China (61179026)and the Fundamental Research of the Central Universi-ties of China Civil Aviation University of Science special(ZXH2012k003)

References

[1] E N Gilbert F J MacWilliams and N J A Sloane ldquoCodeswhich detect deceptionrdquoThe Bell System Technical Journal vol53 pp 405ndash424 1974

[2] Y Desmedt Y Frankel and M Yung ldquoMulti-receivermulti-sender network security efficient authenticated multicastfeedbackrdquo in Proceedings of the the 11th Annual Conference of theIEEEComputer andCommunications Societies (Infocom rsquo92) pp2045ndash2054 May 1992

[3] K Martin and R Safavi-Naini ldquoMultisender authenticationschemes with unconditional securityrdquo in Information and Com-munications Security vol 1334 of Lecture Notes in ComputerScience pp 130ndash143 Springer Berlin Germany 1997

[4] W Ma and X Wang ldquoSeveral new constructions of multitrasmitters authentication codesrdquo Acta Electronica Sinica vol28 no 4 pp 117ndash119 2000

[5] G J Simmons ldquoMessage authentication with arbitration oftransmitterreceiver disputesrdquo in Advances in CryptologymdashEUROCRYPT rsquo87 Workshop on theTheory and Application of ofCryptographic Techniques vol 304 of Lecture Notes in ComputerScience pp 151ndash165 Springer 1988

[6] S Cheng and L Chang ldquoTwo constructions of multi-senderauthentication codes with arbitration based linear codes to bepublished in rdquoWSEAS Transactions on Mathematics vol 11 no12 2012

Journal of Applied Mathematics 7

[7] R Safavi-Naini and H Wang ldquoNew results on multi-receiver authentication codesrdquo in Advances in CryptologymdashEUROCRYPT rsquo98 (Espoo) vol 1403 of Lecture Notes in ComputSci pp 527ndash541 Springer Berlin Germany 1998

[8] R Aparna and B B Amberker ldquoMulti-sender multi-receiverauthentication for dynamic secure group communicationrdquoInternational Journal of Computer Science andNetwork Securityvol 7 no 10 pp 47ndash63 2007

[9] R Safavi-Naini and H Wang ldquoBounds and constructions formultireceiver authentication codesrdquo inAdvances in cryptologymdashASIACRYPTrsquo98 (Beijing) vol 1514 of Lecture Notes in ComputSci pp 242ndash256 Springer Berlin Germany 1998

[10] S Shen and L Chen Information and Coding Theory Sciencepress in China 2002

[11] J J Rotman Advanced Modern Algebra High Education Pressin China 2004

[12] Z Wan Geometry of Classical Group over Finite Field SciencePress in Beijing New York NY USA 2002

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 435730 8 pageshttpdxdoiorg1011552013435730

Research ArticleGeometrical and Spectral Properties of the OrthogonalProjections of the Identity

Luis Gonzaacutelez Antonio Suaacuterez and Dolores Garciacutea

Department of Mathematics Research Institute SIANI University of Las Palmas de Gran Canaria Campus de Tafira35017 Las Palmas de Gran Canaria Spain

Correspondence should be addressed to Luis Gonzalez luisglezdmaulpgces

Received 28 December 2012 Accepted 10 April 2013

Academic Editor K Sivakumar

Copyright copy 2013 Luis Gonzalez et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We analyze the best approximation119860119873 (in the Frobenius sense) to the identitymatrix in an arbitrarymatrix subspace119860119878 (119860 isin R119899times119899

nonsingular 119878 being any fixed subspace of R119899times119899) Some new geometrical and spectral properties of the orthogonal projection 119860119873are derived In particular new inequalities for the trace and for the eigenvalues of matrix119860119873 are presented for the special case that119860119873 is symmetric and positive definite

1 Introduction

The set of all 119899 times 119899 real matrices is denoted by R119899times119899 and119868 denotes the identity matrix of order 119899 In the following119860119879 and tr(119860) denote as usual the transpose and the trace

of matrix 119860 isin R119899times119899 The notations ⟨sdot sdot⟩119865and sdot

119865stand

for the Frobenius inner product and matrix norm definedon the matrix space R119899times119899 Throughout this paper the termsorthogonality angle and cosine will be used in the sense ofthe Frobenius inner product

Our starting point is the linear system

119860119909 = 119887 119860 isin R119899times119899

119909 119887 isin R119899

(1)

where 119860 is a large nonsingular and sparse matrix Theresolution of this system is usually performed by iterativemethods based on Krylov subspaces (see eg [1 2]) Thecoefficient matrix 119860 of the system (1) is often extremelyill-conditioned and highly indefinite so that in this caseKrylov subspace methods are not competitive without agood preconditioner (see eg [2 3]) Then to improve theconvergence of these Krylov methods the system (1) canbe preconditioned with an adequate nonsingular precondi-tioning matrix 119873 transforming it into any of the equivalentsystems

119873119860119909 = 119873119887

119860119873119910 = 119887 119909 = 119873119910

(2)

the so-called left and right preconditioned systems respec-tively In this paper we address only the case of the right-handside preconditioned matrices 119860119873 but analogous results canbe obtained for the left-hand side preconditioned matrices119873119860

The preconditioning of the system (1) is often performedin order to get a preconditioned matrix 119860119873 as close aspossible to the identity in some sense and the preconditioner119873 is called an approximate inverse of119860 The closeness of119860119873to 119868may bemeasured by using a suitablematrix norm like forinstance the Frobenius norm [4] In this way the problemof obtaining the best preconditioner 119873 (with respect to theFrobenius norm) of the system (1) in an arbitrary subspace119878 of R119899times119899 is equivalent to the minimization problem see forexample [5]

min119872isin119878

119860119872 minus 119868119865= 119860119873 minus 119868

119865 (3)

The solution 119873 to the problem (3) will be referred to asthe ldquooptimalrdquo or the ldquobestrdquo approximate inverse of matrix 119860in the subspace 119878 Since matrix119860119873 is the best approximationto the identity in subspace 119860119878 it will be also referred toas the orthogonal projection of the identity matrix onto thesubspace 119860119878 Although many of the results presented in thispaper are also valid for the case thatmatrix119873 is singular fromnow on we assume that the optimal approximate inverse 119873(and thus also the orthogonal projection119860119873) is a nonsingular

2 Journal of Applied Mathematics

matrix The solution 119873 to the problem (3) has been studiedas a natural generalization of the classical Moore-Penroseinverse in [6] where it has been referred to as the 119878-Moore-Penrose inverse of matrix 119860

The main goal of this paper is to derive new geometricaland spectral properties of the best approximations119860119873 (in thesense of formula (3)) to the identity matrix Such propertiescould be used to analyze the quality and theoretical effective-ness of the optimal approximate inverse119873 as preconditionerof the system (1) However it is important to highlightthat the purpose of this paper is purely theoretical and weare not looking for immediate numerical or computationalapproaches (although our theoretical results could be poten-tially applied to the preconditioning problem) In particularthe term ldquooptimal (or best) approximate inverserdquo is used inthe sense of formula (3) and not in any other sense of thisexpression

Among the many different works dealing with practicalalgorithms that can be used to compute approximate inverseswe refer the reader to for example [4 7ndash9] and to thereferences therein In [4] the author presents an exhaustivesurvey of preconditioning techniques and in particulardescribes several algorithms for computing sparse approxi-mate inverses based on Frobenius norm minimization likefor instance the well-known SPAI and FSAI algorithms Adifferent approach (which is also focused on approximateinverses based on minimizing 119860119872 minus 119868

119865) can be found

in [7] where an iterative descent-type method is used toapproximate each column of the inverse and the iterationis done with ldquosparse matrix by sparse vectorrdquo operationsWhen the system matrix is expressed in block-partitionedform some preconditioning options are explored in [8] In[9] the idea of ldquotargetrdquo matrix is introduced in the contextof sparse approximate inverse preconditioners and the gen-eralized Frobenius norms 1198612

119865119867= tr(119861119867119861119879) (119867 symmetric

positive definite) are used for minimization purposes as analternative to the classical Frobenius norm

The last results of our work are devoted to the special casethat matrix 119860119873 is symmetric and positive definite In thissense let us recall that the cone of symmetric and positivedefinite matrices has a rich geometrical structure and in thiscontext the angle that any symmetric and positive definitematrix forms with the identity plays a very important role[10] In this paper the authors extend this geometrical pointof view and analyze the geometrical structure of the subspaceof symmetric matrices of order 119899 including the location of allorthogonal matrices not only the identity matrix

This paper has been organized as follows In Section 2we present some preliminary results required to make thepaper self-contained Sections 3 and 4 are devoted to obtainnew geometrical and spectral relations respectively for theorthogonal projections 119860119873 of the identity matrix FinallySection 5 closes the paper with its main conclusions

2 Some Preliminaries

Now we present some preliminary results concerning theorthogonal projection 119860119873 of the identity onto the matrix

subspace119860119878 sub R119899times119899 For more details about these results andfor their proofs we refer the reader to [5 6 11]

Taking advantage of the prehilbertian character of thematrix Frobenius norm the solution 119873 to the problem (3)can be obtained using the orthogonal projection theoremMore precisely the matrix product 119860119873 is the orthogonalprojection of the identity onto the subspace119860119878 and it satisfiesthe conditions stated by the following lemmas see [5 11]

Lemma 1 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

0 le 1198601198732

119865= tr (119860119873) le 119899 (4)

0 le 119860119873 minus 1198682

119865= 119899 minus tr (119860119873) le 119899 (5)

An explicit formula for matrix119873 can be obtained by ex-pressing the orthogonal projection119860119873 of the identity matrixonto the subspace 119860119878 by its expansion with respect to anorthonormal basis of 119860119878 [5] This is the idea of the followinglemma

Lemma 2 Let 119860 isin R119899times119899 be nonsingular Let 119878 be a linearsubspace of R119899times119899 of dimension 119889 and 119872

1 119872

119889 a basis of 119878

such that 1198601198721 119860119872

119889 is an orthogonal basis of 119860119878 Then

the solution119873 to the problem (3) is

119873 =

119889

sum

119894=1

tr (119860119872119894)

10038171003817100381710038171198601198721198941003817100381710038171003817

2

119865

119872119894 (6)

and the minimum (residual) Frobenius norm is

119860119873 minus 1198682

119865= 119899 minus

119889

sum

119894=1

[tr (119860119872119894)]2

10038171003817100381710038171198601198721198941003817100381710038171003817

2

119865

(7)

Let us mention two possible options both taken from [5]for choosing in practice the subspace 119878 and its correspondingbasis 119872

119894119889

119894=1 The first example consists of considering the

subspace 119878 of 119899times119899matrices with a prescribed sparsity patternthat is

119878 = 119872 isin R119899times119899

119898119894119895= 0 forall (119894 119895) notin 119870

119870 sub 1 2 119899 times 1 2 119899

(8)

Then denoting by119872119894119895 the 119899 times 119899matrix whose only nonzero

entry is 119898119894119895

= 1 a basis of subspace 119878 is clearly 119872119894119895

(119894 119895) isin 119870 and then 119860119872119894119895

(119894 119895) isin 119870 will be a basisof subspace 119860119878 (since we have assumed that matrix 119860 isnonsingular) In general this basis of 119860119878 is not orthogonalso that we only need to use the Gram-Schmidt procedureto obtain an orthogonal basis of 119860119878 in order to apply theorthogonal expansion (6)

For the second example consider a linearly independentset of 119899 times 119899 real symmetric matrices 119875

1 119875

119889 and the

corresponding subspace

1198781015840

= span 1198751119860119879

119875119889119860119879

(9)

Journal of Applied Mathematics 3

which clearly satisfies

1198781015840

sube 119872 = 119875119860119879

119875 isin R119899times119899 119875119879 = 119875

= 119872 isin R119899times119899 (119860119872)119879

= 119860119872

(10)

Hence we can explicitly obtain the solution 119873 to theproblem (3) for subspace 1198781015840 from its basis 119875

1119860119879

119875119889119860119879

as follows If 119860119875

1119860119879

119860119875119889119860119879

is an orthogonal basis ofsubspace 1198601198781015840 then we just use the orthogonal expansion (6)for obtaining119873 Otherwise we use again the Gram-Schmidtprocedure to obtain an orthogonal basis of subspace1198601198781015840 andthenwe apply formula (6)The interest of this second examplestands in the possibility of using the conjugate gradientmethod for solving the preconditioned linear system whenthe symmetric matrix 119860119873 is positive definite For a moredetailed exposition of the computational aspects related tothese two examples we refer the reader to [5]

Now we present some spectral properties of the orthog-onal projection 119860119873 From now on we denote by 120582

119894119899

119894=1and

120590119894119899

119894=1the sets of eigenvalues and singular values respectively

of matrix119860119873 arranged as usual in nonincreasing order thatis

100381610038161003816100381612058211003816100381610038161003816 ge

100381610038161003816100381612058221003816100381610038161003816 ge sdot sdot sdot ge

10038161003816100381610038161205821198991003816100381610038161003816 gt 0

1205901ge 1205902ge sdot sdot sdot ge 120590

119899gt 0

(11)

The following lemma [11] provides some inequalitiesinvolving the eigenvalues and singular values of the precon-ditioned matrix 119860119873

Lemma 3 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Then

119899

sum

119894=1

1205822

119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816

2

le

119899

sum

119894=1

1205902

119894= 119860119873

2

119865

= tr (119860119873) =

119899

sum

119894=1

120582119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816 le

119899

sum

119894=1

120590119894

(12)

The following fact [11] is a direct consequence ofLemma 3

Lemma 4 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Then the smallest singular value and thesmallest eigenvaluersquos modulus of the orthogonal projection 119860119873of the identity onto the subspace 119860119878 are never greater than 1That is

0 lt 120590119899le1003816100381610038161003816120582119899

1003816100381610038161003816 le 1 (13)

The following theorem [11] establishes a tight connectionbetween the closeness of matrix 119860119873 to the identity matrixand the closeness of 120590

119899(|120582119899|) to the unity

Theorem 5 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

(1 minus1003816100381610038161003816120582119899

1003816100381610038161003816)2

le (1 minus 120590119899)2

le 119860119873 minus 1198682

119865

le 119899 (1 minus1003816100381610038161003816120582119899

1003816100381610038161003816

2

) le 119899 (1 minus 1205902

119899)

(14)

Remark 6 Theorem 5 states that the closer the smallestsingular value 120590

119899of matrix 119860119873 is to the unity the closer

matrix 119860119873 will be to the identity that is the smaller119860119873 minus 119868

119865will be and conversely The same happens with

the smallest eigenvaluersquos modulus |120582119899| of matrix119860119873 In other

words we get a good approximate inverse 119873 of 119860 when 120590119899

(|120582119899|) is sufficiently close to 1

To finish this section let us mention that recently lowerand upper bounds on the normalized Frobenius conditionnumber of the orthogonal projection 119860119873 of the identityonto the subspace 119860119878 have been derived in [12] In additionthis work proposes a natural generalization (related to anarbitrary matrix subspace 119878 of R119899times119899) of the normalizedFrobenius condition number of the nonsingular matrix 119860

3 Geometrical Properties

In this section we present some new geometrical propertiesfor matrix 119860119873 119873 being the optimal approximate inverse ofmatrix 119860 defined by (3) Our first lemma states some basicproperties involving the cosine of the angle between matrix119860119873 and the identity that is

cos (119860119873 119868) =⟨119860119873 119868⟩

119865

119860119873119865119868119865

=tr (119860119873)

119860119873119865radic119899

(15)

Lemma 7 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

cos (119860119873 119868) =tr (119860119873)

119860119873119865radic119899

=119860119873

119865

radic119899=radictr (119860119873)

radic119899 (16)

0 le cos (119860119873 119868) le 1 (17)

119860119873 minus 1198682

119865= 119899 (1 minus cos2 (119860119873 119868)) (18)

Proof First using (15) and (4) we immediately obtain (16)As a direct consequence of (16) we derive that cos(119860119873 119868) isalways nonnegative Finally using (5) and (16) we get

119860119873 minus 1198682

119865= 119899 minus tr (119860119873) = 119899 (1 minus cos2 (119860119873 119868)) (19)

and the proof is concluded

Remark 8 In [13] the authors consider an arbitrary approxi-mate inverse119876 of matrix119860 and derive the following equality

119860119876 minus 1198682

119865= (119860119876

119865minus 119868119865)2

+ 2 (1 minus cos (119860119876 119868)) 119860119876119865119868119865

(20)

4 Journal of Applied Mathematics

that is the typical decomposition (valid in any inner productspace) of the strong convergence into the convergence ofthe norms (119860119876

119865minus 119868119865)2 and the weak convergence (1 minus

cos(119860119876 119868))119860119876119865119868119865 Note that for the special case that 119876

is the optimal approximate inverse119873 defined by (3) formula(18) has stated that the strong convergence is reduced justto the weak convergence and indeed just to the cosinecos(119860119873 119868)

Remark 9 More precisely formula (18) states that the closercos(119860119873 119868) is to the unity (ie the smaller the angle ang(119860119873 119868)

is) the smaller 119860119873 minus 119868119865will be and conversely This gives

us a new measure of the quality (in the Frobenius sense) ofthe approximate inverse 119873 of matrix 119860 by comparing theminimum residual norm 119860119873 minus 119868

119865with the cosine of the

angle between 119860119873 and the identity instead of with tr(119860119873)119860119873

119865(Lemma 1) or 120590

119899 |120582119899| (Theorem 5) So for a fixed

nonsingular matrix 119860 isin R119899times119899 and for different subspaces119878 sub R119899times119899 we have

tr (119860119873) 119899 lArrrArr 119860119873119865 radic119899 lArrrArr 120590

119899 1 lArrrArr

10038161003816100381610038161205821198991003816100381610038161003816 1

lArrrArr cos (119860119873 119868) 1 lArrrArr 119860119873 minus 119868119865 0

(21)

Obviously the optimal theoretical situation corresponds tothe case

tr (119860119873) = 119899 lArrrArr 119860119873119865= radic119899 lArrrArr 120590

119899= 1 lArrrArr 120582

119899= 1

lArrrArr cos (119860119873 119868) = 1 lArrrArr 119860119873 minus 119868119865= 0

lArrrArr 119873 = 119860minus1

lArrrArr 119860minus1

isin 119878

(22)

Remark 10 Note that the ratio between cos(119860119873 119868) andcos(119860 119868) is independent of the order 119899 of matrix 119860 Indeedassuming that tr(119860) = 0 andusing (16) we immediately obtain

cos (119860119873 119868)

cos (119860 119868)=

119860119873119865 radic119899

tr (119860) 119860119865radic119899

= 119860119873119865

119860119865

tr (119860) (23)

The following lemma compares the trace and the Frobe-nius norm of the orthogonal projection 119860119873

Lemma 11 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

tr (119860119873) le 119860119873119865lArrrArr 119860119873

119865le 1 lArrrArr tr (119860119873) le 1

lArrrArr cos (119860119873 119868) le1

radic119899

(24)

tr (119860119873) ge 119860119873119865lArrrArr 119860119873

119865ge 1 lArrrArr tr (119860119873) ge 1

lArrrArr cos (119860119873 119868) ge1

radic119899

(25)

Proof Using (4) we immediately obtain the four leftmostequivalences Using (16) we immediately obtain the tworightmost equivalences

The next lemma provides us with a relationship betweenthe Frobenius norms of the inverses of matrices119860 and its bestapproximate inverse119873 in subspace 119878

Lemma 12 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

10038171003817100381710038171003817119860minus110038171003817100381710038171003817119865

10038171003817100381710038171003817119873minus110038171003817100381710038171003817119865

ge 1 (26)

Proof Using (4) we get

119860119873119865le radic119899 = 119868

119865=10038171003817100381710038171003817(119860119873) (119860119873)

minus110038171003817100381710038171003817119865

le 119860119873119865

10038171003817100381710038171003817(119860119873)minus110038171003817100381710038171003817119865

997904rArr10038171003817100381710038171003817(119860119873)minus110038171003817100381710038171003817119865

ge 1

(27)

and hence10038171003817100381710038171003817119860minus110038171003817100381710038171003817119865

10038171003817100381710038171003817119873minus110038171003817100381710038171003817119865

ge10038171003817100381710038171003817119873minus1

119860minus110038171003817100381710038171003817119865

=10038171003817100381710038171003817(119860119873)minus110038171003817100381710038171003817119865

ge 1 (28)

and the proof is concluded

The following lemma compares the minimum residualnorm 119860119873 minus 119868

119865with the distance (with respect to the

Frobenius norm) 119860minus1 minus 119873119865between the inverse of 119860 and

the optimal approximate inverse 119873 of 119860 in any subspace119878 sub R119899times119899 First note that for any two matrices 119860 119861 isin R119899times119899

(119860 nonsingular) from the submultiplicative property of theFrobenius norm we immediately get

119860119861 minus 1198682

119865=10038171003817100381710038171003817119860 (119861 minus 119860

minus1

)10038171003817100381710038171003817

2

119865

le 1198602

119865

10038171003817100381710038171003817119861 minus 119860

minus110038171003817100381710038171003817

2

119865

(29)

However for the special case that 119861 = 119873 (the solution tothe problem (3)) we also get the following inequality

Lemma 13 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

119860119873 minus 1198682

119865le 119860

119865

10038171003817100381710038171003817119873 minus 119860

minus110038171003817100381710038171003817119865 (30)

Proof Using the Cauchy-Schwarz inequality and (5) we get10038161003816100381610038161003816⟨119860minus1

minus 119873119860119879

⟩119865

10038161003816100381610038161003816le10038171003817100381710038171003817119860minus1

minus 11987310038171003817100381710038171003817119865

1003817100381710038171003817100381711986011987910038171003817100381710038171003817119865

997904rArr10038161003816100381610038161003816tr ((119860minus1 minus 119873)119860)

10038161003816100381610038161003816le10038171003817100381710038171003817119860minus1

minus 11987310038171003817100381710038171003817119865

1003817100381710038171003817100381711986011987910038171003817100381710038171003817119865

997904rArr10038161003816100381610038161003816tr (119860 (119860

minus1

minus 119873))10038161003816100381610038161003816le10038171003817100381710038171003817119873 minus Aminus110038171003817100381710038171003817119865119860119865

997904rArr |tr (119868 minus 119860119873)| le10038171003817100381710038171003817119873 minus 119860

minus110038171003817100381710038171003817119865119860119865

997904rArr 119899 minus tr (119860119873) le10038171003817100381710038171003817119873 minus 119860

minus110038171003817100381710038171003817119865119860119865

997904rArr 119860119873 minus 1198682

119865le 119860

119865

10038171003817100381710038171003817119873 minus 119860

minus110038171003817100381710038171003817119865

(31)

Journal of Applied Mathematics 5

The following extension of the Cauchy-Schwarz inequal-ity in a real or complex inner product space (119867 ⟨sdot sdot⟩) wasobtained by Buzano [14] For all 119886 119909 119887 isin 119867 we have

|⟨119886 119909⟩ sdot ⟨119909 119887⟩| le1

2(119886 119887 + |⟨119886 119887⟩|) 119909

2

(32)

Thenext lemmaprovides uswith lower and upper boundson the inner product ⟨119860119873 119861⟩

119865 for any 119899 times 119899 real matrix 119861

Lemma 14 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then for every 119861 isin R119899times119899 we have

1003816100381610038161003816⟨119860119873 119861⟩119865

1003816100381610038161003816 le1

2(radic119899119861

119865+ |tr (119861)|) (33)

Proof Using (32) for 119886 = 119868 119909 = 119860119873 119887 = 119861 and (4) we get

1003816100381610038161003816⟨119868 119860119873⟩119865sdot ⟨119860119873 119861⟩

119865

1003816100381610038161003816 le1

2(119868119865119861119865+1003816100381610038161003816⟨119868 119861⟩119865

1003816100381610038161003816) 1198601198732

119865

997904rArr1003816100381610038161003816tr (119860119873) sdot ⟨119860119873 119861⟩

119865

1003816100381610038161003816

le1

2(radic119899119861

119865+ |tr (119861)|) 119860119873

2

119865

997904rArr1003816100381610038161003816⟨119860119873 119861⟩

119865

1003816100381610038161003816 le1

2(radic119899119861

119865+ |tr (119861)|)

(34)

The next lemma provides an upper bound on the arith-metic mean of the squares of the 1198992 terms in the orthogonalprojection 119860119873 By the way it also provides us with an upperbound on the arithmetic mean of the 119899 diagonal terms inthe orthogonal projection 119860119873 These upper bounds (validfor any matrix subspace 119878) are independent of the optimalapproximate inverse119873 and thus they are independent of thesubspace 119878 and only depend on matrix 119860

Lemma 15 Let 119860 isin R119899times119899 be nonsingular with tr(119860) = 0 andlet 119878 be a linear subspace of R119899times119899 Let119873 be the solution to theproblem (3) Then

1198601198732

119865

1198992le

1198602

119865

[tr (119860)]2

tr (119860119873)

119899le

1198991198602

119865

[tr (119860)]2

(35)

Proof Using (32) for 119886 = 119860119879 119909 = 119868 and 119887 = 119860119873 and the

Cauchy-Schwarz inequality for ⟨119860119879 119860119873⟩119865and (4) we get

10038161003816100381610038161003816⟨119860119879

119868⟩119865

sdot ⟨119868 119860119873⟩119865

10038161003816100381610038161003816

le1

2(1003817100381710038171003817100381711986011987910038171003817100381710038171003817119865119860119873

119865+10038161003816100381610038161003816⟨119860119879

119860119873⟩119865

10038161003816100381610038161003816) 1198682

119865

997904rArr |tr (119860) sdot tr (119860119873)|

le119899

2(119860119865119860119873

119865+10038161003816100381610038161003816⟨119860119879

119860119873⟩119865

10038161003816100381610038161003816)

le119899

2(119860119865119860119873

119865+1003817100381710038171003817100381711986011987910038171003817100381710038171003817119865119860119873

119865)

= 119899119860119865119860119873

119865

997904rArr |tr (119860)| 1198601198732

119865le 119899119860

119865119860119873

119865

997904rArr119860119873

119865

119899le

119860119865

|tr (119860)|

997904rArr119860119873

2

119865

1198992le

1198602

119865

[tr (119860)]2997904rArr

tr (119860119873)

119899le

1198991198602

119865

[tr (119860)]2

(36)

Remark 16 Lemma 15 has the following interpretation interms of the quality of the optimal approximate inverse119873 ofmatrix 119860 in subspace 119878 The closer the ratio 119899119860

119865| tr(119860)|

is to zero the smaller tr(119860119873) will be and thus due to (5)the larger 119860119873 minus 119868

119865will be and this happens for any matrix

subspace 119878

Remark 17 By the way from Lemma 15 we obtain thefollowing inequality for any nonsingular matrix 119860 isin R119899times119899Consider any matrix subspace 119878 st 119860minus1 isin 119878 Then119873 = 119860

minus1and using Lemma 15 we get

1198601198732

119865

1198992=1198682

119865

1198992=1

nle

1198602

119865

[tr (119860)]2

997904rArr |tr (119860)| le radic119899119860119865

(37)

4 Spectral Properties

In this section we present some new spectral propertiesfor matrix 119860119873 119873 being the optimal approximate inverseof matrix 119860 defined by (3) Mainly we focus on the casethat matrix 119860119873 is symmetric and positive definite This hasbeenmotivated by the following reasonWhen solving a largenonsymmetric linear system (1) by using Krylov methodsa possible strategy consists of searching for an adequateoptimal preconditioner 119873 such that the preconditionedmatrix119860119873 is symmetric positive definite [5]This enables oneto use the conjugate gradientmethod (CG-method) which isin general a computationally efficient method for solving thenew preconditioned system [2 15]

Our starting point is Lemma 3 which has established thatthe sets of eigenvalues and singular values of any orthogonalprojection 119860119873 satisfy

119899

sum

119894=1

1205822

119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816

2

le

119899

sum

119894=1

1205902

119894

=

119899

sum

119894=1

120582119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816 le

119899

sum

119894=1

120590119894

(38)

Let us particularize (38) for some special cases

6 Journal of Applied Mathematics

First note that if 119860119873 is normal (ie for all 1 le 119894 le 119899120590119894= |120582119894| [16]) then (38) becomes

119899

sum

119894=1

1205822

119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816

2

=

119899

sum

119894=1

1205902

119894

=

119899

sum

119894=1

120582119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816 =

119899

sum

119894=1

120590119894

(39)

In particular if 119860119873 is symmetric (120590119894= |120582119894| = plusmn120582

119894isin R) then

(38) becomes119899

sum

119894=1

1205822

119894=

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816

2

=

119899

sum

119894=1

1205902

119894

=

119899

sum

119894=1

120582119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816 =

119899

sum

119894=1

120590119894

(40)

In particular if 119860119873 is symmetric and positive definite (120590119894=

|120582119894| = 120582119894isin R+) then the equality holds in all (38) that is

119899

sum

119894=1

1205822

119894=

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816

2

=

119899

sum

119894=1

1205902

119894

=

119899

sum

119894=1

120582119894=

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816 =

119899

sum

119894=1

120590119894

(41)

The next lemma compares the traces of matrices 119860119873 and(119860119873)2

Lemma 18 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

(i) for any orthogonal projection 119860119873

tr ((119860119873)2

) le 1198601198732

119865= tr (119860119873) (42)

(ii) for any symmetric orthogonal projection 11986011987310038171003817100381710038171003817(119860119873)210038171003817100381710038171003817119865

le tr ((119860119873)2

) = 1198601198732

119865= tr (119860119873) (43)

(iii) for any symmetric positive definite orthogonal projec-tion 119860119873

10038171003817100381710038171003817(119860119873)210038171003817100381710038171003817119865

le tr ((119860119873)2

) = 1198601198732

119865= tr (119860119873) le [tr (119860119873)]

2

(44)

Proof (i) Using (38) we get119899

sum

119894=1

1205822

119894le

119899

sum

119894=1

1205902

119894=

119899

sum

119894=1

120582119894 (45)

(ii) It suffices to use the obvious fact that (119860119873)2

119865

le

1198601198732

119865and the following equalities taken from (40)

119899

sum

119894=1

1205822

119894=

119899

sum

119894=1

1205902

119894=

119899

sum

119894=1

120582119894 (46)

(iii) It suffices to use (43) and the fact that (see eg [1718]) if 119875 and 119876 are symmetric positive definite matrices thentr(119875119876) le tr(119875) tr(119876) for 119875 = 119876 = 119860119873

The rest of the paper is devoted to obtain new propertiesabout the eigenvalues of the orthogonal projection119860119873 for thespecial case that this matrix is symmetric positive definite

First let us recall that the smallest singular value and thesmallest eigenvaluersquos modulus of the orthogonal projection119860119873 are never greater than 1 (see Lemma 4) The followingtheorem establishes the dual result for the largest eigenvalueof matrix 119860119873 (symmetric positive definite)

Theorem 19 Let 119860 isin R119899times119899 be nonsingular and let 119878 be alinear subspace of R119899times119899 Let 119873 be the solution to the problem(3) Suppose thatmatrix119860119873 is symmetric and positive definiteThen the largest eigenvalue of the orthogonal projection119860119873 ofthe identity onto the subspace 119860119878 is never less than 1 That is

1205901= 1205821ge 1 (47)

Proof Using (41) we get

119899

sum

119894=1

1205822

119894=

119899

sum

119894=1

120582119894997904rArr 1205822

119899minus 120582119899=

119899minus1

sum

119894=1

(120582119894minus 1205822

119894) (48)

Now since 120582119899le 1 (Lemma 4) then 1205822

119899minus 120582119899le 0 This implies

that at least one summand in the rightmost sum in (48) mustbe less than or equal to zero Suppose that such summand isthe 119896th one (1 le 119896 le 119899 minus 1) Since 119860119873 is positive definitethen 120582

119896gt 0 and thus

120582119896minus 1205822

119896le 0 997904rArr 120582

119896le 1205822

119896997904rArr 120582

119896ge 1 997904rArr 120582

1ge 1 (49)

and the proof is concluded

InTheorem 19 the assumption that matrix119860119873 is positivedefinite is essential for assuring that |120582

1| ge 1 as the following

simple counterexample shows Moreover from Lemma 4 andTheorem 19 respectively we have that the smallest and largesteigenvalues of 119860119873 (symmetric positive definite) satisfy 120582

119899le

1 and 1205821ge 1 respectively Nothing can be asserted about

the remaining eigenvalues of the symmetric positive definitematrix 119860119873 which can be greater than equal to or less thanthe unity as the same counterexample also shows

Example 20 For 119899 = 3 let

119860119896= (

3 0 0

0 119896 0

0 0 1

) 119896 isin R (50)

let 1198683be identitymatrix of order 3 and let 119878 be the subspace of

all 3times3 scalarmatrices that is 119878 = span1198683Then the solution

119873119896to the problem (3) for subspace 119878 can be immediately

obtained by using formula (6) as follows

119873119896=tr (119860119896)

10038171003817100381710038171198601198961003817100381710038171003817

2

119865

1198683=

119896 + 4

1198962 + 101198683 (51)

and then we get

119860119896119873119896=

119896 + 4

1198962 + 10119860119896=

119896 + 4

1198962 + 10(

3 0 0

0 119896 0

0 0 1

) (52)

Journal of Applied Mathematics 7

Let us arrange the eigenvalues and singular values ofmatrix 119860

119896119873119896 as usual in nonincreasing order (as shown in

(11))On one hand for 119896 = minus2 we have

119860minus2119873minus2

=1

7(

3 0 0

0 minus2 0

0 0 1

) (53)

and then

1205901=10038161003816100381610038161205821

1003816100381610038161003816 =3

7

ge 1205902=10038161003816100381610038161205822

1003816100381610038161003816 =2

7

ge 1205903=10038161003816100381610038161205823

1003816100381610038161003816 =1

7

(54)

Hence 119860minus2119873minus2

is indefinite and 1205901= |1205821| = 37 lt 1

On the other hand for 1 lt 119896 lt 3 we have (see matrix(52))

1205901= 1205821= 3

119896 + 4

1198962 + 10

ge 1205902= 1205822= 119896

119896 + 4

1198962 + 10

ge 1205903= 1205823=

119896 + 4

1198962 + 10

(55)

and then

119896 = 2 1205901= 1205821=9

7gt 1

1205902= 1205822=6

7lt 1

1205903= 1205823=3

7lt 1

119896 =5

2 1205901= 1205821=6

5gt 1

1205902= 1205822= 1

1205903= 1205823=2

5lt 1

119896 =8

3 1205901= 1205821=90

77gt 1

1205902= 1205822=80

77gt 1

1205903= 1205823=30

77lt 1

(56)

Hence for 119860119896119873119896positive definite we have (depending on 119896)

1205822lt 1 120582

2= 1 or 120582

2gt 1

The following corollary improves the lower bound zeroon both tr(119860119873) given in (4) and cos(119860119873 119868) given in (17)

Corollary 21 Let 119860 isin R119899times119899 be nonsingular and let 119878

be a linear subspace of R119899times119899 Let 119873 be the solution to the

problem (3) Suppose thatmatrix119860119873 is symmetric and positivedefinite Then

1 le 119860119873119865le tr (119860119873) = 119860119873

2

119865le 119899 (57)

cos (119860119873 119868) ge1

radic119899 (58)

Proof Denote by sdot 2the spectral norm Using the well-

known inequality sdot 2le sdot

119865[19] Theorem 19 and (4) we

get

119860119873119865ge 119860119873

2= 1205901= 1205821ge 1

997904rArr 1 le 119860119873119865le tr (119860119873) = 119860119873

2

119865le 119899

(59)

Finally (58) follows immediately from (57) and (25)

Let usmention that an upper bound on all the eigenvaluesmoduli and on all singular values of any orthogonal projec-tion 119860119873 can be immediately obtained from (38) and (4) asfollows

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816

2

le

119899

sum

119894=1

1205902

119894= 119860119873

2

119865le 119899

997904rArr1003816100381610038161003816120582119894

1003816100381610038161003816 120590119894le radic119899 forall119894 = 1 2 119899

(60)

Our last theorem improves the upper bound given in(60) for the special case that the orthogonal projection 119860119873

is symmetric positive definite

Theorem 22 Let 119860 isin R119899times119899 be nonsingular and let 119878 be alinear subspace of R119899times119899 Let 119873 be the solution to the problem(3) Suppose thatmatrix119860119873 is symmetric and positive definiteThen all the eigenvalues of matrix 119860119873 satisfy

120590119894= 120582119894le1 + radic119899

2forall119894 = 1 2 119899 (61)

Proof First note that the assertion is obvious for the smallestsingular value since |120582

119899| le 1 for any orthogonal projection

119860119873 (Lemma 4) For any eigenvalue of 119860119873 we use the factthat 119909 minus 119909

2

le 14 for all 119909 gt 0 Then from (41) we get119899

sum

119894=1

1205822

119894=

119899

sum

119894=1

120582119894

997904rArr 1205822

1minus 1205821=

119899

sum

119894=2

(120582119894minus 1205822

119894) le

119899 minus 1

4

997904rArr 1205821le1 + radic119899

2997904rArr 120582

119894le1 + radic119899

2

forall119894 = 1 2 119899

(62)

5 Conclusion

In this paper we have considered the orthogonal projection119860119873 (in the Frobenius sense) of the identity matrix onto an

8 Journal of Applied Mathematics

arbitrary matrix subspace 119860119878 (119860 isin R119899times119899 nonsingular 119878 sub

R119899times119899) Among other geometrical properties of matrix 119860119873we have established a strong relation between the qualityof the approximation 119860119873 asymp 119868 and the cosine of the angleang(119860119873 119868) Also the distance between119860119873 and the identity hasbeen related to the ratio 119899119860

119865| tr(119860)| (which is independent

of the subspace 119878) The spectral analysis has provided lowerand upper bounds on the largest eigenvalue of the symmetricpositive definite orthogonal projections of the identity

Acknowledgments

Theauthors are grateful to the anonymous referee for valuablecomments and suggestions which have improved the earlierversion of this paperThisworkwas partially supported by theldquoMinisterio de Economıa y Competitividadrdquo (Spanish Gov-ernment) and FEDER through Grant Contract CGL2011-29396-C03-01

References

[1] O Axelsson Iterative Solution Methods Cambridge UniversityPress Cambridge Mass USA 1994

[2] Y Saad Iterative Methods for Sparse Linear Systems PWS Pub-lishing Boston Mass USA 1996

[3] S-L Wu F Chen and X-Q Niu ldquoA note on the eigenvalueanalysis of the SIMPLE preconditioning for incompressibleflowrdquo Journal of Applied Mathematics vol 2012 Article ID564132 7 pages 2012

[4] M Benzi ldquoPreconditioning techniques for large linear systemsa surveyrdquo Journal of Computational Physics vol 182 no 2 pp418ndash477 2002

[5] G Montero L Gonzalez E Florez M D Garcıa and ASuarez ldquoApproximate inverse computation using Frobenius in-ner productrdquoNumerical Linear Algebra with Applications vol 9no 3 pp 239ndash247 2002

[6] A Suarez and L Gonzalez ldquoA generalization of the Moore-Penrose inverse related to matrix subspaces of 119862119899times119898rdquo AppliedMathematics andComputation vol 216 no 2 pp 514ndash522 2010

[7] E Chow and Y Saad ldquoApproximate inverse preconditioners viasparse-sparse iterationsrdquo SIAM Journal on Scientific Computingvol 19 no 3 pp 995ndash1023 1998

[8] E Chow and Y Saad ldquoApproximate inverse techniques forblock-partitioned matricesrdquo SIAM Journal on Scientific Com-puting vol 18 no 6 pp 1657ndash1675 1997

[9] R M Holland A J Wathen and G J Shaw ldquoSparse approxi-mate inverses and target matricesrdquo SIAM Journal on ScientificComputing vol 26 no 3 pp 1000ndash1011 2005

[10] R Andreani M Raydan and P Tarazaga ldquoOn the geometricalstructure of symmetric matricesrdquo Linear Algebra and Its Appli-cations vol 438 no 3 pp 1201ndash1214 2013

[11] L Gonzalez ldquoOrthogonal projections of the identity spectralanalysis and applications to approximate inverse precondition-ingrdquo SIAM Review vol 48 no 1 pp 66ndash75 2006

[12] A Suarez and L Gonzalez ldquoNormalized Frobenius conditionnumber of the orthogonal projections of the identityrdquo Journalof Mathematical Analysis and Applications vol 400 no 2 pp510ndash516 2013

[13] J-P Chehab and M Raydan ldquoGeometrical properties of theFrobenius condition number for positive definite matricesrdquo

Linear Algebra and Its Applications vol 429 no 8-9 pp 2089ndash2097 2008

[14] M L Buzano ldquoGeneralizzazione della diseguaglianza di Cau-chy-Schwarzrdquo Rendiconti del Seminario Matematico Universitae Politecnico di Torino vol 31 pp 405ndash409 1974 (Italian)

[15] M R Hestenes and E Stiefel ldquoMethods of conjugate gradientsfor solving linear systemsrdquo Journal of Research of the NationalBureau of Standards vol 49 no 6 pp 409ndash436 1952

[16] R A Horn and C R Johnson Topics in Matrix Analysis Cam-bridge University Press Cambridge UK 1991

[17] E H Lieb andWThirring ldquoInequalities for themoments of theeigenvalues of the Schrodinger Hamiltonian and their relationto Sobolev inequalitiesrdquo in Studies in Mathematical PhysicsEssays in Honor of Valentine Bargmann E Lieb B Simon andA Wightman Eds pp 269ndash303 Princeton University PressPrinceton NJ USA 1976

[18] Z Ulukok and R Turkmen ldquoOn some matrix trace inequali-tiesrdquo Journal of Inequalities and Applications vol 2010 ArticleID 201486 8 pages 2010

[19] R A Horn andC R JohnsonMatrix Analysis CambridgeUni-versity Press Cambridge UK 1985

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 489295 6 pageshttpdxdoiorg1011552013489295

Research ArticleA Parameterized Splitting Preconditioner for GeneralizedSaddle Point Problems

Wei-Hua Luo12 and Ting-Zhu Huang1

1 School of Mathematical Sciences University of Electronic Science and Technology of China Chengdu Sichuan 611731 China2 Key Laboratory of Numerical Simulation of Sichuan Province University Neijiang Normal University Neijiang Sichuan 641112 China

Correspondence should be addressed to Ting-Zhu Huang tingzhuhuang126com

Received 4 January 2013 Revised 31 March 2013 Accepted 1 April 2013

Academic Editor P N Shivakumar

Copyright copy 2013 W-H Luo and T-Z Huang This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

By using Sherman-Morrison-Woodbury formula we introduce a preconditioner based on parameterized splitting idea forgeneralized saddle point problems which may be singular and nonsymmetric By analyzing the eigenvalues of the preconditionedmatrix we find that when 120572 is big enough it has an eigenvalue at 1 with multiplicity at least n and the remaining eigenvalues are alllocated in a unit circle centered at 1 Particularly when the preconditioner is used in general saddle point problems it guaranteeseigenvalue at 1 with the same multiplicity and the remaining eigenvalues will tend to 1 as the parameter 120572 rarr 0 Consequently thiscan lead to a good convergence when some GMRES iterative methods are used in Krylov subspace Numerical results of Stokesproblems and Oseen problems are presented to illustrate the behavior of the preconditioner

1 Introduction

In some scientific and engineering applications such as finiteelement methods for solving partial differential equations [12] and computational fluid dynamics [3 4] we often considersolutions of the generalized saddle point problems of the form

(119860 119861119879

minus119861 119862)(

119909

119910) = (

119891

119892) (1)

where 119860 isin R119899times119899 119861 isin R119898times119899 (119898 le 119899) and 119862 isin R119898times119898

are positive semidefinite 119909 119891 isin R119899 and 119910 119892 isin R119898 When119862 = 0 (1) is a general saddle point problem which is also aresearching object for many authors

It is well known that when the matrices 119860 119861 and 119862

are large and sparse the iterative methods are more efficientand attractive than direct methods assuming that (1) has agood preconditioner In recent years a lot of preconditioningtechniques have arisen for solving linear system for exampleSaad [5] and Chen [6] have comprehensively surveyed someclassical preconditioning techniques including ILU pre-conditioner triangular preconditioner SPAI preconditionermultilevel recursive Schur complements preconditioner and

sparse wavelet preconditioner Particularly many precondi-tioning methods for saddle problems have been presentedrecently such as dimensional splitting (DS) [7] relaxeddimensional factorization (RDF) [8] splitting preconditioner[9] and Hermitian and skew-Hermitian splitting precondi-tioner [10]

Among these results Cao et al [9] have used splitting ideato give a preconditioner for saddle point problems where thematrix 119860 is symmetric and positive definite and 119861 is of fullrow rank According to his preconditioner the eigenvaluesof the preconditioned matrix would tend to 1 when theparameter 119905 rarr infin Consequently just as we have seen fromthose examples of [9] preconditioner has guaranteed a goodconvergence when some iterative methods were used

In this paper being motivated by [9] we use the splittingidea to present a preconditioner for the system (1) where 119860may be nonsymmetric and singular (when rank(119861) lt 119898) Wefind that when the parameter is big enough the precondi-tioned matrix has the eigenvalue at 1 with multiplicity at least119899 and the remaining eigenvalues are all located in a unit circlecentered at 1 Particularly when the precondidtioner is usedin some general saddle point problems (namely 119862 = 0) we

2 Journal of Applied Mathematics

see that the multiplicity of the eigenvalue at 1 is also at least 119899but the remaining eigenvalues will tend to 1 as the parameter120572 rarr 0

The remainder of the paper is organized as followsIn Section 2 we present our preconditioner based on thesplitting idea and analyze the bound of eigenvalues of thepreconditioned matrix In Section 3 we use some numericalexamples to show the behavior of the new preconditionerFinally we draw some conclusions and outline our futurework in Section 4

2 A Parameterized Splitting Preconditioner

Now we consider using splitting idea with a variable param-eter to present a preconditioner for the system (1)

Firstly it is evident that when 120572 = 0 the system (1) isequivalent to

(

119860 119861119879

minus119861

120572

119862

120572

)(119909

119910) = (

119891

119892

120572

) (2)

Let

119872 = (

119860 119861119879

minus119861

120572119868

) 119873 = (

0 0

0 119868 minus119862

120572

)

119883 = (119909

119910) 119865 = (

119891

119892

120572

)

(3)

Then the coefficient matrix of (2) can be expressed by119872minus119873Multiplying both sides of system (2) from the left with matrix119872minus1 we have

(119868 minus119872minus1

119873)119883 = 119872minus1

119865 (4)

Hence we obtain a preconditioned linear system from (1)using the idea of splitting and the corresponding precondi-tioner is

119867 = (

119860 119861119879

minus119861

120572119868

)

minus1

(

119868 0

0119868

120572

) (5)

Nowwe analyze the eigenvalues of the preconditioned system(4)

Theorem 1 The preconditioned matrix 119868 minus 119872minus1

119873 has aneigenvalue at 1 with multiplicity at least 119899 The remainingeigenvalues 120582 satisfy

120582 =1199041+ 1199042

1199041+ 120572

(6)

where 1199041= 120596119879

119861119860minus1

119861119879

120596 1199042= 120596119879

119862120596 and 120596 isin C119898 satisfies

(119861119860minus1

119861119879

120572+119862

120572)120596 = 120582(119868 +

119861119860minus1

119861119879

120572)120596 120596 = 1 (7)

Proof Because

119872minus1

=(

(119860 +119861119879

119861

120572)

minus1

minus119860minus1

119861119879

(119868 +119861119860minus1

119861119879

120572)

minus1

119861

120572(119860 +

119861119879

119861

120572)

minus1

(119868 +119861119860minus1

119861119879

120572)

minus1)

(8)

we can easily get

119868 minus119872minus1

119873 =(

119868 119860minus1

119861119879

(119868 +119861119860minus1

119861119879

120572)

minus1

(119868 minus119862

120572)

0 (119868 +119861119860minus1

119861119879

120572)

minus1

(119861119860minus1

119861119879

120572+119862

120572)

)

(9)

which implies the preconditioned matrix 119868 minus 119872minus1119873 has aneigenvalue at 1 with multiplicity at least 119899

For the remaining eigenvalues let

(119868 +119861119860minus1

119861119879

120572)

minus1

(119861119860minus1

119861119879

120572+119862

120572)120596 = 120582120596 (10)

with 120596 = 1 then we have

(119861119860minus1

119861119879

120572+119862

120572)120596 = 120582(119868 +

119861119860minus1

119861119879

120572)120596 (11)

By multiplying both sides of this equality from the left with120596119879 we can get

120582 =1199041+ 1199042

1199041+ 120572

(12)

This completes the proof of Theorem 1

Remark 2 FromTheorem 1 we can get that when parameter120572 is big enough the modulus of nonnil eigenvalues 120582 will belocated in interval (0 1)

Remark 3 InTheorem 1 if the matrix 119862 = 0 then for nonnileigenvalues we have

lim120572rarr0

120582 = 1 (13)

Figures 1 2 and 3 are the eigenvalues plots of the pre-conditioned matrices obtained with our preconditioner Aswe can see in the following numerical experiments thisgood phenomenon is useful for accelerating convergence ofiterative methods in Krylov subspace

Additionally for the purpose of practically executing ourpreconditioner

119867 =(

(119860 +119861119879

119861

120572)

minus1

minus119860minus1

119861119879

120572(119868 +

119861119860minus1

119861119879

120572)

minus1

119861

120572(119860 +

119861119879

119861

120572)

minus1

1

120572(119868 +

119861119860minus1

119861119879

120572)

minus1)

(14)

Journal of Applied Mathematics 3

0 02 04 06 08 12 141

0

1

2

3

4

minus1

minus2

minus3

times10minus4

= 01 120572 = 00001

(a)

0

002

004

006

008

0 02 04 06 08 12 141minus008

minus006

minus004

minus002

= 01 120572 = norm(119862 fro)20

(b)

Figure 1 Spectrum of the preconditioned steady Oseen matrix with viscosity coefficient V = 01 32 times 32 grid (a) Q2-Q1 FEM 119862 = 0 (b)Q1-P0 FEM 119862 = 0

0

0 02 04 06 08 121

02

04

06

08

1

minus02

minus04

minus06

minus08

minus1

minus02

times10minus3

= 001 120572 = 00001

(a)

0 02 04 06 08 12 141minus02

minus015

minus01

minus005

0

005

01

015

02 = 001 120572 = norm(119862 fro)20

(b)

Figure 2 Spectrum of the preconditioned steady Oseen matrix with viscosity coefficient V = 001 32 times 32 grid (a) Q2-Q1 FEM 119862 = 0 (b)Q1-P0 FEM 119862 = 0

0

2

4

6

8

0 02 04 06 08 12 141

times10minus4

minus8

minus6

minus4

minus2

= 0001 120572 = 00001

(a)

0

002

004

006

008

minus008

minus006

minus004

minus002

0 02 04 06 08 121minus02

= 0001 120572 = norm(119862 fro)20

(b)

Figure 3 Spectrum of the preconditioned steady Oseen matrix with viscosity coefficient V = 0001 32 times 32 grid (a) Q2-Q1 FEM 119862 = 0 (b)Q1-P0 FEM 119862 = 0

4 Journal of Applied Mathematics

Table 1 Preconditioned GMRES(20) on Stokes problems withdifferent grid sizes (uniform grids)

Grid 120572 its LU time its time Total time16 times 16 00001 4 00457 00267 0072432 times 32 00001 6 03912 00813 0472564 times 64 00001 9 54472 05519 59991128 times 128 00001 14 914698 57732 972430

Table 2 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 01

Grid 120572 its LU time its time Total time16 times 16 00001 3 00479 00244 0072332 times 32 00001 3 06312 00696 0700964 times 64 00001 4 132959 03811 136770128 times 128 00001 6 1305463 37727 1343190

Table 3 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 001

Grid 120572 its LU time its time Total time16 times 16 00001 2 00442 00236 0067832 times 32 00001 3 04160 00557 0471764 times 64 00001 3 73645 02623 76268128 times 128 00001 4 1694009 31709 1725718

Table 4 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 0001

Grid 120572 its LU time its time Total time16 times 16 00001 2 00481 00206 0068732 times 32 00001 3 04661 00585 0524664 times 64 00001 3 66240 02728 68969128 times 128 00001 4 1773130 29814 1802944

Table 5 Preconditioned GMRES(20) on Stokes problems withdifferent grid sizes (uniform grids)

Grid 120572 its LU time its time Total time16 times 16 10000 5 00471 00272 0074332 times 32 10000 7 03914 00906 0482064 times 64 10000 9 56107 04145 60252128 times 128 10000 14 927154 48908 976062

Table 6 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 01

Grid 120572 its LU time its time Total time16 times 16 10000 4 00458 00223 0068132 times 32 10000 4 05748 00670 0641864 times 64 10000 5 122642 04179 126821128 times 128 10000 7 1282758 16275 1299033

Table 7 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 001

Grid 120572 its LU time its time Total time16 times 16 10000 3 00458 00224 0068232 times 32 10000 3 04309 00451 0476064 times 64 10000 4 76537 01712 78249128 times 128 10000 5 1751587 34554 1786141

Table 8 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 0001

Grid 120572 its LU time its time Total time16 times 16 10000 3 00507 00212 0071932 times 32 10000 3 04735 00449 0518464 times 64 10000 4 66482 01645 68127128 times 128 10000 4 1720516 21216 1741732

Table 9 Preconditioned GMRES(20) on Stokes problems withdifferent grid sizes (uniform grids)

Grid 120572 its LU time its time Total time16 times 16 norm(119862 fro) 21 00240 00473 0071332 times 32 norm(119862 fro) 20 00890 01497 0238764 times 64 norm(119862 fro) 20 08387 06191 14578128 times 128 norm(119862 fro) 20 68917 30866 99783

Table 10 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 01

Grid 120572 its LU time its time Total time16 times 16 norm(119862 fro)20 10 00250 00302 0055232 times 32 norm(119862 fro)20 10 00816 00832 0164864 times 64 norm(119862 fro)20 12 08466 03648 12114128 times 128 norm(119862 fro)20 14 69019 20398 89417

Table 11 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 001

Grid 120572 its LU time its time Total time16 times 16 norm(119862 fro)20 7 00238 00286 0052432 times 32 norm(119862 fro)20 7 00850 00552 0140264 times 64 norm(119862 fro)20 11 08400 03177 11577128 times 128 norm(119862 fro)20 16 69537 22306 91844

Table 12 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 0001

Grid 120572 its LU time its time Total time16 times 16 norm(119862 fro)20 5 00245 00250 0049532 times 32 norm(119862 fro)20 5 00905 00587 0149264 times 64 norm(119862 fro)20 8 12916 03200 16116128 times 128 norm(119862 fro)20 15 104399 27468 131867

Journal of Applied Mathematics 5

Table 13 Size and number of non-nil elements of the coefficient matrix generalized by using Q2-Q1 FEM in steady Stokes problems

Grid dim(119860) nnz(119860) dim(119861) nnz(119861) nnz(119860 + (1120572)119861119879119861)16 times 16 578 times 578 6178 81 times 578 2318 3690432 times 32 2178 times 2178 28418 289 times 2178 10460 19322064 times 64 8450 times 8450 122206 1089 times 8450 44314 875966128 times 128 33282 times 33282 506376 4225 times 33282 184316 3741110

we should efficiently deal with the computation of (119868 +(119861119860minus1

119861119879

120572))minus1 This can been tackled by the well-known

Sherman-Morrison-Woodbury formula

(1198601+ 11988311198602119883119879

2)minus1

= 119860minus1

1minus 119860minus1

11198831(119860minus1

2+ 119883119879

2119860minus1

11198831)minus1

119883119879

2119860minus1

1

(15)

where 1198601isin R119899times119899 and 119860

2isin R1199031times1199031 are invertible matrices

1198831isin R119899times1199031 and 119883

2isin R119899times1199031 are any matrices and 119899 119903

1are

any positive integersFrom (15) we immediately get

(119868 +119861119860minus1

119861119879

120572)

minus1

= 119868 minus 119861(120572119860 + 119861119879

119861)minus1

119861119879

(16)

In the following numerical examples we will always use (16)to compute (119868 + (119861119860minus1119861119879120572))minus1 in (14)

3 Numerical Examples

In this sectionwe give numerical experiments to illustrate thebehavior of our preconditioner The numerical experimentsare done by using MATLAB 71 The linear systems areobtained by using finite element methods in the Stokes prob-lems and steady Oseen problems and they are respectivelythe cases of

(1) 119862 = 0 which is caused by using Q2-Q1 FEM(2) 119862 = 0 which is caused by using Q1-P0 FEM

Furthermore we compare our preconditionerwith that of[9] in the case of general saddle point problems (namely 119862 =0) For the general saddle point problem [9] has presentedthe preconditioner

= (

119860 + 119905119861119879

119861 0

minus2119861119868

119905

)

minus1

(17)

with 119905 as a parameter and has proved that when 119860 issymmetric positive definite the preconditioned matrix hasan eigenvalue 1 with multiplicity at 119899 and the remainingeigenvalues satisfy

120582 =1199051205902

119894

1 + 1199051205902

119894

(18)

lim119905rarrinfin

120582 = 1 (19)

where 120590119894 119894 = 1 2 119898 are119898 positive singular values of the

matrix 119861119860minus12All these systems can be generalized by

using IFISS software package [11] (this is a freepackage that can be downloaded from the sitehttpwwwmathsmanchesteracuksimdjsifiss) We userestarted GMRES(20) as the Krylov subspace method andwe always take a zero initial guess The stopping criterion is

100381710038171003817100381711990311989610038171003817100381710038172

1003817100381710038171003817119903010038171003817100381710038172

le 10minus6

(20)

where 119903119896is the residual vector at the 119896th iteration

In the whole course of computation we always replace(119868 + (119861119860

minus1

119861119879

120572))minus1 in (14) with (16) and use the 119871119880 factor-

ization of 119860 + (119861119879

119861120572) to tackle (119860 + (119861119879

119861120572))minus1V where

V is a corresponding vector in the iteration Concretely let119860 + (119861

119879

119861120572) = 119871119880 then we complete the matrix-vectorproduct (119860 + (119861

119879

119861120572))minus1V by 119880 119871 V in MATLAB term

In the following tables the denotation norm (119862 fro) meansthe Frobenius form of the matrix 119862 The total time is the sumof LU time and iterative time and the LU time is the time tocompute LU factorization of 119860 + (119861119879119861120572)

Case 1 (for our preconditioner) 119862 = 0 (using Q2-Q1 FEM inStokes problems and steady Oseen problems with differentviscosity coefficients The results are in Tables 1 2 3 4 )

Case 1rsquo (for preconditioner of [9]) 119862 = 0 (using Q2-Q1 FEMin Stokes problems and steady Oseen problems with differentviscosity coefficients The results are in Tables 5 6 7 8)

Case 2 119862 = 0 (using Q1-P0 FEM in Stokes problems andsteady Oseen problems with different viscosity coefficientsThe results are in Tables 9 10 11 12)

From Tables 1 2 3 4 5 6 7 and 8 we can see that theseresults are in agreement with the theoretical analyses (13) and(19) respectively Additionally comparing with the results inTables 9 10 11 and 12 we find that although the iterationsused in Case 1 (either for the preconditioner of [9] or ourpreconditioner) are less than those in Case 2 the time spentby Case 1 ismuchmore than that of Case 2This is because thedensity of the coefficient matrix generalized by Q2-Q1 FEMis much larger than that generalized by Q1-P0 FEMThis canbe partly illustrated by Tables 13 and 14 and the others can beillustrated similarly

6 Journal of Applied Mathematics

Table 14 Size and number of non-nil elements of the coefficient matrix generalized by using Q1-P0 FEM in steady Stokes problems

Grid dim(119860) nnz(119860) dim(119861) nnz(119861) nnz(119860 + (1120572)119861119879119861)16 times 16 578 times 578 3826 256 times 578 1800 707632 times 32 2178 times 2178 16818 1024 times 2178 7688 3187464 times 64 8450 times 8450 70450 4096 times 8450 31752 136582128 times 128 33282 times 33282 288306 16384 times 33282 129032 567192

4 Conclusions

In this paper we have introduced a splitting preconditionerfor solving generalized saddle point systems Theoreticalanalysis showed the modulus of eigenvalues of the precon-ditioned matrix would be located in interval (0 1) when theparameter is big enough Particularly when the submatrix119862 = 0 the eigenvalues will tend to 1 as the parameter 120572 rarr 0These performances are tested by some examples and theresults are in agreement with the theoretical analysis

There are still some future works to be done how to prop-erly choose a parameter 120572 so that the preconditioned matrixhas better propertiesHow to further precondition submatrix(119868 + (119861119860

minus1

119861119879

120572))minus1

((119861119860minus1

119861119879

120572) + (119862120572)) to improve ourpreconditioner

Acknowledgments

The authors would express their great thankfulness to thereferees and the editor Professor P N Shivakumar for theirhelpful suggestions for revising this paperThe authors wouldlike to thank H C Elman A Ramage and D J Silvester fortheir free IFISS software package This research is supportedby Chinese Universities Specialized Research Fund for theDoctoral Program (20110185110020) Sichuan Province Sci ampTech Research Project (2012GZX0080) and the Fundamen-tal Research Funds for the Central Universities

References

[1] F Brezzi andM FortinMixed and Hybrid Finite ElementMeth-ods vol 15 of Springer Series in Computational MathematicsSpringer New York NY USA 1991

[2] Z Chen Q Du and J Zou ldquoFinite element methods withmatching and nonmatching meshes for Maxwell equationswith discontinuous coefficientsrdquo SIAM Journal on NumericalAnalysis vol 37 no 5 pp 1542ndash1570 2000

[3] H C Elman D J Silvester and A J Wathen Finite Elementsand Fast Iterative Solvers With Applications in IncompressibleFluid Dynamics Numerical Mathematics and Scientific Com-putation Oxford University Press New York NY USA 2005

[4] C Cuvelier A Segal and A A van Steenhoven Finite ElementMethods and Navier-Stokes Equations vol 22 of Mathematicsand its Applications D Reidel Dordrecht The Netherlands1986

[5] Y Saad Iterative Methods for Sparse Linear Systems Society forIndustrial and Applied Mathematics Philadelphia Pa USA2nd edition 2003

[6] K Chen Matrix Preconditioning Techniques and Applicationsvol 19 ofCambridgeMonographs onApplied andComputational

Mathematics Cambridge University Press Cambridge UK2005

[7] M Benzi and X-P Guo ldquoA dimensional split preconditionerfor Stokes and linearized Navier-Stokes equationsrdquo AppliedNumerical Mathematics vol 61 no 1 pp 66ndash76 2011

[8] M Benzi M Ng Q Niu and Z Wang ldquoA relaxed dimensionalfactorization preconditioner for the incompressible Navier-Stokes equationsrdquo Journal of Computational Physics vol 230no 16 pp 6185ndash6202 2011

[9] YCaoM-Q Jiang andY-L Zheng ldquoA splitting preconditionerfor saddle point problemsrdquo Numerical Linear Algebra withApplications vol 18 no 5 pp 875ndash895 2011

[10] V Simoncini and M Benzi ldquoSpectral properties of the Her-mitian and skew-Hermitian splitting preconditioner for saddlepoint problemsrdquo SIAM Journal on Matrix Analysis and Applica-tions vol 26 no 2 pp 377ndash389 2004

[11] H C Elman A Ramage and D J Silvester ldquoAlgorithm 886IFISS a Matlab toolbox for modelling incompressible flowrdquoACM Transactions on Mathematical Software vol 33 no 2article 14 2007

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 593549 11 pageshttpdxdoiorg1011552013593549

Research ArticleSolving Optimization Problems on Hermitian Matrix Functionswith Applications

Xiang Zhang and Shu-Wen Xiang

Department of Computer Science and Information Guizhou University Guiyang 550025 China

Correspondence should be addressed to Xiang Zhang zxjnsc163com

Received 17 October 2012 Accepted 20 March 2013

Academic Editor K Sivakumar

Copyright copy 2013 X Zhang and S-W Xiang This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

We consider the extremal inertias and ranks of the matrix expressions 119891(119883 119884) = 1198603minus 119861

3119883 minus (119861

3119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast where

1198603= 119860

lowast

3 119861

3 119862

3 and 119863

3are known matrices and 119884 and 119883 are the solutions to the matrix equations 119860

1119884 = 119862

1 119884119861

1= 119863

1 and

1198602119883 = 119862

2 respectively As applications we present necessary and sufficient condition for the previous matrix function 119891(119883 119884)

to be positive (negative) non-negative (positive) definite or nonsingular We also characterize the relations between the Hermitianpart of the solutions of the above-mentioned matrix equations Furthermore we establish necessary and sufficient conditions forthe solvability of the system of matrix equations 119860

1119884 = 119862

1 119884119861

1= 119863

1 119860

2119883 = 119862

2 and 119861

3119883 + (119861

3119883)

lowast

+ 1198623119884119863

3+ (119862

3119884119863

3)lowast

= 1198603

and give an expression of the general solution to the above-mentioned system when it is solvable

1 Introduction

Throughout we denote the field of complex numbers by Cthe set of all 119898 times 119899 matrices over C by C119898times119899 and the set ofall119898 times119898Hermitian matrices by C119898times119898

ℎ The symbols 119860lowast and

R(119860) stand for the conjugate transpose the column space ofa complex matrix119860 respectively 119868

119899denotes the 119899 times 119899 identity

matrix TheMoore-Penrose inverse [1] 119860dagger of 119860 is the uniquesolution119883 to the four matrix equations

(i) 119860119883119860 = 119860

(ii) 119883119860119883 = 119883

(iii) (119860119883)lowast = 119860119883

(iv) (119883119860)lowast = 119883119860

(1)

Moreover 119871119860and 119877

119860stand for the projectors 119871

119860= 119868 minus

119860dagger

119860 119877119860= 119868 minus 119860119860

dagger induced by 119860 It is well known that theeigenvalues of a Hermitian matrix 119860 isin C119899times119899 are real and theinertia of 119860 is defined to be the triplet

I119899(119860) = 119894

+(119860) 119894

minus(119860) 119894

0(119860) (2)

where 119894+(119860) 119894

minus(119860) and 119894

0(119860) stand for the numbers of

positive negative and zero eigenvalues of119860 respectivelyThe

symbols 119894+(119860) and 119894

minus(119860) are called the positive index and

the negative index of inertia respectively For two Hermitianmatrices 119860 and 119861 of the same sizes we say 119860 ge 119861 (119860 le 119861)in the Lowner partial ordering if 119860 minus 119861 is positive (negative)semidefinite The Hermitian part of 119883 is defined as 119867(119883) =119883 + 119883

lowast We will say that 119883 is Re-nnd (Re-nonnegativesemidefinite) if 119867(119883) ge 0 119883 is Re-pd (Re-positive definite)if119867(119883) gt 0 and119883 is Re-ns if119867(119883) is nonsingular

It is well known that investigation on the solvabilityconditions and the general solution to linearmatrix equationsis very active (eg [2ndash9]) In 1999 Braden [10] gave thegeneral solution to

119861119883 + (119861119883)lowast

= 119860 (3)

In 2007 Djordjevic [11] considered the explicit solution to (3)for linear bounded operators on Hilbert spaces MoreoverCao [12] investigated the general explicit solution to

119861119883119862 + (119861119883119862)lowast

= 119860 (4)

Xu et al [13] obtained the general expression of the solutionof operator equation (4) In 2012 Wang and He [14] studied

2 Journal of Applied Mathematics

some necessary and sufficient conditions for the consistenceof the matrix equation

1198601119883 + (119860

1119883)

lowast

+ 1198611119884119862

1+ (119861

1119884119862

1)lowast

= 1198641

(5)

and presented an expression of the general solution to (5)Note that (5) is a special case of the following system

1198601119884 = 119862

1 119884119861

1= 119863

1 119860

2119883 = 119862

2

1198613119883 + (119861

3119883)

lowast

+ 1198623119884119863

3+ (119862

3119884119863

3)lowast

= 1198603

(6)

To our knowledge there has been little information about (6)One goal of this paper is to give some necessary and sufficientconditions for the solvability of the system of matrix (6) andpresent an expression of the general solution to system (6)when it is solvable

In order to find necessary and sufficient conditions for thesolvability of the system of matrix equations (6) we need toconsider the extremal ranks and inertias of (10) subject to (13)and (11)

There have been many papers to discuss the extremalranks and inertias of the following Hermitian expressions

119901 (119883) = 1198603minus 119861

3119883 minus (119861

3119883)

lowast

(7)

119892 (119884) = 119860 minus 119861119884119862 minus (119861119884119862)lowast

(8)

ℎ (119883 119884) = 1198601minus 119861

1119883119861

lowast

1minus 119862

1119884119862

lowast

1 (9)

119891 (119883 119884) = 1198603minus 119861

3119883 minus (119861

3119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

(10)

Tian has contributed much in this field One of his works[15] considered the extremal ranks and inertias of (7) Heand Wang [16] derived the extremal ranks and inertias of (7)subject to119860

1119883 = 119862

1119860

2119883119861

2= 119862

2 Liu and Tian [17] studied

the extremal ranks and inertias of (8) Chu et al [18] and Liuand Tian [19] derived the extremal ranks and inertias of (9)Zhang et al [20] presented the extremal ranks and inertias of(9) where119883 and 119884 are Hermitian solutions of

1198602119883 = 119862

2 (11)

1198841198612= 119863

2 (12)

respectively He and Wang [16] derived the extremal ranksand inertias of (10)We consider the extremal ranks and iner-tias of (10) subject to (11) and

1198601119884 = 119862

1 119884119861

1= 119863

1 (13)

which is not only the generalization of the abovematrix func-tions but also can be used to investigate the solvability con-ditions for the existence of the general solution to the system(6) Moreover it can be applied to characterize the relationsbetween Hermitian part of the solutions of (11) and (13)

The remainder of this paper is organized as follows InSection 2 we consider the extremal ranks and inertias of(10) subject to (11) and (13) In Section 3 we characterize therelations between the Hermitian part of the solution to (11)and (13) In Section 4 we establish the solvability conditionsfor the existence of a solution to (6) and obtain an expressionof the general solution to (6)

2 Extremal Ranks and Inertias of HermitianMatrix Function (10) with Some Restrictions

In this section we consider formulas for the extremal ranksand inertias of (10) subject to (11) and (13) We begin with thefollowing Lemmas

Lemma 1 (see [21]) (a) Let1198601119862

1 119861

1 and119863

1be givenThen

the following statements are equivalent

(1) system (13) is consistent(2)

1198771198601119862

1= 0 119863

1119871

1198611= 0 119860

1119863

1= 119862

11198611 (14)

(3)

119903 [1198601119862

1] = 119903 (119860

1)

[119863

1

1198611

] = 119903 (1198611)

1198601119863

1= 119862

11198611

(15)

In this case the general solution can be written as

119884 = 119860dagger

1119862

1+ 119871

1198601119863

1119861dagger

1+ 119871

1198601119881119877

1198611 (16)

where 119881 is arbitrary(b) Let 119860

2and 119862

2be given Then the following statements

are equivalent

(1) equation (11) is consistent(2)

1198771198602119862

2= 0 (17)

(3)

119903 [1198602119862

2] = 119903 (119860

2) (18)

In this case the general solution can be written as

119883 = 119860dagger

119862 + 119871119860119882 (19)

where119882 is arbitrary

Lemma 2 ([22 Lemma 15 Theorem 23]) Let 119860 isin C119898times119898

ℎ

119861 isin C119898times119899 and119863 isin C119899times119899

ℎ and denote that

119872 = [119860 119861

119861lowast

0]

119873 = [0 119876

119876lowast

0]

119871 = [119860 119861

119861lowast

119863]

119866 = [119875 119872119871

119873

119871119873119872

lowast

0]

(20)

Journal of Applied Mathematics 3

Then one has the following

(a) the following equalities hold

119894plusmn(119872) = 119903 (119861) + 119894

plusmn(119877

119861119860119877

119861) (21)

119894plusmn(119873) = 119903 (119876) (22)

(b) if R(119861) sube R(119860) then 119894plusmn(119871) = 119894

plusmn(119860) + 119894

plusmn(119863 minus

119861lowast

119860dagger

119861)) Thus 119894plusmn(119871) = 119894

plusmn(119860) if and only if R(119861) sube

R(119860) and 119894plusmn(119863 minus 119861

lowast

119860dagger

119861) = 0(c)

119894plusmn(119866) = [

[

119875 119872 0

119872lowast

0 119873lowast

0 119873 0

]

]

minus 119903 (119873) (23)

Lemma 3 (see [23]) Let 119860 isin C119898times119899 119861 isin C119898times119896 and 119862 isin C119897times119899Then they satisfy the following rank equalities

(a) 119903[119860 119861] = 119903(119860) + 119903(119864119860119861) = 119903(119861) + 119903(119864

119861119860)

(b) 119903 [ 119860119862] = 119903(119860) + 119903(119862119865

119860) = 119903(119862) + 119903(119860119865

119862)

(c) 119903 [ 119860 119861

119862 0] = 119903(119861) + 119903(119862) + 119903(119864

119861119860119865

119862)

(d) 119903[119861 119860119865119862] = 119903 [

119861 119860

0 119862] minus 119903(119862)

(e) 119903 [ 119862

119864119861119860] = 119903 [

119862 0

119860 119861] minus 119903(119861)

(f) 119903 [ 119860 119861119865119863

119864119864119862 0] = 119903 [

119860 119861 0

119862 0 119864

0 119863 0

] minus 119903(119863) minus 119903(119864)

Lemma 4 (see [15]) Let119860 isin C119898times119898

ℎ 119861 isin C119898times119899119862 isin C119899times119899

ℎ119876 isin

C119898times119899 and 119875 isin C119901times119899 be given and 119879 isin C119898times119898 be nonsingularThen one has the following

(1) 119894plusmn(119879119860119879

lowast

) = 119894plusmn(119860)

(2) 119894plusmn[119860 0

0 119862] = 119894

plusmn(119860) + 119894

plusmn(119862)

(3) 119894plusmn[

0 119876

119876lowast

0] = 119903(119876)

(4) 119894plusmn[

119860 119861119871119875

119871119875119861lowast

0] + 119903(119875) = 119894

plusmn[

119860 119861 0

119861lowast

0 119875lowast

0 119875 0

]

Lemma 5 (see [22 Lemma 14]) Let 119878 be a set consisting of(square) matrices over C119898times119898 and let 119867 be a set consisting of(square) matrices overC119898times119898

ℎ ThenThen one has the following

(a) 119878 has a nonsingularmatrix if and only ifmax119883isin119878119903(119883) =

119898(b) any 119883 isin 119878 is nonsingular if and only if min

119883isin119878119903(119883) =

119898(c) 0 isin 119878 if and only ifmin

119883isin119878119903(119883) = 0

(d) 119878 = 0 if and only ifmax119883isin119878119903(119883) = 0

(e) 119867 has a matrix 119883 gt 0 (119883 lt 0) if and only ifmax

119883isin119867119894+(119883) = 119898(max

119883isin119867119894minus(119883) = 119898)

(f) any 119883 isin 119867 satisfies 119883 gt 0 (119883 lt 0) if and only ifmin

119883isin119867119894+(119883) = 119898 (min

119883isin119867119894minus(119883) = 119898)

(g) 119867 has a matrix 119883 ge 0 (119883 le 0) if and only ifmin

119883isin119867119894minus(119883) = 0 (min

119883isin119867119894+(119883) = 0)

(h) any 119883 isin 119867 satisfies 119883 ge 0 (119883 le 0) if and only ifmax

119883isin119867119894minus(119883) = 0 (max

119883isin119867119894+(119883) = 0)

Lemma 6 (see [16]) Let 119901(119883 119884) = 119860minus119861119883minus (119861119883)lowast minus119862119884119863minus(119862119884119863)

lowast where119860119861119862 and119863 are givenwith appropriate sizesand denote that

1198721= [

[

119860 119861 119862

119861lowast

0 0

119862lowast

0 0

]

]

1198722= [

[

119860 119861 119863lowast

119861lowast

0 0

119863 0 0

]

]

1198723= [

119860 119861 119862 119863lowast

119861lowast

0 0 0]

1198724= [

[

119860 119861 119862 119863lowast

119861lowast

0 0 0

119862lowast

0 0 0

]

]

1198725= [

[

119860 119861 119862 119863lowast

119861lowast

0 0 0

119863 0 0 0

]

]

(24)

Then one has the following

(1) the maximal rank of 119901(119883 119884) is

max119883isinC119899times119898 119884

119903 [119901 (119883 119884)] = min 119898 119903 (1198721) 119903 (119872

2) 119903 (119872

3)

(25)

(2) the minimal rank of 119901(119883 119884) is

min119883isinC119899times119898 119884

119903 [119901 (119883 119884)]

= 2119903 (1198723) minus 2119903 (119861)

+max 119906++ 119906

minus V

++ V

minus 119906

++ V

minus 119906

minus+ V

+

(26)

(3) the maximal inertia of 119901(119883 119884) is

max119883isinC119899times119898119884

119894plusmn[119901 (119883 119884)] = min 119894

plusmn(119872

1) 119894

plusmn(119872

2) (27)

(4) the minimal inertias of 119901(119883 119884) is

min119883isinC119899times119898119884

119894plusmn[119901 (119883 119884)] = 119903 (119872

3) minus 119903 (119861)

+max 119894plusmn(119872

1) minus 119903 (119872

4)

119894plusmn(119872

2) minus 119903 (119872

5)

(28)

where

119906plusmn= 119894

plusmn(119872

1) minus 119903 (119872

4) V

plusmn= 119894

plusmn(119872

2) minus 119903 (119872

5) (29)

Now we present the main theorem of this section

4 Journal of Applied Mathematics

Theorem7 Let1198601isin C119898times119899119862

1isin C119898times119896119861

1isin C119896times119897119863

1isin C119899times119897

1198602isin C119905times119902 119862

2isin C119905times119901119860

3isin C

119901times119901

ℎ 119861

3isin C119901times119902 119862

3isin C119901times119899

and1198633isin C119901times119899 be given and suppose that the system of matrix

equations (13) and (11) is consistent respectively Denote the setof all solutions to (13) by 119878 and (11) by 119866 Put

1198641=

[[[[[

[

1198603

1198623119863

lowast

3119862

lowast

11198613119862

lowast

2

119862lowast

30 119860

lowast

10 0

1198621119863

3119860

10 0 0

119861lowast

30 0 0 119860

lowast

2

1198622

0 0 11986020

]]]]]

]

1198642= 119903

[[[[[

[

1198603

119863lowast

3119862

3119863

11198613119862

lowast

2

1198633

0 1198611

0 0

119863lowast

1119862

lowast

3119861lowast

10 0 0

119861lowast

30 0 0 119860

lowast

2

1198622

0 0 11986020

]]]]]

]

1198643=

[[[[[

[

1198603

1198613119862

lowast

3119863

lowast

3119862

lowast

2

119861lowast

30 0 0 119860

lowast

2

119863lowast

1119862

lowast

30 0 119861

lowast

10

1198621119863

30 119860

10 0

1198622

11986020 0 0

]]]]]

]

1198644=

[[[[[[[

[

1198603

1198613119862

3119863

lowast

3119862

lowast

2119863

lowast

3119862

lowast

1

119861lowast

30 0 0 119860

lowast

20

119862lowast

30 0 0 0 119860

lowast

1

0 0 0 119861lowast

10 0

1198621119863

30 119860

10 0 0

1198622

11986020 0 0 0

]]]]]]]

]

1198645=

[[[[[[[

[

1198603

1198613119862

3119863

lowast

3119862

lowast

2119862

3119863

1

119861lowast

30 0 0 119860

lowast

20

1198633

0 0 0 0 1198611

119863lowast

1119862

lowast

30 0 119860

lowast

10 0

0 0 11986010 0 0

1198622

11986020 0 0 0

]]]]]]]

]

(30)

Then one has the following(a) the maximal rank of (10) subject to (13) and (11) is

max119883isin119866119884isin119878

119903 [119891 (119883 119884)]

= min 119901 119903 (1198641) minus 2119903 (119860

1) minus 2119903 (119860

2)

119903 (1198642) minus 2119903 (119861

1) minus 2119903 (119860

2)

119903 (1198643) minus 2119903 (119860

2) minus 119903 (119860

1) minus 119903 (119861

1)

(31)

(b) the minimal rank of (10) subject to (13) and (11) is

min119883isin119866119884isin119878

119903 [119891 (119883 119884)]

= 2119903 (1198643) minus 2119903 [

1198613

1198602

]

+max 119903 (1198641) minus 2119903 (119864

4) 119903 (119864

2) minus 2119903 (119864

5)

119894+(119864

1) + 119894

minus(119864

2) minus 119903 (119864

4) minus 119903 (119864

5)

119894minus(119864

1) + 119894

+(119864

2) minus 119903 (119864

4) minus 119903 (119864

5)

(32)

(c) the maximal inertia of (10) subject to (13) and (11) is

max119883isin119866119884isin119878119909

119894plusmn[119891 (119883 119884)] = min 119894

plusmn(119864

1) minus 119903 (119860

1) minus 119903 (119860

2)

119894plusmn(119864

2) minus 119903 (119861

1) minus 119903 (119860

2)

(33)

(d) the minimal inertia of (10) subject to (13) and (11) is

min119883isin119866119884isin119878

119894plusmn[119891 (119883 119884)] = 119903 (119864

3) minus 119903 [

1198613

1198602

]

+max 119894plusmn(119864

1) minus 119903 (119864

4)

119894plusmn(119864

2) minus 119903 (119864

5)

(34)

Proof By Lemma 1 the general solutions to (13) and (11) canbe written as

119883 = 119860dagger

2119862

2+ 119871

1198602119882

119884 = 119860dagger

1119862

1+ 119871

1198601119863

1119861dagger

1+ 119871

1198601119885119877

1198611

(35)

where119882 and 119885 are arbitrary matrices with appropriate sizesPut

119876 = 1198613119871

1198602 119879 = 119862

3119871

1198601 119869 = 119877

1198611119863

3

119875 = 1198603minus 119861

3119860

dagger

2119862

2minus (119861

3119860

dagger

2119862

2)lowast

minus 1198623(119860

dagger

1119862

1+ 119871

1198601119863

1119861dagger

1)119863

3

minus (1198623(119860

dagger

1119862

1+ 119871

1198601119863

1119861dagger

1)119863

3)lowast

(36)

Substituting (36) into (10) yields

119891 (119883 119884) = 119875 minus 119876119882 minus (119876119882)lowast

minus 119879119885119869 minus (119879119885119869)lowast

(37)

Clearly 119875 is Hermitian It follows from Lemma 6 that

max119883isin119866119884isin119878

119903 [119891 (119883 119884)]

= max119882119885

119903 (119875 minus 119876119882 minus (119876119882)lowast

minus 119879119885119869 minus (119879119885119869)lowast

)

= min 119898 119903 (1198731) 119903 (119873

2) 119903 (119873

3)

(38)

min119883isin119866119884isin119878

119903 [119891 (119883 119884)]

= max119882119885

119903 (119875 minus 119876119882 minus (119876119882)lowast

minus 119879119885119869 minus (119879119885119869)lowast

)

= 2119903 (1198733) minus 2119903 (119876)

+max 119904++ 119904

minus 119905

++ 119905

minus 119904

++ 119905

minus 119904

minus+ 119905

+

(39)

Journal of Applied Mathematics 5

max119883isin119866119884isin119878

119894plusmn[119891 (119883 119884)]

= max119882119885

119903 (119875 minus 119876119882 minus (119876119882)lowast

minus 119879119885119869 minus (119879119885119869)lowast

)

= min 119894plusmn(119873

1) 119894

plusmn(119873

2)

(40)

min119883isin119866119884isin119878

119894plusmn[119891 (119883 119884)]

= max119882119885

119903 (119875 minus 119876119882 minus (119876119882)lowast

minus 119879119885119869 minus (119879119885119869)lowast

)

= 119903 (1198733) minus 119903 (119876) +max 119904

plusmn 119905

plusmn

(41)

where

1198731= [

[

119875 119876 119879

119876lowast

0 0

119879lowast

0 0

]

]

1198732= [

[

119875 119876 119869lowast

119876lowast

0 0

119869 0 0

]

]

1198733= [

119875 119876 119879 119869lowast

119876lowast

0 0 0]

1198734= [

[

119875 119876 119879 119869lowast

119876lowast

0 0 0

119879lowast

0 0 0

]

]

1198735= [

[

119875 119876 119879 119869lowast

119876lowast

0 0 0

119869 0 0 0

]

]

119904plusmn= 119894

plusmn(119873

1) minus 119903 (119873

4) 119905

plusmn= 119894

plusmn(119873

2) minus 119903 (119873

5)

(42)

Now we simplify the ranks and inertias of block matrices in(38)ndash(41)

By Lemma 4 blockGaussian elimination and noting that

119871lowast

119878= (119868 minus 119878

dagger

119878)lowast

= 119868 minus 119878lowast

(119878lowast

)dagger

= 119877119878lowast (43)

we have the following

119903 (1198731) = 119903[

[

119875 119876 119879

119876lowast

0 0

119879lowast

0 0

]

]

= 119903

[[[[[

[

1198603

1198623119863

lowast

3119862

lowast

11198613119862

lowast

2

119862lowast

30 119860

lowast

10 0

1198621119863

3119860

10 0 0

119861lowast

30 0 0 119860

lowast

2

1198622

0 0 11986020

]]]]]

]

minus 2119903 (1198601) minus 2119903 (119860

2)

(44)

By 11986211198611= 119860

1119863

1 we obtain

119903 (1198732) = 119903

[[

[

119875 119876 119869lowast

119876lowast

0 0

119869 0 0

]]

]

= 119903

[[[[[[[[

[

1198603

119863lowast

3119862

3119863

11198613119862

lowast

2

1198633

0 1198611

0 0

119863lowast

1119862

lowast

3119861lowast

10 0 0

119861lowast

30 0 0 119860

lowast

2

1198622

0 0 11986020

]]]]]]]]

]

minus 2119903 (1198611) minus 2119903 (119860

2)

119903 (1198733) = 119903 [

119875 119876 119879 119869lowast

119876lowast

0 0 0]

= 119903

[[[[[[[[

[

1198603

1198613119862

lowast

3119863

lowast

3119862

lowast

2

119861lowast

30 0 0 119860

lowast

2

119863lowast

1119862

lowast

30 0 119861

lowast

10

1198621119863

30 119860

10 0

1198622

11986020 0 0

]]]]]]]]

]

minus 119903 (1198611) minus 2119903 (119860

2) minus 119903 (119860

1)

119903 (1198734) = 119903

[[

[

119875 119876 119879 119869lowast

119876lowast

0 0 0

119879lowast

0 0 0

]]

]

= 119903

[[[[[[[

[

1198603

1198613119862

3119863

lowast

3119862

lowast

2119863

lowast

3119862

lowast

1

119861lowast

30 0 0 119860

lowast

20

119862lowast

30 0 0 0 119860

lowast

1

0 0 0 119861lowast

10 0

1198621119863

30 119860

10 0 0

1198622

11986020 0 0 0

]]]]]]]

]

minus 119903 (1198611) minus 2119903 (119860

2) minus 2119903 (119860

1)

119903 (1198735) = 119903[

[

119875 119876 119879 119869lowast

119876lowast

0 0 0

119879lowast

0 0 0

]

]

= 119903

[[[[[[[

[

1198603

1198613119862

3119863

lowast

3119862

lowast

2119862

3119863

1

119861lowast

30 0 0 119860

lowast

20

1198633

0 0 0 0 1198611

119863lowast

1119862

lowast

30 0 119860

lowast

10 0

0 0 11986010 0 0

1198622

11986020 0 0 0

]]]]]]]

]

minus 2119903 (1198611) minus 2119903 (119860

2) minus 119903 (119860

1)

(45)

By Lemma 2 we can get the following

119894plusmn(119873

1) = 119894

plusmn

[

[

119875 119876 119879

119876lowast

0 0

119879lowast

0 0

]

]

6 Journal of Applied Mathematics

= 119894plusmn

[[[[[

[

1198603

1198623119863

lowast

3119862

lowast

11198613119862

lowast

2

119862lowast

30 119860

lowast

10 0

1198621119863

3119860

10 0 0

119861lowast

30 0 0 119860

lowast

2

1198622

0 0 11986020

]]]]]

]

minus 119903 (1198601) minus 119903 (119860

2)

(46)

119894plusmn(119873

2) = 119894

plusmn

[

[

119875 119876 119869lowast

119876lowast

0 0

119869 0 0

]

]

= 119894plusmn

[[[[[

[

1198603

119863lowast

3119862

3119863

11198613119862

lowast

2

1198633

0 1198611

0 0

119863lowast

1119862

lowast

3119861lowast

10 0 0

119861lowast

30 0 0 119860

lowast

2

1198622

0 0 11986020

]]]]]

]

minus 119903 (1198611) minus 119903 (119860

2)

(47)

Substituting (44)-(47) into (38) and (41) yields (31)ndash(34)respectively

Corollary 8 Let 1198601 119862

1 119861

1 119863

1 119860

2 119862

2 119860

3 119861

3 119862

3 119863

3

and 119864119894 (119894 = 1 2 5) be as in Theorem 7 and suppose

that the system of matrix equations (13) and (11) is consistentrespectively Denote the set of all solutions to (13) by 119878 and (11)by 119866 Then one has the following

(a) there exist 119883 isin 119866 and 119884 isin 119878 such that 1198603minus 119861

3119883 minus

(1198613119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

gt 0 if and only if

119894+(119864

1) minus 119903 (119860

1) minus 119903 (119860

2) ge 119901

119894+(119864

2) minus 119903 (119861

1) minus 119903 (119860

2) ge 119901

(48)

(b) there exist 119883 isin 119866 and 119884 isin 119878 such that 1198603minus 119861

3119883 minus

(1198613119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

lt 0 if and only if

119894minus(119864

1) minus 119903 (119860

1) minus 119903 (119860

2) ge 119901

119894minus(119864

2) minus 119903 (119861

1) minus 119903 (119860

2) ge 119901

(49)

(c) there exist 119883 isin 119866 and 119884 isin 119878 such that 1198603minus 119861

3119883 minus

(1198613119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

ge 0 if and only if

119903 (1198643) minus 119903 [

1198613

1198602

] + 119894minus(119864

1) minus 119903 (119864

4) le 0

119903 (1198643) minus 119903 [

1198613

1198602

] + 119894minus(119864

2) minus 119903 (119864

5) le 0

(50)

(d) there exist 119883 isin 119866 and 119884 isin 119878 such that 1198603minus 119861

3119883 minus

(1198613119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

le 0 if and only if

119903 (1198643) minus 119903 [

1198613

1198602

] + 119894+(119864

1) minus 119903 (119864

4) le 0

119903 (1198643) minus 119903 [

1198613

1198602

] + 119894+(119864

2) minus 119903 (119864

5) le 0

(51)

(e) 1198603minus 119861

3119883 minus (119861

3119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

gt 0 for all119883 isin 119866 and 119884 isin 119878 if and only if

119903 (1198643) minus 119903 [

1198613

1198602

] + 119894+(119864

1) minus 119903 (119864

4) = 119901

119900119903 119903 (1198643) minus 119903 [

1198613

1198602

] + 119894+(119864

2) minus 119903 (119864

5) = 119901

(52)

(f) 1198603minus 119861

3119883 minus (119861

3119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

lt 0 for all119883 isin 119866 and 119884 isin 119878 if and only if

119903 (1198643) minus 119903 [

1198613

1198602

] + 119894minus(119864

1) minus 119903 (119864

4) = 119901

119900119903 119903 (1198643) minus 119903 [

1198613

1198602

] + 119894minus(119864

2) minus 119903 (119864

5) = 119901

(53)

(g) 1198603minus 119861

3119883 minus (119861

3119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

ge 0 for all119883 isin 119866 and 119884 isin 119878 if and only if

119894minus(119864

1) minus 119903 (119860

1) minus 119903 (119860

2) le 0

119900119903 119894minus(119864

2) minus 119903 (119861

1) minus 119903 (119860

2) le 0

(54)

(h) 1198603minus 119861

3119883 minus (119861

3119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

le 0 for all119883 isin 119866 and 119884 isin 119878 if and only if

119894+(119864

1) minus 119903 (119860

1) minus 119903 (119860

2) le 0

119900119903 119894+(119864

2) minus 119903 (119861

1) minus 119903 (119860

2) le 0

(55)

(i) there exist 119883 isin 119866 and 119884 isin 119878 such that 1198603minus 119861

3119883 minus

(1198613119883)

lowast

minus1198623119884119863

3minus(119862

3119884119863

3)lowast is nonsingular if and only

if

119903 (1198641) minus 2119903 (119860

1) minus 2119903 (119860

2) ge 119901

119903 (1198642) minus 2119903 (119861

1) minus 2119903 (119860

2) ge 119901

119903 (1198643) minus 2119903 (119860

2) minus 119903 (119860

1) minus 119903 (119861

1) ge 119901

(56)

3 Relations between the Hermitian Part ofthe Solutions to (13) and (11)

Now we consider the extremal ranks and inertias of thedifference between the Hermitian part of the solutions to (13)and (11)

Theorem 9 Let 1198601isin C119898times119901 119862

1isin C119898times119901 119861

1isin C119901times119897 119863

1isin

C119901times119897 1198602isin C119905times119901 and 119862

2isin C119905times119901 be given Suppose that

the system of matrix equations (13) and (11) is consistent

Journal of Applied Mathematics 7

respectively Denote the set of all solutions to (13) by 119878 and (11)by 119866 Put

1198671=

[[[[[

[

0 119868 119862lowast

1minus119868 119862

lowast

2

119868 0 119860lowast

10 0

1198621119860

10 0 0

minus119868 0 0 0 119860lowast

2

11986220 0 119860

20

]]]]]

]

1198672= 119903

[[[[[

[

0 119868 1198631minus119868 119862

lowast

2

119868 0 1198611

0 0

119863lowast

1119861lowast

10 0 0

minus119868 0 0 0 119860lowast

2

11986220 0 119860

20

]]]]]

]

1198673=

[[[[[

[

0 minus119868 119868 119868 119862lowast

2

minus119868 0 0 0 119860lowast

2

119863lowast

10 0 119861

lowast

10

1198621

0 11986010 0

1198622119860

20 0 0

]]]]]

]

1198674=

[[[[[[[

[

0 minus119868 119868 119868 119862lowast

2119862

lowast

1

minus119868 0 0 0 119860lowast

20

119868 0 0 0 0 119860lowast

1

0 0 0 119861lowast

10 0

11986210 119860

10 0 0

1198622119860

20 0 0 0

]]]]]]]

]

1198675=

[[[[[[[

[

0 minus119868 119868 119868 119862lowast

2119863

1

minus119868 0 0 0 119860lowast

20

119868 0 0 0 0 1198611

119863lowast

10 0 119860

lowast

10 0

0 0 11986010 0 0

1198622119860

20 0 0 0

]]]]]]]

]

(57)

Then one has the following

max119883isin119866119884isin119878

119903 [(119883 + 119883lowast

) minus (119884 + 119884lowast

)]

= min 119901 119903 (1198671) minus 2119903 (119860

1) minus 2119903 (119860

2) 119903 (119867

2) minus 2119903 (119861

1)

minus2119903 (1198602) 119903 (119867

3) minus 2119903 (119860

2) minus 119903 (119860

1) minus 119903 (119861

1)

min119883isin119866119884isin119878

119903 [(119883 + 119883lowast

) minus (119884 + 119884lowast

)]

= 2119903 (1198673) minus 2119901

+max 119903 (1198671) minus 2119903 (119867

4) 119903 (119867

2) minus 2119903 (119867

5)

119894+(119867

1) + 119894

minus(119867

2) minus 119903 (119867

4) minus 119903 (119867

5)

119894minus(119867

1) + 119894

+(119867

2) minus 119903 (119867

4) minus 119903 (119867

5)

max119883isin119866119884isin119878

119894plusmn[(119883 + 119883

lowast

) minus (119884 + 119884lowast

)]

= min 119894plusmn(119867

1) minus 119903 (119860

1) minus 119903 (119860

2)

119894plusmn(119867

2) minus 119903 (119861

1) minus 119903 (119860

2)

min119883isin119866119884isin119878

119894plusmn[(119883 + 119883

lowast

) minus (119884 + 119884lowast

)]

= 119903 (1198643) minus 119901 +max 119894

plusmn(119867

1) minus 119903 (119867

4) 119894

plusmn(119867

2) minus 119903 (119867

5)

(58)

Proof By letting 1198603= 0 119861

3= minus119868 119862

3= 119868 and 119863

3= 119868 in

Theorem 7 we can get the results

Corollary 10 Let 1198601 119862

1 119861

1 119863

1 119860

2 119862

2 and 119867

119894 (119894 =

1 2 5) be as in Theorem 9 and suppose that the system ofmatrix equations (13) and (11) is consistent respectively Denotethe set of all solutions to (13) by 119878 and (11) by 119866 Then one hasthe following

(a) there exist 119883 isin 119866 and 119884 isin 119878 such that (119883 + 119883lowast

) gt

(119884 + 119884lowast

) if and only if

119894+(119867

1) minus 119903 (119860

1) minus 119903 (119860

2) ge 119901

119894+(119867

2) minus 119903 (119861

1) minus 119903 (119860

2) ge 119901

(59)

(b) there exist 119883 isin 119866 and 119884 isin 119878 such that (119883 + 119883lowast

) lt

(119884 + 119884lowast

) if and only if

119894minus(119867

1) minus 119903 (119860

1) minus 119903 (119860

2) ge 119901

119894minus(119867

2) minus 119903 (119861

1) minus 119903 (119860

2) ge 119901

(60)

(c) there exist 119883 isin 119866 and 119884 isin 119878 such that (119883 + 119883lowast

) ge

(119884 + 119884lowast

) if and only if

119903 (1198673) minus 119901 + 119894

minus(119867

1) minus 119903 (119867

4) le 0

119903 (1198673) minus 119901 + 119894

minus(119867

1) minus 119903 (119867

4) le 0

(61)

(d) there exist 119883 isin 119866 and 119884 isin 119878 such that (119883 + 119883lowast

) le

(119884 + 119884lowast

) if and only if

119903 (1198673) minus 119901 + 119894

+(119867

1) minus 119903 (119867

4) le 0

119903 (1198673) minus 119901 + 119894

+(119867

2) minus 119903 (119867

5) le 0

(62)

(e) (119883 + 119883lowast

) gt (119884 + 119884lowast

) for all 119883 isin 119866 and 119884 isin 119878 if andonly if

119903 (1198673) minus 119901 + 119894

+(119867

1) minus 119903 (119867

4) = 119901

119900119903 119903 (1198673) minus 119901 + 119894

+(119867

2) minus 119903 (119867

5) = 119901

(63)

(f) (119883 + 119883lowast

) lt (119884 + 119884lowast

) for all 119883 isin 119866 and 119884 isin 119878 if andonly if

119903 (1198673) minus 119901 + 119894

minus(119867

1) minus 119903 (119867

4) = 119901

119900119903 119903 (1198673) minus 119901 + 119894

minus(119867

2) minus 119903 (119867

5) = 119901

(64)

(g) (119883 + 119883lowast

) ge (119884 + 119884lowast

) for all 119883 isin 119866 and 119884 isin 119878 if andonly if

119894minus(119867

1) minus 119903 (119860

1) minus 119903 (119860

2) le 0

119900119903 119894minus(119867

2) minus 119903 (119861

1) minus 119903 (119860

2) le 0

(65)

8 Journal of Applied Mathematics

(h) (119883 + 119883lowast

) le (119884 + 119884lowast

) for all 119883 isin 119866 and 119884 isin 119878 if andonly if

119894+(119867

1) minus 119903 (119860

1) minus 119903 (119860

2) le 0

119900119903 119894+(119867

2) minus 119903 (119861

1) minus 119903 (119860

2) le 0

(66)

(i) there exist119883 isin 119866 and 119884 isin 119878 such that (119883 +119883lowast

) minus (119884 +

119884lowast

) is nonsingular if and only if119903 (119867

1) minus 2119903 (119860

1) minus 2119903 (119860

2) ge 119901

119903 (1198672) minus 2119903 (119861

1) minus 2119903 (119860

2) ge 119901

119903 (1198673) minus 2119903 (119860

2) minus 119903 (119860

1) minus 119903 (119861

1) ge 119901

(67)

4 The Solvability Conditions and the GeneralSolution to System (6)

We now turn our attention to (6) We in this section useTheorem 9 to give some necessary and sufficient conditionsfor the existence of a solution to (6) and present an expressionof the general solution to (6) We begin with a lemma whichis used in the latter part of this section

Lemma 11 (see [14]) Let 1198601isin C119898times1198991 119861

1isin C119898times1198992 119862

1isin

C119902times119898 and 1198641isin C119898times119898

ℎbe given Let 119860 = 119877

11986011198611 119861 = 119862

1119877

1198601

119864 = 1198771198601119864

1119877

1198601119872 = 119877

119860119861lowast119873 = 119860

lowast

119871119861 and 119878 = 119861lowast

119871119872 Then

the following statements are equivalent(1) equation (5) is consistent(2)119877

119872119877

119860119864 = 0 119877

119860119864119877

119860= 0 119871

119861119864119871

119861= 0 (68)

(3)

119903 [119864

11198611119862

lowast

1119860

1

119860lowast

10 0 0

] = 119903 [1198611119862

lowast

1119860

1] + 119903 (119860

1)

119903 [

[

11986411198611119860

1

119860lowast

10 0

119861lowast

10 0

]

]

= 2119903 [1198611119860

1]

119903 [

[

1198641119862

lowast

1119860

1

119860lowast

10 0

11986210 0

]

]

= 2119903 [119862lowast

1119860

1]

(69)

In this case the general solution of (5) can be expressed as

119884 =1

2[119860

dagger

119864119861dagger

minus 119860dagger

119861lowast

119872dagger

119864119861dagger

minus 119860dagger

119878(119861dagger

)lowast

119864119873dagger

119860lowast

119861dagger

+119860dagger

119864(119872dagger

)lowast

+ (119873dagger

)lowast

119864119861dagger

119878dagger

119878] + 1198711198601198811+ 119881

2119877

119861

+ 1198801119871

119878119871

119872+ 119877

119873119880

lowast

2119871

119872minus 119860

dagger

1198781198802119877

119873119860

lowast

119861dagger

119883 = 119860dagger

1[119864

1minus 119861

1119884119862

1minus (119861

1119884119862

1)lowast

]

minus1

2119860

dagger

1[119864

1minus 119861

1119884119862

1minus (119861

1119884119862

1)lowast

] 1198601119860

dagger

1

minus 119860dagger

1119882

1119860

lowast

1+119882

lowast

1119860

1119860

dagger

1+ 119871

1198601119882

2

(70)

where 1198801 119880

2 119881

1 119881

2119882

1 and119882

2are arbitrary matrices over

C with appropriate sizes

Now we give the main theorem of this section

Theorem 12 Let 119860119894 119862

119894 (119894 = 1 2 3) 119861

119895 and 119863

119895 (119895 = 1 3) be

given Set

119860 = 1198613119871

1198602 119861 = 119862

3119871

1198601

119862 = 1198771198611119863

3 119865 = 119877

119860119861

119866 = 119862119877119860 119872 = 119877

119865119866

lowast

119873 = 119865lowast

119871119866 119878 = 119866

lowast

119871119872

(71)

119863 = 1198603minus 119861

3119860

dagger

2119862

2minus (119861

3119860

dagger

2119862

2)lowast

minus 1198623(119860

dagger

1119862

1+ 119871

1198601119863

1119861dagger

1)119863

3

minus 119863lowast

3(119860

dagger

1119862

1+ 119871

1198601119863

1119861dagger

1)lowast

119862lowast

3

(72)

119864 = 119877119860119863119877

119860 (73)

Then the following statements are equivalent

(1) system (6) is consistent

(2) the equalities in (14) and (17) hold and

119877119872119877

119865119864 = 0 119877

119865119864119877

119865= 0 119871

119866119864119871

119866= 0 (74)

(3) the equalities in (15) and (18) hold and

119903

[[[[[

[

1198603

1198623119863

lowast

31198613119862

lowast

2

119861lowast

30 0 0 119860

lowast

2

1198621119863

3119860

10 0 0

119863lowast

1119862

lowast

30 119861

lowast

10 0

1198622

0 0 11986020

]]]]]

]

= 119903

[[[

[

1198623119863

lowast

31198613

11986010 0

0 119861lowast

10

0 0 1198602

]]]

]

+ 119903 [119860

2

1198613

]

119903

[[[[[

[

1198603

11986231198613119863

lowast

3119862

lowast

1119862

lowast

2

119862lowast

30 0 119860

lowast

10

119861lowast

30 0 0 119860

lowast

2

1198621119863

3119860

10 0 0

1198622

0 1198602

0 0

]]]]]

]

= 2119903[

[

11986231198613

11986010

0 1198602

]

]

119903

[[[[[

[

1198603

119863lowast

31198613119862

3119863

1119862

lowast

2

1198633

0 0 1198611

0

119861lowast

30 0 0 119860

lowast

2

119863lowast

1119862

lowast

3119861lowast

10 0 0

1198622

0 1198602

0 0

]]]]]

]

= 2119903[

[

119863lowast

31198613

119861lowast

10

0 1198602

]

]

(75)

In this case the general solution of system (6) can be expressedas

119883 = 119860dagger

2119862

2+ 119871

1198602119880

119884 = 119860dagger

1119862

1+ 119871

1198601119863

1119861dagger

1+ 119871

1198601119881119877

1198611

(76)

Journal of Applied Mathematics 9

where

119881 =1

2[119865

dagger

119864119866dagger

minus 119865dagger

119866lowast

119872dagger

119864119866dagger

minus 119865dagger

119878(119866dagger

)lowast

119864119873dagger

119865lowast

119866dagger

+119865dagger

119864(119872dagger

)lowast

+ (119873dagger

)lowast

119864119866dagger

119878dagger

119878] + 1198711198651198811

+ 1198812119877

119866+ 119880

1119871

119878119871

119872+ 119877

119873119880

lowast

2119871

119872minus 119865

dagger

1198781198802119877

119873119865

lowast

119866dagger

119880 = 119860dagger

[119863 minus 119861119881119862 minus (119861119881119862)lowast

]

minus1

2119860

dagger

[119863 minus 119861119881119862 minus (119861119881119862)lowast

] 119860119860dagger

minus 119860dagger

1198821119860

lowast

+119882lowast

1119860119860

dagger

+ 119871119860119882

2

(77)

where 1198801 119880

2 119881

1 119881

2119882

1 and119882

2are arbitrary matrices over

C with appropriate sizes

Proof (2)hArr (3) Applying Lemma 3 and Lemma 11 gives

119877119872119877

119865119864 = 0 lArrrArr 119903 (119877

119872119877

119865119864)

= 0 lArrrArr 119903[119863 119861 119862

lowast

119860

119860lowast

0 0 0]

= 119903 [119861 119862lowast

119860] + 119903 (119860)

lArrrArr 119903

[[[[[

[

119863 1198623119863

lowast

31198613

0

119861lowast

30 0 0 119860

lowast

2

0 11986010 0 0

0 0 119861lowast

10 0

0 0 0 11986020

]]]]]

]

= 119903

[[[

[

1198623119863

lowast

31198613

11986010 0

0 119861lowast

10

0 0 1198602

]]]

]

+ 119903 [119860

2

1198613

]

lArrrArr 119903

[[[[[

[

1198603

1198623119863

lowast

31198613119862

lowast

2

119861lowast

30 0 0 119860

lowast

2

1198621119863

3119860

10 0 0

119863lowast

1119862

lowast

30 119861

lowast

10 0

1198622

0 0 11986020

]]]]]

]

= 119903

[[[

[

1198623119863

lowast

31198613

11986010 0

0 119861lowast

10

0 0 1198602

]]]

]

+ 119903 [119860

2

1198613

]

(78)

By a similar approach we can obtain that

119877119865119864119877

119865= 0 lArrrArr 119903

[[[[[

[

1198603

11986231198613119863

lowast

3119862

lowast

1119862

lowast

2

119862lowast

30 0 119860

lowast

10

119861lowast

30 0 0 119860

lowast

2

1198621119863

3119860

10 0 0

1198622

0 1198602

0 0

]]]]]

]

= 2119903[

[

11986231198613

11986010

0 1198602

]

]

119871119866119864119871

119866= 0 lArrrArr 119903

[[[[[[[[

[

1198603

119863lowast

31198613119862

3119863

1119862

lowast

2

1198633

0 0 1198611

0

119861lowast

30 0 0 119860

lowast

2

119863lowast

1119862

lowast

3119861lowast

10 0 0

1198622

0 1198602

0 0

]]]]]]]]

]

= 2119903[[

[

119863lowast

31198613

119861lowast

10

0 1198602

]]

]

(79)

(1)hArr (2) We separate the four equations in system (6) intothree groups

1198601119884 = 119862

1 119884119861

1= 119863

1 (80)

1198602119883 = 119862

2 (81)

1198613119883 + (119861

3119883)

lowast

+ 1198623119884119863

3+ (119862

3119884119863

3)lowast

= 1198603 (82)

By Lemma 1 we obtain that system (80) is solvable if andonly if (14) (81) is consistent if and only if (17) The generalsolutions to system (80) and (81) can be expressed as (16) and(19) respectively Substituting (16) and (19) into (82) yields

119860119880 + (119860119880)lowast

+ 119861119881119862 + (119861119881119862)lowast

= 119863 (83)

Hence the system (5) is consistent if and only if (80) (81) and(83) are consistent respectively It follows fromLemma 11 that(83) is solvable if and only if

119877119872119877

119865119864 = 0 119877

119865119864119877

119865= 0 119871

119866119864119871

119866= 0 (84)

We know by Lemma 11 that the general solution of (83) canbe expressed as (77)

InTheorem 12 let 1198601and119863

1vanishThen we can obtain

the general solution to the following system

1198602119883 = 119862

2 119884119861

1= 119863

1

1198613119883 + (119861

3119883)

lowast

+ 1198623119884119863

3+ (119862

3119884119863

3)lowast

= 1198603

(85)

Corollary 13 Let 1198602 119862

2 119861

1 119863

1 119861

3 119862

3 119863

3 and 119860

3= 119860

lowast

3

be given Set

119860 = 1198613119871

1198602 119862 = 119877

1198611119863

3

119865 = 119877119860119862

3 119866 = 119862119877

119860

119872 = 119877119865119866

lowast

119873 = 119865lowast

119871119866

119878 = 119866lowast

119871119872

119863 = 1198603minus 119861

3119860

dagger

2119862

2minus (119861

3119860

dagger

2119862

2)lowast

minus 1198623119863

1119861dagger

1119863

3minus (119862

3119863

1119861dagger

1119863

3)lowast

119864 = 119877119860119863119877

119860

(86)

10 Journal of Applied Mathematics

Then the following statements are equivalent

(1) system (85) is consistent(2)

1198771198602119862

2= 0 119863

1119871

1198611= 0 119877

119872119877

119865119864 = 0 (87)

119877119865119864119877

119865= 0 119871

119866119864119871

119866= 0 (88)

(3)

119903 [1198602119862

2] = 119903 (119860

2) [

1198631

1198611

] = 119903 (1198611)

119903

[[[

[

1198603

1198623119863

lowast

31198613119862

lowast

2

119861lowast

30 0 0 119860

lowast

2

119863lowast

1119862

lowast

30 119861

lowast

10 0

1198622

0 0 11986020

]]]

]

= 119903[

[

1198623119863

lowast

31198613

0 119861lowast

10

0 0 1198602

]

]

+ 119903 [119860

2

1198613

]

119903

[[[

[

1198603119862

31198613119862

lowast

2

119862lowast

30 0 0

119861lowast

30 0 119860

lowast

2

11986220 119860

20

]]]

]

= 2119903 [119862

31198613

0 1198602

]

119903

[[[[[

[

1198603

119863lowast

31198613119862

3119863

1119862

lowast

2

1198633

0 0 1198611

0

119861lowast

30 0 0 119860

lowast

2

119863lowast

1119862

lowast

3119861lowast

10 0 0

1198622

0 1198602

0 0

]]]]]

]

= 2119903[

[

119863lowast

31198613

119861lowast

10

0 1198602

]

]

(89)

In this case the general solution of system (6) can be expressedas

119883 = 119860dagger

2119862

2+ 119871

1198602119880

119884 = 1198631119861dagger

1+ 119881119877

1198611

(90)

where

119881 =1

2[119865

dagger

119864119866dagger

minus 119865dagger

119866lowast

119872dagger

119864119866dagger

minus 119865dagger

119878(119866dagger

)lowast

119864119873dagger

119865lowast

119866dagger

+119865dagger

119864(119872dagger

)lowast

+ (119873dagger

)lowast

119864119866dagger

119878dagger

119878] + 1198711198651198811

+ 1198812119877

119866+ 119880

1119871

119878119871

119872+ 119877

119873119880

lowast

2119871

119872minus 119865

dagger

1198781198802119877

119873119865

lowast

119866dagger

119880 = 119860dagger

[119863 minus 1198623119881119862 minus (119862

3119881119862)

lowast

]

minus1

2119860

dagger

[119863 minus 1198623119881119862 minus (119862

3119881119862)

lowast

] 119860119860dagger

minus 119860dagger

1198821119860

lowast

+119882lowast

1119860119860

dagger

+ 119871119860119882

2

(91)

where 1198801 119880

2 119881

1 119881

2119882

1 and119882

2are arbitrary matrices over

C with appropriate sizes

Acknowledgments

The authors would like to thank Dr Mohamed Dr Sivaku-mar and a referee very much for their valuable suggestions

and comments which resulted in a great improvement of theoriginal paper This research was supported by the Grantsfrom the National Natural Science Foundation of China(NSFC (11161008)) and Doctoral Program Fund of Ministryof Education of PRChina (20115201110002)

References

[1] H Zhang ldquoRank equalities for Moore-Penrose inverse andDrazin inverse over quaternionrdquo Annals of Functional Analysisvol 3 no 2 pp 115ndash127 2012

[2] L Bao Y Lin and YWei ldquoA new projectionmethod for solvinglarge Sylvester equationsrdquo Applied Numerical Mathematics vol57 no 5ndash7 pp 521ndash532 2007

[3] MDehghan andMHajarian ldquoOn the generalized reflexive andanti-reflexive solutions to a system of matrix equationsrdquo LinearAlgebra and Its Applications vol 437 no 11 pp 2793ndash2812 2012

[4] F O Farid M S Moslehian Q-W Wang and Z-C Wu ldquoOnthe Hermitian solutions to a system of adjointable operatorequationsrdquo Linear Algebra and Its Applications vol 437 no 7pp 1854ndash1891 2012

[5] Z H He and Q W Wang ldquoA real quaternion matrix equationwith with applicationsrdquo Linear Multilinear Algebra vol 61 pp725ndash740 2013

[6] Q-W Wang and Z-C Wu ldquoCommon Hermitian solutionsto some operator equations on Hilbert 119862119909-modulesrdquo LinearAlgebra and Its Applications vol 432 no 12 pp 3159ndash3171 2010

[7] Q-W Wang and C-Z Dong ldquoPositive solutions to a systemof adjointable operator equations over Hilbert 119862119909-modulesrdquoLinear Algebra and Its Applications vol 433 no 7 pp 1481ndash14892010

[8] Q-W Wang J W van der Woude and H-X Chang ldquoA systemof real quaternion matrix equations with applicationsrdquo LinearAlgebra and Its Applications vol 431 no 12 pp 2291ndash2303 2009

[9] Q W Wang H-S Zhang and G-J Song ldquoA new solvable con-dition for a pair of generalized Sylvester equationsrdquo ElectronicJournal of Linear Algebra vol 18 pp 289ndash301 2009

[10] H W Braden ldquoThe equations 119860119879

119883 plusmn 119883119879

119860 = 119861rdquo SIAM Journalon Matrix Analysis and Applications vol 20 no 2 pp 295ndash3021999

[11] D S Djordjevic ldquoExplicit solution of the operator equation119860

lowast

119883 + 119883plusmn

119860 = 119861rdquo Journal of Computational and Applied Math-ematics vol 200 no 2 pp 701ndash704 2007

[12] W Cao ldquoSolvability of a quaternion matrix equationrdquo AppliedMathematics vol 17 no 4 pp 490ndash498 2002

[13] Q Xu L Sheng and Y Gu ldquoThe solutions to some operatorequationsrdquo Linear Algebra and Its Applications vol 429 no 8-9pp 1997ndash2024 2008

[14] Q-W Wang and Z-H He ldquoSome matrix equations with appli-cationsrdquo Linear and Multilinear Algebra vol 60 no 11-12 pp1327ndash1353 2012

[15] Y Tian ldquoMaximization and minimization of the rank andinertia of the Hermitian matrix expression 119860 minus 119861119883 minus (119861119883)

119909rdquoLinear Algebra and Its Applications vol 434 no 10 pp 2109ndash2139 2011

[16] Z-H He and Q-WWang ldquoSolutions to optimization problemson ranks and inertias of a matrix function with applicationsrdquoAppliedMathematics andComputation vol 219 no 6 pp 2989ndash3001 2012

[17] Y Liu andY Tian ldquoMax-min problems on the ranks and inertiasof the matrix expressions 119860 minus 119861119883119862 plusmn (119861119883119862)

lowastrdquo Journal of

Journal of Applied Mathematics 11

Optimization Theory and Applications vol 148 no 3 pp 593ndash622 2011

[18] DChu Y SHung andH JWoerdeman ldquoInertia and rank cha-racterizations of some matrix expressionsrdquo SIAM Journal onMatrix Analysis and Applications vol 31 no 3 pp 1187ndash12262009

[19] Y Liu and Y Tian ldquoA simultaneous decomposition of a mat-rix triplet with applicationsrdquoNumerical Linear Algebra with Ap-plications vol 18 no 1 pp 69ndash85 2011

[20] X Zhang Q-W Wang and X Liu ldquoInertias and ranks of someHermitian matrix functions with applicationsrdquo Central Euro-pean Journal of Mathematics vol 10 no 1 pp 329ndash351 2012

[21] Q-W Wang ldquoThe general solution to a system of real quater-nion matrix equationsrdquo Computers amp Mathematics with Appli-cations vol 49 no 5-6 pp 665ndash675 2005

[22] Y Tian ldquoEqualities and inequalities for inertias of Hermitianmatrices with applicationsrdquo Linear Algebra and Its Applicationsvol 433 no 1 pp 263ndash296 2010

[23] G Marsaglia and G P H Styan ldquoEqualities and inequalities forranks of matricesrdquo Linear and Multilinear Algebra vol 2 pp269ndash292 1974

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 230408 6 pageshttpdxdoiorg1011552013230408

Research ArticleLeft and Right Inverse Eigenpairs Problemfor 120581-Hermitian Matrices

Fan-Liang Li1 Xi-Yan Hu2 and Lei Zhang2

1 Institute of Mathematics and Physics School of Sciences Central South University of Forestry and TechnologyChangsha 410004 China

2 College of Mathematics and Econometrics Hunan University Changsha 410082 China

Correspondence should be addressed to Fan-Liang Li lfl302tomcom

Received 6 December 2012 Accepted 20 March 2013

Academic Editor Panayiotis J Psarrakos

Copyright copy 2013 Fan-Liang Li et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Left and right inverse eigenpairs problem for 120581-hermitian matrices and its optimal approximate problem are considered Based onthe special properties of 120581-hermitian matrices the equivalent problem is obtained Combining a new inner product of matricesthe necessary and sufficient conditions for the solvability of the problem and its general solutions are derived Furthermore theoptimal approximate solution and a calculation procedure to obtain the optimal approximate solution are provided

1 Introduction

Throughout this paper we use some notations as follows Let119862119899times119898 be the set of all 119899times119898 complex matrices119880119862

119899times119899119867119862119899times119899

119878119867119862119899times119899 denote the set of all 119899times119899 unitary matrices hermitian

matrices skew-hermitian matrices respectively Let 119860 119860119867and119860

+ be the conjugate conjugate transpose and theMoore-Penrose generalized inverse of 119860 respectively For 119860 119861 isin

119862119899times119898 ⟨119860 119861⟩ = re(tr(119861119867119860)) where re(tr(119861119867119860)) denotes the

real part of tr(119861119867119860) the inner product of matrices 119860 and 119861The induced matrix norm is called Frobenius norm That is119860 = ⟨119860 119860⟩

12

= (tr(119860119867119860))12

Left and right inverse eigenpairs problem is a specialinverse eigenvalue problem That is giving partial left andright eigenpairs (eigenvalue and corresponding eigenvector)(120582119894 119909119894) 119894 = 1 ℎ (120583

119895 119910119895) 119895 = 1 119897 a special matrix set

119878 finding a matrix 119860 isin 119878 such that

119860119909119894= 120582119894119909119894 119894 = 1 ℎ

119910119879

119895119860 = 120583

119895119910119879

119895 119895 = 1 119897

(1)

This problem which usually arises in perturbation analysisof matrix eigenvalues and in recursive matters has profoundapplication background [1ndash6] When the matrix set 119878 isdifferent it is easy to obtain different left and right inverse

eigenpairs problem For example we studied the left andright inverse eigenpairs problem of skew-centrosymmetricmatrices and generalized centrosymmetric matrices respec-tively [5 6] Based on the special properties of left and righteigenpairs of these matrices we derived the solvability condi-tions of the problem and its general solutions In this papercombining the special properties of 120581-hermitianmatrices anda new inner product ofmatrices we first obtain the equivalentproblem then derive the necessary and sufficient conditionsfor the solvability of the problem and its general solutions

Hill and Waters [7] introduced the following matrices

Definition 1 Let 120581 be a fixed product of disjoint transposi-tions and let119870 be the associated permutation matrix that is119870 = 119870

119867

= 119870 1198702 = 119868119899 a matrix 119860 isin 119862

119899times119899 is said to be 120581-hermitian matrices (skew 120581-hermitian matrices) if and onlyif 119886119894119895= 119886119896(119895)119896(119894)

(119886119894119895= minus119886119896(119895)119896(119894)

) 119894 119895 = 1 119899 We denote theset of 120581-hermitian matrices (skew 120581-hermitian matrices) by119870119867119862119899times119899

(119878119870119867119862119899times119899

)

FromDefinition 1 it is easy to see that hermitianmatricesand perhermitian matrices are special cases of 120581-hermitianmatrices with 119896(119894) = 119894 and 119896(119894) = 119899 minus 119894 + 1 respectivelyHermitian matrices and perhermitian matrices which areone of twelve symmetry patterns of matrices [8] are appliedin engineering statistics and so on [9 10]

2 Journal of Applied Mathematics

From Definition 1 it is also easy to prove the followingconclusions

(1) 119860 isin 119870119867119862119899times119899 if and only if 119860 = 119870119860

119867

119870(2) 119860 isin 119878119870119867119862

119899times119899 if and only if 119860 = minus119870119860119867

119870(3) If 119870 is a fixed permutation matrix then 119870119867119862

119899times119899 and119878119870119867119862

119899times119899 are the closed linear subspaces of 119862119899times119899 andsatisfy

119862119899times119899

= 119870119867119862119899times119899

⨁119878119870119867119862119899times119899

(2)

The notation 1198811oplus 1198812stands for the orthogonal direct sum of

linear subspace 1198811and 119881

2

(4) 119860 isin 119870119867119862119899times119899 if and only if there is a matrix isin

119867119862119899times119899 such that = 119870119860

(5) 119860 isin 119878119870119867119862119899times119899 if and only if there is a matrix isin

119878119867119862119899times119899 such that = 119870119860

Proof (1) From Definition 1 if 119860 = (119886119894119895) isin 119870119867119862

119899times119899 then119886119894119895= 119886119896(119895)119896(119894)

this implies119860 = 119870119860119867

119870 for119870119860119867

119870 = (119886119896(119895)119896(119894)

)(2)With the samemethod we can prove (2) So the proof

is omitted

(3) (a) For any 119860 isin 119862119899times119899 there exist 119860

1isin 119870119867119862

119899times1198991198602isin 119878119870119867119862

119899times119899 such that

119860 = 1198601+ 1198602 (3)

where 1198601= (12)(A + 119870119860

119867

119870) 1198602= (12)(119860 minus

119870119860119867

119870)(b) If there exist another 119860

1isin 119870119867119862

119899times119899 1198602

isin

119878119870119867119862119899times119899 such that

119860 = 1198601+ 1198602 (4)

(3)-(4) yields

1198601minus 1198601= minus (119860

2minus 1198602) (5)

Multiplying (5) on the left and on the right by119870 respectively and according to (1) and (2) weobtain

1198601minus 1198601= 1198602minus 1198602 (6)

Combining (5) and (6) gives1198601= 11986011198602= 1198602

(c) For any1198601isin 119870119867119862

119899times1198991198602isin 119878119870119867119862

119899times119899 we have

⟨1198601 1198602⟩ = re (tr (119860119867

21198601)) = re (tr (119870119860

119867

21198701198701198601119870))

= re (tr (minus11986011986721198601)) = minus ⟨119860

1 1198602⟩

(7)

This implies ⟨1198601 1198602⟩ = 0 Combining (a) (b)

and (c) gives (3)

(4) Let = 119870119860 if 119860 isin 119870119867119862119899times119899 then

119867

= isin 119867119862119899times119899

If 119867 = isin 119867119862119899timesn then 119860 = 119870 and 119870119860

119867

119870 = 119870119867

119870119870 =

119870 = 119860 isin 119870119867119862119899times119899

(5)With the samemethod we can prove (5) So the proofis omitted

In this paper we suppose that 119870 is a fixed permutationmatrix and assume (120582

119894 119909119894) 119894 = 1 ℎ be right eigenpairs

of 119860 (120583119895 119910119895) 119895 = 1 119897 be left eigenpairs of 119860 If we let

119883 = (1199091 119909

ℎ) isin 119862

119899timesℎ Λ = diag (1205821 120582

ℎ) isin 119862

ℎtimesℎ119884 = (119910

1 119910

119897) isin 119862119899times119897 Γ = diag(120583

1 120583

119897) isin 119862119897times119897 then the

problems studied in this paper can be described as follows

Problem 2 Giving 119883 isin 119862119899timesℎ Λ = diag(120582

1 120582

ℎ) isin 119862

ℎtimesℎ119884 isin 119862

119899times119897 Γ = diag(1205831 120583

119897) isin 119862119897times119897 find 119860 isin 119870119867119862

119899times119899 suchthat

119860119883 = 119883Λ

119884119879

119860 = Γ119884119879

(8)

Problem 3 Giving 119861 isin 119862119899times119899 find isin 119878

119864such that

10038171003817100381710038171003817119861 minus

10038171003817100381710038171003817= minforall119860isin119878119864

119861 minus 119860 (9)

where 119878119864is the solution set of Problem 2

This paper is organized as follows In Section 2 we firstobtain the equivalent problemwith the properties of119870119867119862

119899times119899

and then derive the solvability conditions of Problem 2 and itsgeneral solutionrsquos expression In Section 3 we first attest theexistence and uniqueness theorem of Problem 3 then presentthe unique approximation solution Finally we provide acalculation procedure to compute the unique approximationsolution and numerical experiment to illustrate the resultsobtained in this paper correction

2 Solvability Conditions of Problem 2

We first discuss the properties of119870119867119862119899times119899

Lemma 4 Denoting 119872 = 119870119864119870119866119864 and 119864 isin 119867119862119899times119899 one has

the following conclusions(1) If 119866 isin 119870119867119862

119899times119899 then119872 isin 119870119867119862119899times119899

(2) If 119866 isin 119878119870119867119862119899times119899 then119872 isin 119878119870119867119862

119899times119899(3) If 119866 = 119866

1+ 1198662 where 119866

1isin 119870119867119862

119899times119899 1198662isin 119878119870119867119862

119899times119899then 119872 isin 119870119867119862

119899times119899 if and only if 1198701198641198701198662119864 = 0 In

addition one has119872 = 1198701198641198701198661119864

Proof (1) 119870119872119867

119870 = 119870119864119866119867

119870119864119870119870 = 119870119864(119870119866119870)119870119864 =

119870119864119870119866119864 = 119872Hence we have119872 isin 119870119867119862

119899times119899(2) 119870119872

119867

119870 = 119870119864119866119867

119870119864119870119870 = 119870119864(minus119870119866119870)119870119864 =

minus119870119864119870119866119864 = minus119872Hence we have119872 isin 119878119870119867119862

119899times119899(3) 119872 = 119870119864119870(119866

1+ 1198662)119864 = 119870119864119870119866

1119864 + 119870119864119870119866

2119864 we

have 1198701198641198701198661119864 isin 119870119867119862

119899times119899 1198701198641198701198662119864 isin 119878119870119867119862

119899times119899 from (1)and (2) If 119872 isin 119870119867119862

119899times119899 then 119872 minus 1198701198641198701198661119864 isin 119870119867119862

119899times119899while119872minus119870119864119870119866

1119864 = 119870119864119870119866

2119864 isin 119878119870119867119862

119899times119899Therefore fromthe conclusion (3) of Definition 1 we have119870119864119870119866

2119864 = 0 that

is119872 = 1198701198641198701198661119864 On the contrary if119870119864119870119866

2119864 = 0 it is clear

that119872 = 1198701198641198701198661119864 isin 119870119867119862

119899times119899 The proof is completed

Lemma 5 Let 119860 isin 119870119867119862119899times119899 if (120582 119909) is a right eigenpair of 119860

then (120582 119870119909) is a left eigenpair of 119860

Journal of Applied Mathematics 3

Proof If (120582 119909) is a right eigenpair of 119860 then we have

119860119909 = 120582119909 (10)

From the conclusion (1) of Definition 1 it follows that

119870119860119867

119870119909 = 120582119909 (11)

This implies

(119870119909)119879

119860 = 120582(119870119909)119879

(12)

So (120582 119870119909) is a left eigenpair of 119860

From Lemma 5 without loss of the generality we mayassume that Problem 2 is as follows

119883 isin 119862119899timesℎ

Λ = diag (1205821 120582

ℎ) isin 119862ℎtimesℎ

119884 = 119870119883 isin 119862119899timesℎ

Γ = Λ isin 119862ℎtimesℎ

(13)

Combining (13) and the conclusion (4) of Definition 1 itis easy to derive the following lemma

Lemma 6 If 119883 Λ 119884 Γ are given by (13) then Problem 2 isequivalent to the following problem If 119883 Λ 119884 Γ are given by(13) find 119870119860 isin 119867119862

119899times119899 such that

119870119860119883 = 119870119883Λ (14)

Lemma 7 (see [11]) If giving119883 isin 119862119899timesℎ119861 isin 119862

119899timesℎ thenmatrixequation 119883 = 119861 has solution isin 119867119862

119899times119899 if and only if

119861 = 119861119883+

119883 119861119867

119883 = 119883119867

119861 (15)

Moreover the general solution can be expressed as

= 119861119883+

+ (119861119883+

)119867

(119868119899minus 119883119883

+

)

+ (119868119899minus 119883119883

+

) (119868119899minus 119883119883

+

) forall isin 119867119862119899times119899

(16)

Theorem 8 If119883Λ119884 Γ are given by (13) then Problem 2 hasa solution in 119870119867119862

119899times119899 if and only if

119883119867

119870119883Λ = Λ119883119867

119870119883 119883Λ = 119883Λ119883+

119883 (17)

Moreover the general solution can be expressed as

119860 = 1198600+ 119870119864119870119866119864 forall119866 isin 119870119867119862

119899times119899

(18)

where

1198600= 119883Λ119883

+

+ 119870(119883Λ119883+

)119867

119870119864 119864 = 119868119899minus 119883119883

+

(19)

Proof Necessity If there is a matrix 119860 isin 119870119867119862119899times119899 such that

(119860119883 = 119883Λ 119884119879119860 = Γ119884119879

) then from Lemma 6 there exists amatrix 119870119860 isin 119867119862

119899times119899 such that 119870119860119883 = 119870119883Λ and accordingto Lemma 7 we have

119870119883Λ = 119870119883Λ119883+

119883 (119870119883Λ)119867

119883 = 119883119867

(119870119883Λ) (20)

It is easy to see that (20) is equivalent to (17)

Sufficiency If (17) holds then (20) holds Hence matrixequation119870119860119883 = 119870119883Λ has solution119870119860 isin 119867119862

119899times119899 Moreoverthe general solution can be expressed as follows

119870119860 = 119870119883Λ119883+

+ (119870119883Λ119883+

)119867

(119868119899minus 119883119883

+

)

+ (119868119899minus 119883119883

+

) (119868119899minus 119883119883

+

) forall isin 119867119862119899times119899

(21)

Let

1198600= 119883Λ119883

+

+ 119870(119883Λ119883+

)119867

119870119864 119864 = 119868119899minus 119883119883

+

(22)

This implies 119860 = 1198600+ 119870119864119864 Combining the definition of

119870 119864 and the first equation of (17) we have

119870119860119867

0119870 = 119870(119883Λ119883

+

)119867

119870 + 119883Λ119883+

minus 119870(119883119883+

)119867

119870(119883Λ119883+

)

= 119883Λ119883+

+ 119870(119883Λ119883+

)119867

119870 minus 119870(119883Λ119883+

)119879

119870119883119883+

= 1198600

(23)

Hence1198600isin 119870119867119862

119899times119899 Combining the definition of119870 119864 (13)and (17) we have

1198600119883 = 119883Λ119883

+

119883 + 119870(119883Λ119883+

)119867

119870(119868119899minus 119883119883

+

)119883 = 119883Λ

119884119879

1198600= 119883119867

119870119883Λ119883+

+ 119883119867

119870119870(119883Λ119883+

)119867

119870119864

= Λ119883119867

119870119883119883+

+ (119870119883Λ119883+

119883)119867

(119868119899minus 119883119883

+

)

= Λ119883119867

119870119883119883+

+ Λ119883119867

119870(119868119899minus 119883119883

+

)

= Λ119883119867

119870 = Γ119884119879

(24)

Therefore 1198600is a special solution of Problem 2 Combining

the conclusion (4) of Definition 1 Lemma 4 and 119864 = 119868119899minus

119883119883+

isin 119867119862119899times119899 it is easy to prove that 119860 = 119860

0+ 119870119864119870119866119864 isin

119870119867119862119899times119899 if and only if 119866 isin 119870119867119862

119899times119899 Hence the solution setof Problem 2 can be expressed as (18)

3 An Expression of the Solution of Problem 3

From (18) it is easy to prove that the solution set 119878119864of

Problem 2 is a nonempty closed convex set if Problem 2 hasa solution in 119870119867119862

119899times119899 We claim that for any given 119861 isin 119877119899times119899

there exists a unique optimal approximation for Problem 3

Theorem9 Giving119861 isin 119862119899times119899 if the conditions of119883119884Λ Γ are

the same as those in Theorem 8 then Problem 3 has a uniquesolution isin 119878

119864 Moreover can be expressed as

= 1198600+ 119870119864119870119861

1119864 (25)

where 1198600 119864 are given by (19) and 119861

1= (12)(119861 + 119870119861

119867

119870)

Proof Denoting1198641= 119868119899minus119864 it is easy to prove thatmatrices119864

and1198641are orthogonal projectionmatrices satisfying119864119864

1= 0

It is clear that matrices 119870119864119870 and 1198701198641119870 are also orthogonal

4 Journal of Applied Mathematics

projection matrices satisfying (119870119864119870)(1198701198641119870) = 0 According

to the conclusion (3) of Definition 1 for any 119861 isin 119862119899times119899 there

exists unique

1198611isin 119870119867119862

119899times119899

1198612isin 119878119870119867119862

119899times119899 (26)

such that

119861 = 1198611+ 1198612 ⟨119861

1 1198612⟩ = 0 (27)

where

1198611=

1

2(119861 + 119870119861

119867

119870) 1198612=

1

2(119861 minus 119870119861

119867

119870) (28)

CombiningTheorem 8 for any 119860 isin 119878119864 we have

119861 minus 1198602

=1003817100381710038171003817119861 minus 119860

0minus 119870119864119870119866119864

1003817100381710038171003817

2

=10038171003817100381710038171198611 + 119861

2minus 1198600minus 119870119864119870119866119864

1003817100381710038171003817

2

=10038171003817100381710038171198611 minus 119860

0minus 119870119864119870119866119864

1003817100381710038171003817

2

+10038171003817100381710038171198612

1003817100381710038171003817

2

=1003817100381710038171003817(1198611 minus 119860

0) (119864 + 119864

1) minus 119870119864119870119866119864

1003817100381710038171003817

2

+10038171003817100381710038171198612

1003817100381710038171003817

2

=1003817100381710038171003817(1198611 minus 119860

0) 119864 minus 119870119864119870119866119864

1003817100381710038171003817

2

+1003817100381710038171003817(1198611 minus 119860

0) 1198641

1003817100381710038171003817

2

+10038171003817100381710038171198612

1003817100381710038171003817

2

=1003817100381710038171003817119870 (119864 + 119864

1)119870 (119861

1minus 1198600) 119864 minus 119870119864119870119866119864

1003817100381710038171003817

2

+1003817100381710038171003817(1198611 minus 119860

0) 1198641

1003817100381710038171003817

2

+10038171003817100381710038171198612

1003817100381710038171003817

2

=1003817100381710038171003817119870119864119870 (119861

1minus 1198600) 119864 minus 119870119864119870119866119864

1003817100381710038171003817

2

+10038171003817100381710038171198701198641119870(1198611minus 1198600) 119864

1003817100381710038171003817

2

+1003817100381710038171003817(1198611 minus 119860

0)1198641

1003817100381710038171003817

2

+10038171003817100381710038171198612

1003817100381710038171003817

2

(29)

It is easy to prove that 1198701198641198701198600119864 = 0 according to the

definitions of 1198600 119864 So we have

119861 minus 1198602

=1003817100381710038171003817119870119864119870119861

1119864 minus 119870119864119870119866119864

1003817100381710038171003817

2

+10038171003817100381710038171198701198641119870(1198611minus 1198600) 119864

1003817100381710038171003817

2

+1003817100381710038171003817(1198611 minus 119860

0)1198641

1003817100381710038171003817

2

+10038171003817100381710038171198612

1003817100381710038171003817

2

(30)

Obviously min119860isin119878119864

119861 minus 119860 is equivalent to

min119866isin119870119867119862

119899times119899

10038171003817100381710038171198701198641198701198611119864 minus 119870119864119870119866119864

1003817100381710038171003817 (31)

Since 1198641198641

= 0 (119870119864119870)(1198701198641119870) = 0 it is clear that 119866 =

1198611+ 11987011986411198701198641 for any isin 119870119867119862

119899times119899 is a solution of (31)Substituting this result to (18) we can obtain (25)

Algorithm 10 (1) Input 119883 Λ 119884 Γ according to (13) (2)Compute 119883

119867

119870119883Λ Λ119883119867

119870119883 119883Λ119883+

119883 119883Λ if (17) holdsthen continue otherwise stop (3) Compute 119860

0according to

(19) and compute 1198611according to (28) (4) According to (25)

calculate

Example 11 (119899 = 8 ℎ = 119897 = 4)

119883 =

((((

(

05661 minus02014 minus 01422119894 01446 + 02138119894 0524

minus02627 + 01875119894 05336 minus02110 minus 04370119894 minus00897 + 03467119894

minus04132 + 02409119894 00226 minus 00271119894 minus01095 + 02115119894 minus03531 minus 00642119894

minus00306 + 02109119894 minus03887 minus 00425119894 02531 + 02542119894 00094 + 02991119894

00842 minus 01778119894 minus00004 minus 03733119894 03228 minus 01113119894 01669 + 01952119894

00139 minus 03757119894 minus02363 + 03856119894 02583 + 00721119894 01841 minus 02202119894

00460 + 03276119894 minus01114 + 00654119894 minus00521 minus 02556119894 minus02351 + 03002119894

00085 minus 01079119894 00974 + 03610119894 05060 minus02901 minus 00268119894

))))

)

Λ = (

minus03967 minus 04050119894 0 0 0

0 minus03967 + 04050119894 0 0

0 0 00001 0

0 0 0 minus00001119894

)

119870 =

((((

(

0 0 0 0 0 0 0 1

0 0 0 0 0 0 1 0

0 0 0 0 0 1 0 0

0 0 0 1 0 0 0 0

0 0 0 0 1 0 0 0

0 0 1 0 0 0 0 0

0 1 0 0 0 0 0 0

1 0 0 0 0 0 0 0

))))

)

119884 = 119870119883 Γ = Λ

(32)

Journal of Applied Mathematics 5

119861 =

From the first column to the fourth column

((((

(

minus05218 + 00406119894 02267 minus 00560119894 minus01202 + 00820119894 minus00072 minus 03362119894

03909 minus 03288119894 02823 minus 02064119894 minus00438 minus 00403119894 02707 + 00547119894

02162 minus 01144119894 minus04307 + 02474119894 minus00010 minus 00412119894 02164 minus 01314119894

minus01872 minus 00599119894 minus00061 + 04698119894 03605 minus 00247119894 04251 + 01869119894

minus01227 minus 00194119894 02477 minus 00606119894 03918 + 06340119894 01226 + 00636119894

minus00893 + 04335119894 00662 + 00199119894 minus00177 minus 01412119894 04047 + 02288119894

01040 minus 02015119894 01840 + 02276119894 02681 minus 03526119894 minus05252 + 01022119894

01808 + 02669119894 02264 + 03860119894 minus01791 + 01976119894 minus00961 minus 00117119894

(33)

From the fifth column to the eighth column

minus02638 minus 04952119894 minus00863 minus 01664119894 02687 + 01958119894 minus02544 minus 01099119894

minus02741 minus 01656119894 minus00227 + 02684119894 01846 + 02456119894 minus00298 + 05163119894

minus01495 minus 03205119894 01391 + 02434119894 01942 minus 05211119894 minus03052 minus 01468119894

minus02554 + 02690119894 minus04222 minus 01080119894 02232 + 00774119894 00965 minus 00421119894

03856 minus 00619119894 01217 minus 00270119894 01106 minus 03090119894 minus01122 + 02379119894

minus01130 + 00766119894 07102 minus 00901119894 01017 + 01397119894 minus00445 + 00038119894

01216 + 00076119894 02343 minus 01772119894 05242 minus 00089119894 minus00613 + 00258119894

minus00750 minus 03581119894 00125 + 00964119894 00779 minus 01074119894 06735 minus 00266119894

))))

)

(34)

It is easy to see that matrices 119883 Λ 119884 Γ satisfy (17) Hencethere exists the unique solution for Problem 3 Using the

software ldquoMATLABrdquo we obtain the unique solution ofProblem 3

From the first column to the fourth column

((((

(

minus01983 + 00491119894 01648 + 00032119894 00002 + 01065119894 01308 + 02690119894

01071 minus 02992119894 02106 + 02381119894 minus00533 minus 03856119894 minus00946 + 00488119894

01935 minus 01724119894 minus00855 minus 00370119894 00200 minus 00665119894 minus02155 minus 00636119894

00085 minus 02373119894 minus00843 minus 01920119894 00136 + 00382119894 minus00328 minus 00000119894

minus00529 + 01703119894 01948 minus 00719119894 01266 + 01752119894 00232 minus 02351119894

00855 + 01065119894 00325 minus 02068119894 02624 + 00000119894 00136 minus 00382119894

01283 minus 01463119894 minus00467 + 00000119894 00325 + 02067119894 minus00843 + 01920119894

02498 + 00000119894 01283 + 01463119894 00855 minus 01065119894 00086 + 02373119894

(35)

From the fifth column to the eighth column

02399 minus 01019119894 01928 minus 01488119894 minus03480 minus 02574119894 01017 minus 00000119894

minus01955 + 00644119894 02925 + 01872119894 03869 + 00000119894 minus03481 + 02574119894

00074 + 00339119894 minus03132 minus 00000119894 02926 minus 01872119894 01928 + 01488119894

00232 + 02351119894 minus02154 + 00636119894 minus00946 minus 00489119894 01309 minus 02691119894

minus00545 minus 00000119894 00074 minus 00339119894 minus01955 minus 00643119894 02399 + 01019119894

01266 minus 01752119894 00200 + 00665119894 minus00533 + 03857119894 00002 minus 01065119894

01949 + 00719119894 minus00855 + 00370119894 02106 minus 02381119894 01648 minus 00032119894

minus00529 minus 01703119894 01935 + 01724119894 01071 + 02992119894 minus01983 minus 00491119894

))))

)

(36)

6 Journal of Applied Mathematics

Conflict of Interests

There is no conflict of interests between the authors

Acknowledgments

This research was supported by National Natural ScienceFoundation of China (31170532)The authors are very gratefulto the referees for their valuable comments and also thank theeditor for his helpful suggestions

References

[1] J H Wilkinson The Algebraic Eigenvalue Problem OxfordUniversity Press Oxford UK 1965

[2] A Berman and R J Plemmons Nonnegative Matrices in theMathematical Sciences vol 9 ofClassics in AppliedMathematicsSIAM Philadelphia Pa USA 1994

[3] MAravDHershkowitz VMehrmann andH Schneider ldquoTherecursive inverse eigenvalue problemrdquo SIAM Journal on MatrixAnalysis and Applications vol 22 no 2 pp 392ndash412 2000

[4] R Loewy and V Mehrmann ldquoA note on the symmetricrecursive inverse eigenvalue problemrdquo SIAM Journal on MatrixAnalysis and Applications vol 25 no 1 pp 180ndash187 2003

[5] F L Li X YHu and L Zhang ldquoLeft and right inverse eigenpairsproblem of skew-centrosymmetric matricesrdquo Applied Mathe-matics and Computation vol 177 no 1 pp 105ndash110 2006

[6] F L Li X Y Hu and L Zhang ldquoLeft and right inverseeigenpairs problem of generalized centrosymmetric matricesand its optimal approximation problemrdquo Applied Mathematicsand Computation vol 212 no 2 pp 481ndash487 2009

[7] R D Hill and S R Waters ldquoOn 120581-real and 120581-Hermitianmatricesrdquo Linear Algebra and its Applications vol 169 pp 17ndash29 1992

[8] S P Irwin ldquoMatrices with multiple symmetry propertiesapplications of centro-Hermitian and per-Hermitian matricesrdquoLinear Algebra and its Applications vol 284 no 1ndash3 pp 239ndash258 1998

[9] L C Biedenharn and J D Louck Angular Momentum inQuantum Physics vol 8 of Encyclopedia of Mathematics and itsApplications Addison-Wesley Reading Mass USA 1981

[10] WM Gibson and B R Pollard Symmetry Principles in Elemen-tary Particle Physics Cambridge University Press CambridgeUK 1976

[11] W F Trench ldquoHermitian Hermitian R-symmetric and Her-mitian R-skew symmetric Procrustes problemsrdquo Linear Algebraand its Applications vol 387 pp 83ndash98 2004

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 271978 5 pageshttpdxdoiorg1011552013271978

Research ArticleCompleting a 2 times 2 Block Matrix of Real Quaternions witha Partial Specified Inverse

Yong Lin12 and Qing-Wen Wang1

1 Department of Mathematics Shanghai University Shanghai 200444 China2 School of Mathematics and Statistics Suzhou University Suzhou 234000 China

Correspondence should be addressed to Qing-Wen Wang wqw858yahoocomcn

Received 4 December 2012 Revised 23 February 2013 Accepted 20 March 2013

Academic Editor K Sivakumar

Copyright copy 2013 Y Lin and Q-W Wang This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

This paper considers a completion problem of a nonsingular 2 times 2 block matrix over the real quaternion algebra H Let1198981 1198982 1198991 1198992be nonnegative integers119898

1+ 1198982= 1198991+ 1198992= 119899 gt 0 and 119860

12isin H1198981times1198992 119860

21isin H1198982times1198991 119860

22isin H1198982times1198992 119861

11isin H1198991times1198981

be given We determine necessary and sufficient conditions so that there exists a variant block entry matrix11986011isin H1198981times1198991 such that

119860 = (11986011 11986012

11986021 11986022) isin H119899times119899 is nonsingular and 119861

11is the upper left block of a partitioning of 119860minus1 The general expression for 119860

11is also

obtained Finally a numerical example is presented to verify the theoretical findings

1 Introduction

The problem of completing a block-partitioned matrix of aspecified type with some of its blocks given has been studiedby many authors Fiedler and Markham [1] considered thefollowing completion problem over the real number field RSuppose119898

1 1198982 1198991 1198992are nonnegative integers119898

1+1198982=

1198991+1198992= 119899 gt 0 119860

11isin R1198981times1198991 119860

12isin R1198981times1198992 119860

21isin R1198982times1198991

and 11986122isin R1198992times1198982 Determine a matrix 119860

22isin R1198982times1198992 such

that

119860 = (11986011

11986012

11986021

11986022

) (1)

is nonsingular and 11986122is the lower right block of a partition-

ing of 119860minus1 This problem has the form of

(11986011

11986012

11986021

)

minus1

= (

11986122

) (2)

and the solution and the expression for 11986022were obtained in

[1] Dai [2] considered this form of completion problemswithsymmetric and symmetric positive definite matrices over R

Some other particular forms for 2times2 block matrices overR have also been examined (see eg [3]) such as

(11986011

11986012

11986021

)

minus1

= (11986111

)

(11986011

)

minus1

= (

11986122

)

(11986011

11986022

)

minus1

= ( 11986112

11986121

)

(3)

The real quaternion matrices play a role in computerscience quantum physics and so on (eg [4ndash6]) Quaternionmatrices are receiving much attention as witnessed recently(eg [7ndash9]) Motivated by the work of [1 10] and keepingsuch applications of quaternionmatrices in view in this paperwe consider the following completion problem over the realquaternion algebra

H = 1198860+ 1198861119894 + 1198862119895 + 1198863119896 |

1198942

= 1198952

= 1198962

= 119894119895119896 = minus1 and 1198860 1198861 1198862 1198863isin R

(4)

Problem 1 Suppose 1198981 1198982 1198991 1198992are nonnegative inte-

gers 1198981+ 1198982= 1198991+ 1198992= 119899 gt 0 and 119860

12isin H1198981times1198992

2 Journal of Applied Mathematics

11986021isin H1198982times1198991 119860

22isin 1198771198982times1198992 119861

11isin H1198991times1198981 Find a matrix

11986011isin H1198981times1198991 such that

119860 = (11986011

11986012

11986021

11986022

) isin H119899times119899 (5)

is nonsingular and11986111is the upper left block of a partitioning

of 119860minus1 That is

( 119860

12

11986021

11986022

)

minus1

= (11986111

) (6)

where H119898times119899 denotes the set of all 119898 times 119899matrices over H and119860minus1 denotes the inverse matrix of 119860

Throughout over the real quaternion algebra H wedenote the identity matrix with the appropriate size by 119868 thetranspose of 119860 by 119860119879 the rank of 119860 by 119903(119860) the conjugatetranspose of119860 by119860lowast = (119860)119879 a reflexive inverse of a matrix119860over H by 119860+ which satisfies simultaneously 119860119860+119860 = 119860 and119860+

119860119860+

= 119860+Moreover119871

119860= 119868minus119860

+

119860 119877119860= 119868minus119860119860

+ where119860+ is an arbitrary but fixed reflexive inverse of 119860 Clearly 119871

119860

and119877119860are idempotent and each is a reflexive inverse of itself

R(119860) denotes the right column space of the matrix 119860The rest of this paper is organized as follows In Section 2

we establish some necessary and sufficient conditions to solveProblem 1 over H and the general expression for 119860

11is also

obtained In Section 3 we present a numerical example toillustrate the developed theory

2 Main Results

In this section we begin with the following lemmas

Lemma 1 (singular-value decomposition [9]) Let 119860 isin H119898times119899

be of rank 119903 Then there exist unitary quaternion matrices 119880 isin

H119898times119898 and 119881 isin H119899times119899 such that

119880119860119881 = (1198631199030

0 0) (7)

where 119863119903= diag(119889

1 119889

119903) and the 119889

119895rsquos are the positive

singular values of 119860

Let H119899119888denote the collection of column vectors with 119899

components of quaternions and 119860 be an 119898 times 119899 quaternionmatrix Then the solutions of 119860119909 = 0 form a subspace of H119899

119888

of dimension 119899(119860) We have the following lemma

Lemma 2 Let

(11986011

11986012

11986021

11986022

) (8)

be a partitioning of a nonsingular matrix 119860 isin H119899times119899 and let

(11986111

11986112

11986121

11986122

) (9)

be the corresponding (ie transpose) partitioning of 119860minus1 Then119899(11986011) = 119899(119861

22)

Proof It is readily seen that

(11986122

11986121

11986112

11986111

)

(11986022

11986021

11986012

11986011

)

(10)

are inverse to each other so we may suppose that 119899(11986011) lt

119899(11986122)

If 119899(11986122) = 0 necessarily 119899(119860

11) = 0 and we are finished

Let 119899(11986122) = 119888 gt 0 then there exists a matrix 119865 with 119888 right

linearly independent columns such that 11986122119865 = 0 Then

using

1198601111986112+ 1198601211986122= 0 (11)

we have

1198601111986112119865 = 0 (12)

From

1198602111986112+ 1198602211986122= 119868 (13)

we have

1198602111986112119865 = 119865 (14)

It follows that the rank 119903(11986112119865) ge 119888 In view of (12) this

implies

119899 (11986011) ge 119903 (119861

12119865) ge 119888 = 119899 (119861

22) (15)

Thus

119899 (11986011) = 119899 (119861

22) (16)

Lemma 3 (see [10]) Let 119860 isin H119898times119899 119861 isin H119901times119902 119863 isin H119898times119902 beknown and 119883 isin H119899times119901 unknown Then the matrix equation

119860119883119861 = 119863 (17)

is consistent if and only if

119860119860+

119863119861+

119861 = 119863 (18)

In that case the general solution is

119883 = 119860+

119863119861+

+ 1198711198601198841+ 1198842119877119861 (19)

where11988411198842are any matrices with compatible dimensions over

H

By Lemma 1 let the singular value decomposition of thematrix 119860

22and 119861

11in Problem 1 be

11986022= 119876(

Λ 0

0 0)119877lowast

(20)

11986111= 119880(

Σ 0

0 0)119881lowast

(21)

Journal of Applied Mathematics 3

where Λ = diag(1205821 1205822 120582

119904) is a positive diagonal matrix

120582119894= 0 (119894 = 1 119904) are the singular values of 119860

22 119904 =

119903(11986022) Σ = diag(120590

1 1205902 120590

119903) is a positive diagonal matrix

120590119894= 0 (119894 = 1 119903) are the singular values of 119861

11and 119903 =

119903(11986111)119876 = (119876

11198762) isin H1198982times1198982 119877 = (119877

11198772) isin H1198992times1198992

119880 = (11988011198802) isin H1198991times1198991 119881 = (119881

11198812) isin H1198981times1198981 are unitary

quaternion matrices where 1198761isin H1198982times119904 119877

1isin H1198992times119904 119880

1isin

H1198991times119903 and 1198811isin H1198981times119903

Theorem 4 Problem 1 has a solution if and only if thefollowing conditions are satisfied

(a) 119903 ( 1198601211986022

) = 1198992

(b) 1198992minus 119903(119860

22) = 119898

1minus 119903(11986111) that is 119899

2minus 119904 = 119898

1minus 119903

(c) R(1198602111986111) subR(119860

22)

(d) R(119860lowast

12119861lowast

11) subR(119860

lowast

22)

In that case the general solution has the form of

11986011= 119861+

11+ 11986012119877(

Λminus1

119876lowast

1119860211198801Σ 0

119867 minus(119881lowast

2119860121198772)minus1)

times 119881lowast

119861+

11+ 119884 minus 119884119861

11119861+

11

(22)

where 119867 is an arbitrary matrix in H(1198992minus119904)times119903 and 119884 is anarbitrary matrix in H1198981times1198991

Proof If there exists an 1198981times 1198991matrix 119860

11such that 119860 is

nonsingular and 11986111is the corresponding block of 119860minus1 then

(a) is satisfied From 119860119861 = 119861119860 = 119868 we have that

1198602111986111+ 1198602211986121= 0

1198611111986012+ 1198611211986022= 0

(23)

so that (c) and (d) are satisfiedBy (11) we have

119903 (11986022) + 119899 (119860

22) = 1198992 119903 (119861

11) + 119899 (119861

11) = 119898

1 (24)

From Lemma 2 Notice that ( 11986011 1198601211986021 11986022

) is the correspondingpartitioning of 119861minus1 we have

119899 (11986111) = 119899 (119860

22) (25)

implying that (b) is satisfiedConversely from (c) we know that there exists a matrix

119870 isin H1198992times1198981 such that

1198602111986111= 11986022119870 (26)

Let

11986121= minus119870 (27)

From (20) (21) and (26) we have

11986021119880(

Σ 0

0 0)119881lowast

= 119876(Λ 0

0 0)119877lowast

119870 (28)

It follows that

119876lowast

11986021119880(

Σ 0

0 0)119881lowast

119881 = 119876lowast

119876(Λ 0

0 0)119877lowast

119870119881 (29)

This implies that

(

119876lowast

1119860211198801119876lowast

1119860211198802

119876lowast

2119860211198801119876lowast

2119860211198802

)(Σ 0

0 0)

= (Λ 0

0 0)(

119877lowast

11198701198811119877lowast

11198701198812

119877lowast

21198701198811119877lowast

21198701198812

)

(30)

Comparing corresponding blocks in (30) we obtain

119876lowast

2119860211198801= 0 (31)

Let 119877lowast119870119881 = From (29) (30) we have

= (Λminus1

119876lowast

1119860211198801Σ 0

119867 11987022

)

119867 isin H(1198992minus119904)times119903 119870

22isin H(1198992minus119904)times(1198981minus119903)

(32)

In the same way from (d) we can obtain

119881lowast

1119860121198772= 0 (33)

Notice that ( 1198601211986022

) in (a) is a full column rank matrix By (20)(21) and (33) we have

(0 119876lowast

119881lowast

0)(

11986012

11986022

)119877 = (

Λ 0

0 0

119881lowast

1119860121198771119881lowast

1119860121198772

119881lowast

2119860121198771119881lowast

2119860121198772

) (34)

so that

1198992= 119903(

11986012

11986022

) = 119903((0 119876lowast

119881lowast

0)(

11986012

11986022

)119877)

= 119903(

Λ 0

0 0

119881lowast

1119860121198771119881lowast

1119860121198772

119881lowast

2119860121198771119881lowast

2119860121198772

)

= 119903 (Λ) + 119903 (119881lowast

2119860121198772)

= 119904 + 119903 (119881lowast

2119860121198772)

(35)

It follows from (b) and (35) that 1198811198792119860121198772is a full column

rank matrix so it is nonsingularFrom 119860119861 = 119868 we have the following matrix equation

1198601111986111+ 1198601211986121= 119868 (36)

that is

1198601111986111= 119868 minus 119860

1211986121 119868 isin H

1198981times1198981 (37)

4 Journal of Applied Mathematics

where 11986111 11986012

were given 11986121

= minus119870 (from (27)) ByLemma 3 the matrix equation (37) has a solution if and onlyif

(119868 minus 1198601211986121) 119861+

1111986111= 119868 minus 119860

1211986121 (38)

By (21) (27) (32) and (33) we have that (38) is equivalent to

(119868 + 11986012119870)119881(

Σminus1

0

0 0)119880lowast

119880(Σ 0

0 0)119881lowast

= 119868 + 11986012119870 (39)

We simplify the equation aboveThe left hand side reduces to(119868 + 119860

12119870)1198811119881lowast

1and so we have

119860121198701198811119881lowast

1minus 11986012119870 = 119868 minus 119881

1119881lowast

1 (40)

So

11986012119877119881lowast

1198811119881lowast

1minus 11986012119877119881lowast

= (11988111198812) (119881lowast

1

119881lowast

2

) minus 1198811119881lowast

1

(41)

This implies that

11986012119877(

119881lowast

11198811

119881lowast

21198811

)119881lowast

1minus 11986012119877(

119881lowast

1

119881lowast

2

) = 1198812119881lowast

2 (42)

so that

11986012119877(

119868

0)119881lowast

1minus 11986012119877(

119881lowast

1

119881lowast

2

) = 1198812119881lowast

2 (43)

So

minus11986012119877(

0

119881lowast

2

) = 1198812119881lowast

2 (44)

and hence

minus (119860121198771119860121198772) (Λminus1

119876lowast

1119860211198801Σ 0

119867 11987022

)(0

119881lowast

2

) = 1198812119881lowast

2

(45)

Finally we obtain

11986012119877211987022119881lowast

2= minus1198812119881lowast

2 (46)

Multiplying both sides of (46) by 119881lowast from the left consider-ing (33) and the fact that 119881lowast

2119860121198772is nonsingular we have

11987022= minus(119881

lowast

2119860121198772)minus1

(47)

From Lemma 3 (38) (47) Problem 1 has a solution and thegeneral solution is

11986011= 119861+

11+ 11986012119877(

Λminus1

119876lowast

1119860211198801Σ 0

119867 minus(119881lowast

2119860121198772)minus1)

times 119881lowast

119861+

11+ 119884 minus 119884119861

11119861+

11

(48)

where 119867 is an arbitrary matrix in H(1198992minus119904)times119903 and 119884 is anarbitrary matrix in H1198981times1198991

3 An Example

In this section we give a numerical example to illustrate thetheoretical results

Example 5 Consider Problem 1 with the parameter matricesas follows

11986012= (

2 + 1198951

2119896

minus119896 1 +1

2119895

)

11986021= (

3

2+1

2119894 minus

1

2119895 minus

1

2119896

1

2119895 +

1

2119896

3

2+1

2119894

)

11986022= (

2 119894

2119895 119896) 119861

11= (

1 119894

119895 119896)

(49)

It is easy to show that (c) (d) are satisfied and that

1198992= 119903(

11986012

11986022

) = 2

1198992minus 119903 (119860

22) = 119898

1minus 119903 (119861

11) = 0

(50)

so (a) (b) are satisfied too Therefore we have

119861+

11= (

1

2minus1

2119895

minus1

2119894 minus

1

2119896

)

11986022= 119876(

Λ 0

0 0)119877lowast

11986111= 119880(

Σ 0

0 0)119881lowast

(51)

where

119876 =1

radic2

(1 119894

119895 119896) Λ = (

2radic2 0

0 radic2)

119877 = (1 0

0 1) 119880 =

1

radic2

(1 119894

119895 119896)

Σ = (radic2 0

0 radic2) 119881 = (

1 0

0 1)

(52)

We also have

1198761=

1

radic2

(1 119894

119895 119896) 119877

1= (

1 0

0 1)

1198801=

1

radic2

(1 119894

119895 119896) 119881

1= (

1 0

0 1)

(53)

Journal of Applied Mathematics 5

By Theorem 4 for an arbitrary matrices 119884 isin H2times2 wehave

11986011= 119861+

11+ 11986012119877 (Λminus1

119876lowast

1119860211198801Σ)119881lowast

119861+

11+ 119884 minus 119884119861

11119861+

11

= (

3

2+1

4119895 +

1

4119896

3

4+1

4119894 minus

3

2119895

1

2minus 119894 +

1

4119895 minus

1

41198961

4minus3

4119894 minus

1

2119895 minus 119896

)

(54)

it follows that

119860 =((

(

3

2

+

1

4

119895 +

1

4

119896

3

4

+

1

4

119894 minus

3

2

119895 2 + 119895

1

2

119896

1

2

minus 119894 +

1

4

119895 minus

1

4

119896

1

4

minus

3

4

119894 minus

1

2

119895 minus 119896 minus119896 1 +

1

2

119895

3

2

+

1

2

119894 minus

1

2

119895 minus

1

2

119896 2 119894

1

2

119895 +

1

2

119896

3

2

+

1

2

119894 2119895 119896

))

)

119860minus1

=(

1 119894 minus1 minus1

119895 119896 0 minus1

minus1 03

4

1

2minus3

4119895

minus1 minus11

2minus 119894

1

2minus1

2119894 minus

1

2119895 minus 119896

)

(55)

The results verify the theoretical findings of Theorem 4

Acknowledgments

The authors would like to give many thanks to the refereesand Professor K C Sivakumar for their valuable suggestionsand comments which resulted in a great improvement ofthe paper This research was supported by Grants from theKey Project of Scientific Research Innovation Foundation ofShanghai Municipal Education Commission (13ZZ080) theNational Natural Science Foundation of China (11171205) theNatural Science Foundation of Shanghai (11ZR1412500) theDiscipline Project at the corresponding level of Shanghai (A13-0101-12-005) and Shanghai Leading Academic DisciplineProject (J50101)

References

[1] M Fiedler and T L Markham ldquoCompleting a matrix whencertain entries of its inverse are specifiedrdquo Linear Algebra andIts Applications vol 74 pp 225ndash237 1986

[2] H Dai ldquoCompleting a symmetric 2 times 2 block matrix and itsinverserdquo Linear Algebra and Its Applications vol 235 pp 235ndash245 1996

[3] W W Barrett M E Lundquist C R Johnson and H JWoerdeman ldquoCompleting a block diagonal matrix with apartially prescribed inverserdquoLinearAlgebra and Its Applicationsvol 223224 pp 73ndash87 1995

[4] S L Adler Quaternionic Quantum Mechanics and QuantumFields Oxford University Press New York NY USA 1994

[5] A Razon and L P Horwitz ldquoUniqueness of the scalar productin the tensor product of quaternion Hilbert modulesrdquo Journalof Mathematical Physics vol 33 no 9 pp 3098ndash3104 1992

[6] J W Shuai ldquoThe quaternion NN model the identification ofcolour imagesrdquo Acta Comput Sinica vol 18 no 5 pp 372ndash3791995 (Chinese)

[7] F Zhang ldquoOn numerical range of normal matrices of quater-nionsrdquo Journal of Mathematical and Physical Sciences vol 29no 6 pp 235ndash251 1995

[8] F Z Zhang Permanent Inequalities and Quaternion Matrices[PhD thesis] University of California Santa Barbara CalifUSA 1993

[9] F Zhang ldquoQuaternions and matrices of quaternionsrdquo LinearAlgebra and Its Applications vol 251 pp 21ndash57 1997

[10] Q-W Wang ldquoThe general solution to a system of real quater-nion matrix equationsrdquo Computers amp Mathematics with Appli-cations vol 49 no 5-6 pp 665ndash675 2005

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 178209 7 pageshttpdxdoiorg1011552013178209

Research ArticleFuzzy Approximate Solution of Positive Fully Fuzzy LinearMatrix Equations

Xiaobin Guo1 and Dequan Shang2

1 College of Mathematics and Statistics Northwest Normal University Lanzhou 730070 China2Department of Public Courses Gansu College of Traditional Chinese Medicine Lanzhou 730000 China

Correspondence should be addressed to Dequan Shang gxbglz163com

Received 22 November 2012 Accepted 19 January 2013

Academic Editor Panayiotis J Psarrakos

Copyright copy 2013 X Guo and D ShangThis is an open access article distributed under theCreativeCommonsAttribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

The fuzzy matrix equations otimesotimes = in which and are119898times119898 119899times119899 and119898times119899 nonnegative LR fuzzy numbers matricesrespectively are investigated The fuzzy matrix systems is extended into three crisp systems of linear matrix equations accordingto arithmetic operations of LR fuzzy numbers Based on pseudoinverse of matrix the fuzzy approximate solution of original fuzzysystems is obtained by solving the crisp linear matrix systems In addition the existence condition of nonnegative fuzzy solution isdiscussed Two examples are calculated to illustrate the proposed method

1 Introduction

Since many real-world engineering systems are too com-plex to be defined in precise terms imprecision is ofteninvolved in any engineering design process Fuzzy systemshave an essential role in this fuzzy modeling which canformulate uncertainty in actual environment Inmanymatrixequations some or all of the system parameters are vagueor imprecise and fuzzy mathematics is a better tool thancrisp mathematics for modeling these problems and hencesolving a fuzzy matrix equation is becomingmore importantThe concept of fuzzy numbers and arithmetic operationswith these numbers were first introduced and investigatedby Zadeh [1 2] Dubois and Prade [3] and Nahmias [4]A different approach to fuzzy numbers and the structureof fuzzy number spaces was given by Puri and Ralescu [5]Goetschel Jr and Voxman [6] and Wu and Ma [7 8]

In the past decades many researchers have studied thefuzzy linear equations such as fuzzy linear systems (FLS) dualfuzzy linear systems (DFLS) general fuzzy linear systems(GFLS) fully fuzzy linear systems (FFLS) dual fully fuzzylinear systems (DFFLS) and general dual fuzzy linear systems(GDFLS) These works were performed mainly by Friedmanet al [9 10] Allahviranloo et al [11ndash17] Abbasbandy et al

[18ndash21]Wang andZheng [22 23] andDehghan et al [24 25]The general method they applied is the fuzzy linear equationswere converted to a crisp function system of linear equationswith high order according to the embedding principlesand algebraic operations of fuzzy numbers Then the fuzzysolution of the original fuzzy linear systems was derived fromsolving the crisp function linear systemsHowever for a fuzzymatrix equationwhich always has awide use in control theoryand control engineering few works have been done in thepast In 2009 Allahviranloo et al [26] discussed the fuzzy lin-ear matrix equations (FLME) of the form 119860119861 = in whichthematrices119860 and119861 are known119898times119898 and 119899times119899 real matricesrespectively is a given 119898 times 119899 fuzzy matrix By using theparametric form of fuzzy number they derived necessaryand sufficient conditions for the existence condition of fuzzysolutions and designed a numerical procedure for calculatingthe solutions of the fuzzy matrix equations In 2011 Guoet al [27ndash29] investigated a class of fuzzy matrix equations119860 = by means of the block Gaussian elimination methodand studied the least squares solutions of the inconsistentfuzzy matrix equation 119860 = by using generalized inversesof the matrix and discussed fuzzy symmetric solutions offuzzy matrix equations 119860 = What is more there are two

2 Journal of Applied Mathematics

shortcomings in the above fuzzy systems The first is that inthese fuzzy linear systems and fuzzy matrix systems the fuzzyelements were denoted by triangular fuzzy numbers so theextended model equations always contain parameter 119903 0 le

119903 le 1 which makes their computation especially numericalimplementation inconvenient in some sense The other oneis that the weak fuzzy solution of fuzzy linear systems119860 =

does not exist sometimes see [30]To make the multiplication of fuzzy numbers easy and

handle the fully fuzzy systems Dubois and Prade [3] intro-duced the LR fuzzy number in 1978 We know that triangularfuzzy numbers and trapezoidal fuzzy numbers [31] are justspecious cases of LR fuzzy numbers In 2006 Dehghan etal [25] discussed firstly computational methods for fullyfuzzy linear systems = whose coefficient matrix andthe right-hand side vector are LR fuzzy numbers In 2012Allahviranloo et al studied LR fuzzy linear systems [32]by the linear programming with equality constraints Otadiand Mosleh considered the nonnegative fuzzy solution [33]of fully fuzzy matrix equations = by employinglinear programming with equality constraints at the sameyear Recently Guo and Shang [34] investigated the fuzzyapproximate solution of LR fuzzy Sylvester matrix equations119860 + 119861 = In this paper we propose a general model forsolving the fuzzy linearmatrix equation otimesotimes = where and are119898times119898 119899times119899 and119898times119899 nonnegative LR fuzzynumbersmatrices respectivelyThemodel is proposed in thisway that is we extend the fuzzy linear matrix system intoa system of linear matrix equations according to arithmeticoperations of LR fuzzy numbers The LR fuzzy solution ofthe original matrix equation is derived from solving crispsystems of linearmatrix equationsThe structure of this paperis organized as follows

In Section 2 we recall the LR fuzzy numbers andpresent the concept of fully fuzzy linear matrix equationThecomputing model to the positive fully fuzzy linear matrixequation is proposed in detail and the fuzzy approximatesolution of the fuzzy linear matrix equation is obtained byusing pseudo-inverse in Section 3 Some examples are givento illustrate our method in Section 4 and the conclusion isdrawn in Section 5

2 Preliminaries

21 The LR Fuzzy Number

Definition 1 (see [1]) A fuzzy number is a fuzzy set like 119906

119877 rarr 119868 = [0 1] which satisfies the following

(1) 119906 is upper semicontinuous(2) 119906 is fuzzy convex that is 119906(120582119909 + (1 minus 120582)119910) ge

min119906(119909) 119906(119910) for all 119909 119910 isin 119877 120582 isin [0 1](3) 119906 is normal that is there exists 119909

0isin 119877 such that

119906(1199090) = 1

(4) supp 119906 = 119909 isin 119877 | 119906(119909) gt 0 is the support of the 119906and its closure cl(supp 119906) is compact

Let 1198641 be the set of all fuzzy numbers on 119877

Definition 2 (see [3]) A fuzzy number is said to be an LRfuzzy number if

120583(119909) =

119871(119898 minus 119909

120572) 119909 le 119898 120572 gt 0

119877(119909 minus 119898

120573) 119909 ge 119898 120573 gt 0

(1)

where 119898 120572 and 120573 are called the mean value left and rightspreads of respectively The function 119871(sdot) which is calledleft shape function satisfies

(1) 119871(119909) = 119871(minus119909)(2) 119871(0) = 1 and 119871(1) = 0(3) 119871(119909) is nonincreasing on [0infin)

The definition of a right shape function 119877(sdot) is similar tothat of 119871(sdot)

Clearly = (119898 120572 120573)LR is positive (negative) if and onlyif119898 minus 120572 gt 0 (119898 + 120573 lt 0)

Also two LR fuzzy numbers = (119898 120572 120573)LR and =

(119899 120574 120575)LR are said to be equal if and only if 119898 = 119899 120572 = 120574and 120573 = 120575

Definition 3 (see [5]) For arbitrary LR fuzzy numbers =

(119898 120572 120573)LR and = (119899 120574 120575)LR we have the following

(1) Addition

oplus = (119898 120572 120573)LR oplus (119899 120574 120575)LR = (119898 + 119899 120572 + 120574 120573 + 120575)LR

(2)

(2) Multiplication(i) If gt 0 and gt 0 then

otimes = (119898 120572 120573)LR otimes (119899 120574 120575)LR

cong (119898119899119898120574 + 119899120572119898120575 + 119899120573)LR(3)

(ii) if lt 0 and gt 0 then

otimes = (119898 120572 120573)RL otimes (119899 120574 120575)LR

cong (119898119899 119899120572 minus 119898120575 119899120573 minus 119898120574)RL(4)

(iii) if lt 0 and lt 0 then

otimes = (119898 120572 120573)RL otimes (119899 120574 120575)LR

cong (119898119899 minus119898120575 minus 119899120573 minus119898120574 minus 119899120572)RL(5)

(3) Scalar multiplication

120582 times = 120582 times (119898 120572 120573)LR

= (120582119898 120582120572 120582120573)LR 120582 ge 0

(120582119898 minus120582120573 minus120582120572)RL 120582 lt 0

(6)

Journal of Applied Mathematics 3

22 The LR Fuzzy Matrix

Definition 4 Amatrix = (119894119895) is called an LR fuzzy matrix

if each element 119894119895of is an LR fuzzy number

will be positive (negative) and denoted by gt 0 ( lt

0) if each element 119894119895of is positive (negative) where

119894119895=

(119886119894119895 119886119897

119894119895 119886119903

119894119895)LR Up to the rest of this paper we use positive

LR fuzzy numbers and formulas given in Definition 3 Forexample we represent 119898 times 119899 LR fuzzy matrix = (

119894119895) that

119894119895= (119886119894119895 120572119894119895 120573119894119895)LR with new notation = (119860119872119873) where

119860 = (119886119894119895) 119872 = (120572

119894119895) and 119873 = (120573

119894119895) are three 119898 times 119899 crisp

matrices In particular an 119899 dimensions LR fuzzy numbersvector can be denoted by (119909 119909119897 119909119903) where 119909 = (119909

119894) 119909119897 =

(119909119897

119894) and 119909119903 = (119909

119903

119894) are three 119899 dimensions crisp vectors

Definition 5 Let = (119894119895) and = (

119894119895) be two 119898 times 119899 and

119899 times 119901 fuzzy matrices we define otimes = = (119894119895) which is an

119898 times 119901 fuzzy matrix where

119894119895=

oplus

sum

119896=12119899

119894119896otimes 119887119896119895 (7)

23 The Fully Fuzzy Linear Matrix Equation

Definition 6 Thematrix system

(

11

12

sdot sdot sdot 1119898

21

22

sdot sdot sdot 2119898

1198981

1198982

sdot sdot sdot 119898119898

)otimes(

11

12

sdot sdot sdot 1119899

21

22

sdot sdot sdot 2119899

1198981

1198982

sdot sdot sdot 119898119899

)

otimes(

11

11988712

sdot sdot sdot 1198871119899

21

11988722

sdot sdot sdot 1198872119899

1198991

1198992

sdot sdot sdot 119899119899

)

=(

11

12

sdot sdot sdot 1119899

21

22

sdot sdot sdot 2119899

1198981

1198982

sdot sdot sdot 119898119899

)

(8)

where 119894119895 1 le 119894 119895 le 119898

119894119895 1 le 119894 119895 le 119899 and

119894119895 1 le 119894 le 119898

1 le 119895 le 119899 are LR fuzzy numbers is called an LR fully fuzzylinear matrix equation (FFLME)

Using matrix notation we have

otimes otimes = (9)

A fuzzy numbers matrix

= (119909119894119895 119910119894119895 119911119894119895)LR 1 le 119894 le 119898 1 le 119895 le 119899 (10)

is called an LR fuzzy approximate solution of fully fuzzy linearmatrix equation (8) if satisfies (9)

Up to the rest of this paper we will discuss the nonnega-tive solution = (119883 119884 119885) ge 0 of FFLME otimes otimes =

where = (119860119872119873) ge 0 = (119861 119864 119865) ge 0 and =

(119862 119866119867) ge 0

3 Method for Solving FFLME

First we extend the fuzzy linear matrix system (9) into threesystems of linear matrix equations according to the LR fuzzynumber and its arithmetic operations

Theorem 7 The fuzzy linear matrix system (9) can beextended into the following model

119860119883119861 = 119862

119860119883119864 + 119860119884119861 +119872119883119861 = 119866

119860119883119865 + 119860119885119861 + 119873119883119861 = 119867

(11)

Proof We denote = (119860119872119873) = (119861 119864 119865) and =

(119862 119866119867) and assume = (119883 119884 119885) ge 0 then

= (119860119872119873) otimes (119883 119884 119885) otimes (119861 119864 119865)

= (119860119883119860119884 +119872119883119860119885 + 119873119883) otimes (119861 119864 119865)

= (119860119883119861119860119883119864 + 119860119884119861 +119872119883119861119860119883119865 + 119860119885119861 + 119873119883119861)

= (119862 119866119867)

(12)

according tomultiplication of nonnegative LR fuzzy numbersof Definition 2 Thus we obtain a model for solving FFLME(9) as follows

119860119883119861 = 119862

119860119883119864 + 119860119884119861 +119872119883119861 = 119866

119860119883119865 + 119860119885119861 + 119873119883119861 = 119867

(13)

Secondly in order to solve the fuzzy linear matrix equa-tion (9) we need to consider the crisp systems of linearmatrix equation (11) Supposing119860 and119861 are nonsingular crispmatrices we have

119883 = 119860minus1

119862119861minus1

119884 = 119860minus1

119866 minus 119883119864119861minus1

minus 119860minus1

119872119883

119885 = 119860minus1

119867 minus 119883119865119861minus1

minus 119860minus1

119873119883

(14)

that is

119883 = 119860minus1

119862119861minus1

119884 = 119860minus1

119866 minus 119860minus1

119862119861minus1

119864119861minus1

minus 119860minus1

119872119860minus1

119862119861minus1

119885 = 119860minus1

119867 minus 119860minus1

119862119861minus1

119865119861minus1

minus 119860minus1

119873119860minus1

119862119861minus1

(15)

Definition 8 Let = (119883 119884 119885) be an LR fuzzy matrix If(119883 119884 119885) is an exact solution of (11) such that 119883 ge 0 119884 ge 0119885 ge 0 and 119883 minus 119884 ge 0 we call = (119883 119884 119885) a nonnegativeLR fuzzy approximate solution of (9)

4 Journal of Applied Mathematics

Theorem 9 Let = (119860119872119873) = (119861 119864 119865) and =

(119862 119866119867) be three nonnegative fuzzymatrices respectively andlet 119860 and 119861 be the product of a permutation matrix by adiagonal matrix with positive diagonal entries Moreover let119866119861 ge 119862119861

minus1

119864 + 119872119860minus1

119862 119867119861 ge 119862119861minus1

119865 + 119873119860minus1

119862 and119862 + 119862119861

minus1

119864 +119872119860minus1

119862 ge 119866119861 Then the systems otimes otimes =

has nonnegative fuzzy solutions

Proof Our hypotheses on 119860 and 119861 imply that 119860minus1 and119861minus1 exist and they are nonnegative matrices Thus 119883 =

119860minus1

119862119861minus1

ge 0On the other hand because 119866119861 ge 119862119861

minus1

119864 + 119872119860minus1

119862 and119867119861 ge 119862119861

minus1

119865 + 119873119860minus1

119862 so with 119884 = 119860minus1

(119866119861 minus 119862119861minus1

119864 minus

119872119860minus1

119862)119861minus1 and 119885 = 119860

minus1

(119867119861 minus 119862119861minus1

119865 minus 119873119860minus1

119862)119861minus1 we

have 119884 ge 0 and 119885 ge 0 Thus = (119883 119884 119885) is a fuzzy matrixwhich satisfies otimes otimes = Since119883 minus 119884 = 119860

minus1

(119862 minus 119866119861 +

119862119861minus1

119864 + 119872119860minus1

119862)119861minus1 the positivity property of can be

obtained from the condition 119862+119862119861minus1119864+119872119860minus1

119862 ge 119866119861

When 119860 or 119861 is a singular crisp matrix the followingresult is obvious

Theorem 10 (see [35]) For linearmatrix equations119860119883119861 = 119862where 119860 isin 119877

119898times119898 119861 isin 119877119899times119899 and 119862 isin 119877

119898times119899 Then

119883 = 119860dagger

119862119861dagger (16)

is its minimal norm least squares solutionBy the pseudoinverse of matrices we solve model (11) and

obtain its minimal norm least squares solution as follows

119883 = 119860dagger

119862119861dagger

119884 = 119860dagger

119866 minus 119883119864119861dagger

minus 119860dagger

119872119883

119885 = 119860dagger

119867 minus 119883119865119861dagger

minus 119860dagger

119873119883

(17)

that is

119883 = 119860dagger

119862119861dagger

119884 = 119860dagger

119866 minus 119860dagger

119862119861dagger

119864119861dagger

minus 119860dagger

119872119860dagger

119862119861dagger

119885 = 119860dagger

119867 minus 119860dagger

119862119861dagger

119865119861dagger

minus 119860dagger

119873119860dagger

119862119861dagger

(18)

Definition 11 Let = (119883 119884 119885) be an LR fuzzy matrix If(119883 119884 119885) is a minimal norm least squares solution of (11) suchthat119883 ge 0119884 ge 0119885 ge 0 and119883minus119884 ge 0 we call = (119883 119884 119885)

a nonnegative LR fuzzy minimal norm least squares solutionof (9)

At last we give a sufficient condition for nonnegativefuzzy minimal norm least squares solution of FFLME (9) inthe same way

Theorem 12 Let 119860dagger and 119861dagger be nonnegative matrices More-over let 119866119861 ge 119862119861

dagger

119864 + 119872119860dagger

119862 119867119861 ge 119862119861dagger

119865 + 119873119860dagger

119862 and119862+119862119861

dagger

119864+119872119860dagger

119862 ge 119866119861 Then the systems otimes otimes = hasnonnegative fuzzy minimal norm least squares solutions

Proof Since 119860dagger and 119861dagger are nonnegative matrices we have

119883 = 119860dagger

119862119861dagger

ge 0Now that 119866119861 ge 119862119861

dagger

119864+119872119860dagger

119862 and119867119861 ge 119862119861dagger

119865+119873119860dagger

119862therefore with 119884 = 119860

dagger

(119866119861 minus 119862119861dagger

119864 minus 119872119860dagger

119862)119861dagger and 119885 =

119860dagger

(119867119861 minus 119862119861dagger

119865 minus 119873119860dagger

119862)119861dagger we have 119884 ge 0 and 119885 ge 0 Thus

= (119883 119884 119885) is a fuzzy matrix which satisfies otimes otimes =

Since 119883 minus 119884 = 119860dagger

(119862 minus 119866119861 + 119862119861dagger

119864 + 119872119860dagger

119862)119861dagger

the nonnegativity property of can be obtained from thecondition 119862 + 119862119861

dagger

119864 +119872119860dagger

119862 ge 119866119861

The following Theorems give some results for such 119878minus1

and 119878dagger to be nonnegative As usual (sdot)⊤ denotes the transposeof a matrix (sdot)

Theorem 13 (see [36]) The inverse of a nonnegative matrix119860 is nonnegative if and only if 119860 is a generalized permutationmatrix

Theorem 14 (see [37]) Let119860 be an119898 times119898 nonnegativematrixwith rank 119903 Then the following assertions are equivalent

(a) 119860dagger ge 0

(b) There exists a permutation matrix 119875 such that 119875119860 hasthe form

119875119860 =(

1198781

1198782

119878119903

119874

) (19)

where each 119878119894has rank 1 and the rows of 119878

119894are

orthogonal to the rows of 119878119895 whenever 119894 = 119895 the zero

matrix may be absent

(c) 119860dagger = (119870119875⊤119870119876⊤

119870119875⊤119870119875⊤ ) for some positive diagonal matrix 119870

In this case

(119875 + 119876)dagger

= 119870(119875 + 119876)⊤

(119875 minus 119876)dagger

= 119870(119875 minus 119876)⊤

(20)

4 Numerical Examples

Example 15 Consider the fully fuzzy linear matrix system

((2 1 1)LR (1 0 1)LR(1 0 0)LR (2 1 0)LR

) otimes (11

12

13

21

22

23

)

otimes (

(1 0 1)LR (2 1 1)LR (1 1 0)LR(2 1 0)LR (1 0 0)LR (2 1 1)LR(1 0 0)LR (2 1 1)LR (1 0 1)LR

)

= ((17 12 23)LR (22 22 29)LR (17 16 28)LR(19 15 13)LR (23 23 16)LR (19 16 17)LR

)

(21)

Journal of Applied Mathematics 5

ByTheorem 7 the model of the above fuzzy linear matrixsystem is made of the following three crisp systems of linearmatrix equations

(2 1

1 2)(

11990911

11990912

11990913

11990921

11990922

11990923

)(

1 2 1

2 1 2

1 2 1

) = (17 22 17

19 23 19)

(2 1

1 2)(

11990911

11990912

11990913

11990921

11990922

11990923

)(

0 1 1

1 0 1

0 1 0

)

+ (2 1

1 2)(

11991011

11991012

11991013

11991021

11991022

11991023

)(

1 2 1

2 1 2

1 2 1

)

+ (1 0

0 1)(

11990911

11990912

11990913

11990921

11990922

11990923

)(

1 2 1

2 1 2

1 2 1

) = (12 22 16

15 23 20)

(2 1

1 2)(

11990911

11990912

11990913

11990921

11990922

11990923

)(

1 1 0

0 0 1

0 1 1

)

+ (2 1

1 2)(

11991111

11991112

11991113

11991121

11991122

11991123

)(

1 2 1

2 1 2

1 2 1

)

+ (1 1

0 0)(

11990911

11990912

11990913

11990921

11990922

11990923

)(

1 2 1

2 1 2

1 2 1

) = (23 29 28

13 16 17)

(22)

Now that the matrix 119861 is singular according to formula(18) the solutions of the above three systems of linear matrixequations are as follows

119883 = 119860dagger

119862119861dagger

= (2 1

1 2)

dagger

(17 22 17

19 23 19)(

1 2 1

2 1 2

1 2 1

)

dagger

= (15000 10000 15000

15000 20000 15000)

119884 = 119860dagger

119866 minus 119883119864119861dagger

minus 119860dagger

119872119883 = (2 1

1 2)

dagger

(12 22 16

19 23 19)

minus 119883(

0 1 1

1 0 1

0 1 0

)(

1 2 1

2 1 2

1 2 1

)

dagger

minus (2 1

1 2)

dagger

(1 0

0 1)119883

= (10333 06667 08725

11250 16553 12578)

119885 = 119860dagger

119867 minus 119883119865119861dagger

minus 119860dagger

119873119883 = (2 1

1 2)

dagger

(23 29 28

13 16 17)

minus 119883(

1 1 0

0 0 1

0 1 1

)(

1 2 1

2 1 2

1 2 1

)

dagger

minus (2 1

1 2)

dagger

(1 1

0 0)119883

= (83333 116667 93333

14167 13333 24167)

(23)

By Definition 11 we know that the original fuzzy linearmatrix equations have a nonnegative LR fuzzy solution

= (11

12

13

21

22

23

) = ((15000 10333 83333)LR (10000 06667 116667)LR (15000 08725 93333)LR(15000 11250 14167)LR (20000 16553 13333)LR (15000 12578 24167)LR

) (24)

since119883 ge 0 119884 ge 0 119885 ge 0 and119883 minus 119884 ge 0

Example 16 Consider the following fuzzy matrix system

((1 1 0)LR (2 0 1)LR(2 0 1)LR (1 0 0)LR

)(11

12

21

22

)((1 0 1)LR (2 1 0)LR(2 1 1)LR (3 1 2)LR

)

= ((15 14 18)LR (25 24 24)LR(12 10 12)LR (20 17 15)LR

)

(25)

ByTheorem 7 the model of the above fuzzy linear matrixsystem is made of following three crisp systems of linear

matrix equations

(2 1

1 2)(

11990911

11990912

11990921

11990922

)(1 2

2 3) = (

15 25

10 20)

(2 1

1 2)(

11990911

11990912

11990921

11990922

)(0 1

1 1) + (

2 1

1 2)(

11991011

11991012

11991021

11991022

)(1 2

2 3)

+ (1 0

0 0)(

11990911

11990912

11990921

11990922

)(1 2

2 3) = (

14 24

10 17)

(2 1

1 2)(

11990911

11990912

11990921

11990922

)(1 0

1 2) + (

2 1

1 2)(

11991111

11991112

11991121

11991122

)(1 2

2 3)

+ (0 1

1 0)(

11990911

11990912

11990921

11990922

)(1 2

2 3) = (

18 24

12 15)

(26)

6 Journal of Applied Mathematics

By the same way we obtain that the solutions of the abovethree systems of linear matrix equations are as follows

119883 = 119860minus1

119862119861minus1

= (1 1

2 2)

119884 = 119860minus1

119866 minus 119883119864119861minus1

minus 119860minus1

119872119883 = (0 1

0 1)

119885 = 119860minus1

119867 minus 119883119865119861minus1

minus 119860minus1

119873119883 = (0 0

1 0)

(27)

Since 119883 ge 0 119884 ge 0 119885 ge 0 and 119883 minus 119884 ge 0 we know thatthe original fuzzy linear matrix equations have a nonnegativeLR fuzzy solution given by

= (11

12

21

22

)

= ((1000 1000 0000)LR (1000 1000 0000)LR(2000 0000 1000)LR (2000 1000 0000)LR

)

(28)

5 Conclusion

In this work we presented a model for solving fuzzy linearmatrix equations otimes otimes = in which and are119898 times 119898 and 119899 times 119899 fuzzy matrices respectively and isan 119898 times 119899 arbitrary LR fuzzy numbers matrix The modelwas made of three crisp systems of linear equations whichdetermined the mean value and the left and right spreads ofthe solution The LR fuzzy approximate solution of the fuzzylinear matrix equation was derived from solving the crispsystems of linear matrix equations In addition the existencecondition of strong LR fuzzy solution was studied Numericalexamples showed that ourmethod is feasible to solve this typeof fuzzy matrix equations Based on LR fuzzy numbers andtheir operations we can investigate all kinds of fully fuzzymatrix equations in future

Acknowledgments

The work is supported by the Natural Scientific Funds ofChina (no 71061013) and the Youth Research Ability Projectof Northwest Normal University (NWNU-LKQN-1120)

References

[1] L A Zadeh ldquoFuzzy setsrdquo Information and Computation vol 8pp 338ndash353 1965

[2] L A Zadeh ldquoThe concept of a linguistic variable and itsapplication to approximate reasoning Irdquo Information Sciencevol 8 pp 199ndash249 1975

[3] D Dubois and H Prade ldquoOperations on fuzzy numbersrdquoInternational Journal of Systems Science vol 9 no 6 pp 613ndash626 1978

[4] S Nahmias ldquoFuzzy variablesrdquo Fuzzy Sets and Systems AnInternational Journal in Information Science and Engineeringvol 1 no 2 pp 97ndash110 1978

[5] M L Puri and D A Ralescu ldquoDifferentials of fuzzy functionsrdquoJournal of Mathematical Analysis and Applications vol 91 no 2pp 552ndash558 1983

[6] R Goetschel Jr and W Voxman ldquoElementary fuzzy calculusrdquoFuzzy Sets and Systems vol 18 no 1 pp 31ndash43 1986

[7] C X Wu and M Ma ldquoEmbedding problem of fuzzy numberspace Irdquo Fuzzy Sets and Systems vol 44 no 1 pp 33ndash38 1991

[8] C X Wu and M Ma ldquoEmbedding problem of fuzzy numberspace IIIrdquo Fuzzy Sets and Systems vol 46 no 2 pp 281ndash2861992

[9] M Friedman M Ming and A Kandel ldquoFuzzy linear systemsrdquoFuzzy Sets and Systems vol 96 no 2 pp 201ndash209 1998

[10] M Ma M Friedman and A Kandel ldquoDuality in fuzzy linearsystemsrdquo Fuzzy Sets and Systems vol 109 no 1 pp 55ndash58 2000

[11] T Allahviranloo and M Ghanbari ldquoOn the algebraic solutionof fuzzy linear systems based on interval theoryrdquo AppliedMathematical Modelling vol 36 no 11 pp 5360ndash5379 2012

[12] T Allahviranloo and S Salahshour ldquoFuzzy symmetric solutionsof fuzzy linear systemsrdquo Journal of Computational and AppliedMathematics vol 235 no 16 pp 4545ndash4553 2011

[13] T Allahviranloo and M Ghanbari ldquoA new approach to obtainalgebraic solution of interval linear systemsrdquo Soft Computingvol 16 no 1 pp 121ndash133 2012

[14] T Allahviranloo ldquoNumericalmethods for fuzzy systemof linearequationsrdquo Applied Mathematics and Computation vol 155 no2 pp 493ndash502 2004

[15] T Allahviranloo ldquoSuccessive over relaxation iterative methodfor fuzzy system of linear equationsrdquo Applied Mathematics andComputation vol 162 no 1 pp 189ndash196 2005

[16] T Allahviranloo ldquoThe Adomian decomposition method forfuzzy system of linear equationsrdquo Applied Mathematics andComputation vol 163 no 2 pp 553ndash563 2005

[17] RNuraei T Allahviranloo andMGhanbari ldquoFinding an innerestimation of the so- lution set of a fuzzy linear systemrdquoAppliedMathematical Modelling vol 37 no 7 pp 5148ndash5161 2013

[18] S Abbasbandy R Ezzati and A Jafarian ldquoLU decompositionmethod for solving fuzzy system of linear equationsrdquo AppliedMathematics and Computation vol 172 no 1 pp 633ndash6432006

[19] S Abbasbandy A Jafarian and R Ezzati ldquoConjugate gradientmethod for fuzzy symmetric positive definite system of linearequationsrdquo Applied Mathematics and Computation vol 171 no2 pp 1184ndash1191 2005

[20] S Abbasbandy M Otadi andM Mosleh ldquoMinimal solution ofgeneral dual fuzzy linear systemsrdquo Chaos Solitons amp Fractalsvol 37 no 4 pp 1113ndash1124 2008

[21] B Asady S Abbasbandy and M Alavi ldquoFuzzy general linearsystemsrdquo Applied Mathematics and Computation vol 169 no 1pp 34ndash40 2005

[22] K Wang and B Zheng ldquoInconsistent fuzzy linear systemsrdquoApplied Mathematics and Computation vol 181 no 2 pp 973ndash981 2006

[23] B Zheng and K Wang ldquoGeneral fuzzy linear systemsrdquo AppliedMathematics and Computation vol 181 no 2 pp 1276ndash12862006

[24] M Dehghan and B Hashemi ldquoIterative solution of fuzzy linearsystemsrdquo Applied Mathematics and Computation vol 175 no 1pp 645ndash674 2006

[25] M Dehghan B Hashemi and M Ghatee ldquoSolution of the fullyfuzzy linear systems using iterative techniquesrdquo Chaos Solitonsamp Fractals vol 34 no 2 pp 316ndash336 2007

Journal of Applied Mathematics 7

[26] T Allahviranloo N Mikaeilvand and M Barkhordary ldquoFuzzylinear matrix equationrdquo Fuzzy Optimization and Decision Mak-ing vol 8 no 2 pp 165ndash177 2009

[27] X Guo and Z Gong ldquoBlock Gaussian elimination methodsfor fuzzy matrix equationsrdquo International Journal of Pure andApplied Mathematics vol 58 no 2 pp 157ndash168 2010

[28] Z Gong and X Guo ldquoInconsistent fuzzy matrix equationsand its fuzzy least squares solutionsrdquo Applied MathematicalModelling vol 35 no 3 pp 1456ndash1469 2011

[29] X Guo and D Shang ldquoFuzzy symmetric solutions of fuzzymatrix equationsrdquo Advances in Fuzzy Systems vol 2012 ArticleID 318069 9 pages 2012

[30] T Allahviranloo M Ghanbari A A Hosseinzadeh E Haghiand R Nuraei ldquoA note on lsquoFuzzy linear systemsrsquordquo Fuzzy Sets andSystems vol 177 pp 87ndash92 2011

[31] C-T Yeh ldquoWeighted semi-trapezoidal approximations of fuzzynumbersrdquo Fuzzy Sets and Systems vol 165 pp 61ndash80 2011

[32] T Allahviranloo F H Lotfi M K Kiasari and M KhezerlooldquoOn the fuzzy solution of LR fuzzy linear systemsrdquo AppliedMathematical Modelling vol 37 no 3 pp 1170ndash1176 2013

[33] MOtadi andMMosleh ldquoSolving fully fuzzymatrix equationsrdquoApplied Mathematical Modelling vol 36 no 12 pp 6114ndash61212012

[34] X B Guo and D Q Shang ldquoApproximate solutions of LR fuzzySylvester matrix equationsrdquo Journal of Applied Mathematics Inpress

[35] A Ben-Israel and T N E Greville Generalized Inverses Theoryand Applications Springer-Verlag New York NY USA 2ndedition 2003

[36] A Berman and R J Plemmons Nonnegative Matrices in theMathematical Sciences Academic Press New York NY USA1979

[37] R J Plemmons ldquoRegular nonnegative matricesrdquo Proceedings ofthe American Mathematical Society vol 39 pp 26ndash32 1973

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 514984 9 pageshttpdxdoiorg1011552013514984

Research ArticleRanks of a Constrained Hermitian MatrixExpression with Applications

Shao-Wen Yu

Department of Mathematics East China University of Science and Technology Shanghai 200237 China

Correspondence should be addressed to Shao-Wen Yu yushaowenecusteducn

Received 12 November 2012 Accepted 4 January 2013

Academic Editor Yang Zhang

Copyright copy 2013 Shao-Wen YuThis is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We establish the formulas of the maximal and minimal ranks of the quaternion Hermitian matrix expression 1198624minus 1198604119883119860lowast

4where

119883 is a Hermitian solution to quaternion matrix equations 1198601119883 = 119862

1 1198831198611= 1198622 and 119860

3119883119860lowast

3= 1198623 As applications we give a new

necessary and sufficient condition for the existence of Hermitian solution to the system of matrix equations 1198601119883 = 119862

11198831198611= 1198622

1198603119883119860lowast

3= 1198623 and 119860

4119883119860lowast

4= 1198624 which was investigated by Wang and Wu 2010 by rank equalities In addition extremal ranks of

the generalized Hermitian Schur complement 1198624minus 1198604119860sim

3119860lowast

4with respect to a Hermitian g-inverse 119860sim

3of 1198603 which is a common

solution to quaternion matrix equations 1198601119883 = 119862

1and119883119861

1= 1198622 are also considered

1 Introduction

Throughout this paper we denote the real number field byRthe complex number field by C the set of all 119898 times 119899 matricesover the quaternion algebra

H = 1198860+ 1198861119894 + 1198862119895 + 1198863119896 | 1198942

= 1198952

= 1198962

= 119894119895119896 = minus1 1198860 1198861 1198862 1198863isin R

(1)

by H119898times119899 the identity matrix with the appropriate size by 119868the column right space the row left space of a matrix 119860 overH by R(119860) N(119860) respectively the dimension of R(119860) bydimR(119860) a Hermitian g-inverse of a matrix 119860 by 119883 = 119860

∽

which satisfies 119860119860∽119860 = 119860 and 119883 = 119883lowast and the Moore-

Penrose inverse of matrix119860 overH by119860dagger which satisfies fourPenrose equations 119860119860dagger119860 = 119860 119860

dagger

119860119860dagger

= 119860dagger

(119860119860dagger

)lowast

=

119860119860dagger

and (119860dagger

119860)lowast

= 119860dagger

119860 In this case 119860dagger is unique and(119860dagger

)lowast

= (119860lowast

)dagger Moreover 119877

119860and 119871

119860stand for the two

projectors 119871119860

= 119868 minus 119860dagger

119860 119877119860

= 119868 minus 119860119860dagger induced by 119860

Clearly119877119860and 119871

119860are idempotent Hermitian and119877

119860= 119871119860lowast

By [1] for a quaternion matrix 119860 dimR(119860) = dimN(119860)dimR(119860) is called the rank of a quaternion matrix 119860 anddenoted by 119903(119860)

Mitra [2] investigated the system of matrix equations

1198601119883 = 119862

1 119883119861

1= 1198622 (2)

Khatri and Mitra [3] gave necessary and sufficient con-ditions for the existence of the common Hermitian solutionto (2) and presented an explicit expression for the generalHermitian solution to (2) by generalized inverses Using thesingular value decomposition (SVD) Yuan [4] investigatedthe general symmetric solution of (2) over the real numberfield R By the SVD Dai and Lancaster [5] considered thesymmetric solution of equation

119860119883119860lowast

= 119862 (3)

over R which was motivated and illustrated with an inverseproblem of vibration theory Groszlig [6] Tian and Liu [7]gave the solvability conditions for Hermitian solution and itsexpressions of (3) over C in terms of generalized inversesrespectively Liu Tian and Takane [8] investigated ranksof Hermitian and skew-Hermitian solutions to the matrixequation (3) By using the generalized SVD Chang andWang[9] examined the symmetric solution to the matrix equations

1198603119883119860lowast

3= 1198623 119860

4119883119860lowast

4= 1198624

(4)

2 Journal of Applied Mathematics

over R Note that all the matrix equations mentioned aboveare special cases of

1198601119883 = 119862

1 119883119861

1= 1198622

1198603119883119860lowast

3= 1198623 119860

4119883119860lowast

4= 1198624

(5)

Wang and Wu [10] gave some necessary and sufficientconditions for the existence of the common Hermitiansolution to (5) for operators between Hilbert Clowast-modulesby generalized inverses and range inclusion of matrices Inview of the complicated computations of the generalizedinverses of matrices we naturally hope to establish a morepractical necessary and sufficient condition for system (5)over quaternion algebra to have Hermitian solution by rankequalities

As is known to us solutions tomatrix equations and ranksof solutions to matrix equations have been considered previ-ously by many authors [10ndash34] and extremal ranks of matrixexpressions can be used to characterize their rank invariancenonsingularity range inclusion and solvability conditionsof matrix equations Tian and Cheng [35] investigated themaximal and minimal ranks of 119860 minus 119861119883119862 with respect to 119883with applications Tian [36] gave the maximal and minimalranks of 119860

1minus 11986111198831198621subject to a consistent matrix equation

11986121198831198622= 1198602 Tian and Liu [7] established the solvability

conditions for (4) to have a Hermitian solution over C bythe ranks of coefficientmatricesWang and Jiang [20] derivedextreme ranks of (skew)Hermitian solutions to a quaternionmatrix equation 119860119883119860

lowast

+ 119861119884119861lowast

= 119862 Wang Yu and Lin[31] derived the extremal ranks of 119862

4minus 11986041198831198614subject to a

consistent system of matrix equations

1198601119883 = 119862

1 119883119861

1= 1198622 119860

31198831198613= 1198623

(6)

over H and gave a new solvability condition to system

1198601119883 = 119862

1 119883119861

1= 1198622

11986031198831198613= 1198623 119860

41198831198614= 1198624

(7)

In matrix theory and its applications there are manymatrix expressions that have symmetric patterns or involveHermitian (skew-Hermitian) matrices For example

119860 minus 119861119883119861lowast

119860 minus 119861119883 plusmn 119883lowast

119861lowast

119860 minus 119861119883119861lowast

minus 119862119884119862lowast

119860 minus 119861119883119862 plusmn (119861119883119862)lowast

(8)

where 119860 = plusmn119860lowast

119861 and 119862 are given and 119883 and 119884 arevariable matrices In recent papers [7 8 37 38] Liu and Tianconsidered some maximization and minimization problemson the ranks of Hermitian matrix expressions (8)

Define a Hermitian matrix expression

119891 (119883) = 1198624minus 1198604119883119860lowast

4 (9)

where 1198624= 119862lowast

4 we have an observation that by investigating

extremal ranks of (9) where 119883 is a Hermitian solution to asystem of matrix equations

1198601119883 = 119862

1 119883119861

1= 1198622 119860

3119883119860lowast

3= 1198623 (10)

A new necessary and sufficient condition for system (5) tohave Hermitian solution can be given by rank equalitieswhich is more practical than one given by generalizedinverses and range inclusion of matrices

It is well known that Schur complement is one of themostimportant matrix expressions in matrix theory there havebeen many results in the literature on Schur complementsand their applications [39ndash41] Tian [36 42] has investigatedthe maximal and minimal ranks of Schur complements withapplications

Motivated by the workmentioned above we in this paperinvestigate the extremal ranks of the quaternion Hermitianmatrix expression (9) subject to the consistent system ofquaternion matrix equations (10) and its applications InSection 2 we derive the formulas of extremal ranks of (9)with respect to Hermitian solution of (10) As applications inSection 3 we give a new necessary and sufficient conditionfor the existence of Hermitian solution to system (5) byrank equalities In Section 4 we derive extremal ranks ofgeneralized Hermitian Schur complement subject to (2) Wealso consider the rank invariance problem in Section 5

2 Extremal Ranks of (9) Subject to System (10)Corollary 8 in [10] over Hilbert Clowast-modules can be changedinto the following lemma over H

Lemma 1 Let 1198601 1198621

isin H119898times119899 1198611 1198622

isin H119899times119904 1198603

isin

H119903times119899 1198623isin H119903times119903 be given and 119865 = 119861

lowast

11198711198601 119872 = 119878119871

119865 119878 =

11986031198711198601 119863 = 119862

lowast

2minus119861lowast

1119860dagger

11198621 119869 = 119860

dagger

11198621+119865dagger

119863 119866 = 1198623minus1198603(119869+

1198711198601119871lowast

119865119869lowast

)119860lowast

3 then the following statements are equivalent

(1) the system (10) have a Hermitian solution(2) 1198623= 119862lowast

3

11986011198622= 11986211198611 119860

1119862lowast

1= 1198621119860lowast

1 119861

lowast

11198622= 119862lowast

21198611 (11)

11987711986011198621= 0 119877

119865119863 = 0 119877

119872119866 = 0 (12)

(3) 1198623= 119862lowast

3 the equalities in (11) hold and

119903 [11986011198621] = 119903 (119860

1) 119903 [

11986011198621

119861lowast

1119862lowast

2

] = 119903 [1198601

119861lowast

1

]

119903[[

[

11986011198621119860lowast

3

119861lowast

1119862lowast

2119860lowast

3

1198603

1198623

]]

]

= 119903[

[

1198601

119861lowast

1

1198603

]

]

(13)

In that case the general Hermitian solution of (10) can beexpressed as

119883 = 119869 + 1198711198601119871119865119869lowast

+ 1198711198601119871119865119872dagger

119866(119872dagger

)lowast

1198711198651198711198601

+ 11987111986011198711198651198711198721198811198711198651198711198601+ 1198711198601119871119865119881lowast

1198711198721198711198651198711198601

(14)

where 119881 is Hermitian matrix over H with compatible size

Journal of Applied Mathematics 3

Lemma 2 (see Lemma 24 in [24]) Let 119860 isin H119898times119899 119861 isin

H119898times119896 119862 isin H119897times119899 119863 isin H119895times119896 and 119864 isin H119897times119894 Then the followingrank equalities hold

(a) 119903(119862119871119860) = 119903 [

119860

119862] minus 119903(119860)

(b) 119903 [ 119861 119860119871119862 ] = 119903 [119861 119860

0 119862] minus 119903(119862)

(c) 119903 [ 119862119877119861119860

] = 119903 [119862 0

119860 119861] minus 119903(119861)

(d) 119903 [ 119860 119861119871119863119877119864119862 0

] = 119903 [119860 119861 0

119862 0 119864

0 119863 0

] minus 119903(119863) minus 119903(119864)

Lemma 2 plays an important role in simplifying ranks ofvarious block matrices

Liu and Tian [38] has given the following lemma over afield The result can be generalized to H

Lemma 3 Let 119860 = plusmn119860lowast

isin H119898times119898 119861 isin H119898times119899 and 119862 isin H119901times119898

be given then

max119883isinH119899times119901

119903 [119860 minus 119861119883119862 ∓ (119861119883119862)lowast

]

= min119903 [119860 119861 119862lowast

] 119903 [119860 119861

119861lowast

0] 119903 [

119860 119862lowast

119862 0]

min119883isinH119899times119901

119903 [119860 minus 119861119883119862 ∓ (119861119883119862)lowast

]

= 2119903 [119860 119861 119862lowast

] +max 1199041 1199042

(15)

where

1199041= 119903 [

119860 119861

119861lowast

0] minus 2119903 [

119860 119861 119862lowast

119861lowast

0 0]

1199042= 119903 [

119860 119862lowast

119862 0] minus 2119903 [

119860 119861 119862lowast

119862 0 0]

(16)

IfR(119861) sube R(119862lowast

)

max119883

119903 [119860 minus 119861119883119862 minus (119861119883119862)lowast

] = min119903 [119860 119862lowast

] 119903 [119860 119861

119861lowast

0]

max119883

119903 [119860 minus 119861119883119862 minus (119861119883119862)lowast

] = min119903 [119860 119862lowast

] 119903 [119860 119861

119861lowast

0]

(17)

Now we consider the extremal ranks of the matrixexpression (9) subject to the consistent system (10)

Theorem 4 Let 1198601 1198621 1198611 1198622 1198603 and 119862

3be defined as

Lemma 11198624isin H119905times119905 and 119860

4isin H119905times119899Then the extremal ranks

of the quaternionmatrix expression119891(119883) defined as (9) subjectto system (10) are the following

max 119903 [119891 (119883)] = min 119886 119887 (18)

where

119886 = 119903

[[[[

[

1198624

1198604

119862lowast

2119860lowast

4119861lowast

1

1198621119860lowast

41198601

]]]]

]

minus 119903 [119861lowast

1

1198601

]

119887 = 119903

[[[[[[[[

[

0 119860lowast

4119860lowast

31198611

119860lowast

1

11986041198624

0 0 0

1198603

0 minus1198623

minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]]

]

minus 2119903[

[

1198603

119861lowast

1

1198601

]

]

(19)

min 119903 [119891 (119883)] = 2119903

[[[[

[

1198624

1198604

119862lowast

2119860lowast

4119861lowast

1

1198621119860lowast

41198601

]]]]

]

+ 119903

[[[[[[[[[

[

0 119860lowast

4119860lowast

31198611

119860lowast

1

11986041198624

0 0 0

1198603

0 minus1198623

minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]]]

]

minus 2119903

[[[[[[[[[

[

0 119860lowast

41198611

119860lowast

1

11986041198624

0 0

1198603

0 minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus11986211198611minus1198621119860lowast

1

]]]]]]]]]

]

(20)

Proof By Lemma 1 the general Hermitian solution of thesystem (10) can be expressed as

119883 = 119869 + 1198711198601119871119865119869lowast

+ 1198711198601119871119865119872dagger

119866(119872dagger

)lowast

1198711198651198711198601

+ 11987111986011198711198651198711198721198811198711198651198711198601+ 1198711198601119871119865119881lowast

1198711198721198711198651198711198601

(21)

where 119881 is Hermitian matrix over H with appropriate sizeSubstituting (21) into (9) yields

119891 (119883) = 1198624minus 1198604(119869 + 119871

1198601119871119865119869lowast

+ 1198711198601119871119865119872dagger

119866(119872dagger

)lowast

1198711198651198711198601)119860

lowast

4

minus 119860411987111986011198711198651198711198721198811198711198651198711198601119860lowast

4

minus 11986041198711198601119871119865119881lowast

1198711198721198711198651198711198601119860lowast

4

(22)

4 Journal of Applied Mathematics

Put

1198624minus 1198604(119869 + 119871

1198601119871119865119869lowast

+ 1198711198601119871119865119872dagger

119866(119872dagger

)lowast

1198711198651198711198601)119860lowast

4= 119860

119869 + 1198711198601119871119865119869lowast

+ 1198711198601119871119865119872dagger

119866(119872dagger

)lowast

1198711198651198711198601

= 1198691015840

11986041198711198601119871119865119871119872= 119873

1198711198651198711198601119860lowast

4= 119875

(23)

then

119891 (119883) = 119860 minus 119873119881119875 minus (119873119881119875)lowast

(24)

Note that119860 = 119860lowast andR(119873) sube R(119875

lowast

) Thus applying (17) to(24) we get the following

max 119903 [119891 (119883)] = max119881

119903 (119860 minus 119873119881119875 minus (119873119881119875)lowast

)

= min119903 [119860 119875lowast

] 119903 [119860 119873

119873lowast

0]

min 119903 [119891 (119883)] = min119881

119903 (119860 minus 119873119881119875 minus (119873119881119875)lowast

)

= 2119903 [119860 119875lowast

] + 119903 [119860 119873

119873lowast

0] minus 2119903 [

119860 119873

119875 0]

(25)

Now we simplify the ranks of block matrices in (25)In view of Lemma 2 block Gaussian elimination (11)

(12) and (23) we have the following

119903 (119865) = 119903 (119861lowast

11198711198601) = 119903 [

119861lowast

1

1198601

] minus 119903 (1198601)

119903 (119872) = 119903 (119878119871119865) = 119903 [

119878

119865] minus 119903 (119865)

= 119903 [

11986031198711198601

119861lowast

11198711198601

] minus 119903 (119865)

= 119903[

[

1198603

119861lowast

1

1198601

]

]

minus 119903 (1198601) minus 119903 (119865)

119903 [119860 119875lowast

] = 119903 [1198624minus 1198604119869119860lowast

4119875lowast

]

= 119903 [1198624minus 1198604119869119860lowast

411986041198711198601

0 119865] minus 119903 (119865)

= 119903[

[

1198624minus 1198604119869119860lowast

41198604

0 119861lowast

1

0 1198601

]

]

minus 119903 (119865) minus 119903 (1198601)

= 119903

[[[

[

1198624

1198604

119862lowast

2119860lowast

4119861lowast

1

1198621119860lowast

41198601

]]]

]

minus 119903 [119861lowast

1

1198601

]

119903 [119860 119873

119873lowast

0] = 119903 [

1198624minus 11986041198691015840

119860lowast

411986041198711198601119871119865119871119872

119877119872lowast119877119865lowast119877119860lowast

1

119860lowast

40

]

= 119903

[[[[[[

[

1198624minus 11986041198691015840

119860lowast

41198604

0 0 0

119860lowast

40 119860lowast

31198611119860lowast

1

0 1198603

0 0 0

0 119861lowast

10 0 0

0 1198601

0 0 0

]]]]]]

]

minus 2119903 (119872) minus 2119903 (119865) minus 2119903 (1198601)

= 119903

[[[[[[[[[[[[

[

1198624

1198604

0 0 0

119860lowast

40 119860lowast

31198611119860lowast

1

11986031198691015840

119860lowast

41198603

0 0 0

119861lowast

11198691015840

119860lowast

4119861lowast

10 0 0

11986011198691015840

119860lowast

41198601

0 0 0

]]]]]]]]]]]]

]

minus 2119903[

[

1198603

119861lowast

1

1198601

]

]

= 119903

[[[[[[[[[

[

11986241198604

0 0 0

119860lowast

40 119860

lowast

31198611

119860lowast

1

0 1198603

minus1198623

minus11986031198622minus1198603119862lowast

1

0 119861lowast

1minus119862lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

0 1198601minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]]]

]

minus 2119903[

[

1198603

119861lowast

1

1198601

]

]

= 119903

[[[[[[[[[

[

0 119860lowast

4119860lowast

31198611

119860lowast

1

11986041198624

0 0 0

1198603

0 minus1198623

minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]]]

]

minus 2119903[

[

1198603

119861lowast

1

1198601

]

]

119903 [119860 119873

119875 0] = 119903 [

1198624minus 11986041198691015840

119860lowast

411986041198711198601119871119865119871119872

119877119865lowast119877119860lowast

1

119860lowast

40

]

= 119903

[[[[[[[[[[

[

11986241198604

0 0

119860lowast

40 119861

1119860lowast

1

0 1198603minus11986031198622minus1198603119862lowast

1

0 119861lowast

1minus119862lowast

21198611minus119862lowast

2119860lowast

1

0 1198601minus11986211198611minus1198621119860lowast

1

]]]]]]]]]]

]

minus 119903[

[

1198603

119861lowast

1

1198601

]

]

minus 119903 [119861lowast

1

1198601

]

Journal of Applied Mathematics 5

= 119903

[[[[[[[[[[

[

0 119860lowast

41198611

119860lowast

1

11986041198624

0 0

1198603

0 minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus11986211198611minus1198621119860lowast

1

]]]]]]]]]]

]

minus 119903[[

[

1198603

119861lowast

1

1198601

]]

]

minus 119903 [

119861lowast

1

1198601

]

(26)

Substituting (26) into (25) yields (18) and (20)

In Theorem 4 letting 1198624vanish and 119860

4be 119868 with

appropriate size respectively we have the following

Corollary 5 Assume that 1198601 1198621

isin H119898times119899 11986111198622

isin

H119899times119904 1198603isin H119903times119899 and 119862

3isin H119903times119903 are given then the maximal

and minimal ranks of the Hermitian solution 119883 to the system(10) can be expressed as

max 119903 (119883) = min 119886 119887 (27)

where

119886 = 119899 + 119903 [119862lowast

2

1198621

] minus 119903 [119861lowast

1

1198601

]

119887 = 2119899 + 119903

[[[[

[

1198623

119860311986221198603119862lowast

1

119862lowast

2119860lowast

3119862lowast

21198611119862lowast

2119860lowast

1

1198621119860lowast

3119862111986111198621119860lowast

1

]]]]

]

minus 2119903[

[

1198603

119861lowast

1

1198601

]

]

min 119903 (119883) = 2119903 [

119862lowast

2

1198621

]

+ 119903

[[[

[

1198623

119860311986221198603119862lowast

1

119862lowast

2119860lowast

3119862lowast

21198611119862lowast

2119860lowast

1

1198621119860lowast

3119862111986111198621119860lowast

1

]]]

]

minus 2119903

[[[[

[

119860311986221198603119862lowast

1

119862lowast

21198611119862lowast

2119860lowast

1

119862111986111198621119860lowast

1

]]]]

]

(28)

InTheorem 4 assuming that 1198601 1198611 1198621 and 119862

2vanish

we have the following

Corollary 6 Suppose that the matrix equation 1198603119883119860lowast

3= 1198623

is consistent then the extremal ranks of the quaternion matrixexpression 119891(119883) defined as (9) subject to 119860

3119883119860lowast

3= 1198623are the

following

max 119903 [119891 (119883)]

= min

119903 [11986241198604] 119903 [

[

0 119860lowast

4119860lowast

3

11986041198624

0

1198603

0 minus1198623

]

]

minus 2119903 (1198603)

min 119903 [119891 (119883)] = 2119903 [11986241198604]

+ 119903[

[

0 119860lowast

4119860lowast

3

11986041198624

0

1198603

0 minus1198623

]

]

minus 2119903[

[

0 119860lowast

4

11986041198624

1198603

0

]

]

(29)

3 A Practical Solvability Condition forHermitian Solution to System (5)

In this section we use Theorem 4 to give a necessary andsufficient condition for the existence of Hermitian solutionto system (5) by rank equalities

Theorem 7 Let 1198601 1198621

isin H119898times119899 11986111198622

isin H119899times119904 1198603

isin

H119903times119899 1198623isin H119903times119903 119860

4isin H119905times119899 and 119862

4isin H119905times119905be given then

the system (5) have Hermitian solution if and only if 1198623= 119862lowast

3

(11) (13) hold and the following equalities are all satisfied

119903 [11986041198624] = 119903 (119860

4) (30)

119903

[[[

[

1198624

1198604

119862lowast

2119860lowast

4119861lowast

1

1198621119860lowast

41198601

]]]

]

= 119903[

[

1198604

119861lowast

1

1198601

]

]

(31)

119903

[[[[[[[[

[

0 119860lowast

4119860lowast

31198611

119860lowast

1

11986041198624

0 0 0

1198603

0 minus1198623

minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]]

]

= 2119903

[[[[

[

1198604

1198603

119861lowast

1

1198601

]]]]

]

(32)

Proof It is obvious that the system (5) have Hermitiansolution if and only if the system (10) haveHermitian solutionand

min 119903 [119891 (119883)] = 0 (33)

where 119891(119883) is defined as (9) subject to system (10) Let1198830be

aHermitian solution to the system (5) then1198830is aHermitian

solution to system (10) and1198830satisfies119860

41198830119860lowast

4= 1198624 Hence

Lemma 1 yields 1198623

= 119862lowast

3 (11) (13) and (30) It follows

6 Journal of Applied Mathematics

from

[[[[[[

[

119868 0 0 0 0

0 119868 0 0 0

119860311988300 119868 0 0

119861lowast

111988300 0 119868 0

119860111988300 0 0 119868

]]]]]]

]

times

[[[[[[[

[

0 119860lowast

4119860lowast

31198611

119860lowast

1

11986041198624

0 0 0

1198603

0 minus1198623

minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]

]

times

[[[[[

[

119868 minus1198830119860lowast

40 0 0

0 119868 0 0 0

0 0 119868 0 0

0 0 0 119868 0

0 0 0 0 119868

]]]]]

]

=

[[[[[

[

0 119860lowast

4119860lowast

31198611119860lowast

1

1198604

0 0 0 0

1198603

0 0 0 0

119861lowast

10 0 0 0

1198601

0 0 0 0

]]]]]

]

(34)

that (32) holds Similarly we can obtain (31)Conversely assume that 119862

3= 119862lowast

3 (11) (13) hold then by

Lemma 1 system (10) have Hermitian solution By (20) (31)-(32) and

119903

[[[[[[[

[

0 119860lowast

41198611

119860lowast

1

11986041198624

0 0

1198603

0 minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus11986211198611minus1198621119860lowast

1

]]]]]]]

]

ge 119903

[[[

[

1198604

1198603

119861lowast

1

1198601

]]]

]

+ 119903[

[

1198604

119861lowast

1

1198601

]

]

(35)

we can get

min 119903 [119891 (119883)] le 0 (36)

However

min 119903 [119891 (119883)] ge 0 (37)

Hence (33) holds implying that the system (5) have Hermi-tian solution

ByTheorem 7 we can also get the following

Corollary 8 Suppose that 1198603 1198623 1198604 and 119862

4are those in

Theorem 7 then the quaternion matrix equations 1198603119883119860lowast

3=

1198623and 119860

4119883119860lowast

4= 1198624have common Hermitian solution if and

only if (30) hold and the following equalities are satisfied

119903 [11986031198623] = 119903 (119860

3)

119903 [

[

0 119860lowast

4119860lowast

3

11986041198624

0

1198603

0 minus1198623

]

]

= 2119903 [1198603

1198604

]

(38)

Corollary 9 Suppose that11986011198621isin H119898times119899119861

11198622isin H119899times119904 and

119860 119861 isin H119899times119899 are Hermitian Then 119860 and 119861 have a common

Hermitian g-inverse which is a solution to the system (2) if andonly if (11) holds and the following equalities are all satisfied

119903[[

[

11986011198621119860

119861lowast

1119862lowast

2119860

119860 119860

]]

]

= 119903[

[

1198601

119861lowast

1

119860

]

]

119903[[

[

11986011198621119861

119861lowast

1119862lowast

2119861

119861 119861

]]

]

= 119903[

[

1198601

119861lowast

1

119861

]

]

(39)

119903

[[[[[[[

[

0 119861 119860 1198611

119860lowast

1

119861 119861 0 0 0

119860 0 minus119860 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

2119860 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

1119860 minus119862

11198611minus1198621119860lowast

1

]]]]]]]

]

= 2119903

[[[[

[

119861

119860

119861lowast

1

1198601

]]]]

]

(40)

4 Extremal Ranks of Schur ComplementSubject to (2)

As is well known for a given block matrix

119872 = [119860 119861

119861lowast

119863] (41)

where 119860 and 119863 are Hermitian quaternion matrices withappropriate sizes then the Hermitian Schur complement of119860 in119872 is defined as

119878119860= 119863 minus 119861

lowast

119860sim

119861 (42)

where 119860sim is a Hermitian g-inverse of 119860 that is 119860sim isin 119883 |

119860119883119860 = 119860119883 = 119883lowast

Now we use Theorem 4 to establish the extremal ranks

of 119878119860given by (42) with respect to 119860sim which is a solution to

system (2)

Theorem 10 Suppose 1198601 1198621isin H119898times119899 119861

1 1198622isin H119899times119904 119863 isin

H119905times119905 119861 isin H119899times119905 and 119860 isin H119899times119899 are given and system (2)is consistent then the extreme ranks of 119878

119860given by (42) with

respect to 119860sim which is a solution of (2) are the following

max1198601119860sim=1198621

119860sim1198611=1198622

119903 (119878119860) = min 119886 119887

(43)

Journal of Applied Mathematics 7

where

119886 = 119903[

[

119863 119861lowast

119862lowast

2119861 119861lowast

1

1198621119861 1198601

]

]

minus 119903 [119861lowast

1

1198601

]

119887 = 119903

[[[[[[

[

0 119861 119860 1198611

119860lowast

1

119861lowast

119863 0 0 0

119860 0 minus119860 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

2119860 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

1119860 minus119862

11198611minus1198621119860lowast

1

]]]]]]

]

minus 2119903[

[

119860

119861lowast

1

1198601

]

]

min1198601119860sim=1198621

119860sim1198611=1198622

119903 (119878119860) = 2119903[

[

119863 119861lowast

119862lowast

2119861 119861lowast

1

1198621119861 1198601

]

]

+ 119903

[[[[[[[[

[

0 119861 119860 1198611

119860lowast

1

119861lowast

119863 0 0 0

119860 0 minus119860 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

2119860 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

1119860 minus119862

11198611minus1198621119860lowast

1

]]]]]]]]

]

minus 2119903

[[[[[[[[

[

0 119861 1198611

119860lowast

1

119861lowast

119863 0 0

119860 0 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

11198611minus1198621119860lowast

1

]]]]]]]]

]

(44)

Proof It is obvious that

max1198601119860sim=1198621 119860

sim1198611=1198622

119903 (119863 minus 119861lowast

119860sim

119861)

= max1198601119883=1198621 1198831198611=1198622 119860119883119860=119860

119903 (119863 minus 119861lowast

119883119861)

min1198601119860sim=1198621 119860

sim1198611=1198622

119903 (119863 minus 119861lowast

119860sim

119861)

= min1198601119883=1198621 1198831198611=1198622 119860119883119860=119860

119903 (119863 minus 119861lowast

119883119861)

(45)

Thus in Theorem 4 and its proof letting 1198603= 119860lowast

3= 1198623= 119860

1198604= 119861lowast

and 1198624= 119863 we can easily get the proof

In Theorem 10 let 1198601 1198621 1198611 and 119862

2vanish Then we

can easily get the following

Corollary 11 The extreme ranks of 119878119860given by (42) with

respect to 119860sim are the following

max119860sim119903 (119878119860) = min

119903 [119863 119861lowast

] 119903 [

[

0 119861 119860

119861lowast

119863 0

119860 0 minus119860

]

]

minus 2119903 (119860)

min119860sim119903 (119878119860) = 2119903 [119863 119861

lowast

] + 119903[

[

0 119861 119860

119861lowast

119863 0

119860 0 minus119860

]

]

minus 2119903[

[

0 119861

119861lowast

119863

119860 0

]

]

(46)

5 The Rank Invariance of (9)As another application of Theorem 4 we in this sectionconsider the rank invariance of thematrix expression (9) withrespect to the Hermitian solution of system (10)

Theorem 12 Suppose that (10) have Hermitian solution thenthe rank of 119891(119883) defined by (9) with respect to the Hermitiansolution of (10) is invariant if and only if

119903

[[[[[[[[

[

0 119860lowast

41198611

119860lowast

1

11986041198624

0 0

1198603

0 minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus11986211198611minus1198621119860lowast

1

]]]]]]]]

]

= 119903[[

[

1198624

1198604

119862lowast

2119860lowast

4119861lowast

1

1198621119860lowast

41198601

]]

]

+ 119903[

[

1198603

119861lowast

1

1198601

]

]

119903

[[[[[[[[

[

0 119860lowast

4119860lowast

31198611

119860lowast

1

11986041198624

0 0 0

1198603

0 minus1198623

minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]]

]

+ 119903 [119861lowast

1

1198601

]

= 119903

[[[[[[

[

0 119860lowast

41198611

119860lowast

1

11986041198624

0 0

1198603

0 minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus11986211198611minus1198621119860lowast

1

]]]]]]

]

+ 119903[

[

1198603

119861lowast

1

1198601

]

]

(47)

or

119903

[[[[[[[

[

0 119860lowast

41198611

119860lowast

1

11986041198624

0 0

1198603

0 minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus11986211198611minus1198621119860lowast

1

]]]]]]]

]

= 119903[[

[

1198624

1198604

119862lowast

2119860lowast

4119861lowast

1

1198621119860lowast

41198601

]]

]

+ 119903[[

[

1198603

119861lowast

1

1198601

]]

]

(48)

Proof It is obvious that the rank of 119891(119883) with respect toHermitian solution of system (10) is invariant if and only if

max 119903 [119891 (119883)] minusmin 119903 [119891 (119883)] = 0 (49)

By (49) Theorem 4 and simplifications we can get (47)and (48)

8 Journal of Applied Mathematics

Corollary 13 The rank of 119878119860defined by (42) with respect to

119860sim which is a solution to system (2) is invariant if and only if

119903

[[[[[[

[

0 119861 1198611

119860lowast

1

119861lowast

119863 0 0

119860 0 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

11198611minus1198621119860lowast

1

]]]]]]

]

= 119903[

[

119863 119861lowast

119862lowast

2119861 119861lowast

1

1198621119861 1198601

]

]

+ 119903[

[

119860

119861lowast

1

1198601

]

]

119903

[[[[[[[[

[

0 119861 119860 1198611

119860lowast

1

119861lowast

119863 0 0 0

119860 0 minus119860 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

2119860 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

1119860 minus119862

11198611minus1198621119860lowast

1

]]]]]]]]

]

+ 119903 [119861lowast

1

1198601

]

= 119903

[[[[[[

[

0 119861 1198611

119860lowast

1

119861lowast

119863 0 0

119860 0 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

11198611minus1198621119860lowast

1

]]]]]]

]

+ 119903[

[

119860

119861lowast

1

1198601

]

]

(50)

or

119903

[[[[[[

[

0 119861 1198611

119860lowast

1

119861lowast

119863 0 0

119860 0 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

11198611minus1198621119860lowast

1

]]]]]]

]

= 119903[

[

119863 119861lowast

119862lowast

2119861 119861lowast

1

1198621119861 1198601

]

]

+ 119903[

[

119860

119861lowast

1

1198601

]

]

(51)

Acknowledgments

This research was supported by the National Natural ScienceFoundation of China Tian Yuan Foundation (11226067) theFundamental Research Funds for the Central Universities(WM1214063) and China Postdoctoral Science Foundation(2012M511014)

References

[1] T W Hungerford Algebra Springer New York NY USA 1980[2] S K Mitra ldquoThe matrix equations 119860119883 = 119862 119883119861 = 119863rdquo Linear

Algebra and its Applications vol 59 pp 171ndash181 1984[3] C G Khatri and S K Mitra ldquoHermitian and nonnegative

definite solutions of linear matrix equationsrdquo SIAM Journal onApplied Mathematics vol 31 no 4 pp 579ndash585 1976

[4] Y X Yuan ldquoOn the symmetric solutions of matrix equation(119860119883119883119862) = (119861119863)rdquo Journal of East China Shipbuilding Institutevol 15 no 4 pp 82ndash85 2001 (Chinese)

[5] H Dai and P Lancaster ldquoLinear matrix equations from aninverse problem of vibration theoryrdquo Linear Algebra and itsApplications vol 246 pp 31ndash47 1996

[6] J Groszlig ldquoA note on the general Hermitian solution to 119860119883119860lowast =119861rdquo Malaysian Mathematical Society vol 21 no 2 pp 57ndash621998

[7] Y G Tian and Y H Liu ldquoExtremal ranks of some symmetricmatrix expressions with applicationsrdquo SIAM Journal on MatrixAnalysis and Applications vol 28 no 3 pp 890ndash905 2006

[8] Y H Liu Y G Tian and Y Takane ldquoRanks of Hermitian andskew-Hermitian solutions to the matrix equation 119860119883119860lowast = 119861rdquoLinear Algebra and its Applications vol 431 no 12 pp 2359ndash2372 2009

[9] X W Chang and J S Wang ldquoThe symmetric solution of thematrix equations 119860119883 + 119884119860 = 119862 119860119883119860119879 + 119861119884119861

119879

= 119862 and(119860119879

119883119860 119861119879

119883119861) = (119862119863)rdquo Linear Algebra and its Applicationsvol 179 pp 171ndash189 1993

[10] Q-W Wang and Z-C Wu ldquoCommon Hermitian solutionsto some operator equations on Hilbert 119862lowast-modulesrdquo LinearAlgebra and its Applications vol 432 no 12 pp 3159ndash3171 2010

[11] F O Farid M S Moslehian Q-W Wang and Z-C Wu ldquoOnthe Hermitian solutions to a system of adjointable operatorequationsrdquo Linear Algebra and its Applications vol 437 no 7pp 1854ndash1891 2012

[12] Z-H He and Q-WWang ldquoSolutions to optimization problemson ranks and inertias of a matrix function with applicationsrdquoAppliedMathematics andComputation vol 219 no 6 pp 2989ndash3001 2012

[13] Q-W Wang ldquoThe general solution to a system of real quater-nion matrix equationsrdquo Computers amp Mathematics with Appli-cations vol 49 no 5-6 pp 665ndash675 2005

[14] Q-W Wang ldquoBisymmetric and centrosymmetric solutions tosystems of real quaternion matrix equationsrdquo Computers ampMathematics with Applications vol 49 no 5-6 pp 641ndash6502005

[15] Q W Wang ldquoA system of four matrix equations over vonNeumann regular rings and its applicationsrdquo Acta MathematicaSinica vol 21 no 2 pp 323ndash334 2005

[16] Q-W Wang ldquoA system of matrix equations and a linear matrixequation over arbitrary regular rings with identityrdquo LinearAlgebra and its Applications vol 384 pp 43ndash54 2004

[17] Q-WWang H-X Chang andQNing ldquoThe common solutionto six quaternion matrix equations with applicationsrdquo AppliedMathematics and Computation vol 198 no 1 pp 209ndash2262008

[18] Q-W Wang H-X Chang and C-Y Lin ldquoP-(skew)symmetriccommon solutions to a pair of quaternion matrix equationsrdquoApplied Mathematics and Computation vol 195 no 2 pp 721ndash732 2008

[19] Q W Wang and Z H He ldquoSome matrix equations withapplicationsrdquo Linear and Multilinear Algebra vol 60 no 11-12pp 1327ndash1353 2012

[20] Q W Wang and J Jiang ldquoExtreme ranks of (skew-)Hermitiansolutions to a quaternion matrix equationrdquo Electronic Journal ofLinear Algebra vol 20 pp 552ndash573 2010

[21] Q-W Wang and C-K Li ldquoRanks and the least-norm of thegeneral solution to a system of quaternion matrix equationsrdquoLinear Algebra and its Applications vol 430 no 5-6 pp 1626ndash1640 2009

[22] Q-W Wang X Liu and S-W Yu ldquoThe common bisymmetricnonnegative definite solutions with extreme ranks and inertiasto a pair of matrix equationsrdquo Applied Mathematics and Com-putation vol 218 no 6 pp 2761ndash2771 2011

Journal of Applied Mathematics 9

[23] Q-W Wang F Qin and C-Y Lin ldquoThe common solution tomatrix equations over a regular ring with applicationsrdquo IndianJournal of Pure andAppliedMathematics vol 36 no 12 pp 655ndash672 2005

[24] Q-W Wang G-J Song and C-Y Lin ldquoExtreme ranks of thesolution to a consistent system of linear quaternion matrixequations with an applicationrdquo Applied Mathematics and Com-putation vol 189 no 2 pp 1517ndash1532 2007

[25] Q-W Wang G-J Song and C-Y Lin ldquoRank equalities relatedto the generalized inverse 119860

(2)

119879119878with applicationsrdquo Applied

Mathematics and Computation vol 205 no 1 pp 370ndash3822008

[26] Q W Wang G J Song and X Liu ldquoMaximal and minimalranks of the common solution of some linear matrix equationsover an arbitrary division ring with applicationsrdquo AlgebraColloquium vol 16 no 2 pp 293ndash308 2009

[27] Q-W Wang Z-C Wu and C-Y Lin ldquoExtremal ranks ofa quaternion matrix expression subject to consistent systemsof quaternion matrix equations with applicationsrdquo AppliedMathematics and Computation vol 182 no 2 pp 1755ndash17642006

[28] Q W Wang and S W Yu ldquoRanks of the common solution tosome quaternion matrix equations with applicationsrdquo Bulletinof Iranian Mathematical Society vol 38 no 1 pp 131ndash157 2012

[29] Q W Wang S W Yu and W Xie ldquoExtreme ranks of realmatrices in solution of the quaternion matrix equation 119860119883119861 =

119862with applicationsrdquoAlgebra Colloquium vol 17 no 2 pp 345ndash360 2010

[30] Q-W Wang S-W Yu and Q Zhang ldquoThe real solutions toa system of quaternion matrix equations with applicationsrdquoCommunications in Algebra vol 37 no 6 pp 2060ndash2079 2009

[31] Q-W Wang S-W Yu and C-Y Lin ldquoExtreme ranks of alinear quaternionmatrix expression subject to triple quaternionmatrix equations with applicationsrdquo Applied Mathematics andComputation vol 195 no 2 pp 733ndash744 2008

[32] Q-W Wang and F Zhang ldquoThe reflexive re-nonnegativedefinite solution to a quaternion matrix equationrdquo ElectronicJournal of Linear Algebra vol 17 pp 88ndash101 2008

[33] Q W Wang X Zhang and Z H He ldquoOn the Hermitianstructures of the solution to a pair of matrix equationsrdquo Linearand Multilinear Algebra vol 61 no 1 pp 73ndash90 2013

[34] X Zhang Q-W Wang and X Liu ldquoInertias and ranks ofsome Hermitian matrix functions with applicationsrdquo CentralEuropean Journal of Mathematics vol 10 no 1 pp 329ndash3512012

[35] Y G Tian and S Z Cheng ldquoThemaximal andminimal ranks of119860 minus 119861119883119862 with applicationsrdquo New York Journal of Mathematicsvol 9 pp 345ndash362 2003

[36] Y G Tian ldquoUpper and lower bounds for ranks of matrixexpressions using generalized inversesrdquo Linear Algebra and itsApplications vol 355 pp 187ndash214 2002

[37] Y H Liu and Y G Tian ldquoMore on extremal ranks of thematrix expressions119860minus119861119883plusmn119883

lowast

119861lowast with statistical applicationsrdquo

Numerical Linear Algebra with Applications vol 15 no 4 pp307ndash325 2008

[38] Y H Liu and Y G Tian ldquoMax-min problems on the ranksand inertias of the matrix expressions 119860 minus 119861119883119862 plusmn (119861119883119862)

lowast withapplicationsrdquo Journal of Optimization Theory and Applicationsvol 148 no 3 pp 593ndash622 2011

[39] T Ando ldquoGeneralized Schur complementsrdquo Linear Algebra andits Applications vol 27 pp 173ndash186 1979

[40] D Carlson E Haynsworth and T Markham ldquoA generalizationof the Schur complement by means of the Moore-Penroseinverserdquo SIAM Journal onAppliedMathematics vol 26 pp 169ndash175 1974

[41] M Fiedler ldquoRemarks on the Schur complementrdquo Linear Algebraand its Applications vol 39 pp 189ndash195 1981

[42] Y G Tian ldquoMore on maximal and minimal ranks of Schurcomplements with applicationsrdquo Applied Mathematics andComputation vol 152 no 3 pp 675ndash692 2004

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 217540 14 pageshttpdxdoiorg1011552013217540

Research ArticleAn Efficient Algorithm for the Reflexive Solution of theQuaternion Matrix Equation 119860119883119861 + 119862119883119867119863 = 119865

Ning Li12 Qing-Wen Wang1 and Jing Jiang3

1 Department of Mathematics Shanghai University Shanghai 200444 China2 School of Mathematics and Quantitative Economics Shandong University of Finance and Economics Jinan 250002 China3Department of Mathematics Qilu Normal University Jinan 250013 China

Correspondence should be addressed to Qing-Wen Wang wqw858yahoocomcn

Received 3 October 2012 Accepted 4 December 2012

Academic Editor P N Shivakumar

Copyright copy 2013 Ning Li et al This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We propose an iterative algorithm for solving the reflexive solution of the quaternion matrix equation 119860119883119861 + 119862119883119867

119863 = 119865 Whenthe matrix equation is consistent over reflexive matrix X a reflexive solution can be obtained within finite iteration steps in theabsence of roundoff errors By the proposed iterative algorithm the least Frobenius norm reflexive solution of the matrix equationcan be derived when an appropriate initial iterative matrix is chosen Furthermore the optimal approximate reflexive solution to agiven reflexive matrix119883

0can be derived by finding the least Frobenius norm reflexive solution of a new corresponding quaternion

matrix equation Finally two numerical examples are given to illustrate the efficiency of the proposed methods

1 Introduction

Throughout the paper the notations R119898times119899 and H119898times119899 repre-sent the set of all 119898 times 119899 real matrices and the set of all 119898 times 119899

matrices over the quaternion algebraH = 1198861+ 119886

2119894+ 119886

3119895+ 119886

4119896 |

1198942

= 1198952

= 1198962

= 119894119895119896 = minus1 1198861 119886

2 119886

3 119886

4isin R We denote the

identity matrix with the appropriate size by 119868 We denote theconjugate transpose the transpose the conjugate the tracethe column space the real part the 119898119899 times 1 vector formedby the vertical concatenation of the respective columnsof a matrix 119860 by 119860

119867

119860119879

119860 tr(119860) 119877(119860)Re(119860) and vec(119860)respectivelyThe Frobenius norm of119860 is denoted by 119860 thatis 119860 = radictr(119860119867119860) Moreover119860otimes119861 and119860⊙119861 stand for theKronecker matrix product and Hadmard matrix product ofthe matrices 119860 and 119861

Let 119876 isin H119899times119899 be a generalized reflection matrix that is119876

2

= 119868 and 119876119867

= 119876 A matrix 119860 is called reflexive withrespect to the generalized reflection matrix 119876 if 119860 = 119876119860119876It is obvious that any matrix is reflexive with respect to 119868Let RH119899times119899

(119876) denote the set of order 119899 reflexive matriceswith respect to 119876 The reflexive matrices with respect to ageneralized reflection matrix 119876 have been widely used inengineering and scientific computations [1 2]

In the field of matrix algebra quaternion matrix equa-tions have received much attention Wang et al [3] gavenecessary and sufficient conditions for the existence andthe representations of P-symmetric and P-skew-symmetricsolutions of quaternion matrix equations 119860

119886119883 = 119862

119886and

119860119887119883119861

119887= 119862

119887 Yuan and Wang [4] derived the expressions of

the least squares 120578-Hermitian solution with the least normand the expressions of the least squares anti-120578-Hermitiansolution with the least norm for the quaternion matrixequation 119860119883119861 + 119862119883119863 = 119864 Jiang and Wei [5] derivedthe explicit solution of the quaternion matrix equation 119883 minus

119860119883119861 = 119862 Li and Wu [6] gave the expressions of symmetricand skew-antisymmetric solutions of the quaternion matrixequations 119860

1119883 = 119862

1and 119883119861

3= 119862

3 Feng and Cheng [7]

gave a clear description of the solution set of the quaternionmatrix equation 119860119883 minus 119883119861 = 0

The iterative method is a very important method tosolve matrix equations Peng [8] constructed a finite iterationmethod to solve the least squares symmetric solutions oflinear matrix equation 119860119883119861 = 119862 Also Peng [9ndash11] presentedseveral efficient iteration methods to solve the constrainedleast squares solutions of linear matrix equations 119860119883119861 =

119862 and 119860119883119861 + 119862119884119863 = 119864 by using Paigersquos algorithm [12]

2 Journal of Applied Mathematics

as the frame method Duan et al [13ndash17] proposed iterativealgorithms for the (Hermitian) positive definite solutionsof some nonlinear matrix equations Ding et al proposedthe hierarchical gradient-based iterative algorithms [18] andhierarchical least squares iterative algorithms [19] for solvinggeneral (coupled) matrix equations based on the hierarchi-cal identification principle [20] Wang et al [21] proposedan iterative method for the least squares minimum-normsymmetric solution of 119860X119861 = 119864 Dehghan and Hajarianconstructed finite iterative algorithms to solve several linearmatrix equations over (anti)reflexive [22ndash24] generalizedcentrosymmetric [25 26] and generalized bisymmetric [2728] matrices Recently Wu et al [29ndash31] proposed iterativealgorithms for solving various complex matrix equations

However to the best of our knowledge there has beenlittle information on iterative methods for finding a solutionof a quaternionmatrix equation Due to the noncommutativemultiplication of quaternions the study of quaternionmatrixequations is more complex than that of real and complexequations Motivated by the work mentioned above andkeeping the interests and wide applications of quaternionmatrices in view (eg [32ndash45]) we in this paper consideran iterative algorithm for the following two problems

Problem 1 For given matrices 119860119862 isin H119898times119899

119861 119863 isin

H119899times119901

119865 isin H119898times119901 and the generalized reflectionmatrix119876 find119883 isin RH119899times119899

(119876) such that

119860119883119861 + 119862119883119867

119863 = 119865 (1)

Problem 2 When Problem 1 is consistent let its solutionset be denoted by 119878

119867 For a given reflexive matrix 119883

0isin

RH119899times119899

(119876) find119883 isin RH119899times119899

(119876) such that10038171003817100381710038171003817119883 minus 119883

0

10038171003817100381710038171003817= min

119883isin119878119867

1003817100381710038171003817119883 minus 1198830

1003817100381710038171003817 (2)

The remainder of this paper is organized as followsIn Section 2 we give some preliminaries In Section 3 weintroduce an iterative algorithm for solving Problem 1 Thenwe prove that the given algorithm can be used to obtain areflexive solution for any initial matrix within finite stepsin the absence of roundoff errors Also we prove that theleast Frobenius norm reflexive solution can be obtained bychoosing a special kind of initial matrix In addition theoptimal reflexive solution of Problem 2 by finding the leastFrobenius norm reflexive solution of a new matrix equationis given In Section 4 we give two numerical examples toillustrate our results In Section 5 we give some conclusionsto end this paper

2 Preliminary

In this section we provide some results which will playimportant roles in this paper First we give a real innerproduct for the space H119898times119899 over the real field R

Theorem 3 In the space H119898times119899 over the field R a real innerproduct can be defined as

⟨119860 119861⟩ = Re [tr (119861119867

119860)] (3)

for 119860 119861 isin H119898times119899 This real inner product space is denoted as(H119898times119899

R ⟨sdot sdot⟩)

Proof (1) For 119860 isin H119898times119899 let 119860 = 1198601+ 119860

2119894 + 119860

3119895 + 119860

4119896 then

⟨119860 119860⟩ = Re [tr (119860119867

119860)]

= tr (119860119879

1119860

1+ 119860

119879

2119860

2+ 119860

119879

3119860

3+ 119860

119879

4119860

4)

(4)

It is obvious that ⟨119860 119860⟩ gt 0 and ⟨119860 119860⟩ = 0 hArr 119860 = 0(2) For 119860 119861 isin H119898times119899 let 119860 = 119860

1+ 119860

2119894 + 119860

3119895 + 119860

4119896 and

119861 = 1198611+ 119861

2119894 + 119861

3119895 + 119861

4119896 then we have

⟨119860 119861⟩ = Re [tr (119861119867

119860)]

= tr (119861119879

1119860

1+ 119861

119879

2119860

2+ 119861

119879

3119860

3+ 119861

119879

4119860

4)

= tr (119860119879

1119861

1+ 119860

119879

2119861

2+ 119860

119879

3119861

3+ 119860

119879

4119861

4)

= Re [tr (119860119867

119861)] = ⟨119861 119860⟩

(5)

(3) For 119860 119861 119862 isin H119898times119899

⟨119860 + 119861 119862⟩ = Re tr [119862119867

(119860 + 119861)]

= Re [tr (119862119867

119860 + 119862119867

119861)]

= Re [tr (119862119867

119860)] + Re [tr (119862119867

119861)]

= ⟨119860 119862⟩ + ⟨119861 119862⟩

(6)

(4) For 119860 119861 isin H119898times119899 and 119886 isin R

⟨119886119860 119861⟩ = Re tr [119861119867

(119886119860)] = Re [tr (119886119861119867

119860)]

= 119886Re [tr (119861119867

119860)] = 119886 ⟨119860 119861⟩

(7)

All the above arguments reveal that the space H119898times119899 overfield R with the inner product defined in (3) is an innerproduct space

Let sdot Δrepresent the matrix norm induced by the inner

product ⟨sdot sdot⟩ For an arbitrary quaternion matrix 119860 isin H119898times119899it is obvious that the following equalities hold

119860Δ= radic⟨119860119860⟩ = radicRe [tr (119860119867119860)] = radictr (119860119867119860) = 119860

(8)

which reveals that the induced matrix norm is exactly theFrobenius norm For convenience we still use sdot to denotethe induced matrix norm

Let 119864119894119895

denote the 119898 times 119899 quaternion matrix whose(119894 119895) entry is 1 and the other elements are zeros In innerproduct space (H119898times119899

R ⟨sdot sdot⟩) it is easy to verify that 119864119894119895 119864

119894119895119894

119864119894119895119895 119864

119894119895119896 119894 = 1 2 119898 119895 = 1 2 119899 is an orthonormal

basis which reveals that the dimension of the inner productspace (H119898times119899

R ⟨sdot sdot⟩) is 4119898119899Next we introduce a real representation of a quaternion

matrix

Journal of Applied Mathematics 3

For an arbitrary quaternion matrix 119872 = 1198721+ 119872

2119894 +

1198723119895+ 119872

4119896 a map 120601(sdot) fromH119898times119899 toR4119898times4119899 can be defined

as

120601 (119872) =

[[[

[

1198721

minus1198722

minus1198723

minus1198724

1198722

1198721

minus1198724

1198723

1198723

1198724

1198721

minus1198722

1198724

minus1198723

1198722

1198721

]]]

]

(9)

Lemma4 (see [41]) Let119872 and119873 be two arbitrary quaternionmatrices with appropriate size The map 120601(sdot) defined by (9)satisfies the following properties

(a) 119872 = 119873 hArr 120601(119872) = 120601(119873)

(b) 120601(119872 + 119873) = 120601(119872) + 120601(119873) 120601(119872119873) = 120601(119872)120601(119873)

120601(119896119872) = 119896120601(119872) 119896 isin R

(c) 120601(119872119867

) = 120601119879

(119872)

(d) 120601(119872) = 119879minus1

119898120601(119872)119879

119899= 119877

minus1

119898120601(119872)119877

119899= 119878

minus1

119898120601(119872)119878

119899

where

119879119905=

[[[

[

0 minus119868119905

0 0

119868119905

0 0 0

0 0 0 119868119905

0 0 minus119868119905

0

]]]

]

119877119905= [

0 minus1198682119905

1198682119905

0]

119878119905=

[[[

[

0 0 0 minus119868119905

0 0 119868119905

0

0 minus119868119905

0 0

119868119905

0 0 0

]]]

]

119905 = 119898 119899

(10)

By (9) it is easy to verify that

1003817100381710038171003817120601 (119872)1003817100381710038171003817 = 2 119872 (11)

Finally we introduce the commutation matrixA commutation matrix 119875(119898 119899) is a 119898119899 times 119898119899 matrix

which has the following explicit form

119875 (119898 119899) =

119898

sum

119894=1

119899

sum

119895=1

119864119894119895otimes 119864

119879

119894119895= [119864

119879

119894119895] 119894=1119898

119895=1119899

(12)

Moreover 119875(119898 119899) is a permutation matrix and 119875(119898 119899) =

119875119879

(119899119898) = 119875minus1

(119899119898) We have the following lemmas on thecommutation matrix

Lemma 5 (see [46]) Let 119860 be a 119898 times 119899 matrix There is acommutation matrix 119875(119898 119899) such that

vec (119860119879

) = 119875 (119898 119899) vec (119860) (13)

Lemma 6 (see [46]) Let 119860 be a 119898 times 119899 matrix and 119861 a 119901 times

119902 matrix There exist two commutation matrices 119875(119898 119901) and119875(119899 119902) such that

119861 otimes 119860 = 119875119879

(119898 119901) (119860 otimes 119861) 119875 (119899 119902) (14)

3 Main Results

31 The Solution of Problem 1 In this subsection we willconstruct an algorithm for solving Problem 1 Then somelemmaswill be given to analyse the properties of the proposedalgorithm Using these lemmas we prove that the proposedalgorithm is convergent

Algorithm 7 (Iterative algorithm for Problem 1)

(1) Choose an initial matrix119883(1) isin RH119899times119899

(119876)(2) Calculate

119877 (1) = 119865 minus 119860119883 (1) 119861 minus 119862119883119867

(1)119863

119875 (1) =1

2(119860

119867

119877 (1) 119861119867

+ 119863119877119867

(1) 119862

+ 119876119860119867

119877 (1) 119861119867

119876 + 119876119863119877119867

(1) 119862119876)

119896 = 1

(15)

(3) If 119877(119896) = 0 then stop and 119883(119896) is the solution ofProblem 1 else if 119877(119896) =0 and 119875(119896) = 0 then stopand Problem 1 is not consistent else 119896 = 119896 + 1

(4) Calculate

119883 (119896) = 119883 (119896 minus 1) +119877 (119896 minus 1)

2

119875 (119896 minus 1)2119875 (119896 minus 1)

119877 (119896) = 119877 (119896 minus 1) minus119877 (119896 minus 1)

2

119875 (119896 minus 1)2

times (119860119875 (119896 minus 1) 119861 + 119862119875119867

(119896 minus 1)119863)

119875 (119896) =1

2(119860

119867

119877 (119896) 119861119867

+ 119863119877119867

(119896) 119862

+ 119876119860119867

119877 (119896) 119861119867

119876 +119876119863119877119867

(119896) 119862119876)

+119877 (119896)

2

119877 (119896 minus 1)2119875 (119896 minus 1)

(16)

(5) Go to Step (3)

Lemma 8 Assume that the sequences 119877(119894) and 119875(119894)

are generated by Algorithm 7 then ⟨119877(119894) 119877(119895)⟩ =

0 and ⟨119875(119894) 119875(119895)⟩ = 0 for 119894 119895 = 1 2

119894 =119895

Proof Since ⟨119877(119894) 119877(119895)⟩ = ⟨119877(119895) 119877(119894)⟩ and⟨119875(119894) 119875(119895)⟩ = ⟨119875(119895) 119875(119894)⟩ for 119894 119895 = 1

2 we only need to prove that ⟨119877(119894)

119877(119895)⟩ = 0 and ⟨119875(119894) 119875(119895)⟩ = 0 for 1 le 119894 lt 119895Now we prove this conclusion by induction

Step 1 We show that

⟨119877 (119894) 119877 (119894 + 1)⟩ = 0

⟨119875 (119894) 119875 (119894 + 1)⟩ = 0 for 119894 = 1 2

(17)

4 Journal of Applied Mathematics

We also prove (17) by induction When 119894 = 1 we have

⟨119877 (1) 119877 (2)⟩

= Re tr [119877119867

(2) 119877 (1)]

= Retr[(119877 (1) minus119877 (1)

2

119875 (1)2(119860119875 (1) 119861 + 119862119875

119867

(1)119863))

119867

times 119877 (1) ]

= 119877 (1)2

minus119877 (1)

2

119875 (1)2

times Re tr [119875119867

(1) (119860119867

119877 (1) 119861119867

+ 119863119877119867

(1) 119862)]

= 119877(1)2

minus119877(1)

2

119875(1)2

timesRetr [119875119867

(1) ((119860119867

119877 (1) 119861119867

+119863119877119867

(1) 119862+119876119860119867

119877 (1)

times 119861119867

119876 + 119876119863119877119867

(1) 119862119876) times (2)minus1

)]

= 119877 (1)2

minus119877 (1)

2

119875 (1)2119875 (1)

2

= 0

(18)

Also we can write

⟨119875 (1) 119875 (2)⟩

= Re tr [119875119867

(2) 119875 (1)]

= Retr[(1

2(119860

119867

119877 (2) 119861119867

+ 119863119877119867

(2) 119862

+ 119876119860119867

119877 (2) 119861119867

119876 + 119876119863119877119867

(2) 119862119876)

+119877 (2)

2

119877 (1)2119875 (1))

119867

119875 (1)]

=119877 (2)

2

119877 (1)2119875 (1)

2

+ Re tr [119875119867

(1) times ((119860119867

119877 (2) 119861119867

+ 119863119877119867

(2) 119862

+ 119876119860119867

119877 (2) 119861119867

119876

+119876119863119877119867

(2) 119862119876) times (2)minus1

)]

=119877 (2)

2

119877 (1)2119875 (1)

2

+ Re tr [119875119867

(1) (119860119867

119877 (2) 119861119867

+ 119863119877119867

(2) 119862)]

=119877 (2)

2

119877 (1)2119875 (1)

2

+ Re tr [119877119867

(2) (119860119875 (1) 119861 + 119862119875119867

(1)119863)]

=119877 (2)

2

119877 (1)2119875 (1)

2

+119875 (1)

2

119877 (1)2Re tr [119877119867

(2) (119877 (1) minus 119877 (2))]

=119877 (2)

2

119877 (1)2119875 (1)

2

minus119875 (1)

2

119877 (1)2119877 (2)

2

= 0

(19)

Now assume that conclusion (17) holds for 1 le 119894 le 119904 minus 1 then

⟨119877 (119904) 119877 (119904 + 1)⟩

= Re tr [119877119867

(119904 + 1) 119877 (119904)]

= Retr[(119877 (119904) minus119877 (119904)

2

119875 (119904)2(119860119875 (119904) 119861 + 119862119875

119867

(119904)119863))

119867

times 119877 (119904) ]

= 119877 (s)2 minus119877 (119904)

2

119875 (119904)2

times Re tr [119875119867

(119904) (119860119867

119877 (119904) 119861119867

+ 119863119877119867

(119904) 119862)]

= 119877 (119904)2

minus119877 (119904)

2

119875 (119904)2

times Re tr [119875119867

(119904) times ((119860119867

119877 (119904) 119861119867

+ 119863119877119867

(119904) 119862

+ 119876119860119867

119877 (119904) 119861119867

119876

+ 119876119863119877119867

(119904) 119862119876) times (2)minus1

)]

= 119877 (119904)2

minus119877 (119904)

2

119875 (119904)2

times Retr[119875119867

(119904) (119875 (119904) minus119877 (119904)

2

119877 (119904 minus 1)2119875 (119904 minus 1))]

= 0

(20)

Journal of Applied Mathematics 5

And it can also be obtained that

⟨119875 (119904) 119875 (119904 + 1)⟩

= Re tr [119875119867

(119904 + 1) 119875 (119904)]

= Retr[(((119860119867

119877 (119904 + 1) 119861119867

+ 119863119877119867

(119904 + 1) 119862

+ 119876119860119867

119877 (119904 + 1) 119861119867

119876

+ 119876119863119877119867

(119904 + 1) 119862119876) times (2)minus1

)

+119877 (119904 + 1)

2

119877 (119904)2

119875(119904))

119867

119875 (119904)]

=119877 (119904 + 1)

2

119877 (119904)2

119875 (119904)2

+ Re tr [119875119867

(119904) (119860119867

119877 (119904 + 1) 119861119867

+ 119863119877119867

(119904 + 1) 119862)]

=119877 (119904 + 1)

2

119877 (119904)2

119875 (119904)2

+ Re tr [119877119867

(119904 + 1) (119860119875 (119904) 119861 + 119862119875119867

(119904)119863)]

=119877 (119904 + 1)

2

119877 (119904)2

119875 (119904)2

+119875 (119904)

2

119877 (119904)2Re tr [119877119867

(119904 + 1) (119877 (119904) minus 119877 (119904 + 1))]

= 0

(21)

Therefore the conclusion (17) holds for 119894 = 1 2 Step 2 Assume that ⟨119877(119894) 119877(119894+119903)⟩ = 0 and ⟨119875(119894) 119875(119894+119903)⟩ = 0

for 119894 ge 1 and 119903 ge 1 We will show that

⟨119877 (119894) 119877 (119894 + 119903 + 1)⟩ = 0 ⟨119875 (119894) 119875 (119894 + 119903 + 1)⟩ = 0

(22)

We prove the conclusion (22) in two substepsSubstep 21 In this substep we show that

⟨119877 (1) 119877 (119903 + 2)⟩ = 0 ⟨119875 (1) 119875 (119903 + 2)⟩ = 0 (23)

It follows from Algorithm 7 that

⟨119877 (1) 119877 (119903 + 2)⟩

= Re tr [119877119867

(119903 + 2) 119877 (1)]

= Retr[(119877 (119903 + 1) minus119877 (119903 + 1)

2

119875 (119903 + 1)2

times (119860119875 (119903 + 1) 119861+119862119875119867

(119903 + 1)119863))

119867

119877 (1)]

= Re tr [119877119867

(119903 + 1) 119877 (1)] minus119877 (119903 + 1)

2

119875 (119903 + 1)2

times Re tr [119875119867

(119903 + 1) (119860119867

119877 (1) 119861119867

+ 119863119877119867

(1) 119862)]

= Re tr [119877119867

(119903 + 1) 119877 (1)] minus119877 (119903 + 1)

2

119875 (119903 + 1)2

times Re tr [119875119867

(119903 + 1) ((119860119867

119877 (1) 119861119867

+ 119863119877119867

(1) 119862

+ 119876119860119867

119877 (1) 119861119867

119876

+ 119876119863119877119867

(1) 119862119876) times (2)minus1

)]

= Re tr [119877119867

(119903 + 1) 119877 (1)]

minus119877 (119903 + 1)

2

119875 (119903 + 1)2Re tr [119875119867

(119903 + 1) 119875 (1)]

= 0

(24)

Also we can write

⟨119875 (1) 119875 (119903 + 2)⟩

= Re tr [119875119867

(119903 + 2) 119875 (1)]

= Retr[( ((119860119867

119877 (119903 + 2) 119861119867

+ 119863119877119867

(119903 + 2) 119862

+119876119860119867

119877 (119903 + 2) 119861119867

119876 + 119876119863119877119867

(119903 + 2) 119862119876)

times (2)minus1

) +119877(119903 + 2)

2

119877(119903 + 1)2119875(119903 + 1))

119867

119875 (1)]

= Re tr [(119860119867

119877 (119903 + 2) 119861119867

+ 119863119877119867

(119903 + 2) 119862)119867

119875 (1)]

+119877 (119903 + 2)

2

119877 (119903 + 1)2Re tr [119875119867

(119903 + 1) 119875 (1)]

= Re tr [119877119867

(119903 + 2) (119860119875 (1) 119861 + 119862119875119867

(1)119863)]

+119877 (119903 + 2)

2

119877 (119903 + 1)2Re tr [119875119867

(119903 + 1) 119875 (1)]

6 Journal of Applied Mathematics

=119875 (1)

2

119877 (1)2Re tr [119877119867

(119903 + 2) (119877 (1) minus 119877 (2))]

+119877 (119903 + 2)

2

119877 (119903 + 1)2Re tr [119875119867

(119903 + 1) 119875 (1)]

= 0

(25)

Substep 22 In this substep we prove the conclusion (22) inStep 2 It follows from Algorithm 7 that

⟨119877 (119894) 119877 (119894 + 119903 + 1)⟩

= Re tr [119877119867

(119894 + 119903 + 1) 119877 (119894)]

= Retr[(119877 (119894 + 119903) minus119877 (119894 + 119903)

2

119875 (119894 + 119903)2

times (119860119875 (119894 + 119903) 119861 + 119862119875119867

(119894 + 119903)119863))

119867

119877 (119894) ]

= Re tr [119877119867

(119894 + 119903) 119877 (119894)] minus119877 (119894 + 119903)

2

119875 (119894 + 119903)2

times Re tr [119875119867

(119894 + 119903) (119860119867

119877 (119894) 119861119867

+ 119863119877119867

(119894) 119862)]

= minus119877 (119894 + 119903)

2

119875 (119894 + 119903)2

times Re tr [119875119867

(119894 + 119903) ((119860119867

119877 (119894) 119861119867

+ 119863119877119867

(119894) 119862

+ 119876119860119867

119877 (119894) 119861119867

119876

+ 119876119863119877119867

(119894) 119862119876) times (2)minus1

)]

= minus119877 (119894 + 119903)

2

119875 (119894 + 119903)2

times Retr[119875119867

(119894 + 119903) (119875 (119894) minus119877 (119894)

2

119877 (119894 minus 1)2119875 (119894 minus 1))]

=119877 (119894 + 119903)

2

119877 (119894)2

119875 (119894 + 119903)2

119877 (119894 minus 1)2Re tr [119875119867

(119894 + 119903) 119875 (119894 minus 1)]

⟨119875 (119894) 119875 (119894 + 119903 + 1)⟩

= Re tr [119875119867

(119894 + 119903 + 1) 119875 (119894)]

= Retr[(1

2(119860

119867

119877 (119894 + 119903 + 1) 119861119867

+ 119863119877119867

(119894 + 119903 + 1) 119862

+ 119876119860119867

119877 (119894 + 119903 + 1) 119861119867

119876

+ 119876119863119877119867

(119894 + 119903 + 1) 119862119876)

+119877(119894 + 119903 + 1)

2

119877(119894 + 119903)2

119875(119894 + 119903))

119867

119875 (119894)]

= Re tr [(119860119867

119877 (119894 + 119903 + 1) 119861119867

+119863119877119867

(119894+119903+1) 119862)119867

119875 (119894)]

= Re tr [119877119867

(119894 + 119903 + 1) (119860119875 (119894)B + 119862119875119867

(119894) 119863)]

=119875 (119894)

2

119877 (119894)2Re tr [119877119867

(119894 + 119903 + 1) (119877 (119894) minus 119877 (119894 + 1))]

=119875 (119894)

2

119877 (119894)2Re tr [119877119867

(119894 + 119903 + 1) 119877 (119894)]

minus119875 (119894)

2

119877 (119894)2Re tr [119877119867

(119894 + 119903 + 1) 119877 (119894 + 1)]

=119875 (119894)

2

119877 (119894 + 119903)2

119877 (119894)2

119877 (119894)2

119875 (119894 + 119903)2

119877 (119894 minus 1)2

times Re tr [119875119867

(119894 + 119903) 119875 (119894 minus 1)]

(26)

Repeating the above process (26) we can obtain

⟨119877 (119894) 119877 (119894 + 119903 + 1)⟩ = sdot sdot sdot = 120572Re tr [119875119867

(119903 + 2) 119875 (1)]

⟨119875 (119894) 119875 (119894 + 119903 + 1)⟩ = sdot sdot sdot = 120573Re tr [119875119867

(119903 + 2) 119875 (1)]

(27)

Combining these two relations with (24) and (25) it impliesthat (22) holds So by the principle of induction we knowthat Lemma 8 holds

Lemma 9 Assume that Problem 1 is consistent and let 119883 isin

RH119899times119899

(119876) be its solution Then for any initial matrix 119883(1) isin

RH119899times119899

(119876) the sequences 119877(119894) 119875(119894) and 119883(119894) generatedby Algorithm 7 satisfy

⟨119875 (119894) 119883 minus 119883 (119894)⟩ = 119877 (119894)2

119894 = 1 2 (28)

Proof We also prove this conclusion by inductionWhen 119894 = 1 it follows from Algorithm 7 that

⟨119875 (1) 119883 minus 119883 (1)⟩

= Re tr [(119883 minus 119883 (1))119867

119875 (1)]

= Re tr [(119883 minus 119883 (1))119867

times ((119860119867

119877 (1) 119861119867

+ 119863119877119867

(1) 119862+119876119860119867

119877 (1) 119861119867

119876

+ 119876119863119877119867

(1) 119862119876) times (2)minus1

) ]

Journal of Applied Mathematics 7

= Re tr [(119883 minus 119883 (1))119867

(119860119867

119877 (1) 119861119867

+ 119863119877119867

(1) 119862)]

= Retr[119877119867

(1) (119860 (119883 minus 119883 (1)) 119861+119862 (119883 minus 119883 (1))119867

119863)]

= Re tr [119877119867

(1) 119877 (1)]

= 119877 (1)2

(29)

This implies that (28) holds for 119894 = 1Now it is assumed that (28) holds for 119894 = 119904 that is

⟨119875 (119904) 119883 minus 119883 (119904)⟩ = 119877 (119904)2

(30)

Then when 119894 = 119904 + 1

⟨119875 (119904 + 1) 119883 minus 119883 (119904 + 1)⟩

= Re tr [(119883 minus 119883 (119904 + 1))119867

119875 (119904 + 1)]

= Retr[(119883 minus 119883 (119904 + 1))119867

times (1

2(119860

119867

119877 (119904 + 1) 119861119867

+ 119863119877119867

(119904 + 1) 119862

+ 119876119860119867

119877 (119904 + 1) 119861119867

119876

+ 119876119863119877119867

(119904 + 1) 119862119876)

+119877(119904 + 1)

2

119877(119904)2

119875 (119904))]

= Re tr [(119883 minus 119883 (119904 + 1))119867

times (119860119867

119877 (119904 + 1) 119861119867

+ 119863119877119867

(119904 + 1) 119862) ]

+119877 (119904 + 1)

2

119877 (119904)2

Re tr [(119883 minus 119883 (119904 + 1))119867

119875 (119904)]

= Re tr [119877119867

(119904 + 1)

times (119860 (119883 minus 119883 (119904 + 1)) 119861

+ 119862 (119883 minus 119883 (119904 + 1))119867

119863)]

+119877 (119904 + 1)

2

119877 (119904)2

Re tr [(119883 minus 119883 (119904))119867

119875 (119904)]

minus Re tr [(119883 (119904 + 1) minus 119883 (119904))119867

119875 (119904)]

= 119877 (119904 + 1)2

+119877 (119904 + 1)

2

119877 (119904)2

times 119877 (119904)2

minus119877 (119904)

2

119875 (119904)2Re tr [119875119867

(119904) 119875 (119904)]

= 119877 (119904 + 1)2

(31)

Therefore Lemma 9 holds by the principle of induction

From the above two lemmas we have the followingconclusions

Remark 10 If there exists a positive number 119894 such that119877(119894) =0 and 119875(119894) = 0 then we can get from Lemma 9 thatProblem 1 is not consistent Hence the solvability of Problem1 can be determined by Algorithm 7 automatically in theabsence of roundoff errors

Theorem 11 Suppose that Problem 1 is consistentThen for anyinitial matrix119883(1) isin RH119899times119899

(119876) a solution of Problem 1 can beobtained within finite iteration steps in the absence of roundofferrors

Proof In Section 2 it is known that the inner productspace (H119898times119901

119877 ⟨sdot sdot⟩) is 4119898119901-dimensional According toLemma 9 if 119877(119894) =0 119894 = 1 2 4119898119901 then we have119875(119894) =0 119894 = 1 2 4119898119901 Hence 119877(4119898119901+1) and 119875(4119898119901+1)

can be computed From Lemma 8 it is not difficult to get

⟨119877 (119894) 119877 (119895)⟩ = 0 for 119894 119895 = 1 2 4119898119901 119894 =119895 (32)

Then 119877(1) 119877(2) 119877(4119898119901) is an orthogonal basis of theinner product space (H119898times119901

119877 ⟨sdot sdot⟩) In addition we can getfrom Lemma 8 that

⟨119877 (119894) 119877 (4119898119901 + 1)⟩ = 0 for 119894 = 1 2 4119898119901 (33)

It follows that 119877(4119898119901+1) = 0 which implies that119883(4119898119901+1)

is a solution of Problem 1

32 The Solution of Problem 2 In this subsection firstly weintroduce some lemmas Then we will prove that the leastFrobenius norm reflexive solution of (1) can be derived bychoosing a suitable initial iterative matrix Finally we solveProblem 2 by finding the least Frobenius norm reflexivesolution of a new-constructed quaternion matrix equation

Lemma 12 (see [47]) Assume that the consistent system oflinear equations119872119910 = 119887 has a solution 119910

0isin 119877(119872

119879

) then 1199100is

the unique least Frobenius norm solution of the system of linearequations

Lemma 13 Problem 1 is consistent if and only if the system ofquaternion matrix equations

119860119883119861 + 119862119883119867

119863 = 119865

119860119876119883119876119861 + 119862119876119883119867

119876119863 = 119865

(34)

8 Journal of Applied Mathematics

is consistent Furthermore if the solution sets of Problem 1 and(34) are denoted by 119878

119867and 119878

1

119867 respectively then we have 119878

119867sube

1198781

119867

Proof First we assume that Problem 1 has a solution 119883 By119860119883119861 + 119862119883

119867

119863 = 119865 and 119876119883119876 = 119883 we can obtain 119860119883119861 +

119862119883119867

119863 = 119865 and 119860119876119883119876119861 + 119862119876119883119867

119876119863 = 119865 which impliesthat 119883 is a solution of quaternion matrix equations (34) and119878119867

sube 1198781

119867

Conversely suppose (34) is consistent Let119883 be a solutionof (34) Set 119883

119886= (119883 + 119876119883119876)2 It is obvious that 119883

119886isin

RH119899times119899

(119876) Now we can write

119860119883119886119861 + 119862119883

119867

119886119863

=1

2[119860 (119883 + 119876119883119876)119861 + 119862(119883 + 119876119883119876)

119867

119863]

=1

2[119860119883119861 + 119862119883

119867

119863 + 119860119876119883119876119861 + 119862119876119883119867

119876119863]

=1

2[119865 + 119865]

= 119865

(35)

Hence 119883119886is a solution of Problem 1 The proof is completed

Lemma 14 The system of quaternion matrix equations (34) isconsistent if and only if the system of real matrix equations

120601 (119860) [119883119894119895]4 times 4

120601 (119861) + 120601 (119862) [119883119894119895]119879

4 times 4

120601 (119863) = 120601 (119865)

120601 (119860) 120601 (119876) [119883119894119895]4 times 4

120601 (119876) 120601 (119861)

+ 120601 (119862) 120601 (119876) [119883119894119895]119879

4 times 4

120601 (119876) 120601 (119863) = 120601 (119865)

(36)

is consistent where 119883119894119895

isin H119899times119899

119894 119895 = 1 2 3 4 are submatri-ces of the unknown matrix Furthermore if the solution sets of(34) and (36) are denoted by 119878

1

119867and 119878

2

119877 respectively then we

have 120601(1198781

119867) sube 119878

2

119877

Proof Suppose that (34) has a solution

119883 = 1198831+ 119883

2119894 + 119883

3119895 + 119883

4119896 (37)

Applying (119886) (119887) and (119888) in Lemma 4 to (34) yields

120601 (119860) 120601 (119883) 120601 (119861) + 120601 (119862) 120601119879

(119883) 120601 (119863) = 120601 (119865)

120601 (119860) 120601 (119876) 120601 (119883) 120601 (119876) 120601 (119861)

+ 120601 (119862) 120601 (119876) 120601119879

(119883) 120601 (119876) 120601 (119863) = 120601 (119865)

(38)

which implies that 120601(119883) is a solution of (36) and 120601(1198781

119867) sube 119878

2

119877

Conversely suppose that (36) has a solution

119883 = [119883119894119895]4 times 4

(39)

By (119889) in Lemma 4 we have that

119879minus1

119898120601 (119860) 119879

119899119883119879

minus1

119899120601 (119861) 119879

119901

+ 119879minus1

119898120601 (119862) 119879

119899119883

119879

119879minus1

119899120601 (119863)119879

119901= 119879

minus1

119898120601 (119865) 119879

119901

119877minus1

119898120601 (119860) 119877

119899119883119877

minus1

119899120601 (119861) 119877

119901

+ 119877minus1

119898120601 (119862) 119877

119899119883

119879

119877minus1

119899120601 (119863) 119877

119901= 119877

minus1

119898120601 (119865) 119877

119901

119878minus1

119898120601 (119860) 119878

119899119883119878

minus1

119899120601 (119861) 119878

119901

+ Sminus1

119898120601 (119862) 119878

119899119883

119879

119878minus1

119899120601 (119863) 119878

119901= 119878

minus1

119898120601 (119865) 119878

119901

119879minus1

119898120601 (119860) 120601 (119876) 119879

119899119883119879

minus1

119899120601 (119876) 120601 (119861) 119879

119901

+ 119879minus1

119898120601 (119862) 120601 (119876) 119879

119899119883

119879

119879minus1

119899120601 (119876) 120601 (119863) 119879

119901= 119879

minus1

119898120601 (119865) 119879

119901

119877minus1

119898120601 (119860) 120601 (119876) 119877

119899119883119877

minus1

119899120601 (119876) 120601 (119861) 119877

119901

+ 119877minus1

119898120601 (119862) 120601 (119876) 119877

119899119883

119879

119877minus1

119899120601 (119876) 120601 (119863) 119877

119901= 119877

minus1

119898120601 (119865) 119877

119901

119878minus1

119898120601 (119860) 120601 (119876) 119878

119899119883119878

minus1

119899120601 (119876) 120601 (119861) 119878

119901

+ 119878minus1

119898120601 (119862) 120601 (119876) 119878

119899119883

119879

119878minus1

119899120601 (119876) 120601 (119863) 119878

119901= 119878

minus1

119898120601 (119865) 119878

119901

(40)

Hence

120601 (119860) 119879119899119883119879

minus1

119899120601 (119861) + 120601 (119862) (119879

119899119883119879

minus1

119899)119879

120601 (119863) = 120601 (119865)

120601 (119860) 119877119899119883119877

minus1

119899120601 (119861) + 120601 (119862) (R

119899119883119877

minus1

119899)119879

120601 (119863) = 120601 (119865)

120601 (119860) 119878119899119883119878

minus1

119899120601 (119861) + 120601 (119862) (119878

119899119883119878

minus1

119899)119879

120601 (119863) = 120601 (119865)

120601 (119860) 120601 (119876) 119879119899119883119879

minus1

119899120601 (119876) 120601 (119861)

+ 120601 (119862) 120601 (119876) (119879119899119883119879

minus1

119899)119879

120601 (119876) 120601 (119863) = 120601 (119865)

120601 (119860) 120601 (119876) 119877119899119883119877

minus1

119899120601 (119876) 120601 (119861)

+ 120601 (119862) 120601 (119876) (119877119899119883119877

minus1

119899)119879

120601 (119876) 120601 (119863) = 120601 (119865)

120601 (119860) 120601 (119876) 119878119899119883119878

minus1

119899120601 (119876) 120601 (119861)

+ 120601 (119862) 120601 (119876) (119878119899119883119878

minus1

119899)119879

120601 (119876) 120601 (119863) = 120601 (119865)

(41)

which implies that 119879119899119883119879

minus1

119899 119877

119899119883119877

minus1

119899 and 119878

119899119883119878

minus1

119899are also

solutions of (36) Thus

1

4(119883 + 119879

119899119883119879

minus1

119899+ 119877

119899119883119877

minus1

119899+ 119878

119899119883119878

minus1

119899) (42)

Journal of Applied Mathematics 9

is also a solution of (36) where

119883 + 119879119899119883119879

minus1

119899+ 119877

119899119883119877

minus1

119899+ 119878

119899119883119878

minus1

119899

= [119883119894119895]4 times 4

119894 119895 = 1 2 3 4

11988311

= 11988311

+ 11988322

+ 11988333

+ 11988344

11988312

= 11988312

minus 11988321

+ 11988334

minus 11988343

11988313

= 11988313

minus 11988324

minus 11988331

+ 11988342

11988314

= 11988314

+ 11988323

minus 11988332

minus 11988341

11988321

= minus11988312

+ 11988321

minus 11988334

+ 11988343

11988322

= 11988311

+ 11988322

+ 11988333

+ 11988344

11988323

= 11988314

+ 11988323

minus 11988332

minus 11988341

11988324

= minus11988313

+ 11988324

+ 11988331

minus 11988342

11988331

= minus11988313

+ 11988324

+ 11988331

minus 11988342

11988332

= minus11988314

minus 11988323

+ 11988332

+ 11988341

11988333

= 11988311

+ 11988322

+ 11988333

+ 11988344

11988334

= 11988312

minus 11988321

+ 11988334

minus 11988343

11988341

= minus11988314

minus 11988323

+ 11988332

+ 11988341

11988342

= 11988313

minus 11988324

minus 11988331

+ 11988342

11988343

= minus11988312

+ 11988321

minus 11988334

+ 11988343

11988344

= 11988311

+ 11988322

+ 11988333

+ 11988344

(43)

Let

119883 =1

4(119883

11+ 119883

22+ 119883

33+ 119883

44)

+1

4(minus119883

12+ 119883

21minus 119883

34+ 119883

43) 119894

+1

4(minus119883

13+ 119883

24+ 119883

31minus 119883

42) 119895

+1

4(minus119883

14minus 119883

23+ 119883

32+ 119883

41) 119896

(44)

Then it is not difficult to verify that

120601 (119883) =1

4(119883 + 119879

119899119883119879

minus1

119899+ 119877

119899119883119877

minus1

119899+ 119878

119899119883119878

minus1

119899) (45)

We have that 119883 is a solution of (34) by (119886) in Lemma 4 Theproof is completed

0 5 10 15 20

0

2

4

Iteration number k

minus12

minus10

minus8

minus6

minus4

minus2

log10r (k)

Figure 1 The convergence curve for the Frobenius norm of theresiduals from Example 18

0 5 10 15 20

0

2

4

Iteration number k

minus12

minus10

minus8

minus6

minus4

minus2

log10r (k)

Figure 2 The convergence curve for the Frobenius norm of theresiduals from Example 19

Lemma 15 There exists a permutation matrix P(4n4n) suchthat (36) is equivalent to

[120601

119879

(119861) otimes 120601 (119860) + (120601119879

(119863) otimes 120601 (119862)) 119875 (4119899 4119899)

120601119879

(119861) 120601 (119876) otimes 120601 (119860) 120601 (119876) + (120601119879

(119863) 120601 (119876) otimes 120601 (119862) 120601 (119876)) 119875 (4119899 4119899)

]

times vec ([119883119894119895]4 times 4

) = [vec (120601 (119865))

vec (120601 (119865))]

(46)

Lemma 15 is easily proven through Lemma 5 So we omitit here

10 Journal of Applied Mathematics

Theorem 16 When Problem 1 is consistent let its solution setbe denoted by 119878

119867 If

∘

119883 isin 119878119867 and

∘

119883 can be expressed as

∘

119883 = 119860119867

119866119861119867

+ 119863119866119867

119862 + 119876119860119867

119866119861119867

119876

+ 119876119863119866119867

119862119876 119866 isin H119898times119901

(47)

then∘

119883 is the least Frobenius norm solution of Problem 1

Proof By (119886) (119887) and (119888) in Lemmas 4 5 and 6 we have that

vec(120601(

∘

119883))

= vec (120601119879

(119860) 120601 (119866) 120601119879

(119861) + 120601 (119863) 120601119879

(119866) 120601 (119862)

+ 120601 (119876) 120601119879

(119860) 120601 (119866) 120601119879

(119861) 120601 (119876)

+ 120601 (119876) 120601 (119863) 120601119879

(119866) 120601 (119862) 120601 (119876))

= [120601 (119861) otimes 120601119879

(119860) + (120601119879

(119862) otimes 120601 (119863)) 119875 (4119898 4119901)

120601 (119876) 120601 (119861) otimes 120601 (119876) 120601119879

(119860)

+ (120601 (119876) 120601119879

(119862) otimes 120601 (119876) 120601 (119863)) 119875 (4119898 4119901)]

times [vec (120601 (119866))

vec (120601 (119866))]

= [120601

119879

(119861) otimes 120601(119860) + (120601119879

(119863) otimes 120601(119862))119875(4119899 4119899)

120601119879

(119861)120601(119876) otimes 120601(119860)120601(119876) + (120601119879

(119863)120601(119876) otimes 120601(119862)120601(119876))119875(4119899 4119899)

]

119879

times [vec (120601 (119866))

vec (120601 (119866))]

isin 119877 [120601

119879

(119861) otimes 120601 (119860)

120601119879

(119861) 120601 (119876) otimes 120601 (119860) 120601 (119876)

+ (120601119879

(119863) otimes 120601 (119862)) 119875 (4119899 4119899)

+ (120601119879

(119863) 120601 (119876) otimes 120601 (119862) 120601 (119876)) 119875 (4119899 4119899)]

119879

(48)

By Lemma 12 120601(∘

119883) is the least Frobenius norm solution ofmatrix equations (46)

Noting (11) we derive from Lemmas 13 14 and 15 that∘

119883

is the least Frobenius norm solution of Problem 1

From Algorithm 7 it is obvious that if we consider

119883(1) = 119860119867

119866119861119867

+ 119863119866119867

119862 + 119876119860119867

119866119861119867

119876

+ 119876119863119866119867

119862119876 119866 isin H119898times119901

(49)

then all119883(119896) generated by Algorithm 7 can be expressed as

119883 (119896) = 119860119867

119866119896119861

119867

+ 119863119866119867

119896119862 + 119876119860

119867

119866119896119861

119867

119876

+ 119876119863119866119867

119896119862119876 119866

119896isin H

119898times119901

(50)

Using the above conclusion and consideringTheorem 16we propose the following theorem

Theorem 17 Suppose that Problem 1 is consistent Let theinitial iteration matrix be

119883 (1) = 119860119867

119866119861H+ 119863119866

119867

119862 + 119876119860119867

119866119861119867

119876 + 119876119863119866119867

119862119876 (51)

where 119866 is an arbitrary quaternion matrix or especially119883(1) = 0 then the solution 119883

lowast generated by Algorithm 7 isthe least Frobenius norm solution of Problem 1

Now we study Problem 2 When Problem 1 is consistentthe solution set of Problem 1 denoted by 119878

119867is not empty

Then For a given reflexive matrix1198830isin RH119899times119899

(119876)

119860119883119861 + 119862119883119867

119863 = 119865 lArrrArr 119860(119883 minus 1198830) 119861 + 119862(119883 minus 119883

0)119867

119863

= 119865 minus 1198601198830119861 minus 119862119883

119867

0119863

(52)

Let 119883 = 119883 minus 1198830and 119865 = 119865 minus 119860119883

0119861 minus 119862119883

119867

0119863 then Problem

2 is equivalent to finding the least Frobenius norm reflexivesolution of the quaternion matrix equation

119860119883119861 + 119862119883119867

119863 = 119865 (53)

By using Algorithm 7 let the initial iteration matrix 119883(1) =

119860119867

119866119861119867

+ 119863119866119867

119862 + 119876119860119867

119866119861119867

119876 + 119876119863119866119867

119862Q where 119866 is anarbitrary quaternion matrix in H119898times119901 or especially 119883(1) = 0we can obtain the least Frobenius norm reflexive solution119883

lowast

of (53) Then we can obtain the solution of Problem 2 whichis

119883 = 119883lowast

+ 1198830 (54)

Journal of Applied Mathematics 11

4 Examples

In this section we give two examples to illustrate the effi-ciency of the theoretical results

Example 18 Consider the quaternion matrix equation

119860119883119861 + 119862119883119867

119863 = 119865 (55)with

119860 = [1 + 119894 + 119895 + 119896 2 + 2119895 + 2119896 119894 + 3119895 + 3119896 2 + 119894 + 4119895 + 4119896

3 + 2 119894 + 2 119895 + 119896 3 + 3119894 + 119895 1 + 7119894 + 3119895 + 119896 6 minus 7119894 + 2119895 minus 4119896]

119861 =

[[[

[

minus2 + 119894 + 3119895 + 119896 3 + 3119894 + 4119895 + 119896

5 minus 4119894 minus 6119895 minus 5119896 minus1 minus 5119894 + 4119895 + 3119896

minus1 + 2119894 + 119895 + 119896 2 + 2119895 + 6119896

3 minus 2119894 + 119895 + 119896 4 + 119894 minus 2119895 minus 4119896

]]]

]

119862 = [1 + 119894 + 119895 + 119896 2 + 2119894 + 2119895 119896 2 + 2119894 + 2119895 + 119896

minus1 + 119894 + 3119895 + 3119896 minus2 + 3119894 + 4119895 + 2119896 6 minus 2119894 + 6119895 + 4119896 3 + 119894 + 119895 + 5119896]

119863 =

[[[

[

minus1 + 4119895 + 2119896 1 + 2119894 + 3119895 + 3119896

7 + 6119894 + 5119895 + 6119896 3 + 7119894 minus 119895 + 9119896

4 + 119894 + 6119895 + 119896 7 + 2119894 + 9119895 + 119896

1 + 119894 + 3119895 minus 3119896 1 + 3119894 + 2119895 + 2119896

]]]

]

119865 = [1 + 3119894 + 2119895 + 119896 2 minus 2119894 + 3119895 + 3119896

minus1 + 4119894 + 2119895 + 119896 minus2 + 2119894 minus 1119895]

(56)

We apply Algorithm 7 to find the reflexive solution with res-pect to the generalized reflection matrix

119876 =

[[[

[

028 0 096119896 0

0 minus1 00 0

minus096119896 0 minus028 0

0 0 0 minus1

]]]

]

(57)

For the initial matrix119883(1) = 119876 we obtain a solution that is

119883lowast

= 119883 (20)

=[[

[

027257 minus 023500119894 + 0034789119895 + 0054677119896 0085841 minus 0064349119894 + 010387119895 minus 026871119896

0028151 minus 0013246119894 minus 0091097119895 + 0073137119896 minus046775 + 0048363119894 minus 0015746119895 + 021642119896

0043349 + 0079178119894 + 0085167119895 minus 084221119896 035828 minus 013849119894 minus 0085799119895 + 011446119896

013454 minus 0028417119894 minus 0063010119895 minus 010574119896 010491 minus 0039417119894 + 018066119895 minus 0066963119896

minus0043349 + 0079178119894 + 0085167119895 + 084221119896 minus0014337 minus 014868119894 minus 0011146119895 minus 000036397119896

0097516 + 012146119894 minus 0017662119895 minus 0037534119896 minus0064483 + 0070885119894 minus 0089331119895 + 0074969119896

minus021872 + 018532119894 + 0011399119895 + 0029390119896 000048529 + 0014861119894 minus 019824119895 minus 0019117119896

minus014099 + 0084013119894 minus 0037890119895 minus 017938119896 minus063928 minus 0067488119894 + 0042030119895 + 010106119896

]]

]

(58)

with corresponding residual 119877(20) = 22775 times 10minus11 The

convergence curve for the Frobenius norm of the residuals119877(119896) is given in Figure 1 where 119903(119896) = 119877(119896)

Example 19 In this example we choose the matrices119860 119861 119862 119863 119865 and 119876 as same as in Example 18 Let

1198830= [

[

1 minus036119894 minus 075119896 0 minus05625119894 minus 048119896

036119894 128 048119895 minus096119895

0 1 minus 048119895 1 064 minus 075119895

048119896 096119895 064 072

]

]

isin RH119899times119899

(119876)

(59)

In order to find the optimal approximation reflexive solutionto the given matrix 119883

0 let 119883 = 119883 minus 119883

0and 119865 = 119865 minus

1198601198830119861 minus 119862119883

119867

0119863 Now we can obtain the least Frobenius

norm reflexive solution119883lowast of the quaternionmatrix equation

119860119883119861 + 119862 119883119867

119863 = 119865 by choosing the initial matrix 119883(1) = 0that is

12 Journal of Applied Mathematics

119883lowast

= 119883 (20)

=

[[[

[

minus047933 + 0021111119894 + 011703119895 minus 0066934119896 minus052020 minus 019377119894 + 017420119895 minus 059403119896

minus031132 minus 011705119894 minus 033980119895 + 011991119896 minus086390 minus 021715119894 minus 0037089119895 + 040609119896

minus024134 minus 0025941119894 minus 031930119895 minus 026961119896 minus044552 + 013065119894 + 014533119895 + 039015119896

minus011693 minus 027043119894 + 022718119895 minus 00000012004119896 minus044790 minus 031260119894 minus 10271119895 + 027275119896

024134 minus 0025941119894 minus 031930119895 + 026961119896 0089010 minus 067960119894 + 043044119895 minus 0045772119896

minus0089933 minus 025485119894 + 0087788119895 minus 023349119896 minus017004 minus 037655119894 + 080481119895 + 037636119896

minus063660 + 016515119894 minus 013216119895 + 0073845119896 minus0034329 + 032283119894 + 050970119895 minus 0066757119896

000000090029 + 017038119894 + 020282119895 minus 0087697119896 minus044735 + 021846119894 minus 021469119895 minus 015347119896

]]]

]

(60)

with corresponding residual 119877(20) = 43455 times 10minus11 The

convergence curve for the Frobenius norm of the residuals119877(119896) is given in Figure 2 where 119903(119896) = 119877(119896)

Therefore the optimal approximation reflexive solutionto the given matrix119883

0is

119883 = 119883lowast

+ 1198830

=

[[[

[

052067 + 0021111119894 + 011703119895 minus 0066934119896 minus052020 minus 055377119894 + 017420119895 minus 13440119896

minus031132 + 024295119894 minus 033980119895 + 011991119896 041610 minus 021715119894 minus 0037089119895 + 040609119896

minus024134 minus 0025941119894 minus 031930119895 minus 026961119896 055448 + 013065119894 minus 033467119895 + 039015119896

minus011693 minus 027043119894 + 022718119895 + 048000119896 minus044790 minus 031260119894 minus 0067117119895 + 027275119896

024134 minus 0025941119894 minus 031930119895 + 026961119896 0089010 minus 12421119894 + 043044119895 minus 052577119896

minus0089933 minus 025485119894 + 056779119895 minus 023349119896 minus017004 minus 037655119894 minus 015519119895 + 037636119896

036340 + 016515119894 minus 013216119895 + 0073845119896 060567 + 032283119894 minus 024030119895 minus 0066757119896

064000 + 017038119894 + 020282119895 minus 0087697119896 027265 + 021846119894 minus 021469119895 minus 015347119896

]]]

]

(61)

The results show that Algorithm 7 is quite efficient

5 Conclusions

In this paper an algorithm has been presented for solving thereflexive solution of the quaternion matrix equation 119860119883119861 +

119862119883119867

119863 = 119865 By this algorithm the solvability of the problemcan be determined automatically Also when the quaternionmatrix equation 119860119883119861 + 119862119883

119867

119863 = 119865 is consistent overreflexive matrix 119883 for any reflexive initial iterative matrixa reflexive solution can be obtained within finite iterationsteps in the absence of roundoff errors It has been proventhat by choosing a suitable initial iterative matrix we canderive the least Frobenius norm reflexive solution of thequaternion matrix equation 119860119883119861 + 119862119883

119867

119863 = 119865 throughAlgorithm 7 Furthermore by using Algorithm 7 we solvedProblem 2 Finally two numerical examples were given toshow the efficiency of the presented algorithm

Acknowledgments

This research was supported by Grants from InnovationProgram of Shanghai Municipal Education Commission(13ZZ080) the National Natural Science Foundation ofChina (11171205) theNatural Science Foundation of Shanghai

(11ZR1412500) the Discipline Project at the correspondinglevel of Shanghai (A 13-0101-12-005) and Shanghai LeadingAcademic Discipline Project (J50101)

References

[1] H C Chen and A H Sameh ldquoAmatrix decompositionmethodfor orthotropic elasticity problemsrdquo SIAM Journal on MatrixAnalysis and Applications vol 10 no 1 pp 39ndash64 1989

[2] H C Chen ldquoGeneralized reflexive matrices special propertiesand applicationsrdquo SIAM Journal on Matrix Analysis and Appli-cations vol 19 no 1 pp 140ndash153 1998

[3] Q W Wang H X Chang and C Y Lin ldquo119875-(skew)symmetriccommon solutions to a pair of quaternion matrix equationsrdquoApplied Mathematics and Computation vol 195 no 2 pp 721ndash732 2008

[4] S F Yuan and Q W Wang ldquoTwo special kinds of least squaressolutions for the quaternionmatrix equation119860119883119861+119862119883119863 = 119864rdquoElectronic Journal of Linear Algebra vol 23 pp 257ndash274 2012

[5] T S Jiang and M S Wei ldquoOn a solution of the quaternionmatrix equation 119883 minus 119860119883119861 = 119862 and its applicationrdquo ActaMathematica Sinica (English Series) vol 21 no 3 pp 483ndash4902005

[6] Y T Li and W J Wu ldquoSymmetric and skew-antisymmetricsolutions to systems of real quaternion matrix equationsrdquoComputers amp Mathematics with Applications vol 55 no 6 pp1142ndash1147 2008

Journal of Applied Mathematics 13

[7] L G Feng and W Cheng ldquoThe solution set to the quaternionmatrix equation119860119883minus119883119861 = 0rdquo Algebra Colloquium vol 19 no1 pp 175ndash180 2012

[8] Z Y Peng ldquoAn iterative method for the least squares symmetricsolution of the linear matrix equation 119860119883119861 = 119862rdquo AppliedMathematics and Computation vol 170 no 1 pp 711ndash723 2005

[9] Z Y Peng ldquoA matrix LSQR iterative method to solve matrixequation 119860119883119861 = 119862rdquo International Journal of ComputerMathematics vol 87 no 8 pp 1820ndash1830 2010

[10] Z Y Peng ldquoNew matrix iterative methods for constraintsolutions of the matrix equation 119860119883119861 = 119862rdquo Journal ofComputational and Applied Mathematics vol 235 no 3 pp726ndash735 2010

[11] Z Y Peng ldquoSolutions of symmetry-constrained least-squaresproblemsrdquo Numerical Linear Algebra with Applications vol 15no 4 pp 373ndash389 2008

[12] C C Paige ldquoBidiagonalization of matrices and solutions of thelinear equationsrdquo SIAM Journal on Numerical Analysis vol 11pp 197ndash209 1974

[13] X F Duan A P Liao and B Tang ldquoOn the nonlinear matrixequation 119883 minus sum

119898

119894=1119860

119909

2119886119894119883

120575119894119860119894= 119876rdquo Linear Algebra and Its

Applications vol 429 no 1 pp 110ndash121 2008[14] X F Duan and A P Liao ldquoOn the existence of Hermitian

positive definite solutions of thematrix equation119883119904

+119860lowast

119883minus119905

119860 =

119876rdquo Linear Algebra and Its Applications vol 429 no 4 pp 673ndash687 2008

[15] X FDuan andA P Liao ldquoOn the nonlinearmatrix equation119883+

119860lowast

119883minus119902

119860 = 119876(119902 ge 1)rdquo Mathematical and Computer Modellingvol 49 no 5-6 pp 936ndash945 2009

[16] X F Duan and A P Liao ldquoOn Hermitian positive definitesolution of the matrix equation119883minussum

119898

119894=1119860

lowast

119894119883

119903

119860119894= 119876rdquo Journal

of Computational and Applied Mathematics vol 229 no 1 pp27ndash36 2009

[17] X F Duan CM Li and A P Liao ldquoSolutions and perturbationanalysis for the nonlinear matrix equation119883+sum

119898

119894=1119860

lowast

119894119883

minus1

119860119894=

119868rdquo Applied Mathematics and Computation vol 218 no 8 pp4458ndash4466 2011

[18] F Ding P X Liu and J Ding ldquoIterative solutions of thegeneralized Sylvester matrix equations by using the hierarchicalidentification principlerdquo Applied Mathematics and Computa-tion vol 197 no 1 pp 41ndash50 2008

[19] F Ding and T Chen ldquoIterative least-squares solutions ofcoupled Sylvester matrix equationsrdquo Systems amp Control Lettersvol 54 no 2 pp 95ndash107 2005

[20] FDing andTChen ldquoHierarchical gradient-based identificationof multivariable discrete-time systemsrdquo Automatica vol 41 no2 pp 315ndash325 2005

[21] M H Wang M S Wei and S H Hu ldquoAn iterative method forthe least-squares minimum-norm symmetric solutionrdquo CMESComputer Modeling in Engineering amp Sciences vol 77 no 3-4pp 173ndash182 2011

[22] M Dehghan and M Hajarian ldquoAn iterative algorithm for thereflexive solutions of the generalized coupled Sylvester matrixequations and its optimal approximationrdquoAppliedMathematicsand Computation vol 202 no 2 pp 571ndash588 2008

[23] M Dehghan and M Hajarian ldquoFinite iterative algorithmsfor the reflexive and anti-reflexive solutions of the matrixequation119860

1119883

1119861

1+119860

2119883

2119861

2= 119862rdquoMathematical and Computer

Modelling vol 49 no 9-10 pp 1937ndash1959 2009[24] M Hajarian and M Dehghan ldquoSolving the generalized

Sylvester matrix equation119901

sum

119894=1

119860119894119883119861

119894+

119902

sum

119895=1

119862119895119884119863

119895= 119864 over

reflexive and anti-reflexive matricesrdquo International Journal ofControl and Automation vol 9 no 1 pp 118ndash124 2011

[25] M Dehghan and M Hajarian ldquoAn iterative algorithm forsolving a pair of matrix equations 119860119884119861 = 119864 119862119884119863 = 119865

over generalized centro-symmetric matricesrdquo Computers ampMathematics with Applications vol 56 no 12 pp 3246ndash32602008

[26] M Dehghan and M Hajarian ldquoAnalysis of an iterative algo-rithm to solve the generalized coupled Sylvester matrix equa-tionsrdquo Applied Mathematical Modelling vol 35 no 7 pp 3285ndash3300 2011

[27] M Dehghan and M Hajarian ldquoThe general coupled matrixequations over generalized bisymmetric matricesrdquo Linear Alge-bra and Its Applications vol 432 no 6 pp 1531ndash1552 2010

[28] M Dehghan and M Hajarian ldquoAn iterative method for solvingthe generalized coupled Sylvester matrix equations over gener-alized bisymmetric matricesrdquo Applied Mathematical Modellingvol 34 no 3 pp 639ndash654 2010

[29] A G Wu B Li Y Zhang and G R Duan ldquoFinite iterativesolutions to coupled Sylvester-conjugate matrix equationsrdquoApplied Mathematical Modelling vol 35 no 3 pp 1065ndash10802011

[30] A G Wu L L Lv and G R Duan ldquoIterative algorithmsfor solving a class of complex conjugate and transpose matrixequationsrdquo Applied Mathematics and Computation vol 217 no21 pp 8343ndash8353 2011

[31] AGWu L L Lv andMZHou ldquoFinite iterative algorithms forextended Sylvester-conjugate matrix equationsrdquo Mathematicaland Computer Modelling vol 54 no 9-10 pp 2363ndash2384 2011

[32] N L Bihan and J Mars ldquoSingular value decomposition ofmatrices of Quaternions matrices a new tool for vector-sensorsignal processingrdquo Signal Process vol 84 no 7 pp 1177ndash11992004

[33] N L Bihan and S J Sangwine ldquoJacobi method for quaternionmatrix singular value decompositionrdquoAppliedMathematics andComputation vol 187 no 2 pp 1265ndash1271 2007

[34] F O Farid Q W Wang and F Zhang ldquoOn the eigenvalues ofquaternion matricesrdquo Linear and Multilinear Algebra vol 59no 4 pp 451ndash473 2011

[35] S de Leo and G Scolarici ldquoRight eigenvalue equation inquaternionic quantummechanicsrdquo Journal of Physics A vol 33no 15 pp 2971ndash2995 2000

[36] S J Sangwine and N L Bihan ldquoQuaternion singular valuedecomposition based on bidiagonalization to a real or com-plex matrix using quaternion Householder transformationsrdquoApplied Mathematics and Computation vol 182 no 1 pp 727ndash738 2006

[37] C C Took D P Mandic and F Zhang ldquoOn the unitarydiagonalisation of a special class of quaternion matricesrdquoApplied Mathematics Letters vol 24 no 11 pp 1806ndash1809 2011

[38] QWWang H X Chang and Q Ning ldquoThe common solutionto six quaternion matrix equations with applicationsrdquo AppliedMathematics and Computation vol 198 no 1 pp 209ndash2262008

[39] Q W Wang J W van der Woude and H X Chang ldquoA systemof real quaternion matrix equations with applicationsrdquo LinearAlgebra and Its Applications vol 431 no 12 pp 2291ndash2303 2009

[40] Q W Wang S W Yu and W Xie ldquoExtreme ranks of realmatrices in solution of the quaternion matrix equation 119860119883119861 =

119862with applicationsrdquoAlgebra Colloquium vol 17 no 2 pp 345ndash360 2010

14 Journal of Applied Mathematics

[41] Q W Wang and J Jiang ldquoExtreme ranks of (skew-)Hermitiansolutions to a quaternion matrix equationrdquo Electronic Journal ofLinear Algebra vol 20 pp 552ndash573 2010

[42] Q W Wang X Liu and S W Yu ldquoThe common bisymmetricnonnegative definite solutions with extreme ranks and inertiasto a pair of matrix equationsrdquo Applied Mathematics and Com-putation vol 218 no 6 pp 2761ndash2771 2011

[43] Q W Wang Y Zhou and Q Zhang ldquoRanks of the commonsolution to six quaternion matrix equationsrdquo Acta Mathemati-cae Applicatae Sinica (English Series) vol 27 no 3 pp 443ndash4622011

[44] F Zhang ldquoGersgorin type theorems for quaternionic matricesrdquoLinear Algebra and Its Applications vol 424 no 1 pp 139ndash1532007

[45] F Zhang ldquoQuaternions and matrices of quaternionsrdquo LinearAlgebra and Its Applications vol 251 pp 21ndash57 1997

[46] R A Horn and C R Johnson Topics in Matrix AnalysisCambridge University Press Cambridge UK 1991

[47] Y X Peng X Y Hu and L Zhang ldquoAn iterationmethod for thesymmetric solutions and the optimal approximation solutionof the matrix equation 119860119883119861 = 119862rdquo Applied Mathematics andComputation vol 160 no 3 pp 763ndash777 2005

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 961568 6 pageshttpdxdoiorg1011552013961568

Research ArticleThe Optimization on Ranks and Inertias of a QuadraticHermitian Matrix Function and Its Applications

Yirong Yao

Department of Mathematics Shanghai University Shanghai 200444 China

Correspondence should be addressed to Yirong Yao yryaostaffshueducn

Received 3 December 2012 Accepted 9 January 2013

Academic Editor Yang Zhang

Copyright copy 2013 Yirong Yao This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We solve optimization problems on the ranks and inertias of the quadratic Hermitian matrix function 119876 minus 119883119875119883lowast subject to a

consistent system of matrix equations 119860119883 = 119862 and 119883119861 = 119863 As applications we derive necessary and sufficient conditions for thesolvability to the systems of matrix equations and matrix inequalities 119860119883 = 119862119883119861 = 119863 and 119883119875119883lowast = (gt lt ge le)119876 in the Lownerpartial ordering to be feasible respectively The findings of this paper widely extend the known results in the literature

1 Introduction

Throughout this paper we denote the complex number fieldby C The notations C119898times119899 and C119898times119898

ℎstand for the sets

of all 119898 times 119899 complex matrices and all 119898 times 119898 complexHermitian matrices respectivelyThe identity matrix with anappropriate size is denoted by 119868 For a complex matrix 119860the symbols 119860lowast and 119903(119860) stand for the conjugate transposeand the rank of119860 respectivelyTheMoore-Penrose inverse of119860 isin C119898times119899 denoted by119860dagger is defined to be the unique solution119883 to the following four matrix equations

(1) 119860119883119860 = 119860 (2) 119883119860119883 = 119883

(3) (119860119883)lowast

= 119860119883 (4) (119883119860)lowast

= 119883119860

(1)

Furthermore 119871119860and 119877

119860stand for the two projectors 119871

119860=

119868 minus 119860dagger

119860 and 119877119860= 119868 minus 119860119860

dagger induced by 119860 respectively It isknown that 119871

119860= 119871lowast

119860and 119877

119860= 119877lowast

119860 For 119860 isin C119898times119898

ℎ its inertia

I119899(119860) = (119894

+(119860) 119894minus(119860) 1198940(119860)) (2)

is the triple consisting of the numbers of the positive nega-tive and zero eigenvalues of 119860 counted with multiplicitiesrespectively It is easy to see that 119894

+(119860) + 119894

minus(119860) = 119903(119860) For

two Hermitian matrices 119860 and 119861 of the same sizes we say119860 gt 119861 (119860 ge 119861) in the Lowner partial ordering if 119860 minus 119861 ispositive (nonnegative) definite

The investigation on maximal and minimal ranks andinertias of linear and quadratic matrix function is activein recent years (see eg [1ndash24]) Tian [21] considered themaximal and minimal ranks and inertias of the Hermitianquadratic matrix function

ℎ (119883) = 119860119883119861119883lowast

119860lowast

+ 119860119883119862 + 119862lowast

119883lowast

119860lowast

+ 119863 (3)

where 119861 and 119863 are Hermitian matrices Moreover Tian [22]investigated the maximal and minimal ranks and inertias ofthe quadratic Hermitian matrix function

119891 (119883) = 119876 minus 119883119875119883lowast (4)

such that 119860119883 = 119862The goal of this paper is to give the maximal andminimal

ranks and inertias of the matrix function (4) subject to theconsistent system of matrix equations

119860119883 = 119862 119883119861 = 119863 (5)

where 119876 isin C119899times119899ℎ

119875 isin C119901times119901

ℎare given complex matrices

As applications we consider the necessary and sufficient

2 Journal of Applied Mathematics

conditions for the solvability to the systems of matrix equa-tions and inequality

119860119883 = 119862 119883119861 = 119863 119883119875119883lowast

= 119876

119860119883 = 119862 119883119861 = 119863 119883119875119883lowast

gt 119876

119860119883 = 119862 119883119861 = 119863 119883119875119883lowast

lt 119876

119860119883 = 119862 119883119861 = 119863 119883119875119883lowast

ge 119876

119860119883 = 119862 119883119861 = 119863 119883119875119883lowast

le 119876

(6)

in the Lowner partial ordering to be feasible respectively

2 The Optimization on Ranks and Inertiasof (4) Subject to (5)

In this section we consider the maximal and minimal ranksand inertias of the quadratic Hermitian matrix function (4)subject to (5) We begin with the following lemmas

Lemma 1 (see [3]) Let 119860 isin C119898times119898ℎ

119861 isin C119898times119901 and 119862 isin C119902times119898

be given and denote

1198751= [

119860 119861

119861lowast

0] 119875

2= [

119860 119862lowast

119862 0]

1198753= [

119860 119861 119862lowast

119861lowast

0 0] 119875

4= [

119860 119861 119862lowast

119862 0 0]

(7)

Then

max119884isinC119901times119902

119903 [119860 minus 119861119884119862 minus (119861119884119862)lowast

]

= min 119903 [119860 119861 119862lowast

] 119903 (1198751) 119903 (119875

2)

min119884isinC119901times119902

119903 [119860 minus 119861119884119862 minus (119861119884119862)lowast

]

= 2119903 [119860 119861 119862lowast

]

+max 119908++ 119908minus 119892++ 119892minus 119908++ 119892minus 119908minus+ 119892+

max119884isinC119901times119902

119894plusmn[119860 minus 119861119884119862 minus (119861119884119862)

lowast

] = min 119894plusmn(1198751) 119894plusmn(1198752)

min119884isinC119901times119902

119894plusmn[119860 minus 119861119884119862 minus (119861119884119862)

lowast

]

= 119903 [119860 119861 119862lowast

] +max 119894plusmn(1198751) minus 119903 (119875

3) 119894plusmn(1198752) minus 119903 (119875

4)

(8)

where

119908plusmn= 119894plusmn(1198751) minus 119903 (119875

3) 119892

plusmn= 119894plusmn(1198752) minus 119903 (119875

4) (9)

Lemma 2 (see [4]) Let 119860 isin C119898times119899 119861 isin C119898times119896 119862 isin C119897times119899 119863 isin

C119898times119901 119864 isin C119902times119899 119876 isin C1198981times119896 and 119875 isin C119897times1198991 be given Then

(1) 119903 (119860) + 119903 (119877119860119861) = 119903 (119861) + 119903 (119877

119861119860) = 119903 [119860 119861]

(2) 119903 (119860) + 119903 (119862119871119860) = 119903 (119862) + 119903 (119860119871

119862) = 119903 [

119860

119862]

(3) 119903 (119861) + 119903 (119862) + 119903 (119877119861119860119871119862) = 119903 [

119860 119861

119862 0]

(4) 119903 (119875) + 119903 (119876) + 119903 [119860 119861119871

119876

119877119875119862 0

] = 119903[

[

119860 119861 0

119862 0 119875

0 119876 0

]

]

(5) 119903 [119877119861119860119871119862119877119861119863

119864119871119862

0] + 119903 (119861) + 119903 (119862) = 119903[

[

119860 119863 119861

119864 0 0

119862 0 0

]

]

(10)

Lemma 3 (see [23]) Let 119860 isin C119898times119898ℎ

119861 isin C119898times119899 119862 isin C119899times119899ℎ

119876 isin C119898times119899 and 119875 isin C119901times119899 be given and 119879 isin C119898times119898 benonsingular Then

(1) 119894plusmn(119879119860119879

lowast

) = 119894plusmn(119860)

(2) 119894plusmn[119860 0

0 119862] = 119894plusmn(119860) + 119894

plusmn(119862)

(3) 119894plusmn[0 119876

119876lowast

0] = 119903 (119876)

(4) 119894plusmn[

119860 119861119871119875

119871119875119861lowast

0] + 119903 (119875) = 119894

plusmn

[

[

119860 119861 0

119861lowast

0 119875lowast

0 119875 0

]

]

(11)

Lemma 4 Let 119860 119862 119861 and 119863 be given Then the followingstatements are equivalent

(1) System (5) is consistent

(2) Let

119903 [119860 119862] = 119903 (119860) [119863

119861] = 119903 (119861) 119860119863 = 119862119861 (12)

In this case the general solution can be written as

119883 = 119860dagger

119862 + 119871119860119863119861dagger

+ 119871119860119881119877119861 (13)

where 119881 is an arbitrary matrix over C with appropriate size

Now we give the fundamental theorem of this paper

Journal of Applied Mathematics 3

Theorem5 Let119891(119883) be as given in (4) and assume that119860119883 =

119862 and 119883119861 = 119863 in (5) is consistent Then

max119860119883=119862119883119861=119863

119903 (119876 minus 119883119875119883lowast

)

= min

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861)

minus 119903 (119875) 2119899 + 119903 (119860119876119860lowast

minus 119862119875119862lowast

)

minus2119903 (119860) 119903 [

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 2119903 (119861)

min119860119883=119862119883119861=119863

119903 (119876 minus 119883119875119883lowast

)

= 2119899 + 2119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 2119903 (119860) minus 2119903 (119861) minus 119903 (119875)

+max 119904++ 119904minus 119905++ 119905minus 119904++ 119905minus 119904minus+ 119905+

max119860119883=119862119883119861=119863

119894plusmn(119876 minus 119883119875119883

lowast

)

= min

119899 + 119894plusmn(119860119876119860

lowast

minus 119862119875119862lowast

)

minus119903 (119860) 119894plusmn

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 119903 (119861)

(14)

min119860119883=119862119883119861=119863

119894plusmn(119876 minus 119883119875119883

lowast

)

= 119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861)

minus 119894plusmn(119875) +max 119904

plusmn 119905plusmn

(15)

where

119904plusmn=minus119899+119903 (119860)minus119894

∓(119875)+119894

plusmn(119860119876119860

lowast

minus119862119875119862lowast

)minus119903 [119862119875 119860119876119860

lowast

119861lowast

119863lowast

119860lowast]

119905plusmn=minus119899+119903 (119860)minus119894

∓(119875)+119894

plusmn

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus[

[

0 119875 119861

119860119876 119862119875 0

119863lowast

119861lowast

0

]

]

(16)

Proof It follows from Lemma 4 that the general solution of(4) can be expressed as

119883 = 1198830+ 119871119860119881119877119861 (17)

where 119881 is an arbitrary matrix over C and 1198830is a special

solution of (5) Then

119876 minus 119883119875119883lowast

= 119876 minus (1198830+ 119871119860119881119877119861) 119875(119883

0+ 119871119860119881119877119861)lowast

(18)

Note that

119903 [119876 minus (1198830+ 119871119860119881119877119861) 119875(119883

0+ 119871119860119881119877119861)lowast

]

= 119903 [119876 (119883

0+ 119871119860119881119877119861) 119875

119875(1198830+ 119871119860119881119877119861)lowast

119875] minus 119903 (119875)

= 119903 [[119876 119883

0119875

119875119883lowast

0119875] + [

119871119860

0]119881 [0 119877

119861119875]

+([119871119860

0]119881 [0 119877

119861119875])

lowast

] minus 119903 (119875)

(19)

119894plusmn[119876 minus (119883

0+ 119871119860119881119877119861) 119875(119883

0+ 119871119860119881119877119861)lowast

]

= 119894plusmn[

119876 (1198830+ 119871119860119881119877119861) 119875

119875(1198830+ 119871119860119881119877119861)lowast

119875] minus 119894plusmn(119875)

= 119894plusmn[[

119876 1198830119875

119875119883lowast

0119875] + [

119871119860

0]119881 [0 119877

119861119875]

+([119871119860

0]119881 [0 119877

119861119875])

lowast

] minus 119894plusmn(119875)

(20)

Let

119902 (119881) = [119876 119883

0119875

119875119883lowast

0119875] + [

119871119860

0]119881 [0 119877

119861119875]

+ ([119871119860

0]119881 [0 119877

119861119875])

lowast

(21)

Applying Lemma 1 to (19) and (20) yields

max119881

119903 [119902 (119881)] = min 119903 (119872) 119903 (1198721) 119903 (119872

2)

min119881

119903 [119902 (119881)]

= 2119903 (119872) +max 119904++ 119904minus 119905++ 119905minus 119904++ 119905minus 119904minus+ 119905+

max119881

119894plusmn[119902 (119881)] = min 119894

plusmn(1198721) 119894plusmn(1198722)

min119881

119894plusmn[119902 (119881)] = 119903 (119872) +max 119904

plusmn 119905plusmn

(22)

where

119872=[119876 119883

0119875 119871119860

0

119875119883lowast

0119875 0 119875119877

119861

] 1198721=[

[

119876 1198830119875 119871119860

119875119883lowast

0119875 0

119871119860

0 0

]

]

1198722=[

[

119876 1198830119875 0

119875119883lowast

0119875 119875119877

119861

0 119877119861119875 0

]

]

1198723=[

[

119876 1198830119875 119871119860

0

119875119883lowast

0119875 0 119875119877

119861

119871119860

0 0 0

]

]

1198724= [

[

119876 1198830119875 119871119860

0

119875119883lowast

0119875 0 119875119877

119861

0 119877119861119875 0 0

]

]

119904plusmn= 119894plusmn(1198721) minus 119903 (119872

3) 119905

plusmn= 119894plusmn(1198722) minus 119903 (119872

4)

(23)

4 Journal of Applied Mathematics

Applying Lemmas 2 and 3 elementary matrix operations andcongruence matrix operations we obtain

119903 (119872) = 119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861)

119903 (1198721) = 2119899 + 119903 (119860119876119860

lowast

minus 119862119875119862lowast

) minus 2119903 (119860) + 119903 (119875)

119894plusmn(1198721) = 119899 + 119894

plusmn(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 (119860) + 119894plusmn(119875)

119903 (1198722) = 119903[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 2119903 (119861) + 119903 (119875)

119894plusmn(1198722) = 119894plusmn

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 119903 (119861) + 119894plusmn(119875)

119903 (1198723) = 2119899 + 119903 (119875) minus 2119903 (119860) minus 119903 (119861) + 119903 [

119862119875 119860119876119860lowast

119861lowast

119863lowast

119860lowast]

119903 (1198724) = 119899 + 119903 (119875) + 119903[

[

0 119875 119861

119860119876 119862119875 0

119863lowast

119861lowast

0

]

]

minus 2119903 (119861) minus 119903 (119860)

(24)

Substituting (24) into (22) we obtain the results

Using immediately Theorem 5 we can easily get thefollowing

Theorem 6 Let 119891(119883) be as given in (4) 119904plusmnand let 119905

plusmnbe as

given in Theorem 5 and assume that 119860119883 = 119862 and 119883119861 = 119863 in(5) are consistent Then we have the following

(a) 119860119883 = 119862 and 119883119861 = 119863 have a common solution suchthat 119876 minus 119883119875119883

lowast

ge 0 if and only if

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894minus(119875) + 119904

minusle 0

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894minus(119875) + 119905

minusle 0

(25)

(b) 119860119883 = 119862 and 119883119861 = 119863 have a common solution suchthat 119876 minus 119883119875119883

lowast

le 0 if and only if

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894+(119875) + 119904

+le 0

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894+(119875) + 119905

+le 0

(26)

(c) 119860119883 = 119862 and 119883119861 = 119863 have a common solution suchthat 119876 minus 119883119875119883

lowast

gt 0 if and only if

119894+(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 (119860) ge 0

119894+

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 119903 (119861) ge 119899

(27)

(d) 119860119883 = 119862 and 119883119861 = 119863 have a common solution suchthat 119876 minus 119883119875119883

lowast

lt 0 if and only if

119894minus(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 (119860) ge 0

119894minus

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 119903 (119861) ge 119899

(28)

(e) All common solutions of 119860119883 = 119862 and 119883119861 = 119863 satisfy119876 minus 119883119875119883

lowast

ge 0 if and only if

119899 + 119894minus(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 (119860) = 0

or 119894minus

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 119903 (119861) = 0

(29)

(f) All common solutions of 119860119883 = 119862 and 119883119861 = 119863 satisfy119876 minus 119883119875119883

lowast

le 0 if and only if

119899 + 119894+(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 (119860) = 0

or 119894+

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 119903 (119861) = 0

(30)

(g) All common solutions of 119860119883 = 119862 and 119883119861 = 119863 satisfy119876 minus 119883119875119883

lowast

gt 0 if and only if

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894+(119875) + 119904

+= 119899

(31)or

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894+(119875) + 119905

+= 119899

(32)

(h) All common solutions of 119860119883 = 119862 and 119883119861 = 119863 satisfy119876 minus 119883119875119883

lowast

lt 0 if and only if

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894minus(119875) + 119904

minus= 119899

(33)or

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894minus(119875) + 119905

minus= 119899

(34)

Journal of Applied Mathematics 5

(i) 119860119883 = 119862 119883119861 = 119863 and 119876 = 119883119875119883lowast have a common

solution if and only if

2119899+2119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus2119903 (119860)minus2119903 (119861)minus119903 (119875)+119904++119904minusle0

2119899+2119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus2119903 (119860)minus2119903 (119861)minus119903 (119875)+119905++119905minusle0

2119899+2119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus2119903 (119860)minus2119903 (119861)minus119903 (119875) + 119904++119905minusle0

2119899+2119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus2119903 (119860)minus2119903 (119861)minus119903 (119875) + 119904minus+119905+le0

(35)

Let 119875 = 119868 in Theorem 5 we get the following corollary

Corollary 7 Let 119876 isin C119899times119899 119860 119861 119862 and119863 be given Assumethat (5) is consistent Denote

1198791= [

119862 119860119876

119861lowast

119863lowast] 119879

2= 119860119876119860

lowast

minus 119862119862lowast

1198793= [

119876 119863

119863lowast

119861lowast

119861] 119879

4= [

119862 119860119876119860lowast

119861lowast

119863lowast

119860lowast]

1198795= [

119862119861 119860119876

119861lowast

119861 119863lowast]

(36)

Then

max119860119883=119862119883119861=119863

119903 (119876 minus 119883119883lowast

)

= min 119899 + 119903 (1198791) minus 119903 (119860) minus 119903 (119861) 2119899 + 119903 (119879

2)

minus2119903 (119860) 119899 + 119903 (1198793) minus 2119903 (119861)

min119860119883=119862119883119861=119863

119903 (119876 minus 119883119883lowast

)

= 2119903 (1198791) +max 119903 (119879

2) minus 2119903 (119879

4) minus119899 + 119903 (119879

3)

minus 2119903 (1198795) 119894+(1198792) + 119894minus(1198793)

minus 119903 (1198794) minus 119903 (119879

5) minus119899 + 119894

minus(1198792)

+119894+(1198793) minus 119903 (119879

4) minus 119903 (119879

5)

max119860119883=119862119883119861=119863

119894+(119876 minus 119883119883

lowast

)

= min 119899 + 119894+(1198792) minus 119903 (119860) 119894

+(1198793) minus 119903 (119861)

max119860119883=119862119883119861=119863

119894minus(119876 minus 119883119883

lowast

)

= min 119899 + 119894minus(1198792) minus 119903 (119860) 119899 + 119894

minus(1198793) minus 119903 (119861)

min119860119883=119862119883119861=119863

119894+(119876 minus 119883119883

lowast

)

= 119903 (1198791) +max 119894

+(1198792) minus 119903 (119879

4) 119894+(1198793) minus 119899 minus 119903 (119879

5)

min119860119883=119862119883119861=119863

119894minus(119876 minus 119883119883

lowast

)

= 119903 (1198791) +max 119894

minus(1198792) minus 119903 (119879

4) 119894minus(1198793) minus 119903 (119879

5)

(37)

Remark 8 Corollary 7 is one of the results in [24]

Let 119861 and 119863 vanish in Theorem 5 then we can obtainthe maximal and minimal ranks and inertias of (4) subjectto 119860119883 = 119862

Corollary 9 Let 119891(119883) be as given in (4) and assume that119860119883 = 119862 is consistent Then

max119860119883=119862

119903 (119876 minus 119883119875119883lowast

)

= min 119899 + 119903 [119860119876 119862119875] minus 119903 (119860) minus 119903 (119861)

2119899 + 119903 (119860119876119860lowast

minus 119862119875119862lowast

) minus 2119903 (119860) 119903 (119876) + 119903 (119875)

min119860119883=119862

119903 (119876 minus 119883119875119883lowast

)

= 2119899 + 2119903 [119860119876 119862119875] minus 2119903 (119860)

+max 119904++ 119904minus 119905++ 119905minus 119904++ 119905minus 119904minus+ 119905+

max119860119883=119862

119894plusmn(119876 minus 119883119875119883

lowast

)

= min 119899 + 119894plusmn(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 (119860) 119894plusmn(119876) + 119894

∓(119875)

min119860119883=119862

119894plusmn(119876 minus 119883119875119883

lowast

)

= 119899 + 119903 [119860119876 119862119875] minus 119903 (119860) + 119894∓(119875) +max 119904

plusmn 119905plusmn

(38)

where

119904plusmn= minus 119899 + 119903 (119860) minus 119894

∓(119875)

+ 119894plusmn(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 [119862119875 119860119876119860lowast

]

119905plusmn= minus119899 + 119903 (119860) + 119894

plusmn(119876) minus 119903 (119875) minus [119860119876 119862119875]

(39)

Remark 10 Corollary 9 is one of the results in [22]

References

[1] D L Chu Y S Hung and H J Woerdeman ldquoInertia and rankcharacterizations of somematrix expressionsrdquo SIAM Journal onMatrix Analysis and Applications vol 31 no 3 pp 1187ndash12262009

[2] Z H He and Q W Wang ldquoSolutions to optimization problemson ranks and inertias of a matrix function with applicationsrdquo

6 Journal of Applied Mathematics

AppliedMathematics andComputation vol 219 no 6 pp 2989ndash3001 2012

[3] Y Liu andY Tian ldquoMax-min problems on the ranks and inertiasof the matrix expressions119860minus119861119883119862plusmn(119861119883119862)lowast with applicationsrdquoJournal of OptimizationTheory and Applications vol 148 no 3pp 593ndash622 2011

[4] G Marsaglia and G P H Styan ldquoEqualities and inequalities forranks of matricesrdquo Linear and Multilinear Algebra vol 2 pp269ndash292 197475

[5] X Zhang Q W Wang and X Liu ldquoInertias and ranks ofsome Hermitian matrix functions with applicationsrdquo CentralEuropean Journal of Mathematics vol 10 no 1 pp 329ndash3512012

[6] Q W Wang Y Zhou and Q Zhang ldquoRanks of the commonsolution to six quaternion matrix equationsrdquo Acta Mathemati-cae Applicatae Sinica vol 27 no 3 pp 443ndash462 2011

[7] Q Zhang and Q W Wang ldquoThe (119875 119876)-(skew)symmetricextremal rank solutions to a system of quaternion matrixequationsrdquo Applied Mathematics and Computation vol 217 no22 pp 9286ndash9296 2011

[8] D Z Lian QWWang and Y Tang ldquoExtreme ranks of a partialbanded block quaternion matrix expression subject to somematrix equationswith applicationsrdquoAlgebraColloquium vol 18no 2 pp 333ndash346 2011

[9] H S Zhang andQWWang ldquoRanks of submatrices in a generalsolution to a quaternion system with applicationsrdquo Bulletin ofthe Korean Mathematical Society vol 48 no 5 pp 969ndash9902011

[10] Q W Wang and J Jiang ldquoExtreme ranks of (skew-)Hermitiansolutions to a quaternion matrix equationrdquo Electronic Journal ofLinear Algebra vol 20 pp 552ndash573 2010

[11] I A Khan Q W Wang and G J Song ldquoMinimal ranks ofsome quaternionmatrix expressions with applicationsrdquoAppliedMathematics and Computation vol 217 no 5 pp 2031ndash20402010

[12] Q Wang S Yu and W Xie ldquoExtreme ranks of real matricesin solution of the quaternion matrix equation 119860119883119861 = 119862 withapplicationsrdquo Algebra Colloquium vol 17 no 2 pp 345ndash3602010

[13] Q W Wang H S Zhang and G J Song ldquoA new solvable con-dition for a pair of generalized Sylvester equationsrdquo ElectronicJournal of Linear Algebra vol 18 pp 289ndash301 2009

[14] Q W Wang S W Yu and Q Zhang ldquoThe real solutions toa system of quaternion matrix equations with applicationsrdquoCommunications in Algebra vol 37 no 6 pp 2060ndash2079 2009

[15] Q W Wang and C K Li ldquoRanks and the least-norm of thegeneral solution to a system of quaternion matrix equationsrdquoLinear Algebra and its Applications vol 430 no 5-6 pp 1626ndash1640 2009

[16] Q W Wang H S Zhang and S W Yu ldquoOn solutions to thequaternionmatrix equation119860119883119861+119862119884119863 = 119864rdquoElectronic Journalof Linear Algebra vol 17 pp 343ndash358 2008

[17] Q W Wang G J Song and X Liu ldquoMaximal and minimalranks of the common solution of some linear matrix equationsover an arbitrary division ring with applicationsrdquo AlgebraColloquium vol 16 no 2 pp 293ndash308 2009

[18] Q W Wang G J Song and C Y Lin ldquoRank equalities relatedto the generalized inverse 119860

(2)

119879119878with applicationsrdquo Applied

Mathematics and Computation vol 205 no 1 pp 370ndash3822008

[19] Q W Wang S W Yu and C Y Lin ldquoExtreme ranks of alinear quaternionmatrix expression subject to triple quaternionmatrix equations with applicationsrdquo Applied Mathematics andComputation vol 195 no 2 pp 733ndash744 2008

[20] Q W Wang G J Song and C Y Lin ldquoExtreme ranks ofthe solution to a consistent system of linear quaternion matrixequations with an applicationrdquo Applied Mathematics and Com-putation vol 189 no 2 pp 1517ndash1532 2007

[21] Y Tian ldquoFormulas for calculating the extremum ranks andinertias of a four-term quadratic matrix-valued function andtheir applicationsrdquo Linear Algebra and its Applications vol 437no 3 pp 835ndash859 2012

[22] Y Tian ldquoSolving optimization problems on ranks and inertiasof some constrained nonlinearmatrix functions via an algebraiclinearization methodrdquo Nonlinear Analysis Theory Methods ampApplications vol 75 no 2 pp 717ndash734 2012

[23] Y Tian ldquoMaximization and minimization of the rank andinertia of the Hermitianmatrix expression119860minus119861119883minus(119861119883)lowast withapplicationsrdquo Linear Algebra and its Applications vol 434 no10 pp 2109ndash2139 2011

[24] Q W Wang X Zhang and Z H He ldquoOn the Hermitianstructures of the solution to a pair of matrix equationsrdquo Linearand Multilinear Algebra vol 61 no 1 pp 73ndash90 2013

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2012 Article ID 471573 19 pagesdoi1011552012471573

Research ArticleConstrained Solutions of a System ofMatrix Equations

Qing-Wen Wang1 and Juan Yu1 2

1 Department of Mathematics Shanghai University 99 Shangda Road Shanghai 200444 China2 Department of Basic Mathematics China University of Petroleum Qingdao 266580 China

Correspondence should be addressed to Qing-Wen Wang wqw858yahoocomcn

Received 26 September 2012 Accepted 7 December 2012

Academic Editor Panayiotis J Psarrakos

Copyright q 2012 Q-W Wang and J Yu This is an open access article distributed under theCreative Commons Attribution License which permits unrestricted use distribution andreproduction in any medium provided the original work is properly cited

We derive the necessary and sufficient conditions of and the expressions for the orthogonalsolutions the symmetric orthogonal solutions and the skew-symmetric orthogonal solutions ofthe system of matrix equations AX = B and XC = D respectively When the matrix equationsare not consistent the least squares symmetric orthogonal solutions and the least squares skew-symmetric orthogonal solutions are respectively given As an auxiliary an algorithm is providedto compute the least squares symmetric orthogonal solutions and meanwhile an example ispresented to show that it is reasonable

1 Introduction

Throughout this paper the following notations will be used Rmtimesn OR

ntimesn SRntimesn and ASR

ntimesn

denote the set of all m times n real matrices the set of all n times n orthogonal matrices the set of alln times n symmetric matrices and the set of all n times n skew-symmetric matrices respectively In isthe identity matrix of order n (middot)T and tr(middot) represent the transpose and the trace of the realmatrix respectively middot stands for the Frobenius norm induced by the inner product Thefollowing two definitions will also be used

Definition 11 (see [1]) A real matrix X isin Rntimesn is said to be a symmetric orthogonal matrix if

XT = X and XTX = In

Definition 12 (see [2]) A real matrix X isin R2mtimes2m is called a skew-symmetric orthogonal

matrix if XT = minusX and XTX = In

The set of all n times n symmetric orthogonal matrices and the set of all 2m times 2m skew-symmetric orthogonal matrices are respectively denoted by SOR

ntimesn and SSOR2mtimes2m Since

2 Journal of Applied Mathematics

the linear matrix equation(s) and its optimal approximation problem have great applicationsin structural design biology control theory and linear optimal control and so forth seefor example [3ndash5] there has been much attention paid to the linear matrix equation(s) Thewell-known system of matrix equations

AX = B XC = D (11)

as one kind of linear matrix equations has been investigated by many authors and a seriesof important and useful results have been obtained For instance the system (11) withunknown matrix X being bisymmetric centrosymmetric bisymmetric nonnegative definiteHermitian and nonnegative definite and (PQ)-(skew) symmetric has been respectivelyinvestigated by Wang et al [6 7] Khatri and Mitra [8] and Zhang and Wang [9] Of courseif the solvability conditions of system (11) are not satisfied we may consider its least squaressolution For example Li et al [10] presented the least squares mirrorsymmetric solutionYuan [11] got the least-squares solution Some results concerning the system (11) can also befound in [12ndash18]

Symmetric orthogonal matrices and skew-symmetric orthogonal matrices play impor-tant roles in numerical analysis and numerical solutions of partial differential equationsPapers [1 2] respectively derived the symmetric orthogonal solution X of the matrixequation XC = D and the skew-symmetric orthogonal solution X of the matrix equationAX = B Motivated by the work mentioned above we in this paper will respectively studythe orthogonal solutions symmetric orthogonal solutions and skew-symmetric orthogonalsolutions of the system (11) Furthermore if the solvability conditions are not satisfiedthe least squares skew-symmetric orthogonal solutions and the least squares symmetricorthogonal solutions of the system (11) will be also given

The remainder of this paper is arranged as follows In Section 2 some lemmas areprovided to give the main results of this paper In Sections 3 4 and 5 the necessary andsufficient conditions of and the expression for the orthogonal the symmetric orthogonaland the skew-symmetric orthogonal solutions of the system (11) are respectively obtainedIn Section 6 the least squares skew-symmetric orthogonal solutions and the least squaressymmetric orthogonal solutions of the system (11) are presented respectively In additionan algorithm is provided to compute the least squares symmetric orthogonal solutions andmeanwhile an example is presented to show that it is reasonable Finally in Section 7 someconcluding remarks are given

2 Preliminaries

In this section we will recall some lemmas and the special C-S decomposition which will beused to get the main results of this paper

Lemma 21 (see [1 Lemmas 1 and 2]) Given C isin R2mtimesn D isin R

2mtimesn The matrix equation YC =D has a solution Y isin OR

2mtimes2m if and only if DTD = CTC Let the singular value decompositions ofC and D be respectively

C = U

(Π 00 0

)V T D = W

(Π 00 0

)V T (21)

Journal of Applied Mathematics 3

where

Π = diag(σ1 σk) gt 0 k = rank(C) = rank(D)

U =(U1 U2

) isin OR2mtimes2m U1 isin R

2mtimesk

W =(W1 W2

) isin OR2mtimes2m W1 isin R

2mtimesk V =(V1 V2

) isin ORntimesn V1 isin R

ntimesk

(22)

Then the orthogonal solutions of YC = D can be described as

Y = W

(Ik 00 P

)UT (23)

where P isin OR(2mminusk)times(2mminusk) is arbitrary

Lemma 22 (see [2 Lemmas 1 and 2]) GivenA isin Rntimesm B isin R

ntimesm The matrix equationAX = Bhas a solution X isin OR

mtimesm if and only if AAT = BBT Let the singular value decompositions of Aand B be respectively

A = U

(Σ 00 0

)V T B = U

(Σ 00 0

)QT (24)

where

Σ = diag(δ1 δl) gt 0 l = rank(A) = rank(B) U =(U1 U2

) isin ORntimesn U1 isin R

ntimesl

V =(V1 V2

) isin ORmtimesm V1 isin R

mtimesl Q =(Q1 Q2

) isin ORmtimesm Q1 isin R

mtimesl(25)

Then the orthogonal solutions of AX = B can be described as

X = V

(Il 00 W

)QT (26)

whereW isin OR(mminusl)times(mminusl) is arbitrary

Lemma 23 (see [2 Theorem 1]) If

X =(X11 X12

X21 X22

)isin OR

2mtimes2m X11 isin ASRktimesk (27)

then the C-S decomposition of X can be expressed as

(D1 00 D2

)T(X11 X12

X21 X22

)(D1 00 R2

)=(Σ11 Σ12

Σ21 Σ22

) (28)

4 Journal of Applied Mathematics

where D1 isin ORktimesk D2 R2 isin OR

(2mminusk)times(2mminusk)

Σ11 =

⎛⎜⎝

I 0 00 C 00 0 0

⎞⎟⎠ Σ12 =

⎛⎝0 0 0

0 S 00 0 I

⎞⎠

Σ21 =

⎛⎝0 0 0

0 S 00 0 I

⎞⎠ Σ22 =

⎛⎜⎝

I 0 00 minusCT 00 0 0

⎞⎟⎠

I = diag(I1 Ir1

) I1 = middot middot middot = Ir1 =

(0 1minus1 0

) C = diag(C1 Cl1)

Ci =(

0 ciminusci 0

) i = 1 l1 S = diag(S1 Sl1)

Si =(si 00 si

) ST

i Si + CTi Ci = I2 i = 1 l1

2r1 + 2l1 = rank(X11) Σ21 = ΣT12 k minus 2r1 = rank(X21)

(29)

Lemma 24 (see [1 Theorem 1]) If

K =(K11 K12

K21 K22

)isin OR

mtimesm K11 isin SRltimesl (210)

then the C-S decomposition of K can be described as

(D1 00 D2

)T(K11 K12

K21 K22

)(D1 00 R2

)=(Π11 Π12

Π21 Π22

) (211)

where D1 isin ORltimesl D2 R2 isin OR

(mminusl)times(mminusl)

Π11 =

⎛⎜⎝

I 0 00 C 00 0 0

⎞⎟⎠ Π12 =

⎛⎝0 0 0

0 S 00 0 I

⎞⎠

Π21 =

⎛⎝0 0 0

0 S 00 0 I

⎞⎠ Π22 =

⎛⎜⎝

I 0 00 minusC 00 0 0

⎞⎟⎠

I = diag(i1 ir) ij = plusmn1 j = 1 r

S = diag(s1prime slprime) C = diag(c1prime clprime)

s2iprime =radic

1 minus c2iprime i = 1prime lprime r + lprime = rank(K11) Π21 = ΠT

12

(212)

Journal of Applied Mathematics 5

Remarks 25 In order to know the C-S decomposition of an orthogonal matrix with ak times k leading (skew-) symmetric submatrix for details one can deeply study the proof ofTheorem 1 in [1] and [2]

Lemma 26 Given A isin Rntimesm B isin R

ntimesm Then the matrix equation AX = B has a solution X isinSOR

mtimesm if and only ifAAT = BBT andABT = BAT When these conditions are satisfied the generalsymmetric orthogonal solutions can be expressed as

X = V

(I2lminusr 0

0 G

)QT (213)

where

Q = JQdiag(ID2) isin ORmtimesm V = JV diag(I R2) isin OR

mtimesm

J =

⎛⎝I 0 0

0 0 I0 I 0

⎞⎠

(214)

and G isin SOR(mminus2l+r)times(mminus2l+r) is arbitrary

Proof The Necessity Assume X isin SORmtimesm is a solution of the matrix equation AX = B then

we have

BBT = AXXTAT = AAT

BAT = AXAT = AXTAT = ABT(215)

The Sufficiency Since the equality AAT = BBT holds then by Lemma 22 the singularvalue decompositions of A and B can be respectively expressed as (24) Moreover thecondition ABT = BAT means

U

(Σ 00 0

)V TQ

(Σ 00 0

)UT = U

(Σ 00 0

)QTV

(Σ 00 0

)UT (216)

which can be written as

V T1 Q1 = QT

1V1 (217)

From Lemma 22 the orthogonal solutions of the matrix equation AX = B can be describedas (26) Now we aim to find that X in (26) is also symmetric Suppose that X is symmetricthen we have

(Il 00 WT

)V TQ = QTV

(Il 00 W

) (218)

6 Journal of Applied Mathematics

together with the partitions of the matrices Q and V in Lemma 22 we get

V T1 Q1 = QT

1V1 (219)

QT1V2W = V T

1 Q2 (220)

QT2 V2W = WTV T

2 Q2 (221)

By (217) we can get (219) Now we aim to find the orthogonal solutions of the systemof matrix equations (220) and (221) Firstly we obtain from (220) that QT

1V2(QT1V2)

T =V T

1 Q2(V T1 Q2)

T then by Lemma 22 (220) has an orthogonal solution W By (217) the l times lleading principal submatrix of the orthogonal matrix V TQ is symmetric Then we have fromLemma 24

V T1 Q2 = D1Π12R

T2 (222)

QT1V2 = D1Π12D

T2 (223)

V T2 Q2 = D2Π22R

T2 (224)

From (220) (222) and (223) the orthogonal solution W of (220) is

W = D2

⎛⎝G 0 0

0 I 00 0 I

⎞⎠RT

2 (225)

where G isin OR(mminus2l+r)times(mminus2l+r) is arbitrary Combining (221) (224) and (225) yields GT =

G that is G is a symmetric orthogonal matrix Denote

V = V diag(ID2) Q = Qdiag(I R2) (226)

then the symmetric orthogonal solutions of the matrix equation AX = B can be expressed as

X = V

⎛⎝I 0 0

0 G 00 0 I

⎞⎠QT (227)

Let the partition matrix J be

J =

⎛⎝I 0 0

0 0 I0 I 0

⎞⎠ (228)

Journal of Applied Mathematics 7

compatible with the block matrix

⎛⎝I 0 0

0 G 00 0 I

⎞⎠ (229)

Put

V = JV Q = JQ (230)

then the symmetric orthogonal solutions of the matrix equation AX = B can be described as(213)

Setting A = CT B = DT and X = YT in [2 Theorem 2] and then by Lemmas 21 and23 we can have the following result

Lemma 27 Given C isin R2mtimesn D isin R

2mtimesn Then the equation has a solution Y isin SSOR2mtimes2m if

and only ifDTD = CTC andDTC = minusCTD When these conditions are satisfied the skew-symmetricorthogonal solutions of the matrix equation YC = D can be described as

Y = W

(I 00 H

)UT (231)

where

W = J primeW diag(IminusD2) isin OR2mtimes2m U = J primeUdiag(I R2) isin OR

2mtimes2m

J prime =

⎛⎝I 0 0

0 0 I0 I 0

⎞⎠

(232)

and H isin SSOR2rtimes2r is arbitrary

3 The Orthogonal Solutions of the System (11)

The following theorems give the orthogonal solutions of the system (11)

Theorem 31 Given A B isin Rntimesm and C D isin R

mtimesn suppose the singular value decompositions ofA and B are respectively as (24) Denote

QTC =(C1

C2

) V TD =

(D1

D2

) (31)

8 Journal of Applied Mathematics

where C1 D1 isin Rltimesn and C2 D2 isin R

(mminusl)timesn Let the singular value decompositions of C2 and D2 berespectively

C2 = U

(Π 00 0

)V T D2 = W

(Π 00 0

)V T (32)

where U W isin OR(mminusl)times(mminusl) V isin OR

ntimesn Π isin Rktimesk is diagonal whose diagonal elements are

nonzero singular values of C2 or D2 Then the system (11) has orthogonal solutions if and only if

AAT = BBT C1 = D1 DT2 D2 = CT

2 C2 (33)

In which case the orthogonal solutions can be expressed as

X = V

(Ik+l 00 Gprime

)QT (34)

where

V = V

(Il 00 W

)isin OR

mtimesm Q = Q

(Il 00 U

)isin OR

mtimesm (35)

and Gprime isin OR(mminuskminusl)times(mminuskminusl) is arbitrary

Proof Let the singular value decompositions of A and B be respectively as (24) Since thematrix equation AX = B has orthogonal solutions if and only if

AAT = BBT (36)

then by Lemma 22 its orthogonal solutions can be expressed as (26) Substituting (26) and(31) into the matrix equation XC = D we have C1 = D1 and WC2 = D2 By Lemma 21 thematrix equation WC2 = D2 has orthogonal solution W if and only if

DT2 D2 = CT

2C2 (37)

Let the singular value decompositions of C2 and D2 be respectively

C2 = U

(Π 00 0

)V T D2 = W

(Π 00 0

)V T (38)

where U W isin OR(mminusl)times(mminusl) V isin OR

ntimesn Π isin Rktimesk is diagonal whose diagonal elements are

nonzero singular values of C2 or D2 Then the orthogonal solutions can be described as

W = W

(Ik 00 Gprime

)UT (39)

Journal of Applied Mathematics 9

where Gprime isin OR(mminuskminusl)times(mminuskminusl) is arbitrary Therefore the common orthogonal solutions of the

system (11) can be expressed as

X = V

(Il 00 W

)QT = V

(Il 00 W

)⎛⎝Il 0 0

0 Ik 00 0 Gprime

⎞⎠(Il 00 UT

)QT = V

(Ik+l 00 Gprime

)QT (310)

where

V = V

(Il 00 W

)isin OR

mtimesm Q = Q

(Il 00 U

)isin OR

mtimesm (311)

and Gprime isin OR(mminuskminusl)times(mminuskminusl) is arbitrary

The following theorem can be shown similarly

Theorem 32 GivenA B isin Rntimesm and C D isin R

mtimesn let the singular value decompositions of C andD be respectively as (21) Partition

AW =(A1 A2

) BU =

(B1 B2

) (312)

where A1 B1 isin Rntimesk A2 B2 isin R

ntimes(mminusk) Assume the singular value decompositions of A2 and B2

are respectively

A2 = U

(Σ 00 0

)V T B2 = U

(Σ 00 0

)QT (313)

where V Q isin OR(mminusk)times(mminusk) U isin OR

ntimesn Σ isin Rlprimetimeslprime is diagonal whose diagonal elements are

nonzero singular values of A2 or B2 Then the system (11) has orthogonal solutions if and onlyif

DTD = CTC A1 = B1 A2AT2 = B2B

T2 (314)

In which case the orthogonal solutions can be expressed as

X = W

(Ik+lprime 0

0 H prime

)UT (315)

where

W = W

(Ik 00 W

)isin OR

mtimesm U = U

(Ik 00 U

)isin OR

mtimesm (316)

and H prime isin OR(mminuskminuslprime)times(mminuskminuslprime) is arbitrary

10 Journal of Applied Mathematics

4 The Symmetric Orthogonal Solutions of the System (11)

We now present the symmetric orthogonal solutions of the system (11)

Theorem 41 Given AB isin Rntimesm CD isin R

mtimesn Let the symmetric orthogonal solutions of thematrix equation AX = B be described as in Lemma 26 Partition

QTC =(Cprime

1Cprime

2

) V TD =

(Dprime

1Dprime

2

) (41)

where Cprime1 D

prime1 isin R

(2lminusr)timesn Cprime2 D

prime2 isin R

(mminus2l+r)timesn Then the system (11) has symmetric orthogonalsolutions if and only if

AAT = BBT ABT = BAT Cprime1 = Dprime

1 DprimeT2 Dprime

2 = CprimeT2 Cprime

2 DprimeT2 Cprime

2 = CprimeT2 Dprime

2 (42)

In which case the solutions can be expressed as

X = V

(I 00 Gprimeprime

)QT (43)

where

V = V

(I2lminusr 0

0 W

)isin OR

mtimesm Q = Q

(I2lminusr 0

0 U

)isin OR

mtimesm (44)

and Gprimeprime isin SOR(mminus2l+rminus2lprime+r prime)times(mminus2l+rminus2lprime+r prime) is arbitrary

Proof From Lemma 26 we obtain that the matrix equation AX = B has symmetricorthogonal solutions if and only if AAT = BBT and ABT = BAT When these conditionsare satisfied the general symmetric orthogonal solutions can be expressed as

X = V

(I2lminusr 0

0 G

)QT (45)

where G isin SOR(mminus2l+r)times(mminus2l+r) is arbitrary Q isin OR

mtimesm V isin ORmtimesm Inserting (41) and (45)

into the matrix equation XC = D we get Cprime1 = Dprime

1 and GCprime2 = Dprime

2 By [1 Theorem 2] thematrix equation GCprime

2 = Dprime2 has a symmetric orthogonal solution if and only if

DprimeT2 Dprime

2 = CprimeT2 Cprime

2 DprimeT2 Cprime

2 = CprimeT2 Dprime

2 (46)

In which case the solutions can be described as

G = W

(I2lprimeminusr prime 0

0 Gprimeprime

)UT (47)

Journal of Applied Mathematics 11

where Gprimeprime isin SOR(mminus2l+rminus2lprime+r prime)times(mminus2l+rminus2lprime+r prime) is arbitrary W isin OR

(mminus2l+r)times(mminus2l+r) and U isinOR

(mminus2l+r)times(mminus2l+r) Hence the system (11) has symmetric orthogonal solutions if and onlyif all equalities in (42) hold In which case the solutions can be expressed as

X = V

(I2lminusr 0

0 G

)QT = V

(I2lminusr 0

0 W

)⎛⎝I2lminusr 0

0(I 00 Gprimeprime

)⎞⎠(I2lminusr 0

0 UT

)QT (48)

that is the expression in (43)

The following theorem can also be obtained by the method used in the proof ofTheorem 41

Theorem 42 Given AB isin Rntimesm CD isin R

mtimesn Let the symmetric orthogonal solutions of thematrix equation XC = D be described as

X = M

(I2kminusr 0

0 G

)NT (49)

where M N isin ORmtimesm G isin SOR

(mminus2k+r)times(mminus2k+r) Partition

AM =(M1 M2

) BN =

(N1 N2

) M1N1 isin R

ntimes(2kminusr) M2N2 isin Rntimes(mminus2k+r) (410)

Then the system (11) has symmetric orthogonal solutions if and only if

DTD = CTC DTC = CTD M1 = N1 M2MT2 = N2N

T2 M2N

T2 = N2M

T2

(411)

In which case the solutions can be expressed as

X = M

(I 00 H primeprime

)NT (412)

where

M = M

(I2kminusr 0

0 W1

)isin OR

mtimesm N = N

(I2kminusr 0

0 U1

)isin OR

mtimesm (413)

and H primeprime isin SOR(mminus2k+rminus2kprime+r prime)times(mminus2k+rminus2kprime+r prime) is arbitrary

12 Journal of Applied Mathematics

5 The Skew-Symmetric Orthogonal Solutions of the System (11)

In this section we show the skew-symmetric orthogonal solutions of the system (11)

Theorem 51 Given AB isin Rntimes2m CD isin R

2mtimesn Suppose the matrix equation AX = B has skew-symmetric orthogonal solutions with the form

X = V1

(I 00 H

)QT

1 (51)

whereH isin SSOR2rtimes2r is arbitrary V1 Q1 isin OR

2mtimes2m Partition

QT1C =

(Q1

Q2

) V T

1 D =(V1

V2

) (52)

where Q1 V1 isin R(2mminus2r)timesn Q2 V2 isin R

2rtimesn Then the system (11) has skew-symmetric orthogonalsolutions if and only if

AAT = BBT ABT = minusBAT Q1 = V1 QT2 Q2 = V T

2 V2 QT2V2 = minusV T

2 Q2 (53)

In which case the solutions can be expressed as

X = V1

(I 00 J prime

)QT

1 (54)

where

V = V1

(I2mminus2r 0

0 W

)isin OR

2mtimes2m Q = Q1

(I2mminus2r 0

0 U

)isin OR

2mtimes2m (55)

and J prime isin SSOR2kprimetimes2kprime

is arbitrary

Proof By [2 Theorem 2] the matrix equation AX = B has the skew-symmetric orthogonalsolutions if and only if AAT = BBT and ABT = minusBAT When these conditions are satisfiedthe general skew-symmetric orthogonal solutions can be expressed as (51) Substituting (51)and (52) into the matrix equation XC = D we get Q1 = V1 and HQ2 = V2 From Lemma 27equation HQ2 = V2 has a skew-symmetric orthogonal solution H if and only if

QT2 Q2 = V T

2 V2 QT2V2 = minusV T

2 Q2 (56)

When these conditions are satisfied the solution can be described as

H = W

(I 00 J prime

)UT (57)

Journal of Applied Mathematics 13

where J prime isin SSOR2kprimetimes2kprime

is arbitrary W U isin OR2rtimes2r Inserting (57) into (51) yields that the

system (11) has skew-symmetric orthogonal solutions if and only if all equalities in (53)hold In which case the solutions can be expressed as (54)

Similarly the following theorem holds

Theorem 52 Given AB isin Rntimes2m CD isin R

2mtimesn Suppose the matrix equation XC = D hasskew-symmetric orthogonal solutions with the form

X = W

(I 00 K

)UT (58)

where K isin SSOR2ptimes2p is arbitrary W U isin OR

2mtimes2m Partition

AW =(W1 W2

) BU =

(U1 U2

) (59)

where W1 U1 isin Rntimes(2mminus2p) W2 U2 isin R

ntimes2p Then the system (11) has skew-symmetric orthogonalsolutions if and only if

DTD = CTC DTC = minusCTD W1 = U1 W2WT2 = U2U

T2 U2W

T2 = minusW2U

T2

(510)

In which case the solutions can be expressed as

X = W1

(I 00 J primeprime

)UT

1 (511)

where

W1 = W

(I2mminus2p 0

0 W1

)isin OR

2mtimes2m U1 =

(I2mminus2p 0

0 UT1

)U isin OR

2mtimes2m (512)

and J primeprime isin SSOR2qtimes2q is arbitrary

6 The Least Squares (Skew-) Symmetric Orthogonal Solutions ofthe System (11)

If the solvability conditions of a system of matrix equations are not satisfied it is natural toconsider its least squares solution In this section we get the least squares (skew-) symmetricorthogonal solutions of the system (11) that is seek X isin SSOR

2mtimes2m(SORntimesn) such that

minXisinSSOR2mtimes2m(SORntimesn)

AX minus B2 + XC minusD2 (61)

14 Journal of Applied Mathematics

With the help of the definition of the Frobenius norm and the properties of the skew-symmetric orthogonal matrix we get that

AX minus B2 + XC minusD2 = A2 + B2 + C2 + D2 minus 2 tr(XT(ATB +DCT

)) (62)

Let

ATB +DCT = K

KT minusK

2= T

(63)

Then it follows from the skew-symmetric matrix X that

tr(XT(ATB +DCT

))= tr(XTK

)= tr(minusXK) = tr

(X

(KT minusK

2

))= tr(XT) (64)

Therefore (61) holds if and only if (64) reaches its maximum Now we pay our attention tofind the maximum value of (64) Assume the eigenvalue decomposition of T is

T = E

(Λ 00 0

)ET (65)

with

Λ = diag(Λ1 Λl) Λi =(

0 αi

minusαi 0

) αi gt 0 i = 1 l 2l = rank(T) (66)

Denote

ETXE =(X11 X12

X21 X22

) (67)

partitioned according to

(Λ 00 0

) (68)

then (64) has the following form

tr(XT) = tr(ETXE

(Λ 00 0

))= tr(X11Λ) (69)

Journal of Applied Mathematics 15

Thus by

X11 = I = diag(I1 Il

) (610)

where

Ii =(

0 minus11 0

) i = 1 l (611)

Equation (69) gets its maximum Since ETXE is skew-symmetric it follows from

X = E

(I 00 G

)ET (612)

where G isin SSOR(2mminus2l)times(2mminus2l) is arbitrary that (61) obtains its minimum Hence we have the

following theorem

Theorem 61 Given AB isin Rntimes2m and CD isin R

2mtimesn denote

ATB +DCT = KKT minusK

2= T (613)

and let the spectral decomposition of T be (65) Then the least squares skew-symmetric orthogonalsolutions of the system (11) can be expressed as (612)

If X in (61) is a symmetric orthogonal matrix then by the definition of the Frobeniusnorm and the properties of the symmetric orthogonal matrix (62) holds Let

ATB +DCT = HHT +H

2= N (614)

Then we get that

minXisinSORntimesn

AX minus B2 + XC minusD2 = minXisinSORntimesn

[A2 + B2 + C2 + D2 minus 2 tr(XN)

] (615)

Thus (615) reaches its minimum if and only if tr(XN) obtains its maximum Now we focuson finding the maximum value of tr(XN) Let the spectral decomposition of the symmetricmatrix N be

N = M

(Σ 00 0

)MT (616)

16 Journal of Applied Mathematics

where

Σ =(Σ+ 00 Σminus

) Σ+ = diag(λ1 λs)

λ1 ge λ2 ge middot middot middot ge λs gt 0 Σminus = diag(λs+1 λt)

λt le λtminus1 le middot middot middot le λs+1 lt 0 t = rank(N)

(617)

Denote

MTXM =

(X11 X12

X21 X22

)(618)

being compatible with

(Σ 00 0

) (619)

Then

tr(XN) = tr(MTXM

(Σ 00 0

))= tr

((X11 X12

X21 X22

)(Σ 00 0

))= tr(X11Σ

) (620)

Therefore it follows from

X11 = I =(Is 00 minusItminuss

)(621)

that (620) reaches its maximum Since MTXM is a symmetric orthogonal matrix then whenX has the form

X = M

(I 00 L

)MT (622)

where L isin SOR(nminust)times(nminust) is arbitrary (615) gets its minimum Thus we obtain the following

theorem

Theorem 62 Given AB isin Rntimesm CD isin R

mtimesn denote

ATB +DCT = HHT +H

2= N (623)

and let the eigenvalue decomposition of N be (616) Then the least squares symmetric orthogonalsolutions of the system (11) can be described as (622)

Journal of Applied Mathematics 17

Algorithm 63 Consider the following

Step 1 Input AB isin Rntimesm and CD isin R

mtimesn

Step 2 Compute

ATB +DCT = H

N =HT +H

2

(624)

Step 3 Compute the spectral decomposition of N with the form (616)

Step 4 Compute the least squares symmetric orthogonal solutions of (11) according to(622)

Example 64 Assume

A =

⎛⎜⎜⎜⎜⎜⎝

122 84 minus56 63 94 107118 29 85 69 96 minus78106 23 115 78 67 8936 78 49 119 94 5945 67 78 31 56 116

⎞⎟⎟⎟⎟⎟⎠

B =

⎛⎜⎜⎜⎜⎜⎝

85 94 36 78 63 4727 36 79 94 56 7837 67 86 98 34 29minus43 62 57 74 54 9529 39 minus52 63 78 46

⎞⎟⎟⎟⎟⎟⎠

C =

⎛⎜⎜⎜⎜⎜⎜⎜⎝

54 64 37 56 9736 42 78 minus63 7867 35 minus46 29 28minus27 72 108 37 3819 39 82 56 11289 78 94 79 56

⎞⎟⎟⎟⎟⎟⎟⎟⎠

D =

⎛⎜⎜⎜⎜⎜⎜⎜⎝

79 95 54 28 8687 26 67 84 8157 39 minus29 52 1948 58 18 minus72 5828 79 45 67 9695 41 34 98 39

⎞⎟⎟⎟⎟⎟⎟⎟⎠

(625)

It can be verified that the given matrices ABC and D do not satisfy the solvabilityconditions in Theorem 41 or Theorem 42 So we intend to derive the least squares symmetricorthogonal solutions of the system (11) By Algorithm 63 we have the following results

(1) the least squares symmetric orthogonal solution

X =

⎛⎜⎜⎜⎜⎜⎜⎜⎝

001200 006621 minus024978 027047 091302 minus016218006621 065601 022702 minus055769 024587 037715minus024978 022702 080661 040254 004066 minus026784027047 minus055769 040254 006853 023799 062645091302 024587 004066 023799 minus012142 minus018135minus016218 037715 minus026784 062645 minus0181354 057825

⎞⎟⎟⎟⎟⎟⎟⎟⎠

(626)

(2)

minXisinSORmtimesm

AX minus B2 + XC minusD2 = 198366

∥∥∥XTX minus I6

∥∥∥ = 284882 times 10minus15∥∥∥XT minusX

∥∥∥ = 000000(627)

18 Journal of Applied Mathematics

Remark 65 (1) There exists a unique symmetric orthogonal solution such that (61) holds ifand only if the matrix

N =HT +H

2 (628)

where

H = ATB +DCT (629)

is invertible Example 64 just illustrates it(2) The algorithm about computing the least squares skew-symmetric orthogonal

solutions of the system (11) can be shown similarly we omit it here

7 Conclusions

This paper is devoted to giving the solvability conditions of and the expressions ofthe orthogonal solutions the symmetric orthogonal solutions and the skew-symmetricorthogonal solutions to the system (11) respectively and meanwhile obtaining the leastsquares symmetric orthogonal and skew-symmetric orthogonal solutions of the system (11)In addition an algorithm and an example have been provided to compute its least squaressymmetric orthogonal solutions

Acknowledgments

This research was supported by the grants from Innovation Program of Shanghai MunicipalEducation Commission (13ZZ080) the National Natural Science Foundation of China(11171205) the Natural Science Foundation of Shanghai (11ZR1412500) the DisciplineProject at the corresponding level of Shanghai (A 13-0101-12-005) and Shanghai LeadingAcademic Discipline Project (J50101)

References

[1] C J Meng and X Y Hu ldquoThe inverse problem of symmetric orthogonal matrices and its optimalapproximationrdquo Mathematica Numerica Sinica vol 28 pp 269ndash280 2006 (Chinese)

[2] C J Meng X Y Hu and L Zhang ldquoThe skew-symmetric orthogonal solutions of the matrix equationAX = Brdquo Linear Algebra and Its Applications vol 402 no 1ndash3 pp 303ndash318 2005

[3] D L Chu H C Chan and D W C Ho ldquoRegularization of singular systems by derivative andproportional output feedbackrdquo SIAM Journal on Matrix Analysis and Applications vol 19 no 1 pp21ndash38 1998

[4] K T Joseph ldquoInverse eigenvalue problem in structural designrdquo AIAA Journal vol 30 no 12 pp2890ndash2896 1992

[5] A Jameson and E Kreinder ldquoInverse problem of linear optimal controlrdquo SIAM Journal on Controlvol 11 no 1 pp 1ndash19 1973

[6] Q W Wang ldquoBisymmetric and centrosymmetric solutions to systems of real quaternion matrixequationsrdquo Computers and Mathematics with Applications vol 49 no 5-6 pp 641ndash650 2005

[7] Q W Wang X Liu and S W Yu ldquoThe common bisymmetric nonnegative definite solutions withextreme ranks and inertias to a pair of matrix equationsrdquo Applied Mathematics and Computation vol218 pp 2761ndash2771 2011

Journal of Applied Mathematics 19

[8] C G Khatri and S K Mitra ldquoHermitian and nonnegative definite solutions of linear matrixequationsrdquo SIAM Journal on Applied Mathematics vol 31 pp 578ndash585 1976

[9] Q Zhang and Q W Wang ldquoThe (PQ)-(skew)symmetric extremal rank solutions to a system ofquaternion matrix equationsrdquo Applied Mathematics and Computation vol 217 no 22 pp 9286ndash92962011

[10] F L Li X Y Hu and L Zhang ldquoLeast-squares mirrorsymmetric solution for matrix equations (AX =BXC = D)rdquo Numerical Mathematics vol 15 pp 217ndash226 2006

[11] Y X Yuan ldquoLeast-squares solutions to the matrix equations AX = B and XC = Drdquo AppliedMathematics and Computation vol 216 no 10 pp 3120ndash3125 2010

[12] A Dajic and J J Koliha ldquoPositive solutions to the equations AX = C and XB = D for Hilbert spaceoperatorsrdquo Journal of Mathematical Analysis and Applications vol 333 no 2 pp 567ndash576 2007

[13] F L Li X Y Hu and L Zhang ldquoThe generalized reflexive solution for a class of matrix equations(AX = BXC = D)rdquo Acta Mathematica Scientia vol 28 no 1 pp 185ndash193 2008

[14] S Kumar Mitra ldquoThe matrix equations AX = CXB = Drdquo Linear Algebra and Its Applications vol 59pp 171ndash181 1984

[15] Y Qiu Z Zhang and J Lu ldquoThe matrix equations AX = BXC = D with PX = sXP constraintrdquoApplied Mathematics and Computation vol 189 no 2 pp 1428ndash1434 2007

[16] Y Y Qiu and A D Wang ldquoLeast squares solutions to the equations AX = BXC = B with someconstraintsrdquo Applied Mathematics and Computation vol 204 no 2 pp 872ndash880 2008

[17] Q W Wang X Zhang and Z H He ldquoOn the Hermitian structures of the solution to a pair of matrixequationsrdquo Linear Multilinear Algebra vol 61 no 1 pp 73ndash90 2013

[18] Q X Xu ldquoCommon Hermitian and positive solutions to the adjointable operator equations AX =CXB = Drdquo Linear Algebra and Its Applications vol 429 no 1 pp 1ndash11 2008

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2012 Article ID 398085 14 pagesdoi1011552012398085

Research ArticleOn the Hermitian R-Conjugate Solution ofa System of Matrix Equations

Chang-Zhou Dong1 Qing-Wen Wang2 and Yu-Ping Zhang3

1 School of Mathematics and Science Shijiazhuang University of Economics ShijiazhuangHebei 050031 China

2 Department of Mathematics Shanghai University Shanghai Shanghai 200444 China3 Department of Mathematics Ordnance Engineering College Shijiazhuang Hebei 050003 China

Correspondence should be addressed to Qing-Wen Wang wqw858yahoocomcn

Received 3 October 2012 Accepted 26 November 2012

Academic Editor Yang Zhang

Copyright q 2012 Chang-Zhou Dong et al This is an open access article distributed under theCreative Commons Attribution License which permits unrestricted use distribution andreproduction in any medium provided the original work is properly cited

Let R be an n by n nontrivial real symmetric involution matrix that is R = Rminus1 = RT = In An n times n

complex matrix A is termed R-conjugate if A = RAR where A denotes the conjugate of A Wegive necessary and sufficient conditions for the existence of the Hermitian R-conjugate solutionto the system of complex matrix equations AX = C andXB = D and present an expression ofthe Hermitian R-conjugate solution to this system when the solvability conditions are satisfiedIn addition the solution to an optimal approximation problem is obtained Furthermore the leastsquares Hermitian R-conjugate solution with the least norm to this system mentioned above isconsidered The representation of such solution is also derived Finally an algorithm and numericalexamples are given

1 Introduction

Throughout we denote the complex m times n matrix space by Cmtimesn the real m times n matrix space

by Rmtimesn and the set of all matrices in R

mtimesn with rank r by Rmtimesnr The symbols IAAT Alowast Adagger

and A stand for the identity matrix with the appropriate size the conjugate the transposethe conjugate transpose the Moore-Penrose generalized inverse and the Frobenius norm ofA isin C

mtimesn respectively We use Vn to denote the ntimes n backward matrix having the elements 1along the southwest diagonal and with the remaining elements being zeros

Recall that an n times n complex matrix A is centrohermitian if A = VnAVn Centro-hermitian matrices and related matrices such as k-Hermitian matrices Hermitian Toeplitzmatrices and generalized centrohermitian matrices appear in digital signal processing andothers areas (see [1ndash4]) As a generalization of a centrohermitian matrix and related matrices

2 Journal of Applied Mathematics

Trench [5] gave the definition of R-conjugate matrix A matrix A isin Cntimesn is R-conjugate if

A = RAR where R is a nontrivial real symmetric involution matrix that is R = Rminus1 = RT andR= In At the same time Trench studied the linear equation Az = w for R-conjugate matricesin [5] where zw are known column vectors

Investigating the matrix equation

AX = B (11)

with the unknown matrix X being symmetric reflexive Hermitian-generalized Hamiltonianand repositive definite is a very active research topic [6ndash14] As a generalization of (11) theclassical system of matrix equations

AX = C XB = D (12)

has attracted many authorrsquos attention For instance [15] gave the necessary and sufficientconditions for the consistency of (12) [16 17] derived an expression for the general solutionby using singular value decomposition of a matrix and generalized inverses of matricesrespectively Moreover many results have been obtained about the system (12) with variousconstraints such as bisymmetric Hermitian positive semidefinite reflexive and generalizedreflexive solutions (see [18ndash28]) To our knowledge so far there has been little investigationof the Hermitian R-conjugate solution to (12)

Motivated by the work mentioned above we investigate Hermitian R-conjugatesolutions to (12) We also consider the optimal approximation problem

∥∥∥X minus E∥∥∥ = min

XisinSX

X minus E (13)

where E is a given matrix in Cntimesn and SX the set of all Hermitian R-conjugate solutions to

(12) In many cases the system (12) has not Hermitian R-conjugate solution Hence we needto further study its least squares solution which can be described as follows Let RHC

ntimesn

denote the set of all Hermitian R-conjugate matrices in Cntimesn

SL =X | min

XisinRHCntimesn

(AX minus C2 + XB minusD2

) (14)

Find X isin Cntimesn such that

∥∥∥X∥∥∥ = minXisinSL

X (15)

In Section 2 we present necessary and sufficient conditions for the existence of theHermitian R-conjugate solution to (12) and give an expression of this solution when thesolvability conditions are met In Section 3 we derive an optimal approximation solution to(13) In Section 4 we provide the least squares Hermitian R-conjugate solution to (15) InSection 5 we give an algorithm and a numerical example to illustrate our results

Journal of Applied Mathematics 3

2 R-Conjugate Hermitian Solution to (12)

In this section we establish the solvability conditions and the general expression for theHermitian R-conjugate solution to (12)

We denote RCntimesn and RHC

ntimesn the set of all R-conjugate matrices and Hermitian R-conjugate matrices respectively that is

RCntimesn =

A | A = RAR

HRCntimesn =

A | A = RARA = Alowast

(21)

where R is n times n nontrivial real symmetric involution matrixChang et al in [29] mentioned that for nontrivial symmetric involution matrix R isin

Rntimesn there exist positive integer r and n times n real orthogonal matrix [P Q] such that

R =[P Q

]⎡⎣Ir 0

0 minusInminusr

⎤⎦⎡⎣P

T

QT

⎤⎦ (22)

where P isin Rntimesr Q isin R

ntimes(nminusr) By (22)

RP = P RQ = minusQ PTP = Ir QTQ = Inminusr PTQ = 0 QTP = 0 (23)

Throughout this paper we always assume that the nontrivial symmetric involutionmatrix R is fixed which is given by (22) and (23) Now we give a criterion of judging amatrix is R-conjugate Hermitian matrix

Theorem 21 A matrix K isin HRCntimesn if and only if there exists a symmetric matrix H isin R

ntimesn suchthat K = ΓHΓlowast where

Γ =[P iQ

] (24)

with PQ being the same as (22)

Proof If K isin HRCntimesn then K = RKR By (22)

K = RKR =[P Q

]⎡⎣Ir 0

0 minusInminusr

⎤⎦⎡⎣P

T

QT

⎤⎦K[P Q

]⎡⎣Ir 0

0 minusInminusr

⎤⎦⎡⎣P

T

QT

⎤⎦ (25)

4 Journal of Applied Mathematics

which is equivalent to

⎡⎣P

T

QT

⎤⎦K[P Q

]

=

⎡⎣Ir 0

0 minusInminusr

⎤⎦⎡⎣P

T

QT

⎤⎦K[P Q

]⎡⎣Ir 0

0 minusInminusr

⎤⎦

(26)

Suppose that

⎡⎣P

T

QT

⎤⎦K[P Q

]=

⎡⎣K11 K12

K21 K22

⎤⎦ (27)

Substituting (27) into (26) we obtain

⎡⎣K11 K12

K21 K22

⎤⎦ =

⎡⎣Ir 0

0 minusInminusr

⎤⎦⎡⎣K11 K12

K21 K22

⎤⎦⎡⎣Ir 0

0 minusInminusr

⎤⎦ =

⎡⎣ K11 minusK12

minusK21 K22

⎤⎦ (28)

Hence K11 = K11 K12 = minusK12 K21 = minusK21 K22 = K22 that is K11 iK12 iK21 K22 are realmatrices If we denote M = iK12 N = minusiK21 then by (27)

K =[P Q

]⎡⎣K11 K12

K21 K22

⎤⎦⎡⎣P

T

QT

⎤⎦ =

[P iQ

]⎡⎣K11 M

N K22

⎤⎦⎡⎣ PT

minusiQT

⎤⎦ (29)

Let Γ = [P iQ] and

H =

⎡⎣K11 M

N K22

⎤⎦ (210)

Then K can be expressed as ΓHΓlowast where Γ is unitary matrix and H is a real matrix ByK = Klowast

ΓHTΓlowast = Klowast = K = ΓHΓlowast (211)

we obtain H = HT

Journal of Applied Mathematics 5

Conversely if there exists a symmetric matrix H isin Rntimesn such that K = ΓHΓlowast then it

follows from (23) that

RKR = RΓHΓlowastR = R[P iQ

]H

⎡⎣ PT

minusiQT

⎤⎦R =

[P minusiQ]H

⎡⎣ PT

iQT

⎤⎦ = ΓHΓlowast = K

Klowast = ΓHTΓlowast = ΓHΓlowast = K

(212)

that is K isin HRCntimesn

Theorem 21 implies that an arbitrary complex Hermitian R-conjugate matrix is equiv-alent to a real symmetric matrix

Lemma 22 For any matrix A isin Cmtimesn A = A1 + iA2 where

A1 =A +A

2 A2 =

A minusA

2i (213)

Proof For any matrix A isin Cmtimesn it is obvious that A = A1 + iA2 where A1 A2 are defined

as (213) Now we prove that the decomposition A = A1 + iA2 is unique If there exist B1 B2

such that A = B1 + iB2 then

A1 minus B1 + i(B2 minusA2) = 0 (214)

It follows from A1 A2 B1 andB2 are real matrix that

A1 = B1 A2 = B2 (215)

Hence A = A1 + iA2 holds where A1 A2 are defined as (213)

By Theorem 21 for X isin HRCntimesn we may assume that

X = ΓYΓlowast (216)

where Γ is defined as (24) and Y isin Rntimesn is a symmetric matrix

Suppose that AΓ = A1 + iA2 isin Cmtimesn CΓ = C1 + iC2 isin C

mtimesn ΓlowastB = B1 + iB2 isin Cntimesl and

ΓlowastD = D1 + iD2 isin Cntimesl where

A1 =AΓ +AΓ

2 A2 =

AΓ minusAΓ2i

C1 =CΓ + CΓ

2 C2 =

CΓ minus CΓ2i

B1 =ΓlowastB + ΓlowastB

2 B2 =

ΓlowastB minus ΓlowastB2i

D1 =ΓlowastD + ΓlowastD

2 D2 =

ΓlowastD minus ΓlowastD2i

(217)

6 Journal of Applied Mathematics

Then system (12) can be reduced into

(A1 + iA2)Y = C1 + iC2 Y (B1 + iB2) = D1 + iD2 (218)

which implies that

⎡⎣A1

A2

⎤⎦Y =

⎡⎣C1

C2

⎤⎦ Y

[B1 B2

]=[D1 D2

] (219)

Let

F =

⎡⎣A1

A2

⎤⎦ G =

⎡⎣C1

C2

⎤⎦ K =

[B1 B2

]

L =[D1 D2

] M =

⎡⎣ F

KT

⎤⎦ N =

⎡⎣G

LT

⎤⎦

(220)

Then system (12) has a solution X in HRCntimesn if and only if the real system

MY = N (221)

has a symmetric solution Y in Rntimesn

Lemma 23 (Theorem 1 in [7]) Let A isin Rmtimesn The SVD of matrix A is as follows

A = U

⎡⎣Σ 0

0 0

⎤⎦V T (222)

where U = [U1 U2] isin Rmtimesm and V = [V1 V2] isin R

ntimesn are orthogonal matrices Σ = diag(σ1 σr) σi gt 0 (i = 1 r) r = rank(A) U1 isin R

mtimesr V1 isin Rntimesr Then (11) has a symmetric

solution if and only if

ABT = BAT UT2 B = 0 (223)

In that case it has the general solution

X = V1Σminus1UT1 B + V2V

T2 B

TU1Σminus1V T1 + V2GVT

2 (224)

where G is an arbitrary (n minus r) times (n minus r) symmetric matrix

By Lemma 23 we have the following theorem

Journal of Applied Mathematics 7

Theorem 24 Given A isin Cmtimesn C isin C

mtimesn B isin Cntimesl and D isin C

ntimesl Let A1 A2 C1C2B1 B2 D1 D2 F G K L M andN be defined in (217) (220) respectively Assume thatthe SVD of M isin R

(2m+2l)timesn is as follows

M = U

⎡⎣M1 0

0 0

⎤⎦V T (225)

where U = [U1 U2] isin R(2m+2l)times(2m+2l) and V = [V1 V2] isin R

ntimesn are orthogonal matrices M1 =diag(σ1 σr) σi gt 0 (i = 1 r) r = rank(M) U1 isin R

(2m+2l)timesr V1 isin Rntimesr Then system

(12) has a solution in HRCntimesn if and only if

MNT = NMT UT2 N = 0 (226)

In that case it has the general solution

X = Γ(V1M

minus11 UT

1 N + V2VT2 N

TU1Mminus11 V T

1 + V2GVT2

)Γlowast (227)

where G is an arbitrary (n minus r) times (n minus r) symmetric matrix

3 The Solution of Optimal Approximation Problem (13)

When the set SX of all Hermitian R-conjugate solution to (12) is nonempty it is easy to verifySX is a closed set Therefore the optimal approximation problem (13) has a unique solutionby [30]

Theorem 31 GivenA isin Cmtimesn C isin C

mtimesn B isin Cntimesl D isin C

ntimesl E isin Cntimesn and E1 = (12)(ΓlowastEΓ+

ΓlowastEΓ) Assume SX is nonempty then the optimal approximation problem (13) has a unique solutionX and

X = Γ(V1M

minus11 UT

1 N + V2VT2 N

TU1Mminus11 V T

1 + V2VT2 E1V2V

T2

)Γlowast (31)

Proof Since SX is nonempty X isin SX has the form of (227) By Lemma 22 ΓlowastEΓ can be writtenas

ΓlowastEΓ = E1 + iE2 (32)

where

E1 =12

(ΓlowastEΓ + ΓlowastEΓ

) E2 =

12i

(ΓlowastEΓ minus ΓlowastEΓ

) (33)

8 Journal of Applied Mathematics

According to (32) and the unitary invariance of Frobenius norm

X minus E =∥∥∥Γ(V1M

minus1UT1N + V2V

T2 N

TU1Mminus1V T

1 + V2GVT2

)Γlowast minus E

∥∥∥=∥∥∥(E1 minus V1M

minus1UT1N minus V2V

T2 N

TU1Mminus1V T

1 minus V2GVT2

)+ iE2

∥∥∥(34)

We get

X minus E2 =∥∥∥E1 minus V1M

minus1UT1 N minus V2V

T2 N

TU1Mminus1V T

1 minus V2GVT2

∥∥∥2+ E22 (35)

Then minXisinSXX minus E is consistent if and only if there exists G isin R(nminusr)times(nminusr) such that

min∥∥∥E1 minus V1M

minus1UT1 N minus V2V

T2 N

TU1Mminus1V T

1 minus V2GVT2

∥∥∥ (36)

For the orthogonal matrix V

∥∥∥E1 minus V1Mminus1UT

1N minus V2VT2 N

TU1Mminus1V T

1 minus V2GVT2

∥∥∥2

=∥∥∥V T (E1 minus V1M

minus1UT1 N minus V2V

T2 N

TU1Mminus1V T

1 minus V2GVT2 )V

∥∥∥2

=∥∥∥V T

1

(E1 minus V1M

minus1UT1 N)V1

∥∥∥2+∥∥∥V T

1 (E1 minus V1Mminus1UT

1 N)V2

∥∥∥2

+∥∥∥V T

2

(E1 minus V2V

T2 N

TU1Mminus1V T

1

)V1

∥∥∥2+∥∥∥V T

2

(E1 minus V2GVT

2

)V2

∥∥∥

(37)

Therefore

min∥∥∥E1 minus V1M

minus1UT1 N minus V2V

T2 N

TU1Mminus1V T

1 minus V2GVT2

∥∥∥ (38)

is equivalent to

G = V T2 E1V2 (39)

Substituting (39) into (227) we obtain (31)

4 The Solution of Problem (15)

In this section we give the explicit expression of the solution to (15)

Theorem 41 Given A isin Cmtimesn C isin C

mtimesn B isin Cntimesl and D isin C

ntimesl Let A1 A2 C1 C2B1 B2 D1 D2 F G K L M and N be defined in (217) (220) respectively Assume that the

Journal of Applied Mathematics 9

SVD of M isin R(2m+2l)timesn is as (225) and system (12) has not a solution in HRC

ntimesn Then X isin SL

can be expressed as

X = ΓV

⎡⎣Mminus1

1 UT1NV1 Mminus1

1 UT1NV2

V T2 N

TU1Mminus11 Y22

⎤⎦V TΓlowast (41)

where Y22 isin R(nminusr)times(nminusr) is an arbitrary symmetric matrix

Proof It yields from (217)ndash(221) and (225) that

AX minus C2 + XB minusD2 = MY minusN2

=

∥∥∥∥∥∥U⎡⎣M1 0

0 0

⎤⎦V TY minusN

∥∥∥∥∥∥2

=

∥∥∥∥∥∥⎡⎣M1 0

0 0

⎤⎦V TYV minusUTNV

∥∥∥∥∥∥2

(42)

Assume that

V TYV =

⎡⎣Y11 Y12

Y21 Y22

⎤⎦ Y11 isin R

rtimesr Y22 isin R(nminusr)times(nminusr) (43)

Then we have

AX minus C2 + XB minusD2

=∥∥∥M1Y11 minusUT

1 NV1

∥∥∥2+∥∥∥M1Y12 minusUT

1NV2

∥∥∥2

+∥∥∥UT

2NV1

∥∥∥2+∥∥∥UT

2 NV2

∥∥∥2

(44)

Hence

min(AX minus C2 + XB minusD2

)(45)

is solvable if and only if there exist Y11 Y12 such that

∥∥∥M1Y11 minusUT1 NV1

∥∥∥2= min

∥∥∥M1Y12 minusUT1NV2

∥∥∥2= min

(46)

10 Journal of Applied Mathematics

It follows from (46) that

Y11 = Mminus11 UT

1 NV1

Y12 = Mminus11 UT

1 NV2(47)

Substituting (47) into (43) and then into (216) we can get that the form of elements in SL is(41)

Theorem 42 Assume that the notations and conditions are the same as Theorem 41 Then

∥∥∥X∥∥∥ = minXisinSL

X (48)

if and only if

X = ΓV

⎡⎣Mminus1

1 UT1NV1 Mminus1

1 UT1NV2

V T2 N

TU1Mminus11 0

⎤⎦V TΓlowast (49)

Proof In Theorem 41 it implies from (41) that minXisinSLX is equivalent to X has theexpression (41) with Y22 = 0 Hence (49) holds

5 An Algorithm and Numerical Example

Base on the main results of this paper we in this section propose an algorithm for finding thesolution of the approximation problem (13) and the least squares problem with least norm(15) All the tests are performed by MATLAB 65 which has a machine precision of around10minus16

Algorithm 51 (1) Input A isin Cmtimesn C isin C

mtimesn B isin Cntimesl D isin C

ntimesl(2) Compute A1 A2 C1 C2 B1 B2 D1 D2 F G K L M andN by (217) and

(220)(3) Compute the singular value decomposition of M with the form of (225)(4) If (226) holds then input E isin C

ntimesn and compute the solution X of problem (13)according (31) else compute the solution X to problem (15) by (49)

To show our algorithm is feasible we give two numerical example Let an nontrivialsymmetric involution be

R =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 0 0 0

0 minus1 0 0

0 0 1 0

0 0 0 minus1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦ (51)

Journal of Applied Mathematics 11

We obtain [P Q] in (22) by using the spectral decomposition of R then by (24)

Γ =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 0 0 0

0 0 i 0

0 1 0 0

0 0 0 i

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦ (52)

Example 52 Suppose A isin C2times4 C isin C

2times4 B isin C4times3 D isin C

4times3 and

A =

⎡⎣333 minus 5987i 45i 721 minusi

0 minus066i 7694 1123i

⎤⎦

C =

⎡⎣02679 minus 00934i 00012 + 40762i minus00777 minus 01718i minus12801i

02207 minus01197i 00877 07058i

⎤⎦

B =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

4 + 12i 2369i 4256 minus 5111i

4i 466i 821 minus 5i

0 483i 56 + i

222i minus4666 7i

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

D =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

00616 + 01872i minus00009 + 01756i 16746 minus 00494i

00024 + 02704i 01775 + 04194i 07359 minus 06189i

minus00548 + 03444i 00093 minus 03075i minus04731 minus 01636i

00337i 01209 minus 01864i minus02484 minus 38817i

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

(53)

We can verify that (226) holds Hence system (12) has an Hermitian R-conjugate solutionGiven

E =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

735i 8389i 99256 minus 651i minus46i

155 456i 771 minus 75i i

5i 0 minus4556i minus799

422i 0 51i 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦ (54)

12 Journal of Applied Mathematics

Applying Algorithm 51 we obtain the following

X =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

15597 00194i 28705 00002i

minus00194i 90001 02005i minus39997

28705 minus02005i minus00452 79993i

minus00002i minus39997 minus79993i 56846

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦ (55)

Example 52 illustrates that we can solve the optimal approximation problem withAlgorithm 51 when system (12) have Hermitian R-conjugate solutions

Example 53 Let A B andC be the same as Example 52 and let D in Example 52 be changedinto

D =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

00616 + 01872i minus00009 + 01756i 16746 + 00494i

00024 + 02704i 01775 + 04194i 07359 minus 06189i

minus00548 + 03444i 00093 minus 03075i minus04731 minus 01636i

00337i 01209 minus 01864i minus02484 minus 38817i

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦ (56)

We can verify that (226) does not hold By Algorithm 51 we get

X =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

052 22417i 04914 03991i

minus22417i 86634 01921i minus28232

04914 minus01921i 01406 13154i

minus03991i minus28232 minus13154i 63974

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦ (57)

Example 53 demonstrates that we can get the least squares solution with Algo-rithm 51 when system (12) has not Hermitian R-conjugate solutions

Acknowledgments

This research was supported by the Grants from the Key Project of Scientific ResearchInnovation Foundation of Shanghai Municipal Education Commission (13ZZ080) theNational Natural Science Foundation of China (11171205) the Natural Science Foundationof Shanghai (11ZR1412500) the PhD Programs Foundation of Ministry of Education ofChina (20093108110001) the Discipline Project at the corresponding level of Shanghai (A13-0101-12-005) Shanghai Leading Academic Discipline Project (J50101) the Natural ScienceFoundation of Hebei province (A2012403013) and the Natural Science Foundation of Hebeiprovince (A2012205028) The authors are grateful to the anonymous referees for their helpfulcomments and constructive suggestions

Journal of Applied Mathematics 13

References

[1] R D Hill R G Bates and S R Waters ldquoOn centro-Hermitian matricesrdquo SIAM Journal on MatrixAnalysis and Applications vol 11 no 1 pp 128ndash133 1990

[2] R Kouassi P Gouton and M Paindavoine ldquoApproximation of the Karhunen-Loeve tranformationand Its application to colour imagesrdquo Signal Processing Image Communication vol 16 pp 541ndash5512001

[3] A Lee ldquoCentro-Hermitian and skew-centro-Hermitian matricesrdquo Linear Algebra and Its Applicationsvol 29 pp 205ndash210 1980

[4] Z-Y Liu H-D Cao and H-J Chen ldquoA note on computing matrix-vector products with generalizedcentrosymmetric (centrohermitian) matricesrdquo Applied Mathematics and Computation vol 169 no 2pp 1332ndash1345 2005

[5] W F Trench ldquoCharacterization and properties of matrices with generalized symmetry or skewsymmetryrdquo Linear Algebra and Its Applications vol 377 pp 207ndash218 2004

[6] K-W E Chu ldquoSymmetric solutions of linear matrix equations by matrix decompositionsrdquo LinearAlgebra and Its Applications vol 119 pp 35ndash50 1989

[7] H Dai ldquoOn the symmetric solutions of linear matrix equationsrdquo Linear Algebra and Its Applicationsvol 131 pp 1ndash7 1990

[8] F J H Don ldquoOn the symmetric solutions of a linear matrix equationrdquo Linear Algebra and ItsApplications vol 93 pp 1ndash7 1987

[9] I Kyrchei ldquoExplicit representation formulas for the minimum norm least squares solutions of somequaternion matrix equationsrdquo Linear Algebra and Its Applications vol 438 no 1 pp 136ndash152 2013

[10] Y Li Y Gao and W Guo ldquoA Hermitian least squares solution of the matrix equation AXB = Csubject to inequality restrictionsrdquo Computers amp Mathematics with Applications vol 64 no 6 pp 1752ndash1760 2012

[11] Z-Y Peng and X-Y Hu ldquoThe reflexive and anti-reflexive solutions of the matrix equation AX = BrdquoLinear Algebra and Its Applications vol 375 pp 147ndash155 2003

[12] L Wu ldquoThe re-positive definite solutions to the matrix inverse problem AX = Brdquo Linear Algebra andIts Applications vol 174 pp 145ndash151 1992

[13] S F Yuan Q W Wang and X Zhang ldquoLeast-squares problem for the quaternion matrix equationAXB + CYD = E over different constrained matricesrdquo Article ID 722626 International Journal ofComputer Mathematics In press

[14] Z-Z Zhang X-Y Hu and L Zhang ldquoOn the Hermitian-generalized Hamiltonian solutions of linearmatrix equationsrdquo SIAM Journal on Matrix Analysis and Applications vol 27 no 1 pp 294ndash303 2005

[15] F Cecioni ldquoSopra operazioni algebricherdquo Annali della Scuola Normale Superiore di Pisa vol 11 pp17ndash20 1910

[16] K-W E Chu ldquoSingular value and generalized singular value decompositions and the solution oflinear matrix equationsrdquo Linear Algebra and Its Applications vol 88-89 pp 83ndash98 1987

[17] S K Mitra ldquoA pair of simultaneous linear matrix equations A1XB1 = C1 A2XB2 = C2 and a matrixprogramming problemrdquo Linear Algebra and Its Applications vol 131 pp 107ndash123 1990

[18] H-X Chang and Q-W Wang ldquoReflexive solution to a system of matrix equationsrdquo Journal of ShanghaiUniversity vol 11 no 4 pp 355ndash358 2007

[19] A Dajic and J J Koliha ldquoEquations ax = c and xb = d in rings and rings with involution withapplications to Hilbert space operatorsrdquo Linear Algebra and Its Applications vol 429 no 7 pp 1779ndash1809 2008

[20] A Dajic and J J Koliha ldquoPositive solutions to the equations AX = C and XB = D for Hilbert spaceoperatorsrdquo Journal of Mathematical Analysis and Applications vol 333 no 2 pp 567ndash576 2007

[21] C-Z Dong Q-W Wang and Y-P Zhang ldquoThe common positive solution to adjointable operatorequations with an applicationrdquo Journal of Mathematical Analysis and Applications vol 396 no 2 pp670ndash679 2012

[22] X Fang J Yu and H Yao ldquoSolutions to operator equations on Hilbert Clowast-modulesrdquo Linear Algebraand Its Applications vol 431 no 11 pp 2142ndash2153 2009

[23] C G Khatri and S K Mitra ldquoHermitian and nonnegative definite solutions of linear matrixequationsrdquo SIAM Journal on Applied Mathematics vol 31 no 4 pp 579ndash585 1976

[24] I Kyrchei ldquoAnalogs of Cramerrsquos rule for the minimum norm least squares solutions of some matrixequationsrdquo Applied Mathematics and Computation vol 218 no 11 pp 6375ndash6384 2012

[25] F L Li X Y Hu and L Zhang ldquoThe generalized reflexive solution for a class of matrix equations(AX = CXB = D)rdquo Acta Mathematica Scientia vol 28B no 1 pp 185ndash193 2008

14 Journal of Applied Mathematics

[26] Q-W Wang and C-Z Dong ldquoPositive solutions to a system of adjointable operator equations overHilbert Clowast-modulesrdquo Linear Algebra and Its Applications vol 433 no 7 pp 1481ndash1489 2010

[27] Q Xu ldquoCommon Hermitian and positive solutions to the adjointable operator equations AX = CXB = Drdquo Linear Algebra and Its Applications vol 429 no 1 pp 1ndash11 2008

[28] F Zhang Y Li and J Zhao ldquoCommon Hermitian least squares solutions of matrix equations A1XAlowast1 =

B1 and A2XAlowast2 = B2 subject to inequality restrictionsrdquo Computers ampMathematics with Applications vol

62 no 6 pp 2424ndash2433 2011[29] H-X Chang Q-W Wang and G-J Song ldquo(R S)-conjugate solution to a pair of linear matrix

equationsrdquo Applied Mathematics and Computation vol 217 no 1 pp 73ndash82 2010[30] E W Cheney Introduction to Approximation Theory McGraw-Hill New York NY USA 1966

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2012 Article ID 784620 13 pagesdoi1011552012784620

Research ArticlePerturbation Analysis for the Matrix EquationX minussumm

i=1 Alowasti XAi +

sumnj=1 B

lowastj XBj = I

Xue-Feng Duan1 2 and Qing-Wen Wang3

1 School of Mathematics and Computational Science Guilin University of Electronic TechnologyGuilin 541004 China

2 Research Center on Data Science and Social Computing Guilin University of Electronic TechnologyGuilin 541004 China

3 Department of Mathematics Shanghai University Shanghai 200444 China

Correspondence should be addressed to Qing-Wen Wang wqw858yahoocomcn

Received 18 September 2012 Accepted 19 November 2012

Academic Editor Yang Zhang

Copyright q 2012 X-F Duan and Q-W Wang This is an open access article distributed underthe Creative Commons Attribution License which permits unrestricted use distribution andreproduction in any medium provided the original work is properly cited

We consider the perturbation analysis of the matrix equation X minussummi=1 A

lowasti XAi +

sumnj=1 B

lowastj XBj = I

Based on the matrix differentiation we first give a precise perturbation bound for the positivedefinite solution A numerical example is presented to illustrate the sharpness of the perturbationbound

1 Introduction

In this paper we consider the matrix equation

X minusmsumi=1

Alowasti XAi +

nsumj=1

Blowastj XBj = I (11)

where A1 A2 Am B1 B2 Bn are n times n complex matrices I is an n times n identity matrixm n are nonnegative integers and the positive definite solution X is practical interestHere Alowast

i and Blowasti denote the conjugate transpose of the matrices Ai and Bi respectively

Equation (11) arises in solving some nonlinear matrix equations with Newton method Seefor example the nonlinear matrix equation which appears in Sakhnovich [1] Solving these

2 Journal of Applied Mathematics

nonlinear matrix equations gives rise to (11) On the other hand (11) is the general case ofthe generalized Lyapunov equation

MYSlowast + SYMlowast +tsum

k=1

NkYNlowastk + CClowast = 0 (12)

whose positive definite solution is the controllability Gramian of the bilinear control system(see [2 3] for more details)

Mx(t) = Sx(t) +tsum

k=1

Nkx(t)uk(t) + Cu(t) (13)

Set

E =1radic2(M minus S + I) F =

1radic2(M + S + I) G = S minus I Q = CClowast

X = Qminus12YQminus12 Ai = Qminus12NiQ12 i = 1 2 t

At+1 = Qminus12FQ12 At+2 = Qminus12GQ12

B1 = Qminus12EQ12 B2 = Qminus12CQ12

(14)

then (12) can be equivalently written as (11) with m = t + 2 and n = 2Some special cases of (11) have been studied Based on the kronecker product and

fixed point theorem in partially ordered sets Reurings [4] and Ran and Reurings [5 6] gavesome sufficient conditions for the existence of a unique positive definite solution of the linearmatrix equations X minussumm

i=1 Alowasti XAi = I and X +

sumnj=1 B

lowasti XBi = I And the expressions for these

unique positive definite solutions were also derived under some constraint conditions Forthe general linear matrix equation (11) Reurings [[4] Page 61] pointed out that it is hard tofind sufficient conditions for the existence of a positive definite solution because the map

G(X) = I +msumi=1

Alowasti XAi minus

nsumj=1

Blowastj XBj (15)

is not monotone and does not map the set of n times n positive definite matrices into itselfRecently Berzig [7] overcame these difficulties by making use of Bhaskar-Lakshmikanthamcoupled fixed point theorem and gave a sufficient condition for (11) existing a uniquepositive definite solution An iterative method was constructed to compute the uniquepositive definite solution and the error estimation was given too

Recently the matrix equations of the form (11) have been studied by many authors(see [8ndash14]) Some numerical methods for solving the well-known Lyapunov equationX + ATXA = Q such as Bartels-Stewart method and Hessenberg-Schur method have beenproposed in [12] Based on the fixed point theorem the sufficient and necessary conditions forthe existence of a positive definite solution of the matrix equation Xs plusmnAlowastXminustA = Q s t isin Nhave been given in [8 9] The fixed point iterative method and inversion-free iterative method

Journal of Applied Mathematics 3

were developed for solving the matrix equations X plusmn AlowastXminusαA = Q α gt 0 in [13 14] Bymaking use of the fixed point theorem of mixed monotone operator the matrix equationX minussumm

i=1 Alowasti X

δiAi = Q 0 lt |δi| lt 1 was studied in [10] and derived a sufficient condition forthe existence of a positive definite solution Assume that F maps positive definite matriceseither into positive definite matrices or into negative definite matrices the general nonlinearmatrix equation X +AlowastF(X)A = Q was studied in [11] and the fixed point iterative methodwas constructed to compute the positive definite solution under some additional conditions

Motivated by the works and applications in [2ndash6] we continue to study the matrixequation (11) Based on a new mathematical tool (ie the matrix differentiation) we firstlygive a differential bound for the unique positive definite solution of (11) and then use it toderive a precise perturbation bound for the unique positive definite solution A numericalexample is used to show that the perturbation bound is very sharp

Throughout this paper we write B gt 0 (B ge 0) if the matrix B is positive definite(semidefinite) If B minus C is positive definite (semidefinite) then we write B gt C (B ge C) Ifa positive definite matrix X satisfies B le X le C we denote that X isin [BC] The symbolsλ1(B) and λn(B) denote the maximal and minimal eigenvalues of an n times n Hermitian matrixB respectively The symbol Hntimesn stands for the set of n times n Hermitian matrices The symbolB denotes the spectral norm of the matrix B

2 Perturbation Analysis for the Matrix Equation (11)

Based on the matrix differentiation we firstly give a differential bound for the unique positivedefinite solution XU of (11) and then use it to derive a precise perturbation bound for XU inthis section

Definition 21 ([15] Definition 36) Let F = (fij)mtimesn then the matrix differentiation of F isdF = (dfij)mtimesn For example let

F =(

s + t s2 minus 2t2s + t3 t2

) (21)

Then

dF =(

ds + dt 2sds minus 2dt2ds + 3t2dt 2tdt

) (22)

Lemma 22 ([15] Theorem 32) The matrix differentiation has the following properties

(1) d(F1 plusmn F2) = dF1 plusmn dF2

(2) d(kF) = k(dF) where k is a complex number

(3) d(Flowast) = (dF)lowast

(4) d(F1F2F3) = (dF1)F2F3 + F1(dF2)F3 + F1F2(dF3)

(5) dFminus1 = minusFminus1(dF)Fminus1

(6) dF = 0 where F is a constant matrix

4 Journal of Applied Mathematics

Lemma 23 ([7] Theorem 31) If

msumi=1

Alowasti Ai lt

12I

nsumj=1

Blowastj Bj lt

12I (23)

then (11) has a unique positive definite solution XU and

XU isin⎡⎣I minus 2

nsumj=1

Blowastj Bj I + 2

msumi=1

Alowasti Ai

⎤⎦ (24)

Theorem 24 If

msumi=1

Ai2 lt12

nsumj=1

∥∥Bj

∥∥2lt

12 (25)

then (11) has a unique positive definite solution XU and it satisfies

dXU le2(

1 + 2summ

i=1 Ai2)[summ

i=1(AidAi) +sumn

j=1(∥∥Bj

∥∥∥∥dBj

∥∥)]

1 minussummi=1 Ai2 minussumn

j=1

∥∥Bj

∥∥2 (26)

Proof Since

λ1(Alowast

i Ai

) le ∥∥Alowasti Ai

∥∥ le Ai2 i = 1 2 m

λ1

(Blowastj Bj

)le∥∥∥Blowast

j Bj

∥∥∥ le ∥∥Bj

∥∥2 j = 1 2 n

(27)

then

Alowasti Ai le λ1

(Alowast

i Ai

)I le ∥∥Alowast

i Ai

∥∥I le Ai2I i = 1 2 m

Blowastj Bj le λ1

(Blowastj Bj

)I le∥∥∥Blowast

j Bj

∥∥∥ le ∥∥Bj

∥∥2I j = 1 2 n

(28)

consequently

msumi=1

Alowasti Ai le

msumi=1

Ai2I

nsumj=1

Blowastj Bj le

nsumj=1

∥∥Bj

∥∥2I

(29)

Journal of Applied Mathematics 5

Combining (25)ndash(29) we have

msumi=1

Alowasti Ai le

msumi=1

Ai2I lt12I

nsumj=1

Blowastj Bj le

nsumj=1

∥∥Bj

∥∥2I lt

12I

(210)

Then by Lemma 23 we obtain that (11) has a unique positive definite solution XU whichsatisfies

XU isin⎡⎣I minus 2

nsumj=1

Blowastj Bj I + 2

msumi=1

Alowasti Ai

⎤⎦ (211)

Noting that XU is the unique positive definite solution of (11) then

XU minusmsumi=1

Alowasti XUAi +

nsumj=1

Blowastj XUBj = I (212)

It is known that the elements of XU are differentiable functions of the elements of Ai and BiDifferentiating (212) and by Lemma 22 we have

dXU minusmsumi=1

[(dAlowast

i

)XUAi +Alowast

i (dXU)Ai +Alowasti XU(dAi)

]

+nsumj=1

[(dBlowast

j

)XUBj + Blowast

j (dXU)Bj + Blowastj XU

(dBj

)]= 0

(213)

which implies that

dXU minusmsumi=1

Alowasti (dXU)Ai +

nsumj=1

Blowastj (dXU)Bj

=msumi=1

(dAlowast

i

)XUAi +

msumi=1

Alowasti XU(dAi) minus

nsumj=1

(dBlowast

j

)XUBj minus

nsumj=1

Blowastj XU

(dBj

)

(214)

6 Journal of Applied Mathematics

By taking spectral norm for both sides of (214) we obtain that

∥∥∥∥∥∥dXU minusmsumi=1

Alowasti (dXU)Ai +

nsumj=1

Blowastj (dXU)Bj

∥∥∥∥∥∥

=

∥∥∥∥∥∥msumi=1

(dAlowast

i

)XUAi +

msumi=1

Alowasti XU(dAi) minus

nsumj=1

(dBlowast

j

)XUBj minus

nsumj=1

Blowastj XU

(dBj

)∥∥∥∥∥∥

le∥∥∥∥∥

msumi=1

(dAlowast

i

)XUAi

∥∥∥∥∥ +∥∥∥∥∥

msumi=1

Alowasti XU(dAi)

∥∥∥∥∥ +∥∥∥∥∥∥

nsumj=1

(dBlowast

j

)XUBj

∥∥∥∥∥∥ +∥∥∥∥∥∥

nsumj=1

Blowastj XU

(dBj

)∥∥∥∥∥∥

lemsumi=1

∥∥(dAlowasti

)XUAi

∥∥ + msumi=1

∥∥Alowasti XU(dAi)

∥∥ + nsumj=1

∥∥∥(dBlowastj

)XUBj

∥∥∥ +nsumj=1

∥∥∥Blowastj XU

(dBj

)∥∥∥

lemsumi=1

∥∥dAlowasti

∥∥XUAi +msumi=1

∥∥Alowasti

∥∥XUdAi +nsumj=1

∥∥∥dBlowastj

∥∥∥XU∥∥Bj

∥∥ + nsumj=1

∥∥∥Blowastj

∥∥∥XU∥∥dBj

∥∥

= 2msumi=1

(AiXUdAi) + 2nsumj=1

(∥∥Bj

∥∥XU∥∥dBj

∥∥)

= 2

⎡⎣ msum

i=1

(AidAi) +nsumj=1

(∥∥Bj

∥∥∥∥dBj

∥∥)⎤⎦XU

(215)

and noting (211) we obtain that

XU le∥∥∥∥∥I + 2

msumi=1

Alowasti Ai

∥∥∥∥∥ le 1 + 2msumi=1

Ai2 (216)

Then

∥∥∥∥∥∥dXU minusmsumi=1

Alowasti (dXU)Ai +

nsumj=1

Blowastj (dXU)Bj

∥∥∥∥∥∥

le 2

⎡⎣ msum

i=1

(AidAi) +nsumj=1

(∥∥Bj

∥∥∥∥dBj

∥∥)⎤⎦XU

le 2

(1 + 2

msumi=1

Ai2

)⎡⎣ msum

i=1

(AidAi) +nsumj=1

(∥∥Bj

∥∥∥∥dBj

∥∥)⎤⎦

(217)

Journal of Applied Mathematics 7∥∥∥∥∥∥dXU minus

msumi=1

Alowasti (dXU)Ai +

nsumj=1

Blowastj (dXU)Bj

∥∥∥∥∥∥

ge dXU minus∥∥∥∥∥

msumi=1

Alowasti (dXU)Ai

∥∥∥∥∥ minus∥∥∥∥∥∥

nsumj=1

Blowastj (dXU)Bj

∥∥∥∥∥∥

ge dXU minusmsumi=1

∥∥Alowasti (dXU)Ai

∥∥ minus nsumj=1

∥∥∥Blowastj (dXU)Bj

∥∥∥

ge dXU minusmsumi=1

∥∥Alowasti

∥∥dXUAi minusnsumj=1

∥∥∥Blowastj

∥∥∥dXU∥∥Bj

∥∥

=

⎛⎝1 minus

msumi=1

Ai2 minusnsumj=1

∥∥Bj

∥∥2

⎞⎠dXU

(218)

Due to (25) we have

1 minusmsumi=1

Ai2 minusnsumj=1

∥∥Bj

∥∥2gt 0 (219)

Combining (217) (218) and noting (219) we have

⎛⎝1 minus

msumi=1

Ai2 minusnsumj=1

∥∥Bj

∥∥2

⎞⎠dXU

le∥∥∥∥∥∥dXU minus

msumi=1

Alowasti (dXU)Ai +

nsumj=1

Blowastj (dXU)Bj

∥∥∥∥∥∥

le 2

(1 + 2

msumi=1

Ai2

)⎡⎣ msum

i=1

(AidAi) +nsumj=1

(∥∥Bj

∥∥∥∥dBj

∥∥)⎤⎦

(220)

which implies that

dXU le2(

1 + 2summ

i=1 Ai2)[summ

i=1(AidAi) +sumn

j=1(∥∥Bj

∥∥∥∥dBj

∥∥)]

1 minussummi=1 Ai2 minussumn

j=1

∥∥Bj

∥∥2 (221)

8 Journal of Applied Mathematics

Theorem 25 Let A1 A2 Am B1 B2 Bn be perturbed matrices of A1 A2 AmB1 B2 Bn in (11) and Δi = Ai minusAi i = 1 2 m Δj = Bj minus Bj j = 1 2 n If

msumi=1

Ai2 lt12

nsumj=1

∥∥Bj

∥∥2lt

12 (222)

2msumi=1

(AiΔi) +msumi=1

Ai2 lt12minus

msumi=1

Ai2 (223)

2nsumj=1

(∥∥Bj

∥∥∥∥Δj

∥∥) + nsumj=1

∥∥Δj

∥∥2lt

12minus

nsumj=1

∥∥Bj

∥∥2 (224)

then (11) and its perturbed equation

X minusmsumi=1

Alowasti XAi +

nsumj=1

Blowastj XBj = I (225)

have unique positive definite solutions XU and XU respectively which satisfy

∥∥∥XU minusXU

∥∥∥ le Serr (226)

where

Serr =2[1 + 2

summi=1 (Ai + Δi)2

][summi=1 (Ai + Δi)2Δi +

sumnj=1(∥∥Bj

∥∥ + ∥∥Δj

∥∥)2∥∥Δj

∥∥]

1 minussummi=1 (Ai + Δi)2 minussumn

j=1(∥∥Bj

∥∥ + ∥∥Δj

∥∥)2

(227)

Proof By (222) and Theorem 24 we know that (11) has a unique positive definite solutionXU And by (223) we have

msumi=1

∥∥∥Ai

∥∥∥2=

msumi=1

Ai + Δi2 lemsumi=1

(Ai + Δi)2

=msumi=1

(Ai2 + 2AiΔi + Δi2

)

=msumi=1

Ai2 + 2msumi=1

(AiΔi) +msumi=1

Δi2

ltmsumi=1

Ai2 +12minus

msumi=1

Ai2 =12

(228)

Journal of Applied Mathematics 9

similarly by (224) we have

nsumj=1

∥∥∥Bj

∥∥∥2lt

12 (229)

By (228) (229) and Theorem 24 we obtain that the perturbed equation (225) has a uniquepositive definite solution XU

Set

Ai(t) = Ai + tΔi Bj(t) = Bj + tΔj t isin [0 1] (230)

then by (223) we have

msumi=1

Ai(t)2 =msumi=1

Ai + tΔi2 lemsumi=1

(Ai + tΔi)2

=msumi=1

(Ai2 + 2tAiΔi + t2Δi2

)

lemsumi=1

(Ai2 + 2AiΔi + Δi2

)

=msumi=1

Ai2 + 2msumi=1

(AiΔi) +msumi=1

Δi2

ltmsumi=1

Ai2 +12minus

msumi=1

Ai2 =12

(231)

similarly by (224) we have

nsumj=1

∥∥Bj(t)∥∥2 =

msumi=1

∥∥Bj + tΔj

∥∥2lt

12 (232)

Therefore by (231) (232) and Theorem 24 we derive that for arbitrary t isin [0 1] thematrix equation

X minusmsumi=1

Alowasti (t)XAi(t) +

nsumj=1

Blowastj (t)XBj(t) = I (233)

has a unique positive definite solution XU(t) especially

XU(0) = XU XU(1) = XU (234)

10 Journal of Applied Mathematics

From Theorem 24 it follows that

∥∥∥XU minusXU

∥∥∥ = XU(1) minusXU(0) =

∥∥∥∥∥int1

0dXU(t)

∥∥∥∥∥ leint1

0dXU(t)

leint1

0

2(

1 + 2summ

i=1 Ai(t)2)[summ

i=1(Ai(t)dAi(t)) +sumn

j=1(∥∥Bj(t)

∥∥∥∥dBj(t)∥∥)]

1 minussummi=1 Ai(t)2 minussumn

j=1

∥∥Bj(t)∥∥2

leint1

0

2(

1 + 2summ

i=1 Ai(t)2)[summ

i=1(Ai(t)Δidt) +sumn

j=1(∥∥Bj(t)

∥∥∥∥Δj

∥∥dt)]

1 minussummi=1 Ai(t)2 minussumn

j=1

∥∥Bj(t)∥∥2

=int1

0

2(

1 + 2summ

i=1 Ai(t)2)[summ

i=1(Ai(t)Δi) +sumn

j=1(∥∥Bj(t)

∥∥∥∥Δj

∥∥)]

1 minussummi=1 Ai(t)2 minussumn

j=1

∥∥Bj(t)∥∥2

dt

(235)

Noting that

Ai(t) = Ai + tΔi le Ai + tΔi i = 1 2 m∥∥Bj(t)

∥∥ =∥∥Bj + tΔi

∥∥ le ∥∥Bj

∥∥ + tΔi j = 1 2 n(236)

and combining Mean Value Theorem of Integration we have

∥∥∥XU minusXU

∥∥∥

leint1

0

2(

1 + 2summ

i=1 Ai(t)2)[summ

i=1(Ai(t)Δi) +sumn

j=1(∥∥Bj(t)

∥∥∥∥Δj

∥∥)]

1 minussummi=1 Ai(t)2 minussumn

j=1

∥∥Bj(t)∥∥2

dt

leint1

0

2[1 + 2

summi=1 (Ai + tΔi)2

][summi=1 (Ai + tΔi)2Δi +

sumnj=1(∥∥Bj

∥∥ + t∥∥Δj

∥∥)2∥∥Δj

∥∥]

1 minussummi=1 (Ai + tΔi)2 minussumn

j=1(∥∥Bj

∥∥ + t∥∥Δj

∥∥)2dt

=2[1 + 2

summi=1 (Ai + ξΔi)2

][summi=1 (Ai + ξΔi)2Δi +

sumnj=1(∥∥Bj

∥∥ + ξ∥∥Δj

∥∥)2∥∥Δj

∥∥]

1 minussummi=1 (Ai + ξΔi)2 minussumn

j=1(∥∥Bj

∥∥ + ξ∥∥Δj

∥∥)2

times (1 minus 0) ξ isin [0 1]

le2[1 + 2

summi=1 (Ai + Δi)2

][summi=1 (Ai + Δi)2Δi +

sumnj=1(∥∥Bj

∥∥ + ∥∥Δj

∥∥)2∥∥Δj

∥∥]

1 minussummi=1 (Ai + Δi)2 minussumn

j=1(∥∥Bj

∥∥ + ∥∥Δj

∥∥)2= Serr

(237)

Journal of Applied Mathematics 11

3 Numerical Experiments

In this section we use a numerical example to confirm the correctness of Theorem 25 and theprecision of the perturbation bound for the unique positive definite solution XU of (11)

Example 31 Consider the symmetric linear matrix equation

X minusAlowast1XA1 minusAlowast

2XA2 + Blowast1XB1 + Blowast

2XB2 = I (31)

and its perturbed equation

X minus Alowast1XA1 minus Alowast

2XA2 + Blowast1XB1 + Blowast

2XB2 = I (32)

where

A1 =

⎛⎝ 002 minus010 minus002

008 minus010 002minus006 minus012 014

⎞⎠ A2 =

⎛⎝ 008 minus010 minus002

008 minus010 002minus006 minus012 014

⎞⎠

B1 =

⎛⎝ 047 002 004

minus010 036 minus002minus004 001 047

⎞⎠ B2 =

⎛⎝010 010 005

015 0275 0075005 005 0175

⎞⎠

A1 = A1 +

⎛⎝ 05 01 minus02

minus04 02 06minus02 01 minus01

⎞⎠ times 10minusj A2 = A2 +

⎛⎝minus04 01 minus02

05 07 minus1311 09 06

⎞⎠ times 10minusj

B1 = B1 +

⎛⎝ 08 02 005

minus02 012 014minus025 minus02 026

⎞⎠ times 10minusj B2 = B2 +

⎛⎝ 02 02 01

minus03 015 minus01501 minus01 025

⎞⎠ times 10minusj j isin N

(33)

It is easy to verify that the conditions (222)ndash(224) are satisfied then (31) and itsperturbed equation (32) have unique positive definite solutions XU and XU respectivelyFrom Berzig [7] it follows that the sequences Xk and Yk generated by the iterative method

X0 = 0 Y0 = 2I

Xk+1 = I +Alowast1XkA1 +Alowast

2XkA2 minus Blowast1YkB1 minus Blowast

2YkB2

Yk+1 = I +Alowast1YkA1 +Alowast

2YkA2 minus Blowast1XkB1 minus Blowast

2XkB2 k = 0 1 2

(34)

both converge to XU Choose τ = 10 times 10minus15 as the termination scalar that is

R(X) =∥∥X minusAlowast

1XA1 minusAlowast2XA2 + Blowast

1XB1 + Blowast2XB2 minus I

∥∥ le τ = 10 times 10minus15 (35)

By using the iterative method (34) we can get the computed solution Xk of (31) SinceR(Xk) lt 10 times 10minus15 then the computed solution Xk has a very high precision For simplicity

12 Journal of Applied Mathematics

Table 1 Numerical results for the different values of j

j 2 3 4 5 6

XU minusXUXU 213 times 10minus4 616 times 10minus6 523 times 10minus8 437 times 10minus10 845 times 10minus12

SerrXU 342 times 10minus4 713 times 10minus6 663 times 10minus8 512 times 10minus10 977 times 10minus12

we write the computed solution as the unique positive definite solution XU Similarly we canalso get the unique positive definite solution XU of the perturbed equation (32)

Some numerical results on the perturbation bounds for the unique positive definitesolution XU are listed in Table 1

From Table 1 we see that Theorem 25 gives a precise perturbation bound for theunique positive definite solution of (31)

4 Conclusion

In this paper we study the matrix equation (11) which arises in solving some nonlinearmatrix equations and the bilinear control system A new method of perturbation analysisis developed for the matrix equation (11) By making use of the matrix differentiationand its elegant properties we derive a precise perturbation bound for the unique positivedefinite solution of (11) A numerical example is presented to illustrate the sharpness of theperturbation bound

Acknowledgments

This research was supported by the National Natural Science Foundation of China(11101100 11171205 11261014 11226323) the Natural Science Foundation of GuangxiProvince (2012GXNSFBA053006 2011GXNSFA018138) the Key Project of Scientific ResearchInnovation Foundation of Shanghai Municipal Education Commission (13ZZ080) theNatural Science Foundation of Shanghai (11ZR1412500) the PhD Programs Foundation ofMinistry of Education of China (20093108110001) the Discipline Project at the correspondinglevel of Shanghai (A 13-0101-12-005) and Shanghai Leading Academic Discipline Project(J50101) The authors wish to thank Professor Y Zhang and the anonymous referees forproviding very useful suggestions for improving this paper The authors also thank DrXiaoshan Chen for discussing the properties of matrix differentiation

References

[1] L A Sakhnovich Interpolation Theory and Its Applications Mathematics and Its Applications KluwerAcademic Publishers Dordrecht The Netherlands 1997

[2] T Damm ldquoDirect methods and ADI-preconditioned Krylov subspace methods for generalizedLyapunov equationsrdquo Numerical Linear Algebra with Applications vol 15 no 9 pp 853ndash871 2008

[3] L Zhang and J Lam ldquoOn H2 model reduction of bilinear systemsrdquo Automatica vol 38 no 2 pp205ndash216 2002

[4] M C B Reurings Symmetric matrix equations [PhD thesis] Vrije Universiteit Amsterdam TheNetherlands 2003

[5] A C M Ran and M C B Reurings ldquoThe symmetric linear matrix equationrdquo Electronic Journal ofLinear Algebra vol 9 pp 93ndash107 2002

Journal of Applied Mathematics 13

[6] A C M Ran and M C B Reurings ldquoA fixed point theorem in partially ordered sets and someapplications to matrix equationsrdquo Proceedings of the American Mathematical Society vol 132 no 5 pp1435ndash1443 2004

[7] M Berzig ldquoSolving a class of matrix equation via Bhaskar-Lakshmikantham coupled fixed pointtheoremrdquo Applied Mathematics Letters vol 25 no 11 pp 1638ndash1643 2012

[8] J Cai and G Chen ldquoSome investigation on Hermitian positive definite solutions of the matrixequation Xs +AlowastXminustA = Qrdquo Linear Algebra and Its Applications vol 430 no 8-9 pp 2448ndash2456 2009

[9] X Duan and A Liao ldquoOn the existence of Hermitian positive definite solutions of the matrix equationXs +AlowastXminustA = Qrdquo Linear Algebra and its Applications vol 429 no 4 pp 673ndash687 2008

[10] X Duan A Liao and B Tang ldquoOn the nonlinear matrix equation X minus summi=1 A

lowasti X

δiAi = Qrdquo LinearAlgebra and its Applications vol 429 no 1 pp 110ndash121 2008

[11] S M El-Sayed and A C M Ran ldquoOn an iteration method for solving a class of nonlinear matrixequationsrdquo SIAM Journal on Matrix Analysis and Applications vol 23 no 3 pp 632ndash645 2001

[12] Z Gajic and M T J Qureshi Lyapunov Matrix Equation in System Stability and Control Academic PressLondon UK 1995

[13] V I Hasanov ldquoNotes on two perturbation estimates of the extreme solutions to the equations X plusmnAlowastXminus1A = Qrdquo Applied Mathematics and Computation vol 216 no 5 pp 1355ndash1362 2010

[14] Z-Y Peng S M El-Sayed and X-l Zhang ldquoIterative methods for the extremal positive definitesolution of the matrix equation X + AlowastXminusαA = Qrdquo Journal of Computational and Applied Mathematicsvol 200 no 2 pp 520ndash527 2007

[15] B R Fang J D Zhou and Y M Li Matrix Theory Tsinghua University Press Beijing China 2006

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2012 Article ID 483624 11 pagesdoi1011552012483624

Research ArticleRefinements of Kantorovich Inequality forHermitian Matrices

Feixiang Chen

School of Mathematics and Statistics Chongqing Three Gorges University WanzhouChongqing 404000 China

Correspondence should be addressed to Feixiang Chen cfx2002126com

Received 28 August 2012 Accepted 5 November 2012

Academic Editor K C Sivakumar

Copyright q 2012 Feixiang Chen This is an open access article distributed under the CreativeCommons Attribution License which permits unrestricted use distribution and reproduction inany medium provided the original work is properly cited

Some new Kantorovich-type inequalities for Hermitian matrix are proposed in this paper Weconsider what happens to these inequalities when the positive definite matrix is allowed to beinvertible and provides refinements of the classical results

1 Introduction and Preliminaries

We first state the well-known Kantorovich inequality for a positive definite Hermite matrix(see [1 2]) let A isin Mn be a positive definite Hermitian matrix with real eigenvalues λ1 leλ2 le middot middot middot le λn Then

1 le xlowastAxxlowastAminus1x le (λ1 + λn)2

4λ1λn (11)

for any x isin Cn x = 1 where Alowast denotes the conjugate transpose of matrix A A matrix

A isin Mn is Hermitian if A = Alowast An equivalent form of this result is incorporated in

0 le xlowastAxxlowastAminus1x minus 1 le (λn minus λ1)2

4λ1λn (12)

for any x isin Cn x = 1

Attributed to Kantorovich the inequality has built up a considerable literatureThis typically comprises generalizations Examples are [3ndash5] for matrix versions Operator

2 Journal of Applied Mathematics

versions are developed in [6 7] Multivariate versions have been useful in statistics to assessthe robustness of least squares see [8 9] and the references therein

Due to the important applications of the original Kantorovich inequality for matrices[10] in Statistics [8 11 12] and Numerical Analysis [13 14] any new inequality of this typewill have a flow of consequences in the areas of applications

Motivated by the interest in both pure and applied mathematics outlined abovewe establish in this paper some improvements of Kantorovich inequalities The classicalKantorovich-type inequalities are modified to apply not only to positive definite but also toinvertible Hermitian matrices As natural tools in deriving the new results the recent Gruss-type inequalities for vectors in inner product in [6 15ndash19] are utilized

To simplify the proof we first introduce some lemmas

2 Lemmas

Let B be a Hermitian matrix with real eigenvalues μ1 le μ2 le middot middot middot le μn if A minus B is positivesemidefinite we write

A ge B (21)

that is λi ge μi i = 1 2 n On Cn we have the standard inner product defined by 〈x y〉 =sumn

i=1 xiyi where x = (x1 xn)lowast isin C

n and y = (y1 yn)lowast isin C

n

Lemma 21 Let a b c and d be real numbers then one has the following inequality

(a2 minus b2

)(c2 minus d2

)le (ac minus bd)2 (22)

Lemma 22 Let A and B be Hermitian matrices if AB = BA then

AB le (A + B)2

4 (23)

Lemma 23 Let A ge 0 B ge 0 if AB = BA then

AB ge 0 (24)

3 Some Results

The following lemmas can be obtained from [16ndash19] by replacing Hilbert space (H 〈middot middot〉) withinner product spaces C

n so we omit the details

Lemma 31 Let u v and e be vectors in Cn and e = 1 If α β δ and γ are real or complex

numbers such that

Relangβe minus u u minus αe

rang ge 0 Relangδe minus v v minus γe

rang ge 0 (31)

Journal of Applied Mathematics 3

then

|〈u v〉 minus 〈u e〉〈e v〉| le 14(β minus α

)(δ minus γ

) minus [Re

langβe minus u u minus αe

rangRe

langδe minus v v minus γe

rang]12 (32)

Lemma 32 With the assumptions in Lemma 31 one has

|〈u v〉 minus 〈u e〉〈e v〉| le 14(β minus α

)(γ minus δ

) minus∣∣∣∣〈u e〉 minus α + β

2

∣∣∣∣∣∣∣∣〈v e〉 minus γ + δ

2

∣∣∣∣ (33)

Lemma 33 With the assumptions in Lemma 31 if Re(βα) ge 0 Re(δγ) ge 0 one has

|〈u v〉 minus 〈u e〉〈e v〉| le∣∣β minus α

∣∣∣∣δ minus γ∣∣

4[Re

(βα

)Re

(δγ

)]12|〈u e〉〈e v〉| (34)

Lemma 34 With the assumptions in Lemma 33 one has

|〈u v〉 minus 〈u e〉〈e v〉|

le(∣∣α + β

∣∣ minus 2[Re

(βα

)]12)(∣∣δ + γ

∣∣ minus 2[Re

(δγ

)]12)12

[|〈u e〉〈e v〉|]12(35)

4 New Kantorovich Inequalities for Hermitian Matrices

For a Hermitian matrix A as in [6] we define the following transform

C(A) = (λnI minusA)(A minus λ1I) (41)

When A is invertible if λ1λn gt 0 then

C(Aminus1

)=(

1λ1

I minusAminus1)(

Aminus1 minus 1λn

I

) (42)

Otherwise λ1λn lt 0 then

C(Aminus1

)=(

1λk+1

I minusAminus1)(

Aminus1 minus 1λk

I

) (43)

where

λ1 le middot middot middot le λk lt 0 lt λk+1 le middot middot middot le λn (44)

From Lemma 23 we can conclude that C(A) ge 0 and C(Aminus1) ge 0

4 Journal of Applied Mathematics

For two Hermitian matrices A and B and x isin Cn x = 1 we define the following

functional

G(ABx) = 〈AxBx〉 minus 〈Ax x〉〈x Bx〉 (45)

When A = B we denote

G(Ax) = Ax2 minus 〈Ax x〉2 (46)

for x isin Cn x = 1

Lemma 41 With notations above and for x isin Cn x = 1 then

0 le 〈C(A)x x〉 le (λn minus λ1)2

4 (47)

if λnλ1 gt 0

0 lelangC(Aminus1

)x x

rangle (λn minus λ1)

2

4(λnλ1)2 (48)

if λnλ1 lt 0

0 lelangC(Aminus1

)x x

rangle (λk+1 minus λk)

2

4(λk+1λk)2 (49)

Proof From C(A) ge 0 then

〈C(A)x x〉 ge 0 (410)

While from Lemma 22 we can get

C(A) = (λnI minusA)(A minus λ1I) le (λn minus λ1)2

4I (411)

Then 〈C(A)x x〉 le (λn minus λ1)24 is straightforward The proof for C(Aminus1) is similar

Lemma 42 With notations above and for x isin Cn x = 1 then

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le G(Ax)G

(Aminus1x

) (412)

Journal of Applied Mathematics 5

Proof Thus

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2

=∣∣∣xlowast

((xlowastAminus1x

)I minusAminus1

)((xlowastAx)I minusA)x

∣∣∣2

le∥∥∥((xlowastAminus1x

)I minusAminus1

)x∥∥∥2

((xlowastAx)I minusA)x2

(413)

while

((xlowastAx)I minusA)x2 = xlowast((xlowastAx)2I minus 2(xlowastAx)A +A2

)x

= xlowastA2x minus (xlowastAx)2

= Ax2 minus 〈Ax x〉2

= G(Ax)

(414)

Similarly we can get ((xlowastAminus1x)I minusAminus1)x2 = G(Aminus1x) then we complete the proof

Theorem 43 Let A B be two Hermitian matrices and C(A) ge 0 C(B) ge 0 are defined as abovethen

|G(ABx)| le 14(λn minus λ1)

(μn minus μ1

) minus [〈C(A)x x〉〈C(B)x x〉]12

|G(ABx)| le 14(λn minus λ1)

(μn minus μ1

) minus∣∣∣∣lang(

A minus λ1 + λn2

)x x

rang∣∣∣∣∣∣∣∣lang(

B minus μ1 + μn

2

)x x

rang∣∣∣∣(415)

for any x isin Cn x = 1

If λ1λn gt 0 μ1μn gt 0 then

|G(ABx)| le (λn minus λ1)(μn minus μ1

)4[(λnλ1)

(μnμ1

)]12|〈Ax x〉〈Bx x〉|

|G(ABx)| le(

|λ1 + λn| minus 2(λ1λn)12

)(∣∣μ1 + μn

∣∣ minus 2(μ1μn

)12)12

[|〈Ax x〉〈Bx x〉|]12

(416)

for any x isin Cn x = 1

Proof The proof follows by Lemmas 31 32 33 and 34 on choosing u = Ax v = Bx ande = x β = λn α = λ1 δ = μn and γ = μ1 x isin C

n x = 1 respectively

6 Journal of Applied Mathematics

Corollary 44 Let A be a Hermitian matrices and C(A) ge 0 is defined as above then

|G(Ax)| le 14(λn minus λ1)

2 minus 〈C(A)x x〉 (417)

|G(Ax)| le 14(λn minus λ1)

2 minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣2

(418)

for any x isin Cn x = 1

If λ1λn gt 0 then

|G(Ax)| le (λn minus λ1)2

4(λnλ1)|〈Ax x〉|2 (419)

|G(Ax)| le(|λ1 + λn| minus 2

radicλ1λn

)|〈Ax x〉| (420)

for any x isin Cn x = 1

Proof The proof follows by Theorem 43 on choosing A = B respectively

Corollary 45 Let A be a Hermitian matrices and C(Aminus1) ge 0 is defined as above then one has thefollowing

If λ1λn gt 0 then

∣∣∣G(Aminus1x

)∣∣∣ le (λn minus λ1)2

4(λnλ1)2

minuslangC(Aminus1

)x x

rang (421)

∣∣∣G(Aminus1x

)∣∣∣ le (λn minus λ1)2

4(λnλ1)2

minus∣∣∣∣lang(

Aminus1 minus λ1 + λn2λnλ1

I

)x x

rang∣∣∣∣2

(422)

∣∣∣G(Aminus1x

)∣∣∣ le (λn minus λ1)2

4λnλ1

∣∣∣langAminus1x xrang∣∣∣2

(423)

∣∣∣G(Aminus1x

)∣∣∣ le(

|λ1 + λn|λ1λn

minus 21radicλ1λn

)∣∣∣langAminus1x xrang∣∣∣ (424)

for any x isin Cn x = 1

If λ1λn lt 0 then

∣∣∣G(Aminus1x

)∣∣∣ le (λk minus λk+1)2

4(λkλk+1)2

minuslangC(Aminus1

)x x

rang (425)

Journal of Applied Mathematics 7

where C(Aminus1) = ((1λk+1)I minusAminus1)(Aminus1 minus (1λk)I) and

∣∣∣G(Aminus1x

)∣∣∣ le (λk minus λk+1)2

4(λkλk+1)2

minus∣∣∣∣lang(

Aminus1 minus λk+1 + λk2λkλk+1

I

)x x

rang∣∣∣∣2

(426)

for any x isin Cn x = 1

Proof The proof follows by Corollary 44 by replacing A with Aminus1 respectively

Theorem 46 LetA be an ntimesn invertible Hermitian matrix with real eigenvalues λ1 le λ2 le middot middot middot le λnthen one has the following

If λ1λn gt 0 then

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le (λn minus λ1)

2

4(λnλ1)minusradic〈C(A)x x〉langC(Aminus1

)x x

rang (427)

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le (λn minus λ1)

2

4(λnλ1)minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣∣∣∣∣lang(

Aminus1 minus λ1 + λn2λnλ1

I

)x x

rang∣∣∣∣(428)

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le (λn minus λ1)

2

4(λnλ1)2 〈Ax x〉

langAminus1x x

rang (429)

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le

(radic|λn| minus

radic|λ1|

)2

radicλnλ1

radic〈Ax x〉〈Aminus1x x〉 (430)

for any x isin Cn x = 1

If λ1λn lt 0 then

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le (λn minus λ1)(λk+1 minus λk)

4|λkλk+1| minusradic〈C(A)x x〉langC(Aminus1

)x x

rang (431)

where C(Aminus1) = ((1λk+1)I minusAminus1)(Aminus1 minus (1λk)I)

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le (λn minus λ1)(λk+1 minus λk)

4|λkλk+1|

minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣∣∣∣∣lang(

Aminus1 minus λk + λk+1

2λ1λk+1I

)x x

rang∣∣∣∣(432)

Proof Considering

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le G(Ax)G

(Aminus1x

) (433)

8 Journal of Applied Mathematics

When λ1λn gt 0 from (417) and (421) we get

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le

14(λn minus λ1)

2 minus 〈C(A)x x〉

(λn minus λ1)2

4(λnλ1)2

minuslangC(Aminus1

)x x

rang (434)

From (a2 minus b2)(c2 minus d2) le (ac minus bd)2 we have

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le

(λn minus λ1)

2

4(λnλ1)minusradic[〈C(A)x x〉langC(Aminus1

)x x

rang]2

(435)

then the conclusion (427) holdsSimilarly from (418) and (422) we get

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le

14(λn minus λ1)

2 minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣2

times(λn minus λ1)

2

4(λnλ1)2

minus∣∣∣∣lang(

Aminus1 minus λ1 + λn2λnλ1

I

)x x

rang∣∣∣∣2

le(λn minus λ1)

2

4(λnλ1)minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣∣∣∣∣lang(

Aminus1 minus λ1 + λn2λnλ1

I

)x x

rang∣∣∣∣2

(436)

then the conclusion (428) holdsFrom (419) and (423) we get

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le (λn minus λ1)

2

4(λnλ1)|〈Ax x〉|2 (λn minus λ1)

2

4(λnλ1)

∣∣∣langAminus1x xrang∣∣∣2

(437)

then the conclusion (429) holdsFrom (420) and (424) we get

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le

(|λ1 + λn| minus 2

radicλ1λn

)|〈Ax x〉|

(|λ1 + λn|λ1λn

minus 21radicλ1λn

)∣∣∣langAminus1x xrang∣∣∣(438)

then the conclusion (430) holdsWhen λ1λn lt 0 from (417) and (425) we get

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2le

14(λn minus λ1)

2 minus 〈C(A)x x〉

(λk minus λk+1)2

4(λkλk+1)2

minuslangC(Aminus1

)x x

rang (439)

then the conclusion (431) holds

Journal of Applied Mathematics 9

From (418) and (426) we get

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le

14(λn minus λ1)

2 minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣2

times(λk minus λk+1)

2

4(λkλk+1)2

minus∣∣∣∣lang(

Aminus1 minus λk+1 + λk2λkλk+1

I

)x x

rang∣∣∣∣2

le(λn minus λ1)(λk+1 minus λk)

4|λkλk+1|

minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣∣∣∣∣lang(

Aminus1 minus λk + λk+1

2λ1λk+1I

)x x

rang∣∣∣∣2

(440)

then the conclusion (432) holds

Corollary 47 With the notations above for any x isin Cn x = 1 one lets

|G1(Ax)| = 14(λn minus λ1)

2 minus 〈C(A)x x〉

|G2(Ax)| = 14(λn minus λ1)

2 minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣2

(441)

If λ1λn gt 0 one lets

|G3(Ax)| = (λn minus λ1)2

4(λnλ1)|〈Ax x〉|2

|G4(Ax)| =(|λ1 + λn| minus 2

radicλ1λn

)|〈Ax x〉|

(442)

If λ1λn gt 0

∣∣∣G1(Aminus1x

)∣∣∣ = (λn minus λ1)2

4(λnλ1)2

minuslangC(Aminus1

)x x

rang

∣∣∣G2(Aminus1x

)∣∣∣ = (λn minus λ1)2

4(λnλ1)2

minus∣∣∣∣lang(

Aminus1 minus λ1 + λn2λnλ1

I

)x x

rang∣∣∣∣2

∣∣∣G3(Aminus1x

)∣∣∣ = (λn minus λ1)2

4λnλ1

∣∣∣langAminus1x xrang∣∣∣2

∣∣∣G4(Aminus1x

)∣∣∣ =(

|λ1 + λn|λ1λn

minus 21radicλ1λn

)∣∣∣langAminus1x xrang∣∣∣

(443)

10 Journal of Applied Mathematics

If λ1λn lt 0 then

∣∣∣G5(Aminus1x

)∣∣∣ = (λk minus λk+1)2

4(λkλk+1)2

minuslangC(Aminus1

)x x

rang (444)

where C(Aminus1) = ((1λk+1)I minusAminus1)(Aminus1 minus (1λk)I) and

∣∣∣G6(Aminus1x

)∣∣∣ = (λk minus λk+1)2

4(λkλk+1)2

minus∣∣∣∣lang(

Aminus1 minus λk+1 + λk2λkλk+1

I

)x x

rang∣∣∣∣2

(445)

Then one has the followingIf λ1λn gt 0

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le

radicGlowast(Ax)Glowast(Aminus1x

) (446)

where

Glowast(Ax) = minG1(Ax) G2(Ax) G3(Ax) G4(Ax)

Glowast(Ax) = minG1

(Aminus11x

) G2

(Aminus1x

) G3

(Aminus1x

) G4

(Aminus1x

)

(447)

If λ1λn lt 0

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le

radicG(Ax)Gbull(Aminus1x

) (448)

where

G(Ax) = minG1(Ax) G2(Ax)

Gbull(Ax) = minG5

(Aminus1x

) G6

(Aminus1x

)

(449)

Proof The proof follows from that the conclusions in Corollaries 44 and 45 are independent

Remark 48 It is easy to see that if λ1 gt 0 λn gt 0 our result coincides with the inequalityof operator versions in [6] So we conclude that our results give an improvement of theKantorovich inequality [6] that applies to all invertible Hermite matrices

5 Conclusion

In this paper we introduce some new Kantorovich-type inequalities for the invertibleHermitian matrices Inequalities (427) and (431) are the same as [4] but our proof is simpleIn Theorem 46 if λ1 gt 0 λn gt 0 the results are similar to the well-known Kantorovich-typeinequalities for operators in [6] Moreover for any invertible Hermitian matrix there exists asimilar inequality

Journal of Applied Mathematics 11

Acknowledgments

The authors are grateful to the associate editor and two anonymous referees whose commentshave helped to improve the final version of the present work This work was supported by theNatural Science Foundation of Chongqing Municipal Education Commission (no KJ091104)

References

[1] R A Horn and C R Johnson Matrix Analysis Cambridge University Press Cambridge UK 1985[2] A S Householder The Theory of Matrices in Numerical Analysis Blaisdell New York NY USA 1964[3] S Liu and H Neudecker ldquoSeveral matrix Kantorovich-type inequalitiesrdquo Journal of Mathematical

Analysis and Applications vol 197 no 1 pp 23ndash26 1996[4] A W Marshall and I Olkin ldquoMatrix versions of the Cauchy and Kantorovich inequalitiesrdquo

Aequationes Mathematicae vol 40 no 1 pp 89ndash93 1990[5] J K Baksalary and S Puntanen ldquoGeneralized matrix versions of the Cauchy-Schwarz and

Kantorovich inequalitiesrdquo Aequationes Mathematicae vol 41 no 1 pp 103ndash110 1991[6] S S Dragomir ldquoNew inequalities of the Kantorovich type for bounded linear operators in Hilbert

spacesrdquo Linear Algebra and its Applications vol 428 no 11-12 pp 2750ndash2760 2008[7] P G Spain ldquoOperator versions of the Kantorovich inequalityrdquo Proceedings of the American

Mathematical Society vol 124 no 9 pp 2813ndash2819 1996[8] S Liu and H Neudecker ldquoKantorovich inequalities and efficiency comparisons for several classes of

estimators in linear modelsrdquo Statistica Neerlandica vol 51 no 3 pp 345ndash355 1997[9] P Bloomfield and G S Watson ldquoThe inefficiency of least squaresrdquo Biometrika vol 62 pp 121ndash128

1975[10] L V Kantorovic ldquoFunctional analysis and applied mathematicsrdquo Uspekhi Matematicheskikh Nauk vol

3 no 6(28) pp 89ndash185 1948 (Russian)[11] C G Khatri and C R Rao ldquoSome extensions of the Kantorovich inequality and statistical

applicationsrdquo Journal of Multivariate Analysis vol 11 no 4 pp 498ndash505 1981[12] C T Lin ldquoExtrema of quadratic forms and statistical applicationsrdquo Communications in Statistics A

vol 13 no 12 pp 1517ndash1520 1984[13] A Galantai ldquoA study of Auchmutyrsquos error estimaterdquo Computers amp Mathematics with Applications vol

42 no 8-9 pp 1093ndash1102 2001[14] P D Robinson and A J Wathen ldquoVariational bounds on the entries of the inverse of a matrixrdquo IMA

Journal of Numerical Analysis vol 12 no 4 pp 463ndash486 1992[15] S S Dragomir ldquoA generalization of Grussrsquos inequality in inner product spaces and applicationsrdquo

Journal of Mathematical Analysis and Applications vol 237 no 1 pp 74ndash82 1999[16] S S Dragomir ldquoSome Gruss type inequalities in inner product spacesrdquo Journal of Inequalities in Pure

and Applied Mathematics vol 4 no 2 article 42 2003[17] S S Dragomir ldquoOn Bessel and Gruss inequalities for orthonormal families in inner product spacesrdquo

Bulletin of the Australian Mathematical Society vol 69 no 2 pp 327ndash340 2004[18] S S Dragomir ldquoReverses of Schwarz triangle and Bessel inequalities in inner product spacesrdquo Journal

of Inequalities in Pure and Applied Mathematics vol 5 no 3 article 76 2004[19] S S Dragomir ldquoReverses of the Schwarz inequality generalising a Klamkin-McLenaghan resultrdquo

Bulletin of the Australian Mathematical Society vol 73 no 1 pp 69ndash78 2006[20] Z Liu L Lu and K Wang ldquoOn several matrix Kantorovich-type inequalitiesrdquo Journal of Inequalities

and Applications vol 2010 Article ID 571629 5 pages 2010

Advances in Matrices Finite and Infinitewith Applications

Journal of Applied Mathematics

Advances in Matrices Finite and Infinitewith Applications

Guest Editors P N Shivakumar Panayiotis PsarrakosK C Sivakumar and Yang Zhang

Copyright copy 2013 Hindawi Publishing Corporation All rights reserved

This is a special issue published in ldquoJournal of Applied Mathematicsrdquo All articles are open access articles distributed under the CreativeCommons Attribution License which permits unrestricted use distribution and reproduction in any medium provided the originalwork is properly cited

Editorial Board

Saeid Abbasbandy IranMina B Abd-El-Malek EgyptMohamed A Abdou EgyptSubhas Abel IndiaMostafa Adimy FranceCarlos J S Alves PortugalMohamad Alwash USAIgor Andrianov GermanySabri Arik TurkeyFrancis TK Au Hong KongOlivier Bahn CanadaRoberto Barrio SpainAlfredo Bellen ItalyJafar Biazar IranHester Bijl The NetherlandsAnjan Biswas Saudi ArabiaStephane PA Bordas USAJames Robert Buchanan USAAlberto Cabada SpainXiao Chuan Cai USAJinde Cao ChinaAlexandre Carvalho BrazilSong Cen ChinaQianshun S Chang ChinaTai-Ping Chang TaiwanShih-sen Chang ChinaRushan Chen ChinaXinfu Chen USAKe Chen UKEric Cheng Hong KongFrancisco Chiclana UKJen-Tzung Chien TaiwanC S Chien TaiwanHan H Choi Republic of KoreaTin-Tai Chow ChinaM S H Chowdhury MalaysiaCarlos Conca ChileVitor Costa PortugalLivija Cveticanin SerbiaEric de Sturler USAOrazio Descalzi ChileKai Diethelm GermanyVit Dolejsi Czech RepublicBo-Qing Dong ChinaMagdy A Ezzat Egypt

Meng Fan ChinaYa Ping Fang ChinaAntonio J M Ferreira PortugalMichel Fliess FranceM A Fontelos SpainHuijun Gao ChinaBernard J Geurts The NetherlandsJamshid Ghaboussi USAPablo Gonzalez-Vera SpainLaurent Gosse ItalyK S Govinder South AfricaJose L Gracia SpainYuantong Gu AustraliaZhihong GUAN ChinaNicola Guglielmi ItalyFrederico G Guimaraes BrazilVijay Gupta IndiaBo Han ChinaMaoan Han ChinaPierre Hansen CanadaFerenc Hartung HungaryXiaoqiao He Hong KongLuis Javier Herrera SpainJ Hoenderkamp The NetherlandsYing Hu FranceNing Hu JapanZhilong L Huang ChinaKazufumi Ito USATakeshi Iwamoto JapanGeorge Jaiani GeorgiaZhongxiao Jia ChinaTarun Kant IndiaIdo Kanter IsraelAbdul Hamid Kara South AfricaHamid Reza Karimi NorwayJae-Wook Kim UKJong Hae Kim Republic of KoreaKazutake Komori JapanFanrong Kong USAVadim Krysko RussiaJin L Kuang SingaporeMiroslaw Lachowicz PolandHak-Keung Lam UKTak-Wah Lam Hong KongPGL Leach Cyprus

Yongkun Li ChinaWan-Tong Li ChinaJ Liang ChinaChing-Jong Liao TaiwanChong Lin ChinaMingzhu Liu ChinaChein-Shan Liu TaiwanKang Liu USAYansheng Liu ChinaFawang Liu AustraliaShutian Liu ChinaZhijun Liu ChinaJulian Lopez-Gomez SpainShiping Lu ChinaGert Lube GermanyNazim Idrisoglu Mahmudov TurkeyOluwole Daniel Makinde South AfricaFrancisco J Marcellan SpainGuiomar Martın-Herran SpainNicola Mastronardi ItalyMichael McAleer The NetherlandsStephane Metens FranceMichael Meylan AustraliaAlain Miranville FranceRam N Mohapatra USAJaime E Munoz Rivera BrazilJavier Murillo SpainRoberto Natalini ItalySrinivasan Natesan IndiaJiri Nedoma Czech RepublicJianlei Niu Hong KongRoger Ohayon FranceJavier Oliver SpainDonal OrsquoRegan IrelandMartin Ostoja-Starzewski USATurgut Ozis TurkeyClaudio Padra ArgentinaReinaldo Martinez Palhares BrazilFrancesco Pellicano ItalyJuan Manuel Pena SpainRicardo Perera SpainMalgorzata Peszynska USAJames F Peters CanadaMark A Petersen South AfricaMiodrag Petkovic Serbia

Vu Ngoc Phat VietnamAndrew Pickering SpainHector Pomares SpainMaurizio Porfiri USAMario Primicerio ItalyMorteza Rafei The NetherlandsRoberto Reno ItalyJacek Rokicki PolandDirk Roose BelgiumCarla Roque PortugalDebasish Roy IndiaSamir H Saker EgyptMarcelo A Savi BrazilWolfgang Schmidt GermanyEckart Schnack GermanyMehmet Sezer TurkeyNaseer Shahzad Saudi ArabiaFatemeh Shakeri IranJian Hua Shen ChinaHui-Shen Shen ChinaFernando Simoes PortugalTheodore E Simos Greece

Abdel-Maksoud A Soliman EgyptXinyu Song ChinaQiankun Song ChinaYuri N Sotskov BelarusPeter J C Spreij The NetherlandsNiclas Stromberg SwedenRay KL Su Hong KongJitao Sun ChinaWenyu Sun ChinaXianHua Tang ChinaAlexander Timokha NorwayMariano Torrisi ItalyJung-Fa Tsai TaiwanCh Tsitouras GreeceKuppalapalle Vajravelu USAAlvaro Valencia ChileErik Van Vleck USAEzio Venturino ItalyJesus Vigo-Aguiar SpainMichael N Vrahatis GreeceBaolin Wang ChinaMingxin Wang China

Qing-Wen Wang ChinaGuangchen Wang ChinaJunjie Wei ChinaLi Weili ChinaMartin Weiser GermanyFrank Werner GermanyShanhe Wu ChinaDongmei Xiao ChinaGongnan Xie ChinaYuesheng Xu USASuh-Yuh Yang TaiwanBo Yu ChinaJinyun Yuan BrazilAlejandro Zarzo SpainGuisheng Zhai JapanJianming Zhan ChinaZhihua Zhang ChinaJingxin Zhang AustraliaShan Zhao USAChongbin Zhao AustraliaRenat Zhdanov USAHongping Zhu China

Advances in Matrices Finite and Infinite with Applications P N Shivakumar Panayiotis PsarrakosK C Sivakumar and Yang ZhangVolume 2013 Article ID 156136 3 pages

On Nonnegative Moore-Penrose Inverses of Perturbed Matrices Shani Jose and K C SivakumarVolume 2013 Article ID 680975 7 pages

A New Construction of Multisender Authentication Codes from Polynomials over Finite FieldsXiuli WangVolume 2013 Article ID 320392 7 pages

Geometrical and Spectral Properties of the Orthogonal Projections of the Identity Luis GonzalezAntonio Suarez and Dolores GarcıaVolume 2013 Article ID 435730 8 pages

A Parameterized Splitting Preconditioner for Generalized Saddle Point ProblemsWei-Hua Luo and Ting-Zhu HuangVolume 2013 Article ID 489295 6 pages

Solving Optimization Problems on Hermitian Matrix Functions with ApplicationsXiang Zhang and Shu-Wen XiangVolume 2013 Article ID 593549 11 pages

Left and Right Inverse Eigenpairs Problem for κ-Hermitian Matrices Fan-Liang Li Xi-Yan Huand Lei ZhangVolume 2013 Article ID 230408 6 pages

Completing a 2times 2 Block Matrix of Real Quaternions with a Partial Specified InverseYong Lin and Qing-Wen WangVolume 2013 Article ID 271978 5 pages

Fuzzy Approximate Solution of Positive Fully Fuzzy Linear Matrix EquationsXiaobin Guo and Dequan ShangVolume 2013 Article ID 178209 7 pages

Ranks of a Constrained Hermitian Matrix Expression with Applications Shao-Wen YuVolume 2013 Article ID 514984 9 pages

An Efficient Algorithm for the Reflexive Solution of the Quaternion Matrix Equation AXB + CXHD = FNing Li Qing-Wen Wang and Jing JiangVolume 2013 Article ID 217540 14 pages

The Optimization on Ranks and Inertias of a Quadratic Hermitian Matrix Function and Its ApplicationsYirong YaoVolume 2013 Article ID 961568 6 pages

Constrained Solutions of a System of Matrix Equations Qing-Wen Wang and Juan YuVolume 2012 Article ID 471573 19 pages

Contents

On the Hermitian R-Conjugate Solution of a System of Matrix Equations Chang-Zhou DongQing-Wen Wang and Yu-Ping ZhangVolume 2012 Article ID 398085 14 pages

Perturbation Analysis for the Matrix Equation X minussummi=1 A

lowasti XAi +

sumnj=1 B

lowastj XBj = I

Xue-Feng Duan and Qing-Wen WangVolume 2012 Article ID 784620 13 pages

Refinements of Kantorovich Inequality for Hermitian Matrices Feixiang ChenVolume 2012 Article ID 483624 11 pages

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 156136 3 pageshttpdxdoiorg1011552013156136

EditorialAdvances in Matrices Finite and Infinite with Applications

P N Shivakumar1 Panayiotis Psarrakos2 K C Sivakumar3 and Yang Zhang1

1 Department of Mathematics University of Manitoba Winnipeg MB Canada R3T 2N22Department of Mathematics School of Applied Mathematical and Physical SciencesNational Technical Univeristy of Athens Zografou Campus 15780 Athens Greece

3 Department of Mathematics Indian Institute of Technology Madras Chennai 600 036 India

Correspondence should be addressed to P N Shivakumar shivakuccumanitobaca

Received 13 June 2013 Accepted 13 June 2013

Copyright copy 2013 P N Shivakumar et al This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

In the mathematical modeling of many problems includingbut not limited to physical sciences biological sciencesengineering image processing computer graphics medicalimaging and even social sciences lately solution methodsmost frequently involve matrix equations In conjunctionwith powerful computing machines complex numericalcalculations employing matrix methods could be performedsometimes even in real time It is well known that infinitematrices arise more naturally than finite matrices and havea colorful history in development from sequences seriesand quadratic forms Modern viewpoint considers infinitematrices more as operators defined between certain spe-cific infinite-dimensional normed spaces Banach spacesor Hilbert spaces Giant and rapid advancements in thetheory and applications of finite matrices have been madeover the past several decades These developments havebeen well documented in many books monographs andresearch journals The present topics of research includediagonally dominantmatrices theirmany extensions inverseof matrices their recursive computation singular matricesgeneralized inverses inverse positive matrices with specificemphasis on their applications to economics like the Leon-tief input-output models and use of finite difference andfinite element methods in the solution of partial differentialequations perturbationmethods and eigenvalue problems ofinterest and importance to numerical analysts statisticiansphysical scientists and engineers to name a few

The aim of this special issue is to announce certainrecent advancements in matrices finite and infinite andtheir applications For a review of infinite matrices andapplications see P N Shivakumar and K C Sivakumar

(Linear Algebra Appl 2009) Specifically an example of anapplication can be found in the classical problem ldquoshapeof a drumrdquo in P N Shivakumar W Yan and Y Zhang(WSES transactions in Mathematics 2011) The focal themesof this special issue are inverse positivematrices including119872-matrices applications of operator theorymatrix perturbationtheory and pseudospectra matrix functions and generalizedeigenvalue problems and inverse problems including scat-tering and matrices over quaternions In the following wepresent a brief overview of a few of the central topics of thisspecial issue

For119872-matrices let us start by mentioning the most citedbooks byR Berman andR J Plemmons (SIAM 1994) and byR A Horn and C R Johnson (Cambridge 1994) as excellentsources One of the most important properties of invertible119872-matrices is that their inverses are nonnegative There areseveral works that have considered generalizations of someof the important properties of 119872-matrices We only pointout a few of them here The article by D Mishra and K CSivakumar (Oper Matrices 2012) considers generalizationsof inverse nonnegativity to the Moore-Penrose inverse whilemore general classes of matrices are the objects of discussionin D Mishra and K C Sivakumar (Linear Algebra Appl2012) and A N Sushama K Premakumari and K CSivakumar (Electron J Linear Algebra 2012)

A brief description of the papers in the issue is as followsIn the work of Z Zhang the problem of solving cer-

tain optimization problems on Hermitian matrix functionsis studied Specifically the author considers the extremalinertias and ranks of a given function 119891(119883 119884) C119898times119899 rarr

C119898times119899 (which is defined in terms of matrices 119860 119861 119862 and 119863

2 Journal of Applied Mathematics

inC119898times119899) subject to119883 119884 satisfying certain matrix equationsAs applications the author presents necessary and sufficientconditions for 119891 to be positive definite positive semidefiniteand so forth As another application the author presents acharacterization for 119891(119883 119884) = 0 again subject to 119883 119884satisfying certain matrix equations

The task of determining parametrized splitting precon-ditioners for certain generalized saddle point problems isundertaken byW-H Luo and T-Z Huang in their workThewell-known and very widely applicable Sherman-Morrison-Woodbury formula for the sum of matrices is used indetermining a preconditioner for linear equations wherethe coefficient matrix may be singular and nonsymmetricIllustrations of the procedure are provided for Stokes andOseen problems

The work reported by A Gonzalez A Suarez and DGarcia deals with the problem of estimating the Frobeniuscondition of the matrix 119860119873 given a matrix 119860 isin R119899times119899 where119873 satisfies the condition that 119873 = argmin

119883isin119878119868 minus 119860119883

119865

Here 119878 is a certain subspace of R119899 and sdot 119865denotes the

Frobenius norm The authors derive certain spectral andgeometrical properties of the preconditioner119873 as above

Let 119861 119862 119863 and 119864 be (not necessarily square) matriceswith quaternion entries so that the matrix 119872 = (

119860 119861

119862 119863)

is square Y Lin and Q-W Wang present necessary andsufficient conditions so that the top left block 119860 of119872 existssatisfying the conditions that119872 is nonsingular and thematrix119864 is the top left block of119872minus1 A general expression for suchan 119860 when it exists is obtained A numerical illustration ispresented

For a complex hermitian 119860 of order 119899 with eigenvalues1205821le 1205822le sdot sdot sdot le 120582

119899 the celebrated Kantorovich inequality is

1 le 119909lowast

119860119909119909lowast

119860minus1

119909 le (120582119899+1205821)2

41205821120582119899 for all unit vectors 119909 isin

C119899 Equivalently 0 le 119909lowast119860119909119909lowast119860minus1119909minus1 le (120582119899minus1205821)2

41205821120582119899 for

all unit vectors 119909 isin C119899 F Chen in his work on refinements ofKantorovich inequality for hermitian matrices obtains somerefinements of the second inequality where the proofs useonly elementary ideas (httpwwwmathntuagrsimppsarr)

It is pertinent to point out that among other things aninteresting generalization of 119872-matrices is obtained by SJose and K C Sivakumar The authors consider the prob-lem of nonnegativity of the Moore-Penrose inverse of aperturbation of the form 119860minus119883119866119884

119879 given that 119860dagger ge 0 Usinga generalized version of the Sherman-Morrison-Woodburyformula conditions for (119860 minus 119883119866119884

119879

)dagger to be nonnegative

are derived Applications of the results are presented brieflyIterative versions of the results are also studied

It is well known that the concept of quaternions wasintroduced by Hamilton in 1843 and has played an importantrole in contemporary mathematics such as noncommuta-tive algebra analysis and topology Nowadays quaternionmatrices are widely and heavily used in computer sciencequantum physics altitude control robotics signal and colourimage processing and so on Many questions can be reducedto quaternion matrices and solving quaternion matrix equa-tionsThis special issue has provided an excellent opportunityfor researchers to report their recent results

Let H119899times119899 denote the space of all 119899 times 119899 matrices whoseentries are quaternions Let 119876 isin H119899times119899 be Hermitian and self-invertible 119883 isin H119899times119899 is called reflexive (relative to 119876) if 119883 =

119876119883119876 The authors F-L Li X-Y Hu and L Zhang presentan efficient algorithm for finding the reflexive solution of thequaternion matrix equation 119860119883119861 + 119862119883

lowast

119863 = 119865 They alsoshow that given an appropriate initialmatrix their procedureyields the least reflexive solutionwith respect to the Frobeniusnorm

In the work on the ranks of a constrained hermitianmatrix expression S W Yu gives formulas for the maximaland minimal ranks of the quaternion Hermitian matrixexpression 119862

4minus 1198604119883119860lowast

4 where 119883 is a Hermitian solution

to quaternion matrix equations 1198601119883 = 119862

1 1198831198611= 1198622 and

1198603119883119860lowast

3= 1198623 and also applies this result to some special

system of quaternion matrix equationsY Yao reports his findings on the optimization on ranks

and inertias of a quadratic hermitian matrix function Hepresents one solution for optimization problems on the ranksand inertias of the quadratic Hermitian matrix function 119876 minus119883119875119883lowast subject to a consistent system of matrix equations

119860119883 = 119862 and 119883119861 = 119863 As applications he derives necessaryand sufficient conditions for the solvability to the systems ofmatrix equations and matrix inequalities 119860119883 = 119862 119883119861 = 119863and119883119875119883lowast = (gt lt ge le)119876

Let 119877 be a nontrivial real symmetric involution matrixA complex matrix 119860 is called 119877-conjugate if 119860 = 119877119860119877In the work on the hermitian 119877-conjugate solution of asystem of matrix equations C-Z Dong Q-W Wang andY-P Zhang present necessary and sufficient conditions forthe existence of the hermitian 119877-conjugate solution to thesystem of complex matrix equations 119860119883 = 119862 and 119883119861 = 119863An expression of the Hermitian 119877-conjugate solution is alsopresented

A perturbation analysis for the matrix equation 119883 minus

sum119898

119894=1119860lowast

119894119883119860119894+ sum119899

119895=1119861lowast

119895119883119861119895= 119868 is undertaken by X-F Duan

and Q-W Wang They give a precise perturbation bound fora positive definite solution A numerical example is presentedto illustrate the sharpness of the perturbation bound

Linear matrix equations and their optimal approxima-tion problems have great applications in structural designbiology statistics control theory and linear optimal controland as a consequence they have attracted the attention ofseveral researchers for many decades In the work of Q-WWang and J Yu necessary and sufficient conditions of and theexpressions for orthogonal solutions symmetric orthogonalsolutions and skew-symmetric orthogonal solutions of thesystem of matrix equations 119860119883 = 119861 and 119883119862 = 119863 arederived When the solvability conditions are not satisfied theleast squares symmetric orthogonal solutions and the leastsquares skew-symmetric orthogonal solutions are obtainedA methodology is provided to compute the least squaressymmetric orthogonal solutions and an illustrative exampleis given to illustrate the proposed algorithm

Many real-world engineering systems are too complex tobe defined in precise terms and in many matrix equationssome or all of the system parameters are vague or impreciseThus solving a fuzzy matrix equation becomes important In

Journal of Applied Mathematics 3

the work reported by X Guo and D Shang the fuzzy matrixequation 119860 otimes 119883 otimes 119861 = 119862 in which 119860 119861 and 119862 are 119898 times 119898119899 times 119899 and 119898 times 119899 nonnegative LR fuzzy numbers matricesrespectively is investigated This fuzzy matrix system whichhas a wide use in control theory and control engineering isextended into three crisp systems of linear matrix equationsaccording to the arithmetic operations of LR fuzzy numbersBased on the pseudoinversematrix a computingmodel to thepositive fully fuzzy linear matrix equation is provided andthe fuzzy approximate solution of original fuzzy systems isobtained by solving the crisp linear matrix systems In addi-tion the existence condition of nonnegative fuzzy solutionis discussed and numerical examples are given to illustratethe proposed method Overall the technique of LR fuzzynumbers and their operations appears to be a strong tool ininvestigating fully fuzzy matrix equations

Left and right inverse eigenpairs problem arises in anatural way when studying spectral perturbations ofmatricesand relative recursive problems F-L Li X-Y Hu and LZhang consider in their work left and right inverse eigenpairsproblem for 119896-hermitian matrices and its optimal approx-imate problem The class of 119896-hermitian matrices containshermitian and perhermitian matrices and has numerousapplications in engineering and statistics Based on thespecial properties of 119896-hermitian matrices and their left andright eigenpairs an equivalent problem is obtained Thencombining a new inner product of matrices necessary andsufficient conditions for the solvability of the problem aregiven and its general solutions are derived Furthermorethe optimal approximate solution a calculation procedure tocompute the unique optimal approximation and numericalexperiments are provided

Finally let us point out a work on algebraic coding theoryRecall that a communication system in which a receivercan verify the authenticity of a message sent by a group ofsenders is referred to as a multisender authentication codeIn the work reported by X Wang the author constructsmulti-sender authentication codes using polynomials overfinite fields Here certain important parameters and theprobabilities of deceptions of these codes are also computed

Acknowledgments

We wish to thank all the authors for their contributions andthe reviewers for their valuable comments in bringing thespecial issue to fruition

P N ShivakumarPanayiotis Psarrakos

K C SivakumarYang Zhang

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 680975 7 pageshttpdxdoiorg1011552013680975

Research ArticleOn Nonnegative Moore-Penrose Inverses of Perturbed Matrices

Shani Jose and K C Sivakumar

Department of Mathematics Indian Institute of Technology Madras Chennai Tamil Nadu 600036 India

Correspondence should be addressed to K C Sivakumar kcskumariitmacin

Received 22 April 2013 Accepted 7 May 2013

Academic Editor Yang Zhang

Copyright copy 2013 S Jose and K C Sivakumar This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

Nonnegativity of theMoore-Penrose inverse of a perturbation of the form119860minus119883119866119884119879 is consideredwhen119860dagger ge 0 Using a generalizedversion of the Sherman-Morrison-Woodbury formula conditions for (119860 minus 119883119866119884119879)dagger to be nonnegative are derived Applications ofthe results are presented briefly Iterative versions of the results are also studied

1 Introduction

We consider the problem of characterizing nonnegativityof the Moore-Penrose inverse for matrix perturbations ofthe type 119860 minus 119883119866119884

119879 when the Moore-Penrose inverse of119860 is nonnegative Here we say that a matrix 119861 = (119887

119894119895) is

nonnegative and denote it by 119861 ge 0 if 119887119894119895ge 0 forall119894 119895 This

problemwasmotivated by the results in [1] where the authorsconsider an 119872-matrix 119860 and find sufficient conditions forthe perturbed matrix (119860 minus 119883119884119879) to be an119872-matrix Let usrecall that a matrix 119861 = (119887

119894119895) is said to 119885-matrix if 119887

119894119895le 0

for all 119894 119895 119894 = 119895 An119872-matrix is a nonsingular 119885-matrix withnonnegative inverse The authors in [1] use the well-knownSherman-Morrison-Woodbury (SMW) formula as one of theimportant tools to prove their main resultThe SMW formulagives an expression for the inverse of (119860minus119883119884119879) in terms of theinverse of119860 when it exists When119860 is nonsingular119860minus119883119884119879

is nonsingular if and only 119868 minus 119884119879119860minus1119883 is nonsingular In thatcase

(119860 minus 119883119884119879

)minus1

= 119860minus1

minus 119860minus1

119883(119868 minus 119884119879

119860minus1

119883)minus1

119884119879

119860minus1

(1)

Themain objective of the present work is to study certainstructured perturbations 119860 minus 119883119884119879 of matrices 119860 such thattheMoore-Penrose inverse of the perturbation is nonnegativewhenever the Moore-Penrose inverse of 119860 is nonnegativeClearly this class of matrices includes the class of matricesthat have nonnegative inverses especially119872-matrices In ourapproach extensions of SMW formula for singular matrices

play a crucial role Let us mention that this problem has beenstudied in the literature (See for instance [2] formatrices and[3] for operators over Hilbert spaces) We refer the reader tothe references in the latter for other recent extensions

In this paper first we present alternative proofs ofgeneralizations of the SMW formula for the cases of theMoore-Penrose inverse (Theorem 5) and the group inverse(Theorem 6) in Section 3 In Section 4 we characterize thenonnegativity of (119860 minus 119883119866119884119879)dagger This is done in Theorem 9and is one of the main results of the present work As aconsequence we present a result for119872-matrices which seemsnew We present a couple of applications of the main resultin Theorems 13 and 15 In the concluding section we studyiterative versions of the results of the second section Weprove two characterizations for (119860minus119883119884119879)dagger to be nonnegativein Theorems 18 and 21

Before concluding this introductory section let us givea motivation for the work that we have undertaken hereIt is a well-documented fact that 119872-matrices arise quiteoften in solving sparse systems of linear equations Anextensive theory of119872-matrices has been developed relativeto their role in numerical analysis involving the notion ofsplitting in iterativemethods and discretization of differentialequations in the mathematical modeling of an economyoptimization and Markov chains [4 5] Specifically theinspiration for the present study comes from the work of [1]where the authors consider a system of linear inequalitiesarising out of a problem in third generation wireless commu-nication systemsThematrix defining the inequalities there is

2 Journal of Applied Mathematics

an 119872-matrix In the likelihood that the matrix of thisproblem is singular (due to truncation or round-off errors)the earlier method becomes inapplicable Our endeavour isto extend the applicability of these results to more generalmatrices for instance matrices with nonnegative Moore-Penrose inverses Finally as mentioned earlier sincematriceswith nonnegative generalized inverses include in particular119872-matrices it is apparent that our results are expected toenlarge the applicability of themethods presently available for119872-matrices even in a very general framework including thespecific problem mentioned above

2 Preliminaries

Let R R119899 and R119898times119899 denote the set of all real numbers the119899-dimensional real Euclidean space and the set of all 119898 times 119899matrices over R For 119860 isin R119898times119899 let 119877(119860) 119873(119860) 119877(119860)perp and119860119879 denote the range space the null space the orthogonal

complement of the range space and the transpose of thematrix 119860 respectively For x = (119909

1 1199092 119909

119899)119879

isin R119899 wesay that x is nonnegative that is x ge 0 if and only if x

119894ge 0

for all 119894 = 1 2 119899 As mentioned earlier for a matrix 119861 weuse 119861 ge 0 to denote that all the entries of 119861 are nonnegativeAlso we write 119860 le 119861 if 119861 minus 119860 ge 0

Let 120588(119860) denote the spectral radius of the matrix 119860 If120588(119860) lt 1 for 119860 isin R119899times119899 then 119868 minus 119860 is invertible Thenext result gives a necessary and sufficient condition for thenonnegativity of (119868 minus 119860)minus1 This will be one of the results thatwill be used in proving the first main result

Lemma 1 (see [5 Lemma 21 Chapter 6]) Let 119860 isin R119899times119899 benonnegative Then 120588(119860) lt 1 if and only if (119868 minus 119860)minus1 exists and

(119868 minus 119860)minus1

=

infin

sum

119896=0

119860119896

ge 0 (2)

More generally matrices having nonnegative inversesare characterized using a property called monotonicity Thenotion of monotonicity was introduced by Collatz [6] A real119899 times 119899matrix 119860 is called monotone if 119860x ge 0 rArr x ge 0 It wasproved by Collatz [6] that 119860 is monotone if and only if 119860minus1exists and 119860minus1 ge 0

One of the frequently used tools in studying monotonematrices is the notion of a regular splitting We only refer thereader to the book [5] for more details on the relationshipbetween these concepts

The notion of monotonicity has been extended in a vari-ety of ways to singular matrices using generalized inversesFirst let us briefly review the notion of two importantgeneralized inverses

For 119860 isin R119898times119899 the Moore-Penrose inverse is the unique119885 isin R119899times119898 satisfying the Penrose equations 119860119885119860 = 119860119885119860119885 = 119885 (119860119885)119879 = 119860119885 and (119885119860)119879 = 119885119860 The uniqueMoore-Penrose inverse of119860 is denoted by119860dagger and it coincideswith 119860minus1 when 119860 is invertible

The following theorem by Desoer and Whalen whichis used in the sequel gives an equivalent definition for theMoore-Penrose inverse Let us mention that this result wasproved for operators between Hilbert spaces

Theorem 2 (see [7]) Let 119860 isin R119898times119899 Then 119860dagger isin R119899times119898 is theunique matrix119883 isin R119899times119898 satisfying

(i) 119885119860119909 = 119909 forall119909 isin 119877(119860119879)(ii) 119885119910 = 0 forall119910 isin 119873(119860119879)

Now for119860 isin R119899times119899 any119885 isin R119899times119899 satisfying the equations119860119885119860 = 119860 119885119860119885 = 119885 and 119860119885 = 119885119860 is called the groupinverse of 119860 The group inverse does not exist for everymatrix But whenever it exists it is unique A necessary andsufficient condition for the existence of the group inverse of119860 is that the index of119860 is 1 where the index of a matrix is thesmallest positive integer 119896 such that rank (119860119896+1) = rank (119860119896)

Some of the well-known properties of theMoore-Penroseinverse and the group inverse are given as follows 119877(119860119879) =119877(119860dagger

)119873(119860119879) = 119873(119860dagger) 119860dagger119860 = 119875119877(119860119879) and 119860119860dagger = 119875

119877(119860) In

particular x isin 119877(119860119879) if and only if x = 119860dagger119860x Also 119877(119860) =119877(119860

) 119873(119860) = 119873(119860

) and 119860119860 = 119860119860

= 119875119877(119860)119873(119860)

Herefor complementary subspaces 119871 and119872 of R119896 119875

119871119872denotes

the projection ofR119896 onto 119871 along119872 119875119871denotes 119875

119871119872if119872 =

119871perp For details we refer the reader to the book [8]Inmatrix analysis a decomposition (splitting) of amatrix

is considered in order to study the convergence of iterativeschemes that are used in the solution of linear systems of alge-braic equations As mentioned earlier regular splittings areuseful in characterizing matrices with nonnegative inverseswhereas proper splittings are used for studying singularsystems of linear equations Let us next recall this notionFor a matrix 119860 isin R119898times119899 a decomposition 119860 = 119880 minus 119881 iscalled a proper splitting [9] if 119877(119860) = 119877(119880) and 119873(119860) =119873(119880) It is rather well-known that a proper splitting existsfor every matrix and that it can be obtained using a full-rank factorization of the matrix For details we refer to [10]Certain properties of a proper splitting are collected in thenext result

Theorem 3 (see [9 Theorem 1]) Let 119860 = 119880 minus 119881 be a propersplitting of 119860 isin R119898times119899 Then

(a) 119860 = 119880(119868 minus 119880dagger119881)(b) 119868 minus 119880dagger119881 is nonsingular and(c) 119860dagger = (119868 minus 119880dagger119881)minus1119880dagger

The following result by Berman and Plemmons [9] givesa characterization for 119860dagger to be nonnegative when 119860 has aproper splitting This result will be used in proving our firstmain result

Theorem 4 (see [9 Corollary 4]) Let 119860 = 119880 minus 119881 be a propersplitting of 119860 isin R119898times119899 where 119880dagger ge 0 and 119880dagger119881 ge 0 Then119860dagger

ge 0 if and only if 120588(119880dagger119881) lt 1

3 Extensions of the SMW Formula forGeneralized Inverses

The primary objects of consideration in this paper are gen-eralized inverses of perturbations of certain types of a matrix

Journal of Applied Mathematics 3

119860 Naturally extensions of the SMW formula for generalizedinverses are relevant in the proofs Inwhat follows we presenttwo generalizations of the SMW formula for matrices Wewould like to emphasize that our proofs also carry overto infinite dimensional spaces (the proof of the first resultis verbatim and the proof of the second result with slightmodifications applicable to range spaces instead of ranks ofthe operators concerned) However we are confining ourattention to the case of matrices Let us also add that theseresults have been proved in [3] for operators over infinitedimensional spaces We have chosen to include these resultshere as our proofs are different from the ones in [3] and thatour intention is to provide a self-contained treatment

Theorem 5 (see [3 Theorem 21]) Let 119860 isin R119898times119899 119883 isin R119898times119896and 119884 isin R119899times119896 be such that

119877 (119883) sube 119877 (119860) 119877 (119884) sube 119877 (119860119879

) (3)

Let 119866 isin R119896times119896 and Ω = 119860 minus 119883119866119884119879 If

119877 (119883119879

) sube 119877 ((119866dagger

minus 119884119879

119860dagger

119883)119879

)

119877 (119884119879

) sube 119877 (119866dagger

minus 119884119879

119860dagger

119883) 119877 (119883119879

) sube 119877 (119866)

119877 (119884119879

) sube 119877 (119866119879

)

(4)

then

Ωdagger

= 119860dagger

+ 119860dagger

119883(119866dagger

minus 119884119879

119860dagger

119883)dagger

119884119879

119860dagger

(5)

Proof Set 119885 = 119860dagger + 119860dagger119883119867dagger119884119879119860dagger where119867 = 119866dagger

minus 119884119879

119860dagger

119883From the conditions (3) and (4) it follows that 119860119860dagger119883 = 119883119860dagger

119860119884 = 119884119867dagger119867119883119879 = 119883119879119867119867dagger119884119879 = 119884119879119866119866dagger119883119879 = 119883119879 and119866dagger

119866119884119879

= 119884119879

Now

119860dagger

119883119867dagger

119884119879

minus 119860dagger

119883119867dagger

119884119879

119860dagger

119883119866119884119879

= 119860dagger

119883119867dagger

(119866dagger

119866119884119879

minus 119884119879

119860dagger

119883119866119884119879

)

= 119860dagger

119883119867dagger

119867119866119884119879

= 119860dagger

119883119866119884119879

(6)

Thus 119885Ω = 119860dagger

119860 Since 119877(Ω119879) sube 119877(119860119879

) it follows that119885Ω(x) = 119909 forallx isin 119877(Ω119879)

Let y isin 119873(Ω119879

) Then 119860119879y minus 119884119866119879119883119879y = 0 so that119883119879

119860dagger119879

(119860119879

minus 119884119866119879

119883119879

)y = 0 Substituting 119883119879 = 119866119866dagger119883119879 andsimplifying it we get119867119879119866119879119883119879y = 0 Also119860119879y = 119884119866119879119883119879y =119884119867119867dagger

119866119879

119883119879y = 0 and so 119860daggery = 0 Thus 119885y = 0 for

y isin 119873(Ω119879) Hence by Theorem 2 119885 = Ωdagger

The result for the group inverse follows

Theorem 6 Let 119860 isin R119899times119899 be such that 119860 exists Let 119883119884 isin R119899times119896 and 119866 be nonsingular Assume that 119877(119883) sube 119877(119860)119877(119884) sube 119877(119860

119879

) and 119866minus1 minus 119884119879119860119883 is nonsingular Suppose that

rank (119860 minus 119883119866119884119879) = rank (119860) Then (119860 minus 119883119866119884119879) exists andthe following formula holds

(119860 minus 119883119866119884119879

)= 119860

+ 119860

119883(119866minus1

minus 119884119879

119860119883)minus1

119884119879

119860 (7)

Conversely if (119860minus119883119866119884119879) exists then the formula above holdsand we have rank (119860 minus 119883119866119884119879) = rank (119860)

Proof Since 119860 exists 119877(119860) and 119873(119860) are complementarysubspaces of R119899times119899

Suppose that rank (119860 minus 119883119866119884119879) = rank (119860) As 119877(119883) sube119877(119860) it follows that 119877(119860 minus 119883119866119884119879) sube 119877(119860) Thus 119877(119860 minus119883119866119884119879

) = 119877(119860) By the rank-nullity theorem the nullity ofboth 119860 minus 119883119866119884119879 and 119860 are the same Again since 119877(119884) sube119877(119860119879

) it follows that 119873(119860 minus 119883119866119884119879

) = 119873(119860) Thus119877(119860 minus 119883119866119884

119879

) and 119873(119860 minus 119883119866119884119879) are complementary sub-spaces This guarantees the existence of the group inverse of119860 minus 119883119866119884

119879Conversely suppose that (119860 minus 119883119866119884119879) exists It can be

verified by direct computation that 119885 = 119860+ 119860

119883(119866minus1

minus

119884119879

119860119883)minus1

119884119879

119860 is the group inverse of 119860 minus 119883119866119884119879 Also we

have (119860minus119883119866119884119879)(119860minus119883119866119884119879) = 119860119860 so that119877(119860minus119883119866119884119879) =119877(119860) and hence the rank (119860 minus 119883119866119884119879) = rank (119860)

We conclude this section with a fairly old result [2] as aconsequence of Theorem 5

Theorem 7 (see [2 Theorem 15]) Let 119860 isin R119898times119899 of rank 119903119883 isin R119898times119903 and 119884 isin R119899times119903 Let119866 be an 119903times119903 nonsingular matrixAssume that 119877(119883) sube 119877(119860) 119877(119884) sube 119877(119860119879) and 119866minus1 minus 119884119879119860dagger119883is nonsingular Let Ω = 119860 minus 119883119866119884119879 Then

(119860 minus 119883119866119884119879

)dagger

= 119860dagger

+ 119860dagger

119883(119866minus1

minus 119884119879

119860dagger

119883)minus1

119884119879

119860dagger

(8)

4 Nonnegativity of (119860minus119883119866119884119879)dagger

In this section we consider perturbations of the form 119860 minus

119883119866119884119879 and derive characterizations for (119860 minus 119883119866119884119879)dagger to be

nonnegative when 119860dagger ge 0 119883 ge 0 119884 ge 0 and 119866 ge 0 In orderto motivate the first main result of this paper let us recall thefollowing well known characterization of119872-matrices [5]

Theorem 8 Let 119860 be a 119885-matrix with the representation 119860 =119904119868 minus 119861 where 119861 ge 0 and 119904 ge 0 Then the following statementsare equivalent

(a) 119860minus1 exists and 119860minus1 ge 0

(b) There exists 119909 gt 0 such that 119860119909 gt 0

(c) 120588(119861) lt 119904

Let us prove the first result of this article This extendsTheorem 8 to singular matrices We will be interested inextensions of conditions (a) and (c) only

Theorem 9 Let 119860 = 119880 minus 119881 be a proper splitting of 119860 isin R119898times119899

with 119880dagger ge 0 119880dagger119881 ge 0 and 120588(119880dagger119881) lt 1 Let 119866 isin R119896times119896 benonsingular and nonnegative 119883 isin R119898times119896 and 119884 isin R119899times119896 be

4 Journal of Applied Mathematics

nonnegative such that 119877(119883) sube 119877(119860) 119877(119884) sube 119877(119860119879) and119866minus1 minus119884119879

119860dagger

119883 is nonsingular LetΩ = 119860minus119883119866119884119879Then the followingare equivalent

(a) Ωdagger ge 0(b) (119866minus1 minus 119884119879119860dagger119883)minus1 ge 0(c) 120588(119880dagger119882) lt 1 where119882 = 119881 + 119883119866119884

119879

Proof First we observe that since 119877(119883) sube 119877(119860) 119877(119884) sube119877(119860119879

) and 119866minus1 minus 119884119879119860dagger119883 is nonsingular by Theorem 7 wehave

Ωdagger

= 119860dagger

+ 119860dagger

119883(119866minus1

minus 119884119879

119860dagger

119883)minus1

119884119879

119860dagger

(9)

We thus haveΩΩdagger = 119860119860dagger andΩdaggerΩ = 119860dagger119860 Therefore

119877 (Ω) = 119877 (119860) 119873 (Ω) = 119873 (119860) (10)

Note that the first statement also implies that 119860dagger ge 0 byTheorem 4

(a) rArr (b) By taking 119864 = 119860 and 119865 = 119883119866119884119879 we get

Ω = 119864 minus 119865 as a proper splitting for Ω such that 119864dagger = 119860dagger ge 0and 119864dagger119865 = 119860dagger119883119866119884119879 ge 0 (since 119883119866119884119879 ge 0) Since Ωdagger ge 0by Theorem 4 we have 120588(119860dagger119883119866119884119879) = 120588(119864

dagger

119865) lt 1 Thisimplies that 120588(119884119879119860dagger119883119866) lt 1 We also have 119884119879119860dagger119883119866 ge 0Thus by Lemma 1 (119868minus119884119879119860dagger119883119866)minus1 exists and is nonnegativeBut we have 119868 minus 119884119879119860dagger119883119866 = (119866

minus1

minus 119884119879

119860dagger

119883)119866 Now 0 le(119868 minus 119884

119879

119860dagger

119883119866)minus1

= 119866minus1

(119866minus1

minus 119884119879

119860dagger

119883)minus1 This implies that

(119866minus1

minus 119884119879

119860dagger

119883)minus1

ge 0 since 119866 ge 0 This proves (b)(b) rArr (c) We have119880minus119882 = 119880minus119881minus119883119866119884

119879

= 119860minus119883119884119879

=

Ω Also 119877(Ω) = 119877(119860) = 119877(119880) and 119873(Ω) = 119873(119860) = 119873(119880)So Ω = 119880 minus 119882 is a proper splitting Also 119880dagger ge 0 and119880dagger

119882 = 119880dagger

(119881+119883119866119884119879

) ge 119880dagger

119881 ge 0 Since (119866minus1minus119884119879119860dagger119883)minus1 ge0 it follows from (9) that Ωdagger ge 0 (c) now follows fromTheorem 4

(c) rArr (a) Since119860 = 119880minus119881 we haveΩ = 119880minus119881minus119883119866119884119879 =119880minus119882 Also we have119880dagger ge 0119880dagger119881 ge 0 Thus119880dagger119882 ge 0 since119883119866119884119879

ge 0 Now by Theorem 4 we are done if the splittingΩ = 119880 minus119882 is a proper splitting Since 119860 = 119880 minus 119881 is a propersplitting we have 119877(119880) = 119877(119860) and 119873(119880) = 119873(119860) Nowfrom the conditions in (10) we get that 119877(119880) = 119877(Ω) and119873(119880) = 119873(Ω) Hence Ω = 119880 minus 119882 is a proper splitting andthis completes the proof

The following result is a special case of Theorem 9

Theorem 10 Let119860 = 119880minus119881 be a proper splitting of119860 isin R119898times119899

with 119880dagger ge 0 119880dagger119881 ge 0 and 120588(119880dagger119881) lt 1 Let 119883 isin R119898times119896 and119884 isin R119899times119896 be nonnegative such that 119877(119883) sube 119877(119860) 119877(119884) sube119877(119860119879

) and 119868minus119884119879119860dagger119883 is nonsingular LetΩ = 119860minus119883119884119879 Thenthe following are equivalent

(a) Ωdagger ge 0(b) (119868 minus 119884119879119860dagger119883)minus1 ge 0(c) 120588(119880dagger119882) lt 1 where119882 = 119881 + 119883119884

119879

The following consequence of Theorem 10 appears to benew This gives two characterizations for a perturbed 119872-matrix to be an119872-matrix

Corollary 11 Let119860 = 119904119868minus119861where119861 ge 0 and 120588(119861) lt 119904 (ie119860is an119872-matrix) Let 119883 isin R119898times119896 and 119884 isin R119899times119896 be nonnegativesuch that 119868 minus 119884119879119860dagger119883 is nonsingular Let Ω = 119860 minus 119883119884119879 Thenthe following are equivalent

(a) Ωminus1 ge 0(b) (119868 minus 119884119879119860dagger119883)minus1 ge 0(c) 120588(119861(119868 + 119883119884119879)) lt 119904

Proof From the proof ofTheorem 10 since 119860 is nonsingularit follows that ΩΩdagger = 119868 and ΩdaggerΩ = 119868 This shows that Ωis invertible The rest of the proof is omitted as it is an easyconsequence of the previous result

In the rest of this section we discuss two applicationsof Theorem 10 First we characterize the least element ina polyhedral set defined by a perturbed matrix Next weconsider the following Suppose that the ldquoendpointsrdquo of aninterval matrix satisfy a certain positivity property Then allmatrices of a particular subset of the interval also satisfythat positivity condition The problem now is if that we aregiven a specific structured perturbation of these endpointswhat conditions guarantee that the positivity property for thecorresponding subset remains valid

The first result is motivated by Theorem 12 below Let usrecall that with respect to the usual order an element xlowast isinX sube R119899 is called a least element ofX if it satisfies xlowast le x forall x isin X Note that a nonempty set may not have the leastelement and if it exists then it is unique In this connectionthe following result is known

Theorem 12 (see [11 Theorem 32]) For 119860 isin R119898times119899 and b isinR119898 let

Xb = x isin R119899

119860x + y ge b 119875119873(119860)

x = 0

119875119877(119860)

y = 0 119891119900119903 119904119900119898119890 y isin R119898 (11)

Then a vector xlowast is the least element ofXb if and only if xlowast =119860daggerb isin Xb with 119860dagger ge 0

Now we obtain the nonnegative least element of apolyhedral set defined by a perturbed matrix This is animmediate application of Theorem 10

Theorem 13 Let 119860 isin R119898times119899 be such that 119860dagger ge 0 Let 119883 isin

R119898times119896 and let 119884 isin R119899times119896 be nonnegative such that 119877(119883) sube119877(119860) 119877(119884) sube 119877(119860119879) and 119868 minus 119884119879119860dagger119883 is nonsingular Supposethat (119868 minus 119884119879119860dagger119883)minus1 ge 0 For b isin R119898 119887 ge 0 let

Sb = x isin R119899

Ωx + y ge b 119875119873(Ω)

x = 0

119875119877(Ω)

y = 0 119891119900119903 119904119900119898119890 y isin R119898 (12)

where Ω = 119860 minus 119883119884119879 Then xlowast = (119860 minus 119883119884119879)daggerb is the least

element of S119887

Proof From the assumptions using Theorem 10 it fol-lows that Ωdagger ge 0 The conclusion now follows fromTheorem 12

Journal of Applied Mathematics 5

To state and prove the result for interval matrices let usfirst recall the notion of interval matrices For119860 119861 isin R119898 times 119899an interval (matrix) 119869 = [119860 119861] is defined as 119869 = [119860 119861] = 119862 119860 le 119862 le 119861 The interval 119869 = [119860 119861] is said to be range-kernelregular if 119877(119860) = 119877(119861) and 119873(119860) = 119873(119861) The followingresult [12] provides necessary and sufficient conditions for119862dagger

ge 0 for 119862 isin 119870 where 119870 = 119862 isin 119869 119877(119862) = 119877(119860) =

119877(119861)119873(119862) = 119873(119860) = 119873(119861)

Theorem 14 Let 119869 = [119860 119861] be range-kernel regular Then thefollowing are equivalent

(a) 119862dagger ge 0 whenever 119862 isin 119870(b) 119860dagger ge 0 and 119861dagger ge 0

In such a case we have 119862dagger = suminfin119895=0(119861dagger

(119861 minus 119862))119895

119861dagger

Now we present a result for the perturbation

Theorem 15 Let 119869 = [119860 119861] be range-kernel regular with119860dagger ge0 and 119861dagger ge 0 Let 119883

1 1198832 1198841 and 119884

2be nonnegative matrices

such that 119877(1198831) sube 119877(119860) 119877(119884

1) sube 119877(119860

119879

) 119877(1198832) sube 119877(119861) and

119877(1198842) sube 119877(119861

119879

) Suppose that 119868minus1198841198791119860dagger

1198831and 119868minus119884119879

2119860dagger

1198832are

nonsingular with nonnegative inverses Suppose further that1198832119884119879

2le 1198831119884119879

1 Let 119869 = [119864 119865] where 119864 = 119860 minus 119883

1119884119879

1and

119865 = 119861 minus 1198832119884119879

2 Finally let = 119885 isin 119869 119877(119885) = 119877(119864) =

119877(119865) 119886119899119889 119873(119885) = 119873(119864) = 119873(119865) Then

(a) 119864dagger ge 0 and 119865dagger ge 0(b) 119885dagger ge 0 whenever 119885 isin

In that case 119885dagger = suminfin119895=0(119861dagger

119861 minus 119865dagger

119885)119895

119865dagger

Proof It follows fromTheorem 7 that

119864dagger

= 119860dagger

+ 119860dagger

1198831(119868 minus 119884

119879

1119860dagger

1198831)minus1

119884119879

1119860dagger

119865dagger

= 119861dagger

+ 119861dagger

1198832(119868 minus 119884

119879

2119861dagger

1198832)minus1

119884119879

2119861dagger

(13)

Also we have 119864119864dagger = 119860119860dagger 119864dagger119864 = 119860

dagger

119860 119865dagger119865 = 119861dagger

119861and 119865119865dagger = 119861119861

dagger Hence 119877(119864) = 119877(119860) 119873(119864) = 119873(119860)119877(119865) = 119877(119861) and 119873(119865) = 119873(119861) This implies that theinterval 119869 is range-kernel regular Now since 119864 and 119865 satisfythe conditions of Theorem 10 we have 119864dagger ge 0 and 119865dagger ge 0proving (a) Hence by Theorem 14 119885dagger ge 0 whenever 119885 isin Again by Theorem 14 we have 119885dagger = suminfin

119895=0(119865dagger

(119865 minus 119885))119895

119865dagger

=

suminfin

119895=0(119861dagger

119861 minus 119865dagger

119885)119895

119865dagger

5 Iterations That Preserve Nonnegativity ofthe Moore-Penrose Inverse

In this section we present results that typically provideconditions for iteratively defined matrices to have nonneg-ative Moore-Penrose inverses given that the matrices thatwe start with have this property We start with the followingresult about the rank-one perturbation case which is a directconsequence of Theorem 10

Theorem 16 Let 119860 isin R119899times119899 be such that 119860dagger ge 0 Let x isin R119898

and y isin R119899 be nonnegative vectors such that x isin 119877(119860) y isin119877(119860119879

) and 1 minus y119879119860daggerx = 0 Then (119860 minus xy119879)dagger ge 0 if and only ify119879119860daggerx lt 1

Example 17 Let us consider 119860 = (13) (1 minus1 1

minus1 4 minus1

1 minus1 1

) It can be

verified that 119860 can be written in the form 119860 = (1198682

119890119879

1

) (1198682+

e1119890119879

1)minus1

119862minus1

(1198682+ e1119890119879

1)minus1

(1198682e1) where e

1= (1 0)

119879 and 119862 isthe 2 times 2 circulant matrix generated by the row (1 12) Wehave 119860dagger = (12) ( 2 1 21 2 1

2 1 2

) ge 0 Also the decomposition 119860 =

119880minus119881 where119880 = (13) ( 1 minus1 2minus1 4 minus2

1 minus1 2

) and119881 = (13) ( 0 0 10 2 minus10 0 1

) isa proper splitting of 119860 with 119880dagger ge 0 119880dagger119881 ge 0 and 120588(119880dagger119881) lt1

Let x = y = (13 16 13)119879 Then x and let y are

nonnegative x isin 119877(119860) and y isin 119877(119860119879) We have 1 minus y119879119860daggerx =512 gt 0 and 120588(119880dagger119882) lt 1 for119882 = 119881 + xy119879 Also it can beseen that (119860 minus xy119879)dagger = (120) ( 47 28 4728 32 28

47 28 47

) ge 0 This illustratesTheorem 10

Let x1 x2 x

119896isin R119898 and y

1 y2 y

119896isin R119899 be nonneg-

ative Denote 119883119894= (x1 x2 x

119894) and 119884

119894= (y1 y2 y

119894) for

119894 = 1 2 119896 Then119860minus119883119896119884119879

119896= 119860minussum

119896

119894=1x119894y119879119894 The following

theorem is obtained by a recurring application of the rank-one result of Theorem 16

Theorem 18 Let119860 isin R119898times119899 and let119883119894 119884119894be as above Further

suppose that x119894isin 119877(119860 minus 119883

119894minus1119884119879

119894minus1) y119894isin 119877(119860 minus 119883

119894minus1119884119879

119894minus1)119879 and

1 minus y119879119894(119860 minus 119883

119894minus1119884119879

119894minus1)daggerx119894be nonzero Let (119860 minus 119883

119894minus1119884119879

119894minus1)dagger

ge 0for all 119894 = 1 2 119896 Then (119860 minus 119883

119896119884119879

119896)dagger

ge 0 if and only ify119879119894(119860 minus 119883

119894minus1119884119879

119894minus1)daggerx119894lt 1 where 119860 minus 119883

0119884119879

0is taken as 119860

Proof Set 119861119894= 119860 minus 119883

119894119884119879

119894for 119894 = 0 1 119896 where 119861

0is

identified as 119860 The conditions in the theorem can be writtenas

x119894isin 119877 (119861

119894minus1) y

119894isin 119877 (119861

119879

119894minus1)

1 minus y119879119894119861dagger

119894minus1x119894= 0

forall119894 = 1 2 119896

(14)

Also we have 119861dagger119894ge 0 for all 119894 = 0 1 119896 minus 1

Now assume that 119861dagger119896ge 0 Then by Theorem 16 and the

conditions in (14) for 119894 = 119896 we have y119896119861dagger

119896minus1x119896lt 1 Also

since by assumption 119861dagger119894ge 0 for all 119894 = 0 1 119896 minus 1 we have

y119879119894119861dagger

119894minus1x119894lt 1 for all 119894 = 1 2 119896

The converse part can be proved iteratively Then condi-tion y119879

11198610

daggerx1lt 1 and the conditions in (14) for 119894 = 1 imply

that 1198611

dagger

ge 0 Repeating the argument for 119894 = 2 to 119896 proves theresult

The following result is an extension of Lemma 23 in [1]which is in turn obtained as a corollaryThis corollary will beused in proving another characterization for (119860minus119883

119896119884119879

119896)dagger

ge 0

6 Journal of Applied Mathematics

Theorem 19 Let 119860 isin R119899times119899 bc isin R119899 be such that minusb ge 0minusc ge 0 and 120572(isin R) gt 0 Further suppose that b isin 119877(119860) andc isin 119877(119860119879) Let 119860 = ( 119860 b

c119879 120572 ) isin R(119899+1)times(119899+1) Then 119860dagger ge 0 if andonly if 119860dagger ge 0 and c119879119860daggerb lt 120572

Proof Let us first observe that 119860119860daggerb = b and c119879119860dagger119860 = c119879Set

119883 =(

119860dagger

+119860daggerbc119879119860dagger

120572 minus c119879119860daggerbminus119860daggerb

120572 minus c119879119860daggerbminus

c119879119860dagger

120572 minus c119879119860daggerb1

120572 minus c119879119860daggerb) (15)

It then follows that 119860119883 = ( 119860119860dagger 00119879 1 ) and 119883119860 = (119860dagger119860 0

0119879 1 ) Usingthese two equations it can easily be shown that119883 = 119860dagger

Suppose that 119860dagger ge 0 and 120573 = 120572 minus c119879119860daggerb gt 0 Then119860dagger

ge 0 Conversely suppose that 119860dagger ge 0 Then we musthave 120573 gt 0 Let e

1 e2 e

119899+1 denote the standard basis

of R119899+1 Then for 119894 = 1 2 119899 we have 0 le 119860dagger

(e119894) =

(119860daggere119894+(119860daggerbc119879119860daggere

119894120573) minus(c119879119860daggere

119894120573))119879 Since c le 0 it follows

that 119860daggere119894ge 0 for 119894 = 1 2 119899 Thus 119860dagger ge 0

Corollary 20 (see [1 Lemma 23]) Let 119860 isin R119899times119899 benonsingular Let b c isin R119899 be such that minusb ge 0 minusc ge 0

and 120572(isin R) gt 0 Let 119860 = (119860 bc119879 120572 ) Then 119860minus1 isin R(119899+1)times(119899+1)

is nonsingular with 119860minus1 ge 0 if and only if 119860minus1 ge 0 andc119879119860minus1b lt 120572

Now we obtain another necessary and sufficient condi-tion for (119860 minus 119883

119896119884119879

119896)dagger to be nonnegative

Theorem 21 Let 119860 isin R119898times119899 be such that 119860dagger ge 0 Let x119894 y119894

be nonnegative vectors inR119899 andR119898 respectively for every 119894 =1 119896 such that

x119894isin 119877 (119860 minus 119883

119894minus1119884119879

119894minus1) y

119894isin 119877(119860 minus 119883

119894minus1119884119879

119894minus1)119879

(16)

where 119883119894= (x1 x

119894) 119884119894= (y1 y

119894) and 119860 minus 119883

0119884119879

0is

taken as119860 If119867119894= 119868119894minus119884119879

119894119860dagger

119883119894is nonsingular and 1minus y119879

119894119860daggerx119894

is positive for all 119894 = 1 2 119896 then (119860 minus 119883119896119884119879

119896)dagger

ge 0 if andonly if

y119879119894119860daggerx119894lt 1 minus y119879

119894119860dagger

119883119894minus1119867minus1

119894minus1119884119879

119894minus1119860daggerx119894 forall119894 = 1 119896 (17)

Proof The range conditions in (16) imply that 119877(119883119896) sube 119877(119860)

and 119877(119884119896) sube 119877(119860

119879

) Also from the assumptions it followsthat 119867

119896is nonsingular By Theorem 10 (119860 minus 119883

119896119884119879

119896)dagger

ge 0 ifand only if119867minus1

119896ge 0 Now

119867119896= (

119867119896minus1

minus119884119879

119896minus1119860daggerx119896

minusy119879119896119860dagger

119883119896minus1

1 minus y119879119896119860daggerx119896

) (18)

Since 1minusy119879119896119860daggerx119896gt 0 usingCorollary 20 it follows that119867minus1

119896ge

0 if and only if y119879119896119860dagger

119883119896minus1119867minus1

119896minus1119884119879

119896minus1119860daggerx119896lt 1 minus y119879

119896119860daggerx119896and

119867minus1

119896minus1ge 0

Now applying the above argument to the matrix 119867119896minus1

we have that 119867minus1

119896minus1ge 0 holds if and only if 119867minus1

119896minus2ge 0

holds and y119879119896minus1119860dagger

119883119896minus2(119868119896minus2

minus 119884119879

119896minus2119860dagger

119883119896minus2)minus1

119884119879

119896minus2119860daggerx119896minus1 lt

1minus y119879119896minus1119860daggerx119896minus1 Continuing the above argument we get (119860minus

119883119896y119879119896)dagger

ge 0 if and only if y119879119894119860dagger

119883119894minus1119867minus1

119894minus1119884119879

119894minus1119860daggerx119894lt 1minusy119879

119894119860daggerx119894

forall119894 = 1 119896 This is condition (17)

We conclude the paper by considering an extension ofExample 17

Example 22 For a fixed 119899 let 119862 be the circulant matrixgenerated by the row vector (1 (12) (13) (1(119899 minus 1)))Consider

119860 = (119868

119890119879

1

) (119868 + e1119890119879

1)minus1

119862minus1

(119868 + e1119890119879

1)minus1

(119868 e1) (19)

where 119868 is the identity matrix of order 119899 minus 1 e1is (119899 minus 1) times 1

vector with 1 as the first entry and 0 elsewhere Then 119860 isin

R119899times119899 119860dagger is the 119899 times 119899 nonnegative Toeplitz matrix

119860dagger

=

((((((

(

11

119899 minus 1

1

119899 minus 2

1

21

1

21

1

119899 minus 1

1

3

1

2

1

119899 minus 1

1

119899 minus 2

1

119899 minus 3 1

1

119899 minus 1

11

119899 minus 1

1

119899 minus 2

1

21

))))))

)

(20)

Let 119909119894be the first row of 119861dagger

119894minus1written as a column vector

multiplied by 1119899119894 and let 119910119894be the first column of 119861dagger

119894minus1

multiplied by 1119899119894 where 119861119894= 119860 minus 119883

119894119884119879

119894 with 119861

0being

identified with 119860 119894 = 0 2 119896 We then have x119894ge 0

y119894ge 0 119909

1isin 119877(119860) and 119910

1isin 119877(119860

119879

) Now if 119861dagger119894minus1

ge 0 foreach iteration then the vectors 119909

119894and 119910

119894will be nonnegative

Hence it is enough to check that 1 minus 119910119879119894119861dagger

119894minus1119909119894gt 0 in order

to get 119861dagger119894ge 0 Experimentally it has been observed that for

119899 gt 3 the condition 1 minus 119910119879119894119861dagger

119894minus1119909119894gt 0 holds true for any

iteration However for 119899 = 3 it is observed that 119861dagger1ge 0 119861dagger

2ge

0 1 minus 1199101198791119860dagger

1199091gt 0 and 1 minus 119910dagger

2119861dagger

11199092gt 0 But 1 minus 119910119879

3119861dagger

21199093lt 0

and in that case we observe that 119861dagger3is not nonnegative

References

[1] J Ding W Pye and L Zhao ldquoSome results on structured119872-matrices with an application to wireless communicationsrdquoLinear Algebra and its Applications vol 416 no 2-3 pp 608ndash614 2006

[2] S K Mitra and P Bhimasankaram ldquoGeneralized inverses ofpartitionedmatrices and recalculation of least squares estimatesfor data ormodel changesrdquo Sankhya A vol 33 pp 395ndash410 1971

[3] C Y Deng ldquoA generalization of the Sherman-Morrison-Woodbury formulardquo Applied Mathematics Letters vol 24 no9 pp 1561ndash1564 2011

[4] A Berman M Neumann and R J SternNonnegative Matricesin Dynamical Systems JohnWiley amp Sons New York NY USA1989

Journal of Applied Mathematics 7

[5] A Berman and R J Plemmons Nonnegative Matrices inthe Mathematical Sciences Society for Industrial and AppliedMathematics (SIAM) 1994

[6] L Collatz Functional Analysis and Numerical MathematicsAcademic Press New York NY USA 1966

[7] C A Desoer and B H Whalen ldquoA note on pseudoinversesrdquoSociety for Industrial and Applied Mathematics vol 11 pp 442ndash447 1963

[8] A Ben-Israel and T N E Greville Generalized InversesTheoryand Applications Springer New York NY USA 2003

[9] A Berman and R J Plemmons ldquoCones and iterative methodsfor best least squares solutions of linear systemsrdquo SIAM Journalon Numerical Analysis vol 11 pp 145ndash154 1974

[10] D Mishra and K C Sivakumar ldquoOn splittings of matrices andnonnegative generalized inversesrdquo Operators and Matrices vol6 no 1 pp 85ndash95 2012

[11] D Mishra and K C Sivakumar ldquoNonnegative generalizedinverses and least elements of polyhedral setsrdquo Linear Algebraand its Applications vol 434 no 12 pp 2448ndash2455 2011

[12] M R Kannan and K C Sivakumar ldquoMoore-Penrose inversepositivity of interval matricesrdquo Linear Algebra and its Applica-tions vol 436 no 3 pp 571ndash578 2012

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 320392 7 pageshttpdxdoiorg1011552013320392

Research ArticleA New Construction of Multisender Authentication Codes fromPolynomials over Finite Fields

Xiuli Wang

College of Science Civil Aviation University of China Tianjin 300300 China

Correspondence should be addressed to Xiuli Wang xlwangcauceducn

Received 3 February 2013 Accepted 7 April 2013

Academic Editor Yang Zhang

Copyright copy 2013 Xiuli Wang This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Multisender authentication codes allow a group of senders to construct an authenticated message for a receiver such that thereceiver can verify the authenticity of the received message In this paper we construct one multisender authentication code frompolynomials over finite fields Some parameters and the probabilities of deceptions of this code are also computed

1 Introduction

Multisender authentication code was firstly constructed byGilbert et al [1] in 1974 Multisender authentication systemrefers towho a groupof senders cooperatively send amessageto a receiver then the receiver should be able to ascertainthat the message is authentic About this case many scholarsand researchers had made great contributions to multisenderauthentication codes such as [2ndash6]

In the actual computer network communications mul-tisender authentication codes include sequential model andsimultaneous model Sequential model is that each senderuses his own encoding rules to encode a source state orderlythe last sender sends the encoded message to the receiverand the receiver receives themessage and verifies whether themessage is legal or not Simultaneousmodel is that all sendersuse their own encoding rules to encode a source state andeach sender sends the encoded message to the synthesizerrespectively then the synthesizer forms an authenticatedmessage and verifies whether the message is legal or not Inthis paper we will adopt the second model

In a simultaneous model there are four participants agroup of senders 119880 = 119880

1 1198802 119880

119899 the key distribution

center he is responsible for the key distribution to sendersand receiver including solving the disputes between them areceiver 119877 and a synthesizer where he only runs the trustedsynthesis algorithm The code works as follows each senderand receiver has their own Cartesian authentication code

respectively Let (119878 119864119894 119879119894 119891119894) (119894 = 1 2 119899) be the sendersrsquo

Cartesian authentication code (119878 119864119877 119879 119892) be the receiverrsquos

Cartesian authentication code ℎ 1198791times 1198792times sdot sdot sdot times 119879

119899rarr

119879 be the synthesis algorithm and 120587119894

119864 rarr 119864119894be a

subkey generation algorithm where 119864 is the key set of thekey distribution center When authenticating a message thesenders and the receiver should comply with the protocolThe key distribution center randomly selects an encodingrule 119890 isin 119864 and sends 119890

119894= 120587119894(119890) to the 119894th sender 119880

119894(119894 =

1 2 119899) secretly then he calculates 119890119877by 119890 according to

an effective algorithm and secretly sends 119890119877to the receiver

119877 If the senders would like to send a source state 119904 to thereceiver 119877 119880

119894computes 119905

119894= 119891119894(119904 119890119894) (119894 = 1 2 119899) and

sends 119898119894= (119904 119905

119894) (119894 = 1 2 119899) to the synthesizer through

an open channel The synthesizer receives the message 119898119894=

(119904 119905119894) (119894 = 1 2 119899) and calculates 119905 = ℎ(119905

1 1199052 119905

119899) by the

synthesis algorithm ℎ and then sends message 119898 = (119904 119905) tothe receiver he checks the authenticity by verifying whether119905 = 119892(119904 119890

119877) or not If the equality holds the message is

authentic and is accepted Otherwise the message is rejectedWe assume that the key distribution center is credible and

though he know the sendersrsquo and receiverrsquos encoding rules hewill not participate in any communication activities Whentransmitters and receiver are disputing the key distributioncenter settles it At the same time we assume that the systemfollows the Kerckhoff principle in which except the actualused keys the other information of the whole system ispublic

2 Journal of Applied Mathematics

In a multisender authentication system we assume thatthe whole senders are cooperative to form a valid messagethat is all senders as a whole and receiver are reliable Butthere are some malicious senders who together cheat thereceiver the part of senders and receiver are not credible andthey can take impersonation attack and substitution attackIn the whole system we assume that 119880

1 1198802 119880

119899 are

senders 119877 is a receiver 119864119894is the encoding rules set of the

sender 119880119894 and 119864

119877is the decoding rules set of the receiver

119877 If the source state space 119878 and the key space 119864119877of receiver

119877 are according to a uniform distribution then the messagespace119872 and the tag space119879 are determined by the probabilitydistribution of 119878 and 119864

119877 119871 = 119894

1 1198942 119894

119897 sub 1 2 119899

119897 lt 119899 119880119871

= 1198801198941 1198801198942 119880

119894119897 119864119871

= 1198641198801198941

1198641198801198942

119864119880119894119897

Now consider that let us consider the attacks from maliciousgroups of senders Here there are two kinds of attack

The opponentrsquos impersonation attack to receiver119880119871 after

receiving their secret keys encode a message and send it tothe receiver 119880

119871are successful if the receiver accepts it as

legitimate message Denote by 119875119868the largest probability of

some opponentrsquos successful impersonation attack to receiverit can be expressed as

119875119868= max119898isin119872

1003816100381610038161003816119890119877 isin 119864119877| 119890119877sub 119898

10038161003816100381610038161003816100381610038161003816119864119877

1003816100381610038161003816

(1)

The opponentrsquos substitution attack to the receiver 119880119871

replace 119898 with another message 1198981015840 after they observe a

legitimatemessage119898119880119871are successful if the receiver accepts

it as legitimate message it can be expressed as

119875119878= max119898isin119872

max1198981015840=119898isin119872

10038161003816100381610038161003816119890119877isin 119864119877| 119890119877sub 119898119898

1015840

10038161003816100381610038161003816

1003816100381610038161003816119890119877 isin 119864119877| 119890119877sub 119898

1003816100381610038161003816

(2)

There might be 119897 malicious senders who together cheatthe receiver that is the part of senders and the receiverare not credible and they can take impersonation attackLet 119871 = 119894

1 1198942 119894

119897 sub 1 2 119899 119897 lt 119899 and 119864

119871=

1198641198801198941

1198641198801198942

119864119880119894119897

Assume that 119880119871

= 1198801198941 1198801198942 119880

119894119897

after receiving their secret keys send a message 119898 to thereceiver 119877119880

119871are successful if the receiver accepts it as legiti-

mate message Denote by 119875119880(119871) the maximum probability of

success of the impersonation attack to the receiver It can beexpressed as

119875119880(119871)

=max119890119871isin119864119871

max119890119871isin119890119880

max119898isin119872

1003816100381610038161003816119890119877isin119864119877 | 119890119877sub119898 119901 (119890119877 119890119875) =0

10038161003816100381610038161003816100381610038161003816119890119877 isin 119864

119877| 119901 (119890119877 119890119875) =0

1003816100381610038161003816

(3)

Notes119901(119890119877 119890119875) = 0 implies that any information 119904 encoded by

119890119879can be authenticated by 119890

119877

In [2] Desmedt et al gave two constructions for MRA-codes based on polynomials and finite geometries respec-tively To construct multisender or multireceiver authenti-cation by polynomials over finite fields many researchershave done much work for example [7ndash9] There are other

constructions of multisender authentication codes that aregiven in [3ndash6] The construction of authentication codesis combinational design in its nature We know that thepolynomial over finite fields can provide a better algebrastructure and is easy to count In this paper we constructone multisender authentication code from the polynomialover finite fields Some parameters and the probabilities ofdeceptions of this code are also computed We realize thegeneralization and the application of the similar idea andmethod of the paper [7ndash9]

2 Some Results about Finite Field

Let 119865119902be the finite field with 119902 elements where 119902 is a power

of a prime 119901 and 119865 is a field containing 119865119902 denote by 119865

lowast

119902be

the nonzero elements set of 119865119902 In this paper we will use the

following conclusions over finite fields

Conclusion 1 A generator 120572 of 119865lowast

119902is called a primitive

element of 119865119902

Conclusion 2 Let 120572 isin 119865119902 if some polynomials contain 120572 as

their root and their leading coefficient are 1 over 119865119902 then the

polynomial having least degree among all such polynomialsis called a minimal polynomial over 119865

119902

Conclusion 3 Let |119865| = 119902119899 then 119865 is an 119899-dimensional

vector space over 119865119902 Let 120572 be a primitive element of 119865

119902

and 119892(119909) the minimal polynomial about 120572 over 119865119902 then

dim119892(119909) = 119899 and 1 120572 1205722

120572119899minus1 is a basis of 119865 Further-

more 1 120572 1205722 120572119899minus1 is linear independent and it is equalto 120572 120572

2

120572119899minus1

120572119899 (120572 is a primitive element 120572 = 0) is also

linear independent moreover 120572119901 1205721199012

120572119901119899minus1

120572119901119899

is alsolinear independent

Conclusion 4 Consider (1199091+ 1199092+ sdot sdot sdot + 119909

119899)119898

= (1199091)119898

+

(1199092)119898

+ sdot sdot sdot + (119909119899)119898 where 119909

119894isin 119865119902 (1 le 119894 le 119899) and 119898 is a

nonnegative power of character 119901 of 119865119902

Conclusion 5 Let119898 le 119899 Then the number of119898times119899matricesof rank119898 over 119865

119902is 119902119898(119898minus1)2prod119899

119894=119899minus119898+1(119902119894

minus 1)

More results about finite fields can be found in [10ndash12]

3 Construction

Let the polynomial 119901119895(119909) = 119886

1198951119909119901119899

+ 1198861198952119909119901(119899minus1)

+ sdot sdot sdot +

119886119895119899119909119901

(1 le 119895 le 119896) where the coefficient 119886119894119897

isin 119865119902

(1 le 119897 le 119899) and these vectors by the composition of theircoefficient are linearly independent The set of source states119878 = 119865

119902 the set of 119894th transmitterrsquos encoding rules 119864

119880119894=

1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) 119909119894

isin 119865lowast

119902 (1 le 119894 le 119899) the set

of receiverrsquos encoding rules 119864119877

= 1199011(120572) 1199012(120572) 119901

119896(120572)

where 120572 is a primitive element of 119865119902 the set of 119894th transmit-

terrsquos tags 119879119894= 119905119894| 119905119894isin 119865119902 (1 le 119894 le 119899) the set of receiverrsquos

tags 119879 = 119905 | 119905 isin 119865119902

Journal of Applied Mathematics 3

Define the encoding map 119891119894 119878 times 119864

119880119894rarr 119879119894 119891119894(119904 119890119880119894) =

1199041199011(119909119894) + 1199042

1199012(119909119894) + sdot sdot sdot + 119904

119896

119901119896(119909119894) 1 le 119894 le 119899

The decoding map 119891 119878 times 119864119877

rarr 119879 119891(119904 119890119877) = 119904119901

1(120572) +

1199042

1199012(120572) + sdot sdot sdot + 119904

119896

119901119896(120572)

The synthesizing map ℎ 1198791times 1198792times sdot sdot sdot times 119879

119899rarr 119879

ℎ(1199051 1199052 119905

119899) = 1199051+ 1199052+ sdot sdot sdot + 119905

119899

The code works as followsAssume that 119902 is larger than or equal to the number of

the possible message and 119899 le 119902

31 Key Distribution The key distribution center randomlygenerates 119896 (119896 le 119899) polynomials 119901

1(119909) 1199012(119909) 119901

119896(119909)

where 119901119895(119909) = 119886

1198951119909119901119899

+ 1198861198952119909119901(119899minus1)

+ sdot sdot sdot + 119886119895119899119909119901

(1 le 119895 le 119896)and make these vectors by composed of their coefficient islinearly independent it is equivalent to the column vectors

of the matrix (

11988611 11988621 sdotsdotsdot 1198861198961

11988612 11988622 sdotsdotsdot 1198861198962

1198861119899 1198862119899 sdotsdotsdot 119886119896119899

) is linearly independent He

selects 119899 distinct nonzero elements 1199091 1199092 119909

119899isin 119865119902again

and makes 119909119894(1 le 119894 le 119899) secret then he sends privately

1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) to the sender 119880

119894(1 le 119894 le 119899) The

key distribution center also randomly chooses a primitiveelement 120572 of 119865

119902satisfying 119909

1+ 1199092+ sdot sdot sdot + 119909

119899= 120572 and sends

1199011(120572) 1199012(120572) 119901

119896(120572) to the receiver 119877

32 Broadcast If the senderswant to send a source state 119904 isin 119878

to the receiver 119877 the sender 119880119894calculates 119905

119894= 119891119894(119904 119890119880119894) =

119860119904(119909119894) = 119904119901

1(119909119894)+1199042

1199012(119909119894)+ sdot sdot sdot+119904

119896

119901119896(119909119894) 1 le 119894 le 119899 and then

sends 119860119904(119909119894) = 119905119894to the synthesizer

33 Synthesis After the synthesizer receives 1199051 1199052 119905

119899 he

calculates ℎ(1199051 1199052 119905

119899) = 1199051+ 1199052+ sdot sdot sdot + 119905

119899and then sends

119898 = (119904 119905) to the receiver 119877

34 Verification When the receiver 119877 receives 119898 = (119904 119905) hecalculates 1199051015840 = 119892(119904 119890

119877) = 119860

119904(120572) = 119904119901

1(120572) + 119904

2

1199012(120572) + sdot sdot sdot +

119904119896

119901119896(120572) If 119905 = 119905

1015840 he accepts 119905 otherwise he rejects itNext we will show that the above construction is a well

defined multisender authentication code with arbitration

Lemma 1 Let 119862119894= (119878 119864

119875119894 119879119894 119891119894) then the code is an A-code

1 le 119894 le 119899

Proof (1) For any 119890119880119894

isin 119864119880119894 119904 isin 119878 because 119864

119880119894= 1199011(119909119894)

1199012(119909119894) 119901

119896(119909119894) 119909119894isin 119865lowast

119902 so 119905

119894= 1199041199011(119909119894) + 1199042

1199012(119909119894) + sdot sdot sdot +

119904119896

119901119896(119909119894) isin 119879119894= 119865119902 Conversely for any 119905

119894isin 119879119894 choose 119890

119880119894=

1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) 119909119894isin 119865lowast

119902 where 119901

119895(119909) = 119886

1198951119909119901119899

+

1198861198952119909119901(119899minus1)

+ sdot sdot sdot + 119886119895119899119909119901

(1 le 119895 le 119896) and let 119905119894= 119891119894(119904 119890119880119894) =

1199041199011(119909119894) + 1199042

1199012(119909119894) + sdot sdot sdot + 119904

119896

119901119896(119909119894) it is equivalent to

(119909119901119899

119894 119909119901119899minus1

119894 119909

119901

119894)(

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)(

119904

1199042

119904119896

) = 119905119894

(4)

It follows that

(

(

119909119901119899

1119909119901119899minus1

1sdot sdot sdot 119909119901

1

119909119901n

2119909119901119899minus1

2sdot sdot sdot 119909119901

2

119909119901119899

119899119909119901119899minus1

119899sdot sdot sdot 119909119901

119899

)

)

times (

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)(

119904

1199042

119904119896

) = (

1199051

1199052

119905119899

)

(5)

Denote

119860 = (

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)

119883 = (

119909119901119899

1119909119901119899minus1

1sdot sdot sdot 119909119901

1

119909119901119899

2119909119901119899minus1

2sdot sdot sdot 119909119901

2

119909119901119899

119899119909119901119899minus1

119899sdot sdot sdot 119909119901

119899

)

119878 = (

119904

1199042

119904119896

) 119905 = (

1199051

1199052

119905119899

)

(6)

The above linear equation is equivalent to 119883119860119878 = 119905because the column vectors of 119860 are linearly independent119883 is equivalent to a Vandermonde matrix and 119883 is inversetherefore the above linear equation has a unique solution so119904 is only defined that is 119891

119894(1 le 119894 le 119899) is a surjection

(2) If 1199041015840 isin 119878 is another source state satisfying 1199041199011(119909119894) +

1199042

1199012(119909119894)+sdot sdot sdot+119904

119896

119901119896(119909119894) = 1199041015840

1199011(119909119894)+11990410158402

1199012(119909119894)+sdot sdot sdot+119904

1015840119896

119901119896(119909119894) =

119905119894 and it is equivalent to (119904 minus 119904

1015840

)1199011(119909119894) + (1199042

minus 11990410158402

)1199012(119909119894) + sdot sdot sdot +

(119904119896

minus 1199041015840119896

)119901119896(119909119894) = 0 then

(119909119901119899

119894 119909119901119899minus1

119894 119909

119901

119894)

times (

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)(

119904 minus 1199041015840

1199042

minus 11990410158402

119904119896

minus 1199041015840119896

) = 0

(7)

4 Journal of Applied Mathematics

Thus

(

(

119909119901119899

1119909119901119899minus1

1sdot sdot sdot 119909119901

1

119909119901n

2119909119901119899minus1

2sdot sdot sdot 119909119901

2

119909119901119899

119899119909119901119899minus1

119899sdot sdot sdot 119909119901

119899

)

)

times (

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)(

119904 minus 1199041015840

1199042

minus 11990410158402

119904119896

minus 1199041015840119896

) = (

0

0

0

)

(8)

Similar to (1) we know that the homogeneous linear equation119883119860119878 = 0 has a unique solution that is there is only zerosolution so 119904 = 119904

1015840 So 119904 is the unique source state determinedby 119890119880119894and 119905119894 thus 119862

119894(1 le 119894 le 119899) is an A-code

Lemma 2 Let 119862 = (119878 119864119877 119879 119892) then the code is an A-code

Proof (1) For any 119904 isin 119878 119890119877isin 119864119877 from the definition of 119890

119877

we assume that 119864119877

= 1199011(120572) 1199012(120572) 119901

119896(120572) where 120572 is a

primitive element of 119865119902 119892(119904 119890

119877) = 119904119901

1(120572) + 119904

2

1199012(120572) + sdot sdot sdot +

119904119896

119901119896(120572) isin 119879 = 119865

119902 on the other hand for any 119905 isin 119879 choose

119890119877= 1199011(120572) 1199012(120572) 119901

119896(120572) where 120572 is a primitive element

of 119865119902 119892(119904 119890

119877) = 119904119901

1(120572) + 119904

2

1199012(120572) + sdot sdot sdot + 119904

119896

119901119896(120572) = 119905 it is

equivalent to

(120572119901119899

120572119901119899minus1

sdot sdot sdot 120572119901

)

times (

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)(

119904

1199042

119904119896

) = 119905

119860 = (

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)

(9)

that is (120572119901119899

120572119901119899minus1

120572119901

)119860(

119904

1199042

119904119896

) = 119905 From Conclusion

3 we know that (120572119901119899

120572119901119899minus1

120572119901

) is linearly independentand the column vectors of 119860 are also linearly independenttherefore the above linear equation has unique solution so 119904

is only defined that is 119892 is a surjection

(2) If 1199041015840 is another source state satisfying 119905 = 119892(1199041015840

119890119877) then

(120572119901119899

120572119901119899minus1

120572119901

)119860(

1199041015840

11990410158402

1199041015840119896

)

= (120572119901119899

120572119901119899minus1

120572119901

)119860(

119904

1199042

119904119896

)

(10)

that is (120572119901119899

120572119901119899minus1

120572119901

)119860([[

[

1199041015840

11990410158402

1199041015840119896

]]

]

minus [

[

119904

1199042

119904119896

]

]

) = 0 Sim-

ilar to (1) we get that the homogeneous linear equation(120572119901119899

120572119901119899minus1

120572119901

)119860(1198781015840

minus 119878) = 0 has a unique solution thatis there is only zero solution so 119878 = 119878

1015840 that is 119904 = 1199041015840 So

119904 is the unique source state determined by 119890119877and 119905 thus

119862 = (119878 119864119877 119879 119892) is an A-code

At the same time for any valid119898 = (119904 119905) we have knownthat 120572 = 119909

1+ 1199092+ sdot sdot sdot + 119909

119899 and it follows that 1199051015840 = 119904119901

1(120572) +

1199042

1199012(120572)+sdot sdot sdot+119904

119896

119901119896(120572) = 119904119901

1(1199091+1199092+sdot sdot sdot+119909

119899)+1199042

1199012(1199091+1199092+

sdot sdot sdot + 119909119899) + sdot sdot sdot + 119904

119896

119901119896(1199091+ 1199092+ sdot sdot sdot + 119909

119899) We also have known

that 119901119895(119909) = 119886

1198951119909119901119899

+1198861198952119909119901(119899minus1)

+ sdot sdot sdot + 119886119895119899119909119901

(1 le 119895 le 119896) fromConclusion 4 (119909

1+ 1199092+ sdot sdot sdot + 119909

119899)119901119898

= (1199091)119901119898

+ (1199092)119901119898

+ sdot sdot sdot +

(119909119899)119901119898

where119898 is a nonnegative power of character 119901 of 119865119902

andwe get119901119895(1199091+1199092+sdot sdot sdot+119909

119899) = 119901119895(1199091)+119901119895(1199092)+sdot sdot sdot+119901

119895(119909119899)

therefore 1199051015840 = 1199041199011(120572) + 119904

2

1199012(120572) + sdot sdot sdot + 119904

119896

119901119896(120572) = (119904119901

1(1199091) +

1199042

1199012(1199091)+sdot sdot sdot+119904

119896

119901119896(1199091))+(119904119901

1(1199092)+1199042

1199012(1199092)+sdot sdot sdot+119904

119896

119901119896(1199092))+

sdot sdot sdot+(1199041199011(119909119899)+1199042

1199012(119909119899)+ sdot sdot sdot+119904

119896

119901119896(119909119899)) = 1199051+1199052+sdot sdot sdot+119905

119899= 119905

and the receiver 119877 accepts119898

From Lemmas 1 and 2 we know that such construction ofmultisender authentication codes is reasonable and there are119899 senders in this system Next we compute the parametersof this code and the maximum probability of success inimpersonation attack and substitution attack by the group ofsenders

Theorem 3 Some parameters of this construction are|119878| = 119902 |119864

119880119894| = [119902

119896(119896minus1)2

prod119899

119894=119899minus119896+1(119902119894

minus 1)] (119902minus1

1) =

[119902119896(119896minus1)2

prod119899

119894=119899minus119896+1(119902119894

minus 1)](119902 minus 1) (1 le 119894 le 119899) |119879119894| = 119902 (1 le

119894 le 119899) |119864119877| = [119902

119896(119896minus1)2

prod119899

119894=119899minus119896+1(119902119894

minus 1)]120593(119902 minus 1) |119879| = 119902Where 120593(119902 minus 1) is the 119864119906119897119890119903 119891119906119899119888119905119894119900119899 of 119902 minus 1 it representsthe number of primitive element of 119865

119902here

Proof For |119878| = 119902 |119879119894| = 119902 and |119879| = 119902 the results

are straightforward For119864119880119894 because119864

119880119894=1199011(119909119894) 1199012(119909119894)

119901119896(119909119894) 119909119894isin 119865lowast

119902 where 119901

119895(119909) = 119886

1198951119909119901119899

+ 1198861198952119909119901(119899minus1)

+ sdot sdot sdot +

119886119895119899119909119901

(1 le 119895 le 119896) and these vectors by the composition oftheir coefficient are linearly independent it is equivalent to

the columns of 119860 = (

11988611 11988621 sdotsdotsdot 1198861198961

11988612 11988622 sdotsdotsdot 1198861198962

1198861119899 1198862119899 sdotsdotsdot 119886119896119899

) is linear independent

Journal of Applied Mathematics 5

From Conclusion 5 we can conclude that the number of 119860satisfying the condition is 119902

119896(119896minus1)2

prod119899

119894=119899minus119896+1(119902119894

minus 1) On theother hand the number of distinct nonzero elements 119909

119894(1 le

119894 le 119899) in 119865119902is ( 119902minus11

) = 119902 minus 1 so |119864119880119894| = [119902

119896(119896minus1)2

prod119899

119894=119899minus119896+1(119902119894

minus

1)](119902 minus 1) For 119864119877 119864119877

= 1199011(120572) 1199012(120572) 119901

119896(120572) where 120572

is a primitive element of 119865119902 For 120572 from Conclusion 1 a

generator of119865lowast119902is called a primitive element of119865

119902 |119865lowast119902| = 119902minus1

by the theory of the group we know that the number ofgenerator of 119865lowast

119902is 120593(119902minus1) that is the number of 120572 is 120593(119902minus1)

For 1199011(119909) 1199012(119909) 119901

119896(119909) From above we have confirmed

that the number of these polynomials is 119902119896(119896minus1)2prod119899119894=119899minus119896+1

(119902119894

minus

1) therefore |119864119877| = [119902

119896(119896minus1)2

prod119899

119894=119899minus119896+1(119902119894

minus 1)]120593(119902 minus 1)

Lemma 4 For any 119898 isin 119872 the number of 119890119877contained 119898 is

120593(119902 minus 1)

Proof Let 119898 = (119904 119905) isin 119872 119890119877

= 1199011(120572) 1199012(120572) 119901

119896(120572)

where 120572 is a primitive element of 119865119902 isin 119864

119877 If 119890119877

sub 119898then 119904119901

1(120572) + 119904

2

1199012(120572) + sdot sdot sdot + 119904

119896

119901119896(120572) = 119905 hArr (120572

119901119899

120572119901119899minus1

120572119901

)119860(

119904

1199042

119904119896

) = 119905 For any 120572 suppose that there is

another 1198601015840 such that (120572

119901119899

120572119901119899minus1

120572119901

)1198601015840

(

119904

1199042

119904119896

) = 119905

then (120572119901119899

120572119901119899minus1

120572119901

)(119860 minus 1198601015840

)(

119904

1199042

119904119896

) = 0 because

120572119901119899

120572119901119899minus1

120572119901 is linearly independent so (119860minus119860

1015840

)(

119904

1199042

119904119896

) =

0 but (

119904

1199042

119904119896

) is arbitrarily therefore 119860 minus 1198601015840

= 0 that is

119860 = 1198601015840 and it follows that 119860 is only determined by 120572

Therefore as 120572 isin 119864119877 for any given 119904 and 119905 the number of

119890119877contained in119898 is 120593(119902 minus 1)

Lemma 5 For any119898 = (119904 119905) isin 119872 and1198981015840

= (1199041015840

1199051015840

) isin 119872 with119904 = 1199041015840 the number of 119890

119877contained119898 and119898

1015840 is 1

Proof Assume that 119890119877

= 1199011(120572) 1199012(120572) 119901

119896(120572) where

120572 is a primitive element of 119865119902 isin 119864

119877 If 119890119877

sub 119898 and119890119877

sub 1198981015840 then 119904119901

1(120572) + 119904

2

1199012(120572) + sdot sdot sdot + 119904

119896

119901119896(120572) = 119905 hArr

(120572119901119899

120572119901119899minus1

120572119901

)119860(

119904

1199042

119904119896

) = 119905 11990410158401199011(120572) + 119904

10158402

1199012(120572) + sdot sdot sdot +

1199041015840119896

119901119896(120572) = 119905 hArr (120572

119901119899

120572119901119899minus1

120572119901

)119860(

1199041015840

11990410158402

1199041015840119896

) = 119905 It is

equivalent to (120572119901119899

120572119901119899minus1

120572119901

)119860(

119904minus1199041015840

1199042minus11990410158402

119904119896minus1199041015840119896

) = 119905 minus 1199051015840 because

119904 = 1199041015840 so 119905 = 119905

1015840 otherwise we assume that 119905 = 1199051015840 and

since 120572119901119899

120572119901119899minus1

120572119901 and the column vectors of 119860 both

are linearly independent it forces that 119904 = 1199041015840 this is a

contradiction Therefore we get

(119905 minus 1199051015840

)minus1

[[[[

[

(120572119901119899

120572119901119899minus1

120572119901

)119860 (

119904 minus 1199041015840

1199042

minus 11990410158402

119904119896

minus 1199041015840119896

)

]]]]

]

= 1

(lowast)

since 119905 1199051015840 is given (119905 minus 119905

1015840

)minus1 is unique by equation (lowast) for

any given 119904 1199041015840 and 119905 119905

1015840 we obtain that (120572119901119899

120572119901119899minus1

120572119901

)119860 isonly determined thus the number of 119890

119877contained119898 and119898

1015840

is 1

Lemma6 For any fixed 119890119880= 1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) 119909119894isin

119865lowast

119902 (1 le 119894 le 119899) containing a given 119890

119871 then the number of 119890

119877

which is incidence with 119890119880is 120593(119902 minus 1)

Proof For any fixed 119890119880

= 1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) 119909119894isin

119865lowast

119902 (1 le 119894 le 119899) containing a given 119890

119871 we assume that

119901119895(119909119894) = 1198861198951119909119901119899

119894+1198861198952119909119901(119899minus1)

119894+sdot sdot sdot+119886

119895119899119909119901

119894(1 le 119895 le 119896 1 le 119894 le 119899)

119890119877= 1199011(120572) 1199012(120572) 119901

119896(120572) where 120572 is a primitive element

of 119865119902 From the definitions of 119890

119877and 119890119880and Conclusion 4

we can conclude that 119890119877is incidence with 119890

119880if and only if

1199091+ 1199092+ sdot sdot sdot + 119909

119899= 120572 For any 120572 since Rank(1 1 1) =

Rank(1 1 1 120572) = 1 lt 119899 so the equation1199091+1199092+sdot sdot sdot+119909

119899=

120572 always has a solution From the proof of Theorem 3 weknow the number of 119890

119877which is incident with 119890

119880(ie the

number of all 119864119877) is [119902119896(119896minus1)2prod119899

119894=119899minus119896+1(119902119894

minus 1)] 120593(119902 minus 1)

Lemma 7 For any fixed 119890119880= 1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) 119909119894isin

119865lowast

119902 (1 le 119894 le 119899) containing a given 119890

119871and119898 = (119904 119905) the num-

ber of 119890119877which is incidence with 119890

119880and contained in119898 is 1

Proof For any 119904 isin 119878 119890119877

isin 119864119877 we assume that 119890

119877=

1199011(120572) 1199012(120572) 119901

119896(120572) where 120572 is a primitive element

of 119865119902 Similar to Lemma 6 for any fixed 119890

119880=

1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) 119909119894isin 119865lowast

119902 (1 le 119894 le 119899) containing

a given 119890119871 we have known that 119890

119877is incident with 119890

119880if and

only if

1199091+ 1199092+ sdot sdot sdot + 119909

119899= 120572 (11)

Again with 119890119877sub 119898 we can get

1199041199011(120572) + 119904

2

1199012(120572) + sdot sdot sdot + 119904

119896

119901119896(120572) = 119905 (12)

By (11) and (12) and the property of 119901119895(119909) (1 le 119895 le 119896) we

have the following conclusion

1199041199011(

119899

sum

119894=1

119909119894) + 1199042

1199012(

119899

sum

119894=1

119909119894) + sdot sdot sdot + 119904

119896

119901119896(

119899

sum

119894=1

119909119894)

= 119905 lArrrArr (1199011(

119899

sum

119894=1

119909119894) 1199012(

119899

sum

119894=1

119909119894) 119901

119896(

119899

sum

119894=1

119909119894))

6 Journal of Applied Mathematics

times (

119904

1199042

119904119896

)

= 119905 lArrrArr (

119899

sum

119894=1

1199011(119909119894)

119899

sum

119894=1

1199012(119909119894)

119899

sum

119894=1

119901119896(119909119894))(

119904

s2119904119896

)

=119905lArrrArr( (

119899

sum

119894=1

119909119894)

119899

(

119899

sum

119894=1

119909119894)

119899minus1

(

119899

sum

119894=1

119909119894))119860(

119904

1199042

119904119896

)

= 119905 lArrrArr [(

119899

sum

119894=1

1199011(119909119894)

119899

sum

119894=1

1199012(119909119894)

119899

sum

119894=1

119901119896(119909119894))

minus((

119899

sum

119894=1

119909119894)

119899

(

119899

sum

119894=1

119909119894)

119899minus1

(

119899

sum

119894=1

119909119894))119860]

times (

119904

1199042

119904119896

) = 0

(13)

because 119904 is any given Similar to the proof of Lemma 4we can get (sum

119899

119894=11199011(119909119894) sum119899

119894=11199012(119909119894) sum

119899

119894=1119901119896(119909119894)) minus

((sum119899

119894=1119909119894)119899(sum119899119894=1

119909119894)119899minus1

(sum119899

119894=1119909119894))119860=0 that is ((sum119899

119894=1119909119894)119899

(sum119899

119894=1119909119894)119899minus1

(sum119899

119894=1119909119894))119860 = (sum

119899

119894=11199011(119909119894) sum119899

119894=11199012(119909119894)

sum119899

119894=1119901119896(119909119894)) but 119901

1(119909119894) 1199012(119909119894) 119901

119896(119909119894) and 119909

119894(1 le 119894 le 119899)

also are fixed thus 120572 and 119860are only determined so thenumber of 119890

119877which is incident with 119890

119880and contained in 119898

is 1

Theorem 8 In the constructed multisender authenticationcodes if the sendersrsquo encoding rules and the receiverrsquos decodingrules are chosen according to a uniform probability distribu-tion then the largest probabilities of success for different typesof deceptions respectively are

119875119868=

1

119902119896(119896minus1)2prod119899

119894=119899minus119896+1(119902119894 minus 1)

119875119878=

1

120593 (119902 minus 1)

119875119880(119871) =

1

[119902119896(119896minus1)2prod119899

119894=119899minus119896+1(119902119894 minus 1)] 120593 (119902 minus 1)

(14)

Proof By Theorem 3 and Lemma 4 we get

119875119868= max119898isin119872

1003816100381610038161003816119890119877 isin 119864119877| 119890119877sub 119898

10038161003816100381610038161003816100381610038161003816119864119877

1003816100381610038161003816

=120593 (119902 minus 1)

[119902119896(119896minus1)2prod119899

119894=119899minus119896+1(119902119894 minus 1)] 120593 (119902 minus 1)

=1

119902119896(119896minus1)2prod119899

119894=119899minus119896+1(119902119894 minus 1)

(15)

By Lemmas 4 and 5 we get

119875119878= max119898isin119872

max1198981015840=119898isin119872

10038161003816100381610038161003816119890119877isin 119864119877| 119890119877sub 119898119898

1015840

10038161003816100381610038161003816

1003816100381610038161003816119890119877 isin 119864119877| 119890119877sub 119898

1003816100381610038161003816

=1

120593 (119902 minus 1)

(16)

By Lemmas 6 and 7 we get

119875119880(119871)

=max119890119871isin119864119871

max119890119871isin119890119880

max119898isin119872

1003816100381610038161003816119890119877 isin 119864119877| 119890119877sub119898 119901 (119890

119877 119890119875) =0

10038161003816100381610038161003816100381610038161003816119890119877 isin 119864

119877| 119901 (119890119877 119890119875) =0

1003816100381610038161003816

=1

[119902119896(119896minus1)2prod119899

119894=119899minus119896+1(119902119894 minus 1)] 120593 (119902 minus 1)

(17)

Acknowledgments

This paper is supported by the NSF of China (61179026)and the Fundamental Research of the Central Universi-ties of China Civil Aviation University of Science special(ZXH2012k003)

References

[1] E N Gilbert F J MacWilliams and N J A Sloane ldquoCodeswhich detect deceptionrdquoThe Bell System Technical Journal vol53 pp 405ndash424 1974

[2] Y Desmedt Y Frankel and M Yung ldquoMulti-receivermulti-sender network security efficient authenticated multicastfeedbackrdquo in Proceedings of the the 11th Annual Conference of theIEEEComputer andCommunications Societies (Infocom rsquo92) pp2045ndash2054 May 1992

[3] K Martin and R Safavi-Naini ldquoMultisender authenticationschemes with unconditional securityrdquo in Information and Com-munications Security vol 1334 of Lecture Notes in ComputerScience pp 130ndash143 Springer Berlin Germany 1997

[4] W Ma and X Wang ldquoSeveral new constructions of multitrasmitters authentication codesrdquo Acta Electronica Sinica vol28 no 4 pp 117ndash119 2000

[5] G J Simmons ldquoMessage authentication with arbitration oftransmitterreceiver disputesrdquo in Advances in CryptologymdashEUROCRYPT rsquo87 Workshop on theTheory and Application of ofCryptographic Techniques vol 304 of Lecture Notes in ComputerScience pp 151ndash165 Springer 1988

[6] S Cheng and L Chang ldquoTwo constructions of multi-senderauthentication codes with arbitration based linear codes to bepublished in rdquoWSEAS Transactions on Mathematics vol 11 no12 2012

Journal of Applied Mathematics 7

[7] R Safavi-Naini and H Wang ldquoNew results on multi-receiver authentication codesrdquo in Advances in CryptologymdashEUROCRYPT rsquo98 (Espoo) vol 1403 of Lecture Notes in ComputSci pp 527ndash541 Springer Berlin Germany 1998

[8] R Aparna and B B Amberker ldquoMulti-sender multi-receiverauthentication for dynamic secure group communicationrdquoInternational Journal of Computer Science andNetwork Securityvol 7 no 10 pp 47ndash63 2007

[9] R Safavi-Naini and H Wang ldquoBounds and constructions formultireceiver authentication codesrdquo inAdvances in cryptologymdashASIACRYPTrsquo98 (Beijing) vol 1514 of Lecture Notes in ComputSci pp 242ndash256 Springer Berlin Germany 1998

[10] S Shen and L Chen Information and Coding Theory Sciencepress in China 2002

[11] J J Rotman Advanced Modern Algebra High Education Pressin China 2004

[12] Z Wan Geometry of Classical Group over Finite Field SciencePress in Beijing New York NY USA 2002

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 435730 8 pageshttpdxdoiorg1011552013435730

Research ArticleGeometrical and Spectral Properties of the OrthogonalProjections of the Identity

Luis Gonzaacutelez Antonio Suaacuterez and Dolores Garciacutea

Department of Mathematics Research Institute SIANI University of Las Palmas de Gran Canaria Campus de Tafira35017 Las Palmas de Gran Canaria Spain

Correspondence should be addressed to Luis Gonzalez luisglezdmaulpgces

Received 28 December 2012 Accepted 10 April 2013

Academic Editor K Sivakumar

Copyright copy 2013 Luis Gonzalez et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We analyze the best approximation119860119873 (in the Frobenius sense) to the identitymatrix in an arbitrarymatrix subspace119860119878 (119860 isin R119899times119899

nonsingular 119878 being any fixed subspace of R119899times119899) Some new geometrical and spectral properties of the orthogonal projection 119860119873are derived In particular new inequalities for the trace and for the eigenvalues of matrix119860119873 are presented for the special case that119860119873 is symmetric and positive definite

1 Introduction

The set of all 119899 times 119899 real matrices is denoted by R119899times119899 and119868 denotes the identity matrix of order 119899 In the following119860119879 and tr(119860) denote as usual the transpose and the trace

of matrix 119860 isin R119899times119899 The notations ⟨sdot sdot⟩119865and sdot

119865stand

for the Frobenius inner product and matrix norm definedon the matrix space R119899times119899 Throughout this paper the termsorthogonality angle and cosine will be used in the sense ofthe Frobenius inner product

Our starting point is the linear system

119860119909 = 119887 119860 isin R119899times119899

119909 119887 isin R119899

(1)

where 119860 is a large nonsingular and sparse matrix Theresolution of this system is usually performed by iterativemethods based on Krylov subspaces (see eg [1 2]) Thecoefficient matrix 119860 of the system (1) is often extremelyill-conditioned and highly indefinite so that in this caseKrylov subspace methods are not competitive without agood preconditioner (see eg [2 3]) Then to improve theconvergence of these Krylov methods the system (1) canbe preconditioned with an adequate nonsingular precondi-tioning matrix 119873 transforming it into any of the equivalentsystems

119873119860119909 = 119873119887

119860119873119910 = 119887 119909 = 119873119910

(2)

the so-called left and right preconditioned systems respec-tively In this paper we address only the case of the right-handside preconditioned matrices 119860119873 but analogous results canbe obtained for the left-hand side preconditioned matrices119873119860

The preconditioning of the system (1) is often performedin order to get a preconditioned matrix 119860119873 as close aspossible to the identity in some sense and the preconditioner119873 is called an approximate inverse of119860 The closeness of119860119873to 119868may bemeasured by using a suitablematrix norm like forinstance the Frobenius norm [4] In this way the problemof obtaining the best preconditioner 119873 (with respect to theFrobenius norm) of the system (1) in an arbitrary subspace119878 of R119899times119899 is equivalent to the minimization problem see forexample [5]

min119872isin119878

119860119872 minus 119868119865= 119860119873 minus 119868

119865 (3)

The solution 119873 to the problem (3) will be referred to asthe ldquooptimalrdquo or the ldquobestrdquo approximate inverse of matrix 119860in the subspace 119878 Since matrix119860119873 is the best approximationto the identity in subspace 119860119878 it will be also referred toas the orthogonal projection of the identity matrix onto thesubspace 119860119878 Although many of the results presented in thispaper are also valid for the case thatmatrix119873 is singular fromnow on we assume that the optimal approximate inverse 119873(and thus also the orthogonal projection119860119873) is a nonsingular

2 Journal of Applied Mathematics

matrix The solution 119873 to the problem (3) has been studiedas a natural generalization of the classical Moore-Penroseinverse in [6] where it has been referred to as the 119878-Moore-Penrose inverse of matrix 119860

The main goal of this paper is to derive new geometricaland spectral properties of the best approximations119860119873 (in thesense of formula (3)) to the identity matrix Such propertiescould be used to analyze the quality and theoretical effective-ness of the optimal approximate inverse119873 as preconditionerof the system (1) However it is important to highlightthat the purpose of this paper is purely theoretical and weare not looking for immediate numerical or computationalapproaches (although our theoretical results could be poten-tially applied to the preconditioning problem) In particularthe term ldquooptimal (or best) approximate inverserdquo is used inthe sense of formula (3) and not in any other sense of thisexpression

Among the many different works dealing with practicalalgorithms that can be used to compute approximate inverseswe refer the reader to for example [4 7ndash9] and to thereferences therein In [4] the author presents an exhaustivesurvey of preconditioning techniques and in particulardescribes several algorithms for computing sparse approxi-mate inverses based on Frobenius norm minimization likefor instance the well-known SPAI and FSAI algorithms Adifferent approach (which is also focused on approximateinverses based on minimizing 119860119872 minus 119868

119865) can be found

in [7] where an iterative descent-type method is used toapproximate each column of the inverse and the iterationis done with ldquosparse matrix by sparse vectorrdquo operationsWhen the system matrix is expressed in block-partitionedform some preconditioning options are explored in [8] In[9] the idea of ldquotargetrdquo matrix is introduced in the contextof sparse approximate inverse preconditioners and the gen-eralized Frobenius norms 1198612

119865119867= tr(119861119867119861119879) (119867 symmetric

positive definite) are used for minimization purposes as analternative to the classical Frobenius norm

The last results of our work are devoted to the special casethat matrix 119860119873 is symmetric and positive definite In thissense let us recall that the cone of symmetric and positivedefinite matrices has a rich geometrical structure and in thiscontext the angle that any symmetric and positive definitematrix forms with the identity plays a very important role[10] In this paper the authors extend this geometrical pointof view and analyze the geometrical structure of the subspaceof symmetric matrices of order 119899 including the location of allorthogonal matrices not only the identity matrix

This paper has been organized as follows In Section 2we present some preliminary results required to make thepaper self-contained Sections 3 and 4 are devoted to obtainnew geometrical and spectral relations respectively for theorthogonal projections 119860119873 of the identity matrix FinallySection 5 closes the paper with its main conclusions

2 Some Preliminaries

Now we present some preliminary results concerning theorthogonal projection 119860119873 of the identity onto the matrix

subspace119860119878 sub R119899times119899 For more details about these results andfor their proofs we refer the reader to [5 6 11]

Taking advantage of the prehilbertian character of thematrix Frobenius norm the solution 119873 to the problem (3)can be obtained using the orthogonal projection theoremMore precisely the matrix product 119860119873 is the orthogonalprojection of the identity onto the subspace119860119878 and it satisfiesthe conditions stated by the following lemmas see [5 11]

Lemma 1 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

0 le 1198601198732

119865= tr (119860119873) le 119899 (4)

0 le 119860119873 minus 1198682

119865= 119899 minus tr (119860119873) le 119899 (5)

An explicit formula for matrix119873 can be obtained by ex-pressing the orthogonal projection119860119873 of the identity matrixonto the subspace 119860119878 by its expansion with respect to anorthonormal basis of 119860119878 [5] This is the idea of the followinglemma

Lemma 2 Let 119860 isin R119899times119899 be nonsingular Let 119878 be a linearsubspace of R119899times119899 of dimension 119889 and 119872

1 119872

119889 a basis of 119878

such that 1198601198721 119860119872

119889 is an orthogonal basis of 119860119878 Then

the solution119873 to the problem (3) is

119873 =

119889

sum

119894=1

tr (119860119872119894)

10038171003817100381710038171198601198721198941003817100381710038171003817

2

119865

119872119894 (6)

and the minimum (residual) Frobenius norm is

119860119873 minus 1198682

119865= 119899 minus

119889

sum

119894=1

[tr (119860119872119894)]2

10038171003817100381710038171198601198721198941003817100381710038171003817

2

119865

(7)

Let us mention two possible options both taken from [5]for choosing in practice the subspace 119878 and its correspondingbasis 119872

119894119889

119894=1 The first example consists of considering the

subspace 119878 of 119899times119899matrices with a prescribed sparsity patternthat is

119878 = 119872 isin R119899times119899

119898119894119895= 0 forall (119894 119895) notin 119870

119870 sub 1 2 119899 times 1 2 119899

(8)

Then denoting by119872119894119895 the 119899 times 119899matrix whose only nonzero

entry is 119898119894119895

= 1 a basis of subspace 119878 is clearly 119872119894119895

(119894 119895) isin 119870 and then 119860119872119894119895

(119894 119895) isin 119870 will be a basisof subspace 119860119878 (since we have assumed that matrix 119860 isnonsingular) In general this basis of 119860119878 is not orthogonalso that we only need to use the Gram-Schmidt procedureto obtain an orthogonal basis of 119860119878 in order to apply theorthogonal expansion (6)

For the second example consider a linearly independentset of 119899 times 119899 real symmetric matrices 119875

1 119875

119889 and the

corresponding subspace

1198781015840

= span 1198751119860119879

119875119889119860119879

(9)

Journal of Applied Mathematics 3

which clearly satisfies

1198781015840

sube 119872 = 119875119860119879

119875 isin R119899times119899 119875119879 = 119875

= 119872 isin R119899times119899 (119860119872)119879

= 119860119872

(10)

Hence we can explicitly obtain the solution 119873 to theproblem (3) for subspace 1198781015840 from its basis 119875

1119860119879

119875119889119860119879

as follows If 119860119875

1119860119879

119860119875119889119860119879

is an orthogonal basis ofsubspace 1198601198781015840 then we just use the orthogonal expansion (6)for obtaining119873 Otherwise we use again the Gram-Schmidtprocedure to obtain an orthogonal basis of subspace1198601198781015840 andthenwe apply formula (6)The interest of this second examplestands in the possibility of using the conjugate gradientmethod for solving the preconditioned linear system whenthe symmetric matrix 119860119873 is positive definite For a moredetailed exposition of the computational aspects related tothese two examples we refer the reader to [5]

Now we present some spectral properties of the orthog-onal projection 119860119873 From now on we denote by 120582

119894119899

119894=1and

120590119894119899

119894=1the sets of eigenvalues and singular values respectively

of matrix119860119873 arranged as usual in nonincreasing order thatis

100381610038161003816100381612058211003816100381610038161003816 ge

100381610038161003816100381612058221003816100381610038161003816 ge sdot sdot sdot ge

10038161003816100381610038161205821198991003816100381610038161003816 gt 0

1205901ge 1205902ge sdot sdot sdot ge 120590

119899gt 0

(11)

The following lemma [11] provides some inequalitiesinvolving the eigenvalues and singular values of the precon-ditioned matrix 119860119873

Lemma 3 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Then

119899

sum

119894=1

1205822

119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816

2

le

119899

sum

119894=1

1205902

119894= 119860119873

2

119865

= tr (119860119873) =

119899

sum

119894=1

120582119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816 le

119899

sum

119894=1

120590119894

(12)

The following fact [11] is a direct consequence ofLemma 3

Lemma 4 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Then the smallest singular value and thesmallest eigenvaluersquos modulus of the orthogonal projection 119860119873of the identity onto the subspace 119860119878 are never greater than 1That is

0 lt 120590119899le1003816100381610038161003816120582119899

1003816100381610038161003816 le 1 (13)

The following theorem [11] establishes a tight connectionbetween the closeness of matrix 119860119873 to the identity matrixand the closeness of 120590

119899(|120582119899|) to the unity

Theorem 5 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

(1 minus1003816100381610038161003816120582119899

1003816100381610038161003816)2

le (1 minus 120590119899)2

le 119860119873 minus 1198682

119865

le 119899 (1 minus1003816100381610038161003816120582119899

1003816100381610038161003816

2

) le 119899 (1 minus 1205902

119899)

(14)

Remark 6 Theorem 5 states that the closer the smallestsingular value 120590

119899of matrix 119860119873 is to the unity the closer

matrix 119860119873 will be to the identity that is the smaller119860119873 minus 119868

119865will be and conversely The same happens with

the smallest eigenvaluersquos modulus |120582119899| of matrix119860119873 In other

words we get a good approximate inverse 119873 of 119860 when 120590119899

(|120582119899|) is sufficiently close to 1

To finish this section let us mention that recently lowerand upper bounds on the normalized Frobenius conditionnumber of the orthogonal projection 119860119873 of the identityonto the subspace 119860119878 have been derived in [12] In additionthis work proposes a natural generalization (related to anarbitrary matrix subspace 119878 of R119899times119899) of the normalizedFrobenius condition number of the nonsingular matrix 119860

3 Geometrical Properties

In this section we present some new geometrical propertiesfor matrix 119860119873 119873 being the optimal approximate inverse ofmatrix 119860 defined by (3) Our first lemma states some basicproperties involving the cosine of the angle between matrix119860119873 and the identity that is

cos (119860119873 119868) =⟨119860119873 119868⟩

119865

119860119873119865119868119865

=tr (119860119873)

119860119873119865radic119899

(15)

Lemma 7 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

cos (119860119873 119868) =tr (119860119873)

119860119873119865radic119899

=119860119873

119865

radic119899=radictr (119860119873)

radic119899 (16)

0 le cos (119860119873 119868) le 1 (17)

119860119873 minus 1198682

119865= 119899 (1 minus cos2 (119860119873 119868)) (18)

Proof First using (15) and (4) we immediately obtain (16)As a direct consequence of (16) we derive that cos(119860119873 119868) isalways nonnegative Finally using (5) and (16) we get

119860119873 minus 1198682

119865= 119899 minus tr (119860119873) = 119899 (1 minus cos2 (119860119873 119868)) (19)

and the proof is concluded

Remark 8 In [13] the authors consider an arbitrary approxi-mate inverse119876 of matrix119860 and derive the following equality

119860119876 minus 1198682

119865= (119860119876

119865minus 119868119865)2

+ 2 (1 minus cos (119860119876 119868)) 119860119876119865119868119865

(20)

4 Journal of Applied Mathematics

that is the typical decomposition (valid in any inner productspace) of the strong convergence into the convergence ofthe norms (119860119876

119865minus 119868119865)2 and the weak convergence (1 minus

cos(119860119876 119868))119860119876119865119868119865 Note that for the special case that 119876

is the optimal approximate inverse119873 defined by (3) formula(18) has stated that the strong convergence is reduced justto the weak convergence and indeed just to the cosinecos(119860119873 119868)

Remark 9 More precisely formula (18) states that the closercos(119860119873 119868) is to the unity (ie the smaller the angle ang(119860119873 119868)

is) the smaller 119860119873 minus 119868119865will be and conversely This gives

us a new measure of the quality (in the Frobenius sense) ofthe approximate inverse 119873 of matrix 119860 by comparing theminimum residual norm 119860119873 minus 119868

119865with the cosine of the

angle between 119860119873 and the identity instead of with tr(119860119873)119860119873

119865(Lemma 1) or 120590

119899 |120582119899| (Theorem 5) So for a fixed

nonsingular matrix 119860 isin R119899times119899 and for different subspaces119878 sub R119899times119899 we have

tr (119860119873) 119899 lArrrArr 119860119873119865 radic119899 lArrrArr 120590

119899 1 lArrrArr

10038161003816100381610038161205821198991003816100381610038161003816 1

lArrrArr cos (119860119873 119868) 1 lArrrArr 119860119873 minus 119868119865 0

(21)

Obviously the optimal theoretical situation corresponds tothe case

tr (119860119873) = 119899 lArrrArr 119860119873119865= radic119899 lArrrArr 120590

119899= 1 lArrrArr 120582

119899= 1

lArrrArr cos (119860119873 119868) = 1 lArrrArr 119860119873 minus 119868119865= 0

lArrrArr 119873 = 119860minus1

lArrrArr 119860minus1

isin 119878

(22)

Remark 10 Note that the ratio between cos(119860119873 119868) andcos(119860 119868) is independent of the order 119899 of matrix 119860 Indeedassuming that tr(119860) = 0 andusing (16) we immediately obtain

cos (119860119873 119868)

cos (119860 119868)=

119860119873119865 radic119899

tr (119860) 119860119865radic119899

= 119860119873119865

119860119865

tr (119860) (23)

The following lemma compares the trace and the Frobe-nius norm of the orthogonal projection 119860119873

Lemma 11 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

tr (119860119873) le 119860119873119865lArrrArr 119860119873

119865le 1 lArrrArr tr (119860119873) le 1

lArrrArr cos (119860119873 119868) le1

radic119899

(24)

tr (119860119873) ge 119860119873119865lArrrArr 119860119873

119865ge 1 lArrrArr tr (119860119873) ge 1

lArrrArr cos (119860119873 119868) ge1

radic119899

(25)

Proof Using (4) we immediately obtain the four leftmostequivalences Using (16) we immediately obtain the tworightmost equivalences

The next lemma provides us with a relationship betweenthe Frobenius norms of the inverses of matrices119860 and its bestapproximate inverse119873 in subspace 119878

Lemma 12 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

10038171003817100381710038171003817119860minus110038171003817100381710038171003817119865

10038171003817100381710038171003817119873minus110038171003817100381710038171003817119865

ge 1 (26)

Proof Using (4) we get

119860119873119865le radic119899 = 119868

119865=10038171003817100381710038171003817(119860119873) (119860119873)

minus110038171003817100381710038171003817119865

le 119860119873119865

10038171003817100381710038171003817(119860119873)minus110038171003817100381710038171003817119865

997904rArr10038171003817100381710038171003817(119860119873)minus110038171003817100381710038171003817119865

ge 1

(27)

and hence10038171003817100381710038171003817119860minus110038171003817100381710038171003817119865

10038171003817100381710038171003817119873minus110038171003817100381710038171003817119865

ge10038171003817100381710038171003817119873minus1

119860minus110038171003817100381710038171003817119865

=10038171003817100381710038171003817(119860119873)minus110038171003817100381710038171003817119865

ge 1 (28)

and the proof is concluded

The following lemma compares the minimum residualnorm 119860119873 minus 119868

119865with the distance (with respect to the

Frobenius norm) 119860minus1 minus 119873119865between the inverse of 119860 and

the optimal approximate inverse 119873 of 119860 in any subspace119878 sub R119899times119899 First note that for any two matrices 119860 119861 isin R119899times119899

(119860 nonsingular) from the submultiplicative property of theFrobenius norm we immediately get

119860119861 minus 1198682

119865=10038171003817100381710038171003817119860 (119861 minus 119860

minus1

)10038171003817100381710038171003817

2

119865

le 1198602

119865

10038171003817100381710038171003817119861 minus 119860

minus110038171003817100381710038171003817

2

119865

(29)

However for the special case that 119861 = 119873 (the solution tothe problem (3)) we also get the following inequality

Lemma 13 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

119860119873 minus 1198682

119865le 119860

119865

10038171003817100381710038171003817119873 minus 119860

minus110038171003817100381710038171003817119865 (30)

Proof Using the Cauchy-Schwarz inequality and (5) we get10038161003816100381610038161003816⟨119860minus1

minus 119873119860119879

⟩119865

10038161003816100381610038161003816le10038171003817100381710038171003817119860minus1

minus 11987310038171003817100381710038171003817119865

1003817100381710038171003817100381711986011987910038171003817100381710038171003817119865

997904rArr10038161003816100381610038161003816tr ((119860minus1 minus 119873)119860)

10038161003816100381610038161003816le10038171003817100381710038171003817119860minus1

minus 11987310038171003817100381710038171003817119865

1003817100381710038171003817100381711986011987910038171003817100381710038171003817119865

997904rArr10038161003816100381610038161003816tr (119860 (119860

minus1

minus 119873))10038161003816100381610038161003816le10038171003817100381710038171003817119873 minus Aminus110038171003817100381710038171003817119865119860119865

997904rArr |tr (119868 minus 119860119873)| le10038171003817100381710038171003817119873 minus 119860

minus110038171003817100381710038171003817119865119860119865

997904rArr 119899 minus tr (119860119873) le10038171003817100381710038171003817119873 minus 119860

minus110038171003817100381710038171003817119865119860119865

997904rArr 119860119873 minus 1198682

119865le 119860

119865

10038171003817100381710038171003817119873 minus 119860

minus110038171003817100381710038171003817119865

(31)

Journal of Applied Mathematics 5

The following extension of the Cauchy-Schwarz inequal-ity in a real or complex inner product space (119867 ⟨sdot sdot⟩) wasobtained by Buzano [14] For all 119886 119909 119887 isin 119867 we have

|⟨119886 119909⟩ sdot ⟨119909 119887⟩| le1

2(119886 119887 + |⟨119886 119887⟩|) 119909

2

(32)

Thenext lemmaprovides uswith lower and upper boundson the inner product ⟨119860119873 119861⟩

119865 for any 119899 times 119899 real matrix 119861

Lemma 14 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then for every 119861 isin R119899times119899 we have

1003816100381610038161003816⟨119860119873 119861⟩119865

1003816100381610038161003816 le1

2(radic119899119861

119865+ |tr (119861)|) (33)

Proof Using (32) for 119886 = 119868 119909 = 119860119873 119887 = 119861 and (4) we get

1003816100381610038161003816⟨119868 119860119873⟩119865sdot ⟨119860119873 119861⟩

119865

1003816100381610038161003816 le1

2(119868119865119861119865+1003816100381610038161003816⟨119868 119861⟩119865

1003816100381610038161003816) 1198601198732

119865

997904rArr1003816100381610038161003816tr (119860119873) sdot ⟨119860119873 119861⟩

119865

1003816100381610038161003816

le1

2(radic119899119861

119865+ |tr (119861)|) 119860119873

2

119865

997904rArr1003816100381610038161003816⟨119860119873 119861⟩

119865

1003816100381610038161003816 le1

2(radic119899119861

119865+ |tr (119861)|)

(34)

The next lemma provides an upper bound on the arith-metic mean of the squares of the 1198992 terms in the orthogonalprojection 119860119873 By the way it also provides us with an upperbound on the arithmetic mean of the 119899 diagonal terms inthe orthogonal projection 119860119873 These upper bounds (validfor any matrix subspace 119878) are independent of the optimalapproximate inverse119873 and thus they are independent of thesubspace 119878 and only depend on matrix 119860

Lemma 15 Let 119860 isin R119899times119899 be nonsingular with tr(119860) = 0 andlet 119878 be a linear subspace of R119899times119899 Let119873 be the solution to theproblem (3) Then

1198601198732

119865

1198992le

1198602

119865

[tr (119860)]2

tr (119860119873)

119899le

1198991198602

119865

[tr (119860)]2

(35)

Proof Using (32) for 119886 = 119860119879 119909 = 119868 and 119887 = 119860119873 and the

Cauchy-Schwarz inequality for ⟨119860119879 119860119873⟩119865and (4) we get

10038161003816100381610038161003816⟨119860119879

119868⟩119865

sdot ⟨119868 119860119873⟩119865

10038161003816100381610038161003816

le1

2(1003817100381710038171003817100381711986011987910038171003817100381710038171003817119865119860119873

119865+10038161003816100381610038161003816⟨119860119879

119860119873⟩119865

10038161003816100381610038161003816) 1198682

119865

997904rArr |tr (119860) sdot tr (119860119873)|

le119899

2(119860119865119860119873

119865+10038161003816100381610038161003816⟨119860119879

119860119873⟩119865

10038161003816100381610038161003816)

le119899

2(119860119865119860119873

119865+1003817100381710038171003817100381711986011987910038171003817100381710038171003817119865119860119873

119865)

= 119899119860119865119860119873

119865

997904rArr |tr (119860)| 1198601198732

119865le 119899119860

119865119860119873

119865

997904rArr119860119873

119865

119899le

119860119865

|tr (119860)|

997904rArr119860119873

2

119865

1198992le

1198602

119865

[tr (119860)]2997904rArr

tr (119860119873)

119899le

1198991198602

119865

[tr (119860)]2

(36)

Remark 16 Lemma 15 has the following interpretation interms of the quality of the optimal approximate inverse119873 ofmatrix 119860 in subspace 119878 The closer the ratio 119899119860

119865| tr(119860)|

is to zero the smaller tr(119860119873) will be and thus due to (5)the larger 119860119873 minus 119868

119865will be and this happens for any matrix

subspace 119878

Remark 17 By the way from Lemma 15 we obtain thefollowing inequality for any nonsingular matrix 119860 isin R119899times119899Consider any matrix subspace 119878 st 119860minus1 isin 119878 Then119873 = 119860

minus1and using Lemma 15 we get

1198601198732

119865

1198992=1198682

119865

1198992=1

nle

1198602

119865

[tr (119860)]2

997904rArr |tr (119860)| le radic119899119860119865

(37)

4 Spectral Properties

In this section we present some new spectral propertiesfor matrix 119860119873 119873 being the optimal approximate inverseof matrix 119860 defined by (3) Mainly we focus on the casethat matrix 119860119873 is symmetric and positive definite This hasbeenmotivated by the following reasonWhen solving a largenonsymmetric linear system (1) by using Krylov methodsa possible strategy consists of searching for an adequateoptimal preconditioner 119873 such that the preconditionedmatrix119860119873 is symmetric positive definite [5]This enables oneto use the conjugate gradientmethod (CG-method) which isin general a computationally efficient method for solving thenew preconditioned system [2 15]

Our starting point is Lemma 3 which has established thatthe sets of eigenvalues and singular values of any orthogonalprojection 119860119873 satisfy

119899

sum

119894=1

1205822

119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816

2

le

119899

sum

119894=1

1205902

119894

=

119899

sum

119894=1

120582119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816 le

119899

sum

119894=1

120590119894

(38)

Let us particularize (38) for some special cases

6 Journal of Applied Mathematics

First note that if 119860119873 is normal (ie for all 1 le 119894 le 119899120590119894= |120582119894| [16]) then (38) becomes

119899

sum

119894=1

1205822

119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816

2

=

119899

sum

119894=1

1205902

119894

=

119899

sum

119894=1

120582119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816 =

119899

sum

119894=1

120590119894

(39)

In particular if 119860119873 is symmetric (120590119894= |120582119894| = plusmn120582

119894isin R) then

(38) becomes119899

sum

119894=1

1205822

119894=

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816

2

=

119899

sum

119894=1

1205902

119894

=

119899

sum

119894=1

120582119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816 =

119899

sum

119894=1

120590119894

(40)

In particular if 119860119873 is symmetric and positive definite (120590119894=

|120582119894| = 120582119894isin R+) then the equality holds in all (38) that is

119899

sum

119894=1

1205822

119894=

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816

2

=

119899

sum

119894=1

1205902

119894

=

119899

sum

119894=1

120582119894=

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816 =

119899

sum

119894=1

120590119894

(41)

The next lemma compares the traces of matrices 119860119873 and(119860119873)2

Lemma 18 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

(i) for any orthogonal projection 119860119873

tr ((119860119873)2

) le 1198601198732

119865= tr (119860119873) (42)

(ii) for any symmetric orthogonal projection 11986011987310038171003817100381710038171003817(119860119873)210038171003817100381710038171003817119865

le tr ((119860119873)2

) = 1198601198732

119865= tr (119860119873) (43)

(iii) for any symmetric positive definite orthogonal projec-tion 119860119873

10038171003817100381710038171003817(119860119873)210038171003817100381710038171003817119865

le tr ((119860119873)2

) = 1198601198732

119865= tr (119860119873) le [tr (119860119873)]

2

(44)

Proof (i) Using (38) we get119899

sum

119894=1

1205822

119894le

119899

sum

119894=1

1205902

119894=

119899

sum

119894=1

120582119894 (45)

(ii) It suffices to use the obvious fact that (119860119873)2

119865

le

1198601198732

119865and the following equalities taken from (40)

119899

sum

119894=1

1205822

119894=

119899

sum

119894=1

1205902

119894=

119899

sum

119894=1

120582119894 (46)

(iii) It suffices to use (43) and the fact that (see eg [1718]) if 119875 and 119876 are symmetric positive definite matrices thentr(119875119876) le tr(119875) tr(119876) for 119875 = 119876 = 119860119873

The rest of the paper is devoted to obtain new propertiesabout the eigenvalues of the orthogonal projection119860119873 for thespecial case that this matrix is symmetric positive definite

First let us recall that the smallest singular value and thesmallest eigenvaluersquos modulus of the orthogonal projection119860119873 are never greater than 1 (see Lemma 4) The followingtheorem establishes the dual result for the largest eigenvalueof matrix 119860119873 (symmetric positive definite)

Theorem 19 Let 119860 isin R119899times119899 be nonsingular and let 119878 be alinear subspace of R119899times119899 Let 119873 be the solution to the problem(3) Suppose thatmatrix119860119873 is symmetric and positive definiteThen the largest eigenvalue of the orthogonal projection119860119873 ofthe identity onto the subspace 119860119878 is never less than 1 That is

1205901= 1205821ge 1 (47)

Proof Using (41) we get

119899

sum

119894=1

1205822

119894=

119899

sum

119894=1

120582119894997904rArr 1205822

119899minus 120582119899=

119899minus1

sum

119894=1

(120582119894minus 1205822

119894) (48)

Now since 120582119899le 1 (Lemma 4) then 1205822

119899minus 120582119899le 0 This implies

that at least one summand in the rightmost sum in (48) mustbe less than or equal to zero Suppose that such summand isthe 119896th one (1 le 119896 le 119899 minus 1) Since 119860119873 is positive definitethen 120582

119896gt 0 and thus

120582119896minus 1205822

119896le 0 997904rArr 120582

119896le 1205822

119896997904rArr 120582

119896ge 1 997904rArr 120582

1ge 1 (49)

and the proof is concluded

InTheorem 19 the assumption that matrix119860119873 is positivedefinite is essential for assuring that |120582

1| ge 1 as the following

simple counterexample shows Moreover from Lemma 4 andTheorem 19 respectively we have that the smallest and largesteigenvalues of 119860119873 (symmetric positive definite) satisfy 120582

119899le

1 and 1205821ge 1 respectively Nothing can be asserted about

the remaining eigenvalues of the symmetric positive definitematrix 119860119873 which can be greater than equal to or less thanthe unity as the same counterexample also shows

Example 20 For 119899 = 3 let

119860119896= (

3 0 0

0 119896 0

0 0 1

) 119896 isin R (50)

let 1198683be identitymatrix of order 3 and let 119878 be the subspace of

all 3times3 scalarmatrices that is 119878 = span1198683Then the solution

119873119896to the problem (3) for subspace 119878 can be immediately

obtained by using formula (6) as follows

119873119896=tr (119860119896)

10038171003817100381710038171198601198961003817100381710038171003817

2

119865

1198683=

119896 + 4

1198962 + 101198683 (51)

and then we get

119860119896119873119896=

119896 + 4

1198962 + 10119860119896=

119896 + 4

1198962 + 10(

3 0 0

0 119896 0

0 0 1

) (52)

Journal of Applied Mathematics 7

Let us arrange the eigenvalues and singular values ofmatrix 119860

119896119873119896 as usual in nonincreasing order (as shown in

(11))On one hand for 119896 = minus2 we have

119860minus2119873minus2

=1

7(

3 0 0

0 minus2 0

0 0 1

) (53)

and then

1205901=10038161003816100381610038161205821

1003816100381610038161003816 =3

7

ge 1205902=10038161003816100381610038161205822

1003816100381610038161003816 =2

7

ge 1205903=10038161003816100381610038161205823

1003816100381610038161003816 =1

7

(54)

Hence 119860minus2119873minus2

is indefinite and 1205901= |1205821| = 37 lt 1

On the other hand for 1 lt 119896 lt 3 we have (see matrix(52))

1205901= 1205821= 3

119896 + 4

1198962 + 10

ge 1205902= 1205822= 119896

119896 + 4

1198962 + 10

ge 1205903= 1205823=

119896 + 4

1198962 + 10

(55)

and then

119896 = 2 1205901= 1205821=9

7gt 1

1205902= 1205822=6

7lt 1

1205903= 1205823=3

7lt 1

119896 =5

2 1205901= 1205821=6

5gt 1

1205902= 1205822= 1

1205903= 1205823=2

5lt 1

119896 =8

3 1205901= 1205821=90

77gt 1

1205902= 1205822=80

77gt 1

1205903= 1205823=30

77lt 1

(56)

Hence for 119860119896119873119896positive definite we have (depending on 119896)

1205822lt 1 120582

2= 1 or 120582

2gt 1

The following corollary improves the lower bound zeroon both tr(119860119873) given in (4) and cos(119860119873 119868) given in (17)

Corollary 21 Let 119860 isin R119899times119899 be nonsingular and let 119878

be a linear subspace of R119899times119899 Let 119873 be the solution to the

problem (3) Suppose thatmatrix119860119873 is symmetric and positivedefinite Then

1 le 119860119873119865le tr (119860119873) = 119860119873

2

119865le 119899 (57)

cos (119860119873 119868) ge1

radic119899 (58)

Proof Denote by sdot 2the spectral norm Using the well-

known inequality sdot 2le sdot

119865[19] Theorem 19 and (4) we

get

119860119873119865ge 119860119873

2= 1205901= 1205821ge 1

997904rArr 1 le 119860119873119865le tr (119860119873) = 119860119873

2

119865le 119899

(59)

Finally (58) follows immediately from (57) and (25)

Let usmention that an upper bound on all the eigenvaluesmoduli and on all singular values of any orthogonal projec-tion 119860119873 can be immediately obtained from (38) and (4) asfollows

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816

2

le

119899

sum

119894=1

1205902

119894= 119860119873

2

119865le 119899

997904rArr1003816100381610038161003816120582119894

1003816100381610038161003816 120590119894le radic119899 forall119894 = 1 2 119899

(60)

Our last theorem improves the upper bound given in(60) for the special case that the orthogonal projection 119860119873

is symmetric positive definite

Theorem 22 Let 119860 isin R119899times119899 be nonsingular and let 119878 be alinear subspace of R119899times119899 Let 119873 be the solution to the problem(3) Suppose thatmatrix119860119873 is symmetric and positive definiteThen all the eigenvalues of matrix 119860119873 satisfy

120590119894= 120582119894le1 + radic119899

2forall119894 = 1 2 119899 (61)

Proof First note that the assertion is obvious for the smallestsingular value since |120582

119899| le 1 for any orthogonal projection

119860119873 (Lemma 4) For any eigenvalue of 119860119873 we use the factthat 119909 minus 119909

2

le 14 for all 119909 gt 0 Then from (41) we get119899

sum

119894=1

1205822

119894=

119899

sum

119894=1

120582119894

997904rArr 1205822

1minus 1205821=

119899

sum

119894=2

(120582119894minus 1205822

119894) le

119899 minus 1

4

997904rArr 1205821le1 + radic119899

2997904rArr 120582

119894le1 + radic119899

2

forall119894 = 1 2 119899

(62)

5 Conclusion

In this paper we have considered the orthogonal projection119860119873 (in the Frobenius sense) of the identity matrix onto an

8 Journal of Applied Mathematics

arbitrary matrix subspace 119860119878 (119860 isin R119899times119899 nonsingular 119878 sub

R119899times119899) Among other geometrical properties of matrix 119860119873we have established a strong relation between the qualityof the approximation 119860119873 asymp 119868 and the cosine of the angleang(119860119873 119868) Also the distance between119860119873 and the identity hasbeen related to the ratio 119899119860

119865| tr(119860)| (which is independent

of the subspace 119878) The spectral analysis has provided lowerand upper bounds on the largest eigenvalue of the symmetricpositive definite orthogonal projections of the identity

Acknowledgments

Theauthors are grateful to the anonymous referee for valuablecomments and suggestions which have improved the earlierversion of this paperThisworkwas partially supported by theldquoMinisterio de Economıa y Competitividadrdquo (Spanish Gov-ernment) and FEDER through Grant Contract CGL2011-29396-C03-01

References

[1] O Axelsson Iterative Solution Methods Cambridge UniversityPress Cambridge Mass USA 1994

[2] Y Saad Iterative Methods for Sparse Linear Systems PWS Pub-lishing Boston Mass USA 1996

[3] S-L Wu F Chen and X-Q Niu ldquoA note on the eigenvalueanalysis of the SIMPLE preconditioning for incompressibleflowrdquo Journal of Applied Mathematics vol 2012 Article ID564132 7 pages 2012

[4] M Benzi ldquoPreconditioning techniques for large linear systemsa surveyrdquo Journal of Computational Physics vol 182 no 2 pp418ndash477 2002

[5] G Montero L Gonzalez E Florez M D Garcıa and ASuarez ldquoApproximate inverse computation using Frobenius in-ner productrdquoNumerical Linear Algebra with Applications vol 9no 3 pp 239ndash247 2002

[6] A Suarez and L Gonzalez ldquoA generalization of the Moore-Penrose inverse related to matrix subspaces of 119862119899times119898rdquo AppliedMathematics andComputation vol 216 no 2 pp 514ndash522 2010

[7] E Chow and Y Saad ldquoApproximate inverse preconditioners viasparse-sparse iterationsrdquo SIAM Journal on Scientific Computingvol 19 no 3 pp 995ndash1023 1998

[8] E Chow and Y Saad ldquoApproximate inverse techniques forblock-partitioned matricesrdquo SIAM Journal on Scientific Com-puting vol 18 no 6 pp 1657ndash1675 1997

[9] R M Holland A J Wathen and G J Shaw ldquoSparse approxi-mate inverses and target matricesrdquo SIAM Journal on ScientificComputing vol 26 no 3 pp 1000ndash1011 2005

[10] R Andreani M Raydan and P Tarazaga ldquoOn the geometricalstructure of symmetric matricesrdquo Linear Algebra and Its Appli-cations vol 438 no 3 pp 1201ndash1214 2013

[11] L Gonzalez ldquoOrthogonal projections of the identity spectralanalysis and applications to approximate inverse precondition-ingrdquo SIAM Review vol 48 no 1 pp 66ndash75 2006

[12] A Suarez and L Gonzalez ldquoNormalized Frobenius conditionnumber of the orthogonal projections of the identityrdquo Journalof Mathematical Analysis and Applications vol 400 no 2 pp510ndash516 2013

[13] J-P Chehab and M Raydan ldquoGeometrical properties of theFrobenius condition number for positive definite matricesrdquo

Linear Algebra and Its Applications vol 429 no 8-9 pp 2089ndash2097 2008

[14] M L Buzano ldquoGeneralizzazione della diseguaglianza di Cau-chy-Schwarzrdquo Rendiconti del Seminario Matematico Universitae Politecnico di Torino vol 31 pp 405ndash409 1974 (Italian)

[15] M R Hestenes and E Stiefel ldquoMethods of conjugate gradientsfor solving linear systemsrdquo Journal of Research of the NationalBureau of Standards vol 49 no 6 pp 409ndash436 1952

[16] R A Horn and C R Johnson Topics in Matrix Analysis Cam-bridge University Press Cambridge UK 1991

[17] E H Lieb andWThirring ldquoInequalities for themoments of theeigenvalues of the Schrodinger Hamiltonian and their relationto Sobolev inequalitiesrdquo in Studies in Mathematical PhysicsEssays in Honor of Valentine Bargmann E Lieb B Simon andA Wightman Eds pp 269ndash303 Princeton University PressPrinceton NJ USA 1976

[18] Z Ulukok and R Turkmen ldquoOn some matrix trace inequali-tiesrdquo Journal of Inequalities and Applications vol 2010 ArticleID 201486 8 pages 2010

[19] R A Horn andC R JohnsonMatrix Analysis CambridgeUni-versity Press Cambridge UK 1985

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 489295 6 pageshttpdxdoiorg1011552013489295

Research ArticleA Parameterized Splitting Preconditioner for GeneralizedSaddle Point Problems

Wei-Hua Luo12 and Ting-Zhu Huang1

1 School of Mathematical Sciences University of Electronic Science and Technology of China Chengdu Sichuan 611731 China2 Key Laboratory of Numerical Simulation of Sichuan Province University Neijiang Normal University Neijiang Sichuan 641112 China

Correspondence should be addressed to Ting-Zhu Huang tingzhuhuang126com

Received 4 January 2013 Revised 31 March 2013 Accepted 1 April 2013

Academic Editor P N Shivakumar

Copyright copy 2013 W-H Luo and T-Z Huang This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

By using Sherman-Morrison-Woodbury formula we introduce a preconditioner based on parameterized splitting idea forgeneralized saddle point problems which may be singular and nonsymmetric By analyzing the eigenvalues of the preconditionedmatrix we find that when 120572 is big enough it has an eigenvalue at 1 with multiplicity at least n and the remaining eigenvalues are alllocated in a unit circle centered at 1 Particularly when the preconditioner is used in general saddle point problems it guaranteeseigenvalue at 1 with the same multiplicity and the remaining eigenvalues will tend to 1 as the parameter 120572 rarr 0 Consequently thiscan lead to a good convergence when some GMRES iterative methods are used in Krylov subspace Numerical results of Stokesproblems and Oseen problems are presented to illustrate the behavior of the preconditioner

1 Introduction

In some scientific and engineering applications such as finiteelement methods for solving partial differential equations [12] and computational fluid dynamics [3 4] we often considersolutions of the generalized saddle point problems of the form

(119860 119861119879

minus119861 119862)(

119909

119910) = (

119891

119892) (1)

where 119860 isin R119899times119899 119861 isin R119898times119899 (119898 le 119899) and 119862 isin R119898times119898

are positive semidefinite 119909 119891 isin R119899 and 119910 119892 isin R119898 When119862 = 0 (1) is a general saddle point problem which is also aresearching object for many authors

It is well known that when the matrices 119860 119861 and 119862

are large and sparse the iterative methods are more efficientand attractive than direct methods assuming that (1) has agood preconditioner In recent years a lot of preconditioningtechniques have arisen for solving linear system for exampleSaad [5] and Chen [6] have comprehensively surveyed someclassical preconditioning techniques including ILU pre-conditioner triangular preconditioner SPAI preconditionermultilevel recursive Schur complements preconditioner and

sparse wavelet preconditioner Particularly many precondi-tioning methods for saddle problems have been presentedrecently such as dimensional splitting (DS) [7] relaxeddimensional factorization (RDF) [8] splitting preconditioner[9] and Hermitian and skew-Hermitian splitting precondi-tioner [10]

Among these results Cao et al [9] have used splitting ideato give a preconditioner for saddle point problems where thematrix 119860 is symmetric and positive definite and 119861 is of fullrow rank According to his preconditioner the eigenvaluesof the preconditioned matrix would tend to 1 when theparameter 119905 rarr infin Consequently just as we have seen fromthose examples of [9] preconditioner has guaranteed a goodconvergence when some iterative methods were used

In this paper being motivated by [9] we use the splittingidea to present a preconditioner for the system (1) where 119860may be nonsymmetric and singular (when rank(119861) lt 119898) Wefind that when the parameter is big enough the precondi-tioned matrix has the eigenvalue at 1 with multiplicity at least119899 and the remaining eigenvalues are all located in a unit circlecentered at 1 Particularly when the precondidtioner is usedin some general saddle point problems (namely 119862 = 0) we

2 Journal of Applied Mathematics

see that the multiplicity of the eigenvalue at 1 is also at least 119899but the remaining eigenvalues will tend to 1 as the parameter120572 rarr 0

The remainder of the paper is organized as followsIn Section 2 we present our preconditioner based on thesplitting idea and analyze the bound of eigenvalues of thepreconditioned matrix In Section 3 we use some numericalexamples to show the behavior of the new preconditionerFinally we draw some conclusions and outline our futurework in Section 4

2 A Parameterized Splitting Preconditioner

Now we consider using splitting idea with a variable param-eter to present a preconditioner for the system (1)

Firstly it is evident that when 120572 = 0 the system (1) isequivalent to

(

119860 119861119879

minus119861

120572

119862

120572

)(119909

119910) = (

119891

119892

120572

) (2)

Let

119872 = (

119860 119861119879

minus119861

120572119868

) 119873 = (

0 0

0 119868 minus119862

120572

)

119883 = (119909

119910) 119865 = (

119891

119892

120572

)

(3)

Then the coefficient matrix of (2) can be expressed by119872minus119873Multiplying both sides of system (2) from the left with matrix119872minus1 we have

(119868 minus119872minus1

119873)119883 = 119872minus1

119865 (4)

Hence we obtain a preconditioned linear system from (1)using the idea of splitting and the corresponding precondi-tioner is

119867 = (

119860 119861119879

minus119861

120572119868

)

minus1

(

119868 0

0119868

120572

) (5)

Nowwe analyze the eigenvalues of the preconditioned system(4)

Theorem 1 The preconditioned matrix 119868 minus 119872minus1

119873 has aneigenvalue at 1 with multiplicity at least 119899 The remainingeigenvalues 120582 satisfy

120582 =1199041+ 1199042

1199041+ 120572

(6)

where 1199041= 120596119879

119861119860minus1

119861119879

120596 1199042= 120596119879

119862120596 and 120596 isin C119898 satisfies

(119861119860minus1

119861119879

120572+119862

120572)120596 = 120582(119868 +

119861119860minus1

119861119879

120572)120596 120596 = 1 (7)

Proof Because

119872minus1

=(

(119860 +119861119879

119861

120572)

minus1

minus119860minus1

119861119879

(119868 +119861119860minus1

119861119879

120572)

minus1

119861

120572(119860 +

119861119879

119861

120572)

minus1

(119868 +119861119860minus1

119861119879

120572)

minus1)

(8)

we can easily get

119868 minus119872minus1

119873 =(

119868 119860minus1

119861119879

(119868 +119861119860minus1

119861119879

120572)

minus1

(119868 minus119862

120572)

0 (119868 +119861119860minus1

119861119879

120572)

minus1

(119861119860minus1

119861119879

120572+119862

120572)

)

(9)

which implies the preconditioned matrix 119868 minus 119872minus1119873 has aneigenvalue at 1 with multiplicity at least 119899

For the remaining eigenvalues let

(119868 +119861119860minus1

119861119879

120572)

minus1

(119861119860minus1

119861119879

120572+119862

120572)120596 = 120582120596 (10)

with 120596 = 1 then we have

(119861119860minus1

119861119879

120572+119862

120572)120596 = 120582(119868 +

119861119860minus1

119861119879

120572)120596 (11)

By multiplying both sides of this equality from the left with120596119879 we can get

120582 =1199041+ 1199042

1199041+ 120572

(12)

This completes the proof of Theorem 1

Remark 2 FromTheorem 1 we can get that when parameter120572 is big enough the modulus of nonnil eigenvalues 120582 will belocated in interval (0 1)

Remark 3 InTheorem 1 if the matrix 119862 = 0 then for nonnileigenvalues we have

lim120572rarr0

120582 = 1 (13)

Figures 1 2 and 3 are the eigenvalues plots of the pre-conditioned matrices obtained with our preconditioner Aswe can see in the following numerical experiments thisgood phenomenon is useful for accelerating convergence ofiterative methods in Krylov subspace

Additionally for the purpose of practically executing ourpreconditioner

119867 =(

(119860 +119861119879

119861

120572)

minus1

minus119860minus1

119861119879

120572(119868 +

119861119860minus1

119861119879

120572)

minus1

119861

120572(119860 +

119861119879

119861

120572)

minus1

1

120572(119868 +

119861119860minus1

119861119879

120572)

minus1)

(14)

Journal of Applied Mathematics 3

0 02 04 06 08 12 141

0

1

2

3

4

minus1

minus2

minus3

times10minus4

= 01 120572 = 00001

(a)

0

002

004

006

008

0 02 04 06 08 12 141minus008

minus006

minus004

minus002

= 01 120572 = norm(119862 fro)20

(b)

Figure 1 Spectrum of the preconditioned steady Oseen matrix with viscosity coefficient V = 01 32 times 32 grid (a) Q2-Q1 FEM 119862 = 0 (b)Q1-P0 FEM 119862 = 0

0

0 02 04 06 08 121

02

04

06

08

1

minus02

minus04

minus06

minus08

minus1

minus02

times10minus3

= 001 120572 = 00001

(a)

0 02 04 06 08 12 141minus02

minus015

minus01

minus005

0

005

01

015

02 = 001 120572 = norm(119862 fro)20

(b)

Figure 2 Spectrum of the preconditioned steady Oseen matrix with viscosity coefficient V = 001 32 times 32 grid (a) Q2-Q1 FEM 119862 = 0 (b)Q1-P0 FEM 119862 = 0

0

2

4

6

8

0 02 04 06 08 12 141

times10minus4

minus8

minus6

minus4

minus2

= 0001 120572 = 00001

(a)

0

002

004

006

008

minus008

minus006

minus004

minus002

0 02 04 06 08 121minus02

= 0001 120572 = norm(119862 fro)20

(b)

Figure 3 Spectrum of the preconditioned steady Oseen matrix with viscosity coefficient V = 0001 32 times 32 grid (a) Q2-Q1 FEM 119862 = 0 (b)Q1-P0 FEM 119862 = 0

4 Journal of Applied Mathematics

Table 1 Preconditioned GMRES(20) on Stokes problems withdifferent grid sizes (uniform grids)

Grid 120572 its LU time its time Total time16 times 16 00001 4 00457 00267 0072432 times 32 00001 6 03912 00813 0472564 times 64 00001 9 54472 05519 59991128 times 128 00001 14 914698 57732 972430

Table 2 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 01

Grid 120572 its LU time its time Total time16 times 16 00001 3 00479 00244 0072332 times 32 00001 3 06312 00696 0700964 times 64 00001 4 132959 03811 136770128 times 128 00001 6 1305463 37727 1343190

Table 3 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 001

Grid 120572 its LU time its time Total time16 times 16 00001 2 00442 00236 0067832 times 32 00001 3 04160 00557 0471764 times 64 00001 3 73645 02623 76268128 times 128 00001 4 1694009 31709 1725718

Table 4 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 0001

Grid 120572 its LU time its time Total time16 times 16 00001 2 00481 00206 0068732 times 32 00001 3 04661 00585 0524664 times 64 00001 3 66240 02728 68969128 times 128 00001 4 1773130 29814 1802944

Table 5 Preconditioned GMRES(20) on Stokes problems withdifferent grid sizes (uniform grids)

Grid 120572 its LU time its time Total time16 times 16 10000 5 00471 00272 0074332 times 32 10000 7 03914 00906 0482064 times 64 10000 9 56107 04145 60252128 times 128 10000 14 927154 48908 976062

Table 6 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 01

Grid 120572 its LU time its time Total time16 times 16 10000 4 00458 00223 0068132 times 32 10000 4 05748 00670 0641864 times 64 10000 5 122642 04179 126821128 times 128 10000 7 1282758 16275 1299033

Table 7 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 001

Grid 120572 its LU time its time Total time16 times 16 10000 3 00458 00224 0068232 times 32 10000 3 04309 00451 0476064 times 64 10000 4 76537 01712 78249128 times 128 10000 5 1751587 34554 1786141

Table 8 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 0001

Grid 120572 its LU time its time Total time16 times 16 10000 3 00507 00212 0071932 times 32 10000 3 04735 00449 0518464 times 64 10000 4 66482 01645 68127128 times 128 10000 4 1720516 21216 1741732

Table 9 Preconditioned GMRES(20) on Stokes problems withdifferent grid sizes (uniform grids)

Grid 120572 its LU time its time Total time16 times 16 norm(119862 fro) 21 00240 00473 0071332 times 32 norm(119862 fro) 20 00890 01497 0238764 times 64 norm(119862 fro) 20 08387 06191 14578128 times 128 norm(119862 fro) 20 68917 30866 99783

Table 10 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 01

Grid 120572 its LU time its time Total time16 times 16 norm(119862 fro)20 10 00250 00302 0055232 times 32 norm(119862 fro)20 10 00816 00832 0164864 times 64 norm(119862 fro)20 12 08466 03648 12114128 times 128 norm(119862 fro)20 14 69019 20398 89417

Table 11 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 001

Grid 120572 its LU time its time Total time16 times 16 norm(119862 fro)20 7 00238 00286 0052432 times 32 norm(119862 fro)20 7 00850 00552 0140264 times 64 norm(119862 fro)20 11 08400 03177 11577128 times 128 norm(119862 fro)20 16 69537 22306 91844

Table 12 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 0001

Grid 120572 its LU time its time Total time16 times 16 norm(119862 fro)20 5 00245 00250 0049532 times 32 norm(119862 fro)20 5 00905 00587 0149264 times 64 norm(119862 fro)20 8 12916 03200 16116128 times 128 norm(119862 fro)20 15 104399 27468 131867

Journal of Applied Mathematics 5

Table 13 Size and number of non-nil elements of the coefficient matrix generalized by using Q2-Q1 FEM in steady Stokes problems

Grid dim(119860) nnz(119860) dim(119861) nnz(119861) nnz(119860 + (1120572)119861119879119861)16 times 16 578 times 578 6178 81 times 578 2318 3690432 times 32 2178 times 2178 28418 289 times 2178 10460 19322064 times 64 8450 times 8450 122206 1089 times 8450 44314 875966128 times 128 33282 times 33282 506376 4225 times 33282 184316 3741110

we should efficiently deal with the computation of (119868 +(119861119860minus1

119861119879

120572))minus1 This can been tackled by the well-known

Sherman-Morrison-Woodbury formula

(1198601+ 11988311198602119883119879

2)minus1

= 119860minus1

1minus 119860minus1

11198831(119860minus1

2+ 119883119879

2119860minus1

11198831)minus1

119883119879

2119860minus1

1

(15)

where 1198601isin R119899times119899 and 119860

2isin R1199031times1199031 are invertible matrices

1198831isin R119899times1199031 and 119883

2isin R119899times1199031 are any matrices and 119899 119903

1are

any positive integersFrom (15) we immediately get

(119868 +119861119860minus1

119861119879

120572)

minus1

= 119868 minus 119861(120572119860 + 119861119879

119861)minus1

119861119879

(16)

In the following numerical examples we will always use (16)to compute (119868 + (119861119860minus1119861119879120572))minus1 in (14)

3 Numerical Examples

In this sectionwe give numerical experiments to illustrate thebehavior of our preconditioner The numerical experimentsare done by using MATLAB 71 The linear systems areobtained by using finite element methods in the Stokes prob-lems and steady Oseen problems and they are respectivelythe cases of

(1) 119862 = 0 which is caused by using Q2-Q1 FEM(2) 119862 = 0 which is caused by using Q1-P0 FEM

Furthermore we compare our preconditionerwith that of[9] in the case of general saddle point problems (namely 119862 =0) For the general saddle point problem [9] has presentedthe preconditioner

= (

119860 + 119905119861119879

119861 0

minus2119861119868

119905

)

minus1

(17)

with 119905 as a parameter and has proved that when 119860 issymmetric positive definite the preconditioned matrix hasan eigenvalue 1 with multiplicity at 119899 and the remainingeigenvalues satisfy

120582 =1199051205902

119894

1 + 1199051205902

119894

(18)

lim119905rarrinfin

120582 = 1 (19)

where 120590119894 119894 = 1 2 119898 are119898 positive singular values of the

matrix 119861119860minus12All these systems can be generalized by

using IFISS software package [11] (this is a freepackage that can be downloaded from the sitehttpwwwmathsmanchesteracuksimdjsifiss) We userestarted GMRES(20) as the Krylov subspace method andwe always take a zero initial guess The stopping criterion is

100381710038171003817100381711990311989610038171003817100381710038172

1003817100381710038171003817119903010038171003817100381710038172

le 10minus6

(20)

where 119903119896is the residual vector at the 119896th iteration

In the whole course of computation we always replace(119868 + (119861119860

minus1

119861119879

120572))minus1 in (14) with (16) and use the 119871119880 factor-

ization of 119860 + (119861119879

119861120572) to tackle (119860 + (119861119879

119861120572))minus1V where

V is a corresponding vector in the iteration Concretely let119860 + (119861

119879

119861120572) = 119871119880 then we complete the matrix-vectorproduct (119860 + (119861

119879

119861120572))minus1V by 119880 119871 V in MATLAB term

In the following tables the denotation norm (119862 fro) meansthe Frobenius form of the matrix 119862 The total time is the sumof LU time and iterative time and the LU time is the time tocompute LU factorization of 119860 + (119861119879119861120572)

Case 1 (for our preconditioner) 119862 = 0 (using Q2-Q1 FEM inStokes problems and steady Oseen problems with differentviscosity coefficients The results are in Tables 1 2 3 4 )

Case 1rsquo (for preconditioner of [9]) 119862 = 0 (using Q2-Q1 FEMin Stokes problems and steady Oseen problems with differentviscosity coefficients The results are in Tables 5 6 7 8)

Case 2 119862 = 0 (using Q1-P0 FEM in Stokes problems andsteady Oseen problems with different viscosity coefficientsThe results are in Tables 9 10 11 12)

From Tables 1 2 3 4 5 6 7 and 8 we can see that theseresults are in agreement with the theoretical analyses (13) and(19) respectively Additionally comparing with the results inTables 9 10 11 and 12 we find that although the iterationsused in Case 1 (either for the preconditioner of [9] or ourpreconditioner) are less than those in Case 2 the time spentby Case 1 ismuchmore than that of Case 2This is because thedensity of the coefficient matrix generalized by Q2-Q1 FEMis much larger than that generalized by Q1-P0 FEMThis canbe partly illustrated by Tables 13 and 14 and the others can beillustrated similarly

6 Journal of Applied Mathematics

Table 14 Size and number of non-nil elements of the coefficient matrix generalized by using Q1-P0 FEM in steady Stokes problems

Grid dim(119860) nnz(119860) dim(119861) nnz(119861) nnz(119860 + (1120572)119861119879119861)16 times 16 578 times 578 3826 256 times 578 1800 707632 times 32 2178 times 2178 16818 1024 times 2178 7688 3187464 times 64 8450 times 8450 70450 4096 times 8450 31752 136582128 times 128 33282 times 33282 288306 16384 times 33282 129032 567192

4 Conclusions

In this paper we have introduced a splitting preconditionerfor solving generalized saddle point systems Theoreticalanalysis showed the modulus of eigenvalues of the precon-ditioned matrix would be located in interval (0 1) when theparameter is big enough Particularly when the submatrix119862 = 0 the eigenvalues will tend to 1 as the parameter 120572 rarr 0These performances are tested by some examples and theresults are in agreement with the theoretical analysis

There are still some future works to be done how to prop-erly choose a parameter 120572 so that the preconditioned matrixhas better propertiesHow to further precondition submatrix(119868 + (119861119860

minus1

119861119879

120572))minus1

((119861119860minus1

119861119879

120572) + (119862120572)) to improve ourpreconditioner

Acknowledgments

The authors would express their great thankfulness to thereferees and the editor Professor P N Shivakumar for theirhelpful suggestions for revising this paperThe authors wouldlike to thank H C Elman A Ramage and D J Silvester fortheir free IFISS software package This research is supportedby Chinese Universities Specialized Research Fund for theDoctoral Program (20110185110020) Sichuan Province Sci ampTech Research Project (2012GZX0080) and the Fundamen-tal Research Funds for the Central Universities

References

[1] F Brezzi andM FortinMixed and Hybrid Finite ElementMeth-ods vol 15 of Springer Series in Computational MathematicsSpringer New York NY USA 1991

[2] Z Chen Q Du and J Zou ldquoFinite element methods withmatching and nonmatching meshes for Maxwell equationswith discontinuous coefficientsrdquo SIAM Journal on NumericalAnalysis vol 37 no 5 pp 1542ndash1570 2000

[3] H C Elman D J Silvester and A J Wathen Finite Elementsand Fast Iterative Solvers With Applications in IncompressibleFluid Dynamics Numerical Mathematics and Scientific Com-putation Oxford University Press New York NY USA 2005

[4] C Cuvelier A Segal and A A van Steenhoven Finite ElementMethods and Navier-Stokes Equations vol 22 of Mathematicsand its Applications D Reidel Dordrecht The Netherlands1986

[5] Y Saad Iterative Methods for Sparse Linear Systems Society forIndustrial and Applied Mathematics Philadelphia Pa USA2nd edition 2003

[6] K Chen Matrix Preconditioning Techniques and Applicationsvol 19 ofCambridgeMonographs onApplied andComputational

Mathematics Cambridge University Press Cambridge UK2005

[7] M Benzi and X-P Guo ldquoA dimensional split preconditionerfor Stokes and linearized Navier-Stokes equationsrdquo AppliedNumerical Mathematics vol 61 no 1 pp 66ndash76 2011

[8] M Benzi M Ng Q Niu and Z Wang ldquoA relaxed dimensionalfactorization preconditioner for the incompressible Navier-Stokes equationsrdquo Journal of Computational Physics vol 230no 16 pp 6185ndash6202 2011

[9] YCaoM-Q Jiang andY-L Zheng ldquoA splitting preconditionerfor saddle point problemsrdquo Numerical Linear Algebra withApplications vol 18 no 5 pp 875ndash895 2011

[10] V Simoncini and M Benzi ldquoSpectral properties of the Her-mitian and skew-Hermitian splitting preconditioner for saddlepoint problemsrdquo SIAM Journal on Matrix Analysis and Applica-tions vol 26 no 2 pp 377ndash389 2004

[11] H C Elman A Ramage and D J Silvester ldquoAlgorithm 886IFISS a Matlab toolbox for modelling incompressible flowrdquoACM Transactions on Mathematical Software vol 33 no 2article 14 2007

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 593549 11 pageshttpdxdoiorg1011552013593549

Research ArticleSolving Optimization Problems on Hermitian Matrix Functionswith Applications

Xiang Zhang and Shu-Wen Xiang

Department of Computer Science and Information Guizhou University Guiyang 550025 China

Correspondence should be addressed to Xiang Zhang zxjnsc163com

Received 17 October 2012 Accepted 20 March 2013

Academic Editor K Sivakumar

Copyright copy 2013 X Zhang and S-W Xiang This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

We consider the extremal inertias and ranks of the matrix expressions 119891(119883 119884) = 1198603minus 119861

3119883 minus (119861

3119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast where

1198603= 119860

lowast

3 119861

3 119862

3 and 119863

3are known matrices and 119884 and 119883 are the solutions to the matrix equations 119860

1119884 = 119862

1 119884119861

1= 119863

1 and

1198602119883 = 119862

2 respectively As applications we present necessary and sufficient condition for the previous matrix function 119891(119883 119884)

to be positive (negative) non-negative (positive) definite or nonsingular We also characterize the relations between the Hermitianpart of the solutions of the above-mentioned matrix equations Furthermore we establish necessary and sufficient conditions forthe solvability of the system of matrix equations 119860

1119884 = 119862

1 119884119861

1= 119863

1 119860

2119883 = 119862

2 and 119861

3119883 + (119861

3119883)

lowast

+ 1198623119884119863

3+ (119862

3119884119863

3)lowast

= 1198603

and give an expression of the general solution to the above-mentioned system when it is solvable

1 Introduction

Throughout we denote the field of complex numbers by Cthe set of all 119898 times 119899 matrices over C by C119898times119899 and the set ofall119898 times119898Hermitian matrices by C119898times119898

ℎ The symbols 119860lowast and

R(119860) stand for the conjugate transpose the column space ofa complex matrix119860 respectively 119868

119899denotes the 119899 times 119899 identity

matrix TheMoore-Penrose inverse [1] 119860dagger of 119860 is the uniquesolution119883 to the four matrix equations

(i) 119860119883119860 = 119860

(ii) 119883119860119883 = 119883

(iii) (119860119883)lowast = 119860119883

(iv) (119883119860)lowast = 119883119860

(1)

Moreover 119871119860and 119877

119860stand for the projectors 119871

119860= 119868 minus

119860dagger

119860 119877119860= 119868 minus 119860119860

dagger induced by 119860 It is well known that theeigenvalues of a Hermitian matrix 119860 isin C119899times119899 are real and theinertia of 119860 is defined to be the triplet

I119899(119860) = 119894

+(119860) 119894

minus(119860) 119894

0(119860) (2)

where 119894+(119860) 119894

minus(119860) and 119894

0(119860) stand for the numbers of

positive negative and zero eigenvalues of119860 respectivelyThe

symbols 119894+(119860) and 119894

minus(119860) are called the positive index and

the negative index of inertia respectively For two Hermitianmatrices 119860 and 119861 of the same sizes we say 119860 ge 119861 (119860 le 119861)in the Lowner partial ordering if 119860 minus 119861 is positive (negative)semidefinite The Hermitian part of 119883 is defined as 119867(119883) =119883 + 119883

lowast We will say that 119883 is Re-nnd (Re-nonnegativesemidefinite) if 119867(119883) ge 0 119883 is Re-pd (Re-positive definite)if119867(119883) gt 0 and119883 is Re-ns if119867(119883) is nonsingular

It is well known that investigation on the solvabilityconditions and the general solution to linearmatrix equationsis very active (eg [2ndash9]) In 1999 Braden [10] gave thegeneral solution to

119861119883 + (119861119883)lowast

= 119860 (3)

In 2007 Djordjevic [11] considered the explicit solution to (3)for linear bounded operators on Hilbert spaces MoreoverCao [12] investigated the general explicit solution to

119861119883119862 + (119861119883119862)lowast

= 119860 (4)

Xu et al [13] obtained the general expression of the solutionof operator equation (4) In 2012 Wang and He [14] studied

2 Journal of Applied Mathematics

some necessary and sufficient conditions for the consistenceof the matrix equation

1198601119883 + (119860

1119883)

lowast

+ 1198611119884119862

1+ (119861

1119884119862

1)lowast

= 1198641

(5)

and presented an expression of the general solution to (5)Note that (5) is a special case of the following system

1198601119884 = 119862

1 119884119861

1= 119863

1 119860

2119883 = 119862

2

1198613119883 + (119861

3119883)

lowast

+ 1198623119884119863

3+ (119862

3119884119863

3)lowast

= 1198603

(6)

To our knowledge there has been little information about (6)One goal of this paper is to give some necessary and sufficientconditions for the solvability of the system of matrix (6) andpresent an expression of the general solution to system (6)when it is solvable

In order to find necessary and sufficient conditions for thesolvability of the system of matrix equations (6) we need toconsider the extremal ranks and inertias of (10) subject to (13)and (11)

There have been many papers to discuss the extremalranks and inertias of the following Hermitian expressions

119901 (119883) = 1198603minus 119861

3119883 minus (119861

3119883)

lowast

(7)

119892 (119884) = 119860 minus 119861119884119862 minus (119861119884119862)lowast

(8)

ℎ (119883 119884) = 1198601minus 119861

1119883119861

lowast

1minus 119862

1119884119862

lowast

1 (9)

119891 (119883 119884) = 1198603minus 119861

3119883 minus (119861

3119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

(10)

Tian has contributed much in this field One of his works[15] considered the extremal ranks and inertias of (7) Heand Wang [16] derived the extremal ranks and inertias of (7)subject to119860

1119883 = 119862

1119860

2119883119861

2= 119862

2 Liu and Tian [17] studied

the extremal ranks and inertias of (8) Chu et al [18] and Liuand Tian [19] derived the extremal ranks and inertias of (9)Zhang et al [20] presented the extremal ranks and inertias of(9) where119883 and 119884 are Hermitian solutions of

1198602119883 = 119862

2 (11)

1198841198612= 119863

2 (12)

respectively He and Wang [16] derived the extremal ranksand inertias of (10)We consider the extremal ranks and iner-tias of (10) subject to (11) and

1198601119884 = 119862

1 119884119861

1= 119863

1 (13)

which is not only the generalization of the abovematrix func-tions but also can be used to investigate the solvability con-ditions for the existence of the general solution to the system(6) Moreover it can be applied to characterize the relationsbetween Hermitian part of the solutions of (11) and (13)

The remainder of this paper is organized as follows InSection 2 we consider the extremal ranks and inertias of(10) subject to (11) and (13) In Section 3 we characterize therelations between the Hermitian part of the solution to (11)and (13) In Section 4 we establish the solvability conditionsfor the existence of a solution to (6) and obtain an expressionof the general solution to (6)

2 Extremal Ranks and Inertias of HermitianMatrix Function (10) with Some Restrictions

In this section we consider formulas for the extremal ranksand inertias of (10) subject to (11) and (13) We begin with thefollowing Lemmas

Lemma 1 (see [21]) (a) Let1198601119862

1 119861

1 and119863

1be givenThen

the following statements are equivalent

(1) system (13) is consistent(2)

1198771198601119862

1= 0 119863

1119871

1198611= 0 119860

1119863

1= 119862

11198611 (14)

(3)

119903 [1198601119862

1] = 119903 (119860

1)

[119863

1

1198611

] = 119903 (1198611)

1198601119863

1= 119862

11198611

(15)

In this case the general solution can be written as

119884 = 119860dagger

1119862

1+ 119871

1198601119863

1119861dagger

1+ 119871

1198601119881119877

1198611 (16)

where 119881 is arbitrary(b) Let 119860

2and 119862

2be given Then the following statements

are equivalent

(1) equation (11) is consistent(2)

1198771198602119862

2= 0 (17)

(3)

119903 [1198602119862

2] = 119903 (119860

2) (18)

In this case the general solution can be written as

119883 = 119860dagger

119862 + 119871119860119882 (19)

where119882 is arbitrary

Lemma 2 ([22 Lemma 15 Theorem 23]) Let 119860 isin C119898times119898

ℎ

119861 isin C119898times119899 and119863 isin C119899times119899

ℎ and denote that

119872 = [119860 119861

119861lowast

0]

119873 = [0 119876

119876lowast

0]

119871 = [119860 119861

119861lowast

119863]

119866 = [119875 119872119871

119873

119871119873119872

lowast

0]

(20)

Journal of Applied Mathematics 3

Then one has the following

(a) the following equalities hold

119894plusmn(119872) = 119903 (119861) + 119894

plusmn(119877

119861119860119877

119861) (21)

119894plusmn(119873) = 119903 (119876) (22)

(b) if R(119861) sube R(119860) then 119894plusmn(119871) = 119894

plusmn(119860) + 119894

plusmn(119863 minus

119861lowast

119860dagger

119861)) Thus 119894plusmn(119871) = 119894

plusmn(119860) if and only if R(119861) sube

R(119860) and 119894plusmn(119863 minus 119861

lowast

119860dagger

119861) = 0(c)

119894plusmn(119866) = [

[

119875 119872 0

119872lowast

0 119873lowast

0 119873 0

]

]

minus 119903 (119873) (23)

Lemma 3 (see [23]) Let 119860 isin C119898times119899 119861 isin C119898times119896 and 119862 isin C119897times119899Then they satisfy the following rank equalities

(a) 119903[119860 119861] = 119903(119860) + 119903(119864119860119861) = 119903(119861) + 119903(119864

119861119860)

(b) 119903 [ 119860119862] = 119903(119860) + 119903(119862119865

119860) = 119903(119862) + 119903(119860119865

119862)

(c) 119903 [ 119860 119861

119862 0] = 119903(119861) + 119903(119862) + 119903(119864

119861119860119865

119862)

(d) 119903[119861 119860119865119862] = 119903 [

119861 119860

0 119862] minus 119903(119862)

(e) 119903 [ 119862

119864119861119860] = 119903 [

119862 0

119860 119861] minus 119903(119861)

(f) 119903 [ 119860 119861119865119863

119864119864119862 0] = 119903 [

119860 119861 0

119862 0 119864

0 119863 0

] minus 119903(119863) minus 119903(119864)

Lemma 4 (see [15]) Let119860 isin C119898times119898

ℎ 119861 isin C119898times119899119862 isin C119899times119899

ℎ119876 isin

C119898times119899 and 119875 isin C119901times119899 be given and 119879 isin C119898times119898 be nonsingularThen one has the following

(1) 119894plusmn(119879119860119879

lowast

) = 119894plusmn(119860)

(2) 119894plusmn[119860 0

0 119862] = 119894

plusmn(119860) + 119894

plusmn(119862)

(3) 119894plusmn[

0 119876

119876lowast

0] = 119903(119876)

(4) 119894plusmn[

119860 119861119871119875

119871119875119861lowast

0] + 119903(119875) = 119894

plusmn[

119860 119861 0

119861lowast

0 119875lowast

0 119875 0

]

Lemma 5 (see [22 Lemma 14]) Let 119878 be a set consisting of(square) matrices over C119898times119898 and let 119867 be a set consisting of(square) matrices overC119898times119898

ℎ ThenThen one has the following

(a) 119878 has a nonsingularmatrix if and only ifmax119883isin119878119903(119883) =

119898(b) any 119883 isin 119878 is nonsingular if and only if min

119883isin119878119903(119883) =

119898(c) 0 isin 119878 if and only ifmin

119883isin119878119903(119883) = 0

(d) 119878 = 0 if and only ifmax119883isin119878119903(119883) = 0

(e) 119867 has a matrix 119883 gt 0 (119883 lt 0) if and only ifmax

119883isin119867119894+(119883) = 119898(max

119883isin119867119894minus(119883) = 119898)

(f) any 119883 isin 119867 satisfies 119883 gt 0 (119883 lt 0) if and only ifmin

119883isin119867119894+(119883) = 119898 (min

119883isin119867119894minus(119883) = 119898)

(g) 119867 has a matrix 119883 ge 0 (119883 le 0) if and only ifmin

119883isin119867119894minus(119883) = 0 (min

119883isin119867119894+(119883) = 0)

(h) any 119883 isin 119867 satisfies 119883 ge 0 (119883 le 0) if and only ifmax

119883isin119867119894minus(119883) = 0 (max

119883isin119867119894+(119883) = 0)

Lemma 6 (see [16]) Let 119901(119883 119884) = 119860minus119861119883minus (119861119883)lowast minus119862119884119863minus(119862119884119863)

lowast where119860119861119862 and119863 are givenwith appropriate sizesand denote that

1198721= [

[

119860 119861 119862

119861lowast

0 0

119862lowast

0 0

]

]

1198722= [

[

119860 119861 119863lowast

119861lowast

0 0

119863 0 0

]

]

1198723= [

119860 119861 119862 119863lowast

119861lowast

0 0 0]

1198724= [

[

119860 119861 119862 119863lowast

119861lowast

0 0 0

119862lowast

0 0 0

]

]

1198725= [

[

119860 119861 119862 119863lowast

119861lowast

0 0 0

119863 0 0 0

]

]

(24)

Then one has the following

(1) the maximal rank of 119901(119883 119884) is

max119883isinC119899times119898 119884

119903 [119901 (119883 119884)] = min 119898 119903 (1198721) 119903 (119872

2) 119903 (119872

3)

(25)

(2) the minimal rank of 119901(119883 119884) is

min119883isinC119899times119898 119884

119903 [119901 (119883 119884)]

= 2119903 (1198723) minus 2119903 (119861)

+max 119906++ 119906

minus V

++ V

minus 119906

++ V

minus 119906

minus+ V

+

(26)

(3) the maximal inertia of 119901(119883 119884) is

max119883isinC119899times119898119884

119894plusmn[119901 (119883 119884)] = min 119894

plusmn(119872

1) 119894

plusmn(119872

2) (27)

(4) the minimal inertias of 119901(119883 119884) is

min119883isinC119899times119898119884

119894plusmn[119901 (119883 119884)] = 119903 (119872

3) minus 119903 (119861)

+max 119894plusmn(119872

1) minus 119903 (119872

4)

119894plusmn(119872

2) minus 119903 (119872

5)

(28)

where

119906plusmn= 119894

plusmn(119872

1) minus 119903 (119872

4) V

plusmn= 119894

plusmn(119872

2) minus 119903 (119872

5) (29)

Now we present the main theorem of this section

4 Journal of Applied Mathematics

Theorem7 Let1198601isin C119898times119899119862

1isin C119898times119896119861

1isin C119896times119897119863

1isin C119899times119897

1198602isin C119905times119902 119862

2isin C119905times119901119860

3isin C

119901times119901

ℎ 119861

3isin C119901times119902 119862

3isin C119901times119899

and1198633isin C119901times119899 be given and suppose that the system of matrix

equations (13) and (11) is consistent respectively Denote the setof all solutions to (13) by 119878 and (11) by 119866 Put

1198641=

[[[[[

[

1198603

1198623119863

lowast

3119862

lowast

11198613119862

lowast

2

119862lowast

30 119860

lowast

10 0

1198621119863

3119860

10 0 0

119861lowast

30 0 0 119860

lowast

2

1198622

0 0 11986020

]]]]]

]

1198642= 119903

[[[[[

[

1198603

119863lowast

3119862

3119863

11198613119862

lowast

2

1198633

0 1198611

0 0

119863lowast

1119862

lowast

3119861lowast

10 0 0

119861lowast

30 0 0 119860

lowast

2

1198622

0 0 11986020

]]]]]

]

1198643=

[[[[[

[

1198603

1198613119862

lowast

3119863

lowast

3119862

lowast

2

119861lowast

30 0 0 119860

lowast

2

119863lowast

1119862

lowast

30 0 119861

lowast

10

1198621119863

30 119860

10 0

1198622

11986020 0 0

]]]]]

]

1198644=

[[[[[[[

[

1198603

1198613119862

3119863

lowast

3119862

lowast

2119863

lowast

3119862

lowast

1

119861lowast

30 0 0 119860

lowast

20

119862lowast

30 0 0 0 119860

lowast

1

0 0 0 119861lowast

10 0

1198621119863

30 119860

10 0 0

1198622

11986020 0 0 0

]]]]]]]

]

1198645=

[[[[[[[

[

1198603

1198613119862

3119863

lowast

3119862

lowast

2119862

3119863

1

119861lowast

30 0 0 119860

lowast

20

1198633

0 0 0 0 1198611

119863lowast

1119862

lowast

30 0 119860

lowast

10 0

0 0 11986010 0 0

1198622

11986020 0 0 0

]]]]]]]

]

(30)

Then one has the following(a) the maximal rank of (10) subject to (13) and (11) is

max119883isin119866119884isin119878

119903 [119891 (119883 119884)]

= min 119901 119903 (1198641) minus 2119903 (119860

1) minus 2119903 (119860

2)

119903 (1198642) minus 2119903 (119861

1) minus 2119903 (119860

2)

119903 (1198643) minus 2119903 (119860

2) minus 119903 (119860

1) minus 119903 (119861

1)

(31)

(b) the minimal rank of (10) subject to (13) and (11) is

min119883isin119866119884isin119878

119903 [119891 (119883 119884)]

= 2119903 (1198643) minus 2119903 [

1198613

1198602

]

+max 119903 (1198641) minus 2119903 (119864

4) 119903 (119864

2) minus 2119903 (119864

5)

119894+(119864

1) + 119894

minus(119864

2) minus 119903 (119864

4) minus 119903 (119864

5)

119894minus(119864

1) + 119894

+(119864

2) minus 119903 (119864

4) minus 119903 (119864

5)

(32)

(c) the maximal inertia of (10) subject to (13) and (11) is

max119883isin119866119884isin119878119909

119894plusmn[119891 (119883 119884)] = min 119894

plusmn(119864

1) minus 119903 (119860

1) minus 119903 (119860

2)

119894plusmn(119864

2) minus 119903 (119861

1) minus 119903 (119860

2)

(33)

(d) the minimal inertia of (10) subject to (13) and (11) is

min119883isin119866119884isin119878

119894plusmn[119891 (119883 119884)] = 119903 (119864

3) minus 119903 [

1198613

1198602

]

+max 119894plusmn(119864

1) minus 119903 (119864

4)

119894plusmn(119864

2) minus 119903 (119864

5)

(34)

Proof By Lemma 1 the general solutions to (13) and (11) canbe written as

119883 = 119860dagger

2119862

2+ 119871

1198602119882

119884 = 119860dagger

1119862

1+ 119871

1198601119863

1119861dagger

1+ 119871

1198601119885119877

1198611

(35)

where119882 and 119885 are arbitrary matrices with appropriate sizesPut

119876 = 1198613119871

1198602 119879 = 119862

3119871

1198601 119869 = 119877

1198611119863

3

119875 = 1198603minus 119861

3119860

dagger

2119862

2minus (119861

3119860

dagger

2119862

2)lowast

minus 1198623(119860

dagger

1119862

1+ 119871

1198601119863

1119861dagger

1)119863

3

minus (1198623(119860

dagger

1119862

1+ 119871

1198601119863

1119861dagger

1)119863

3)lowast

(36)

Substituting (36) into (10) yields

119891 (119883 119884) = 119875 minus 119876119882 minus (119876119882)lowast

minus 119879119885119869 minus (119879119885119869)lowast

(37)

Clearly 119875 is Hermitian It follows from Lemma 6 that

max119883isin119866119884isin119878

119903 [119891 (119883 119884)]

= max119882119885

119903 (119875 minus 119876119882 minus (119876119882)lowast

minus 119879119885119869 minus (119879119885119869)lowast

)

= min 119898 119903 (1198731) 119903 (119873

2) 119903 (119873

3)

(38)

min119883isin119866119884isin119878

119903 [119891 (119883 119884)]

= max119882119885

119903 (119875 minus 119876119882 minus (119876119882)lowast

minus 119879119885119869 minus (119879119885119869)lowast

)

= 2119903 (1198733) minus 2119903 (119876)

+max 119904++ 119904

minus 119905

++ 119905

minus 119904

++ 119905

minus 119904

minus+ 119905

+

(39)

Journal of Applied Mathematics 5

max119883isin119866119884isin119878

119894plusmn[119891 (119883 119884)]

= max119882119885

119903 (119875 minus 119876119882 minus (119876119882)lowast

minus 119879119885119869 minus (119879119885119869)lowast

)

= min 119894plusmn(119873

1) 119894

plusmn(119873

2)

(40)

min119883isin119866119884isin119878

119894plusmn[119891 (119883 119884)]

= max119882119885

119903 (119875 minus 119876119882 minus (119876119882)lowast

minus 119879119885119869 minus (119879119885119869)lowast

)

= 119903 (1198733) minus 119903 (119876) +max 119904

plusmn 119905

plusmn

(41)

where

1198731= [

[

119875 119876 119879

119876lowast

0 0

119879lowast

0 0

]

]

1198732= [

[

119875 119876 119869lowast

119876lowast

0 0

119869 0 0

]

]

1198733= [

119875 119876 119879 119869lowast

119876lowast

0 0 0]

1198734= [

[

119875 119876 119879 119869lowast

119876lowast

0 0 0

119879lowast

0 0 0

]

]

1198735= [

[

119875 119876 119879 119869lowast

119876lowast

0 0 0

119869 0 0 0

]

]

119904plusmn= 119894

plusmn(119873

1) minus 119903 (119873

4) 119905

plusmn= 119894

plusmn(119873

2) minus 119903 (119873

5)

(42)

Now we simplify the ranks and inertias of block matrices in(38)ndash(41)

By Lemma 4 blockGaussian elimination and noting that

119871lowast

119878= (119868 minus 119878

dagger

119878)lowast

= 119868 minus 119878lowast

(119878lowast

)dagger

= 119877119878lowast (43)

we have the following

119903 (1198731) = 119903[

[

119875 119876 119879

119876lowast

0 0

119879lowast

0 0

]

]

= 119903

[[[[[

[

1198603

1198623119863

lowast

3119862

lowast

11198613119862

lowast

2

119862lowast

30 119860

lowast

10 0

1198621119863

3119860

10 0 0

119861lowast

30 0 0 119860

lowast

2

1198622

0 0 11986020

]]]]]

]

minus 2119903 (1198601) minus 2119903 (119860

2)

(44)

By 11986211198611= 119860

1119863

1 we obtain

119903 (1198732) = 119903

[[

[

119875 119876 119869lowast

119876lowast

0 0

119869 0 0

]]

]

= 119903

[[[[[[[[

[

1198603

119863lowast

3119862

3119863

11198613119862

lowast

2

1198633

0 1198611

0 0

119863lowast

1119862

lowast

3119861lowast

10 0 0

119861lowast

30 0 0 119860

lowast

2

1198622

0 0 11986020

]]]]]]]]

]

minus 2119903 (1198611) minus 2119903 (119860

2)

119903 (1198733) = 119903 [

119875 119876 119879 119869lowast

119876lowast

0 0 0]

= 119903

[[[[[[[[

[

1198603

1198613119862

lowast

3119863

lowast

3119862

lowast

2

119861lowast

30 0 0 119860

lowast

2

119863lowast

1119862

lowast

30 0 119861

lowast

10

1198621119863

30 119860

10 0

1198622

11986020 0 0

]]]]]]]]

]

minus 119903 (1198611) minus 2119903 (119860

2) minus 119903 (119860

1)

119903 (1198734) = 119903

[[

[

119875 119876 119879 119869lowast

119876lowast

0 0 0

119879lowast

0 0 0

]]

]

= 119903

[[[[[[[

[

1198603

1198613119862

3119863

lowast

3119862

lowast

2119863

lowast

3119862

lowast

1

119861lowast

30 0 0 119860

lowast

20

119862lowast

30 0 0 0 119860

lowast

1

0 0 0 119861lowast

10 0

1198621119863

30 119860

10 0 0

1198622

11986020 0 0 0

]]]]]]]

]

minus 119903 (1198611) minus 2119903 (119860

2) minus 2119903 (119860

1)

119903 (1198735) = 119903[

[

119875 119876 119879 119869lowast

119876lowast

0 0 0

119879lowast

0 0 0

]

]

= 119903

[[[[[[[

[

1198603

1198613119862

3119863

lowast

3119862

lowast

2119862

3119863

1

119861lowast

30 0 0 119860

lowast

20

1198633

0 0 0 0 1198611

119863lowast

1119862

lowast

30 0 119860

lowast

10 0

0 0 11986010 0 0

1198622

11986020 0 0 0

]]]]]]]

]

minus 2119903 (1198611) minus 2119903 (119860

2) minus 119903 (119860

1)

(45)

By Lemma 2 we can get the following

119894plusmn(119873

1) = 119894

plusmn

[

[

119875 119876 119879

119876lowast

0 0

119879lowast

0 0

]

]

6 Journal of Applied Mathematics

= 119894plusmn

[[[[[

[

1198603

1198623119863

lowast

3119862

lowast

11198613119862

lowast

2

119862lowast

30 119860

lowast

10 0

1198621119863

3119860

10 0 0

119861lowast

30 0 0 119860

lowast

2

1198622

0 0 11986020

]]]]]

]

minus 119903 (1198601) minus 119903 (119860

2)

(46)

119894plusmn(119873

2) = 119894

plusmn

[

[

119875 119876 119869lowast

119876lowast

0 0

119869 0 0

]

]

= 119894plusmn

[[[[[

[

1198603

119863lowast

3119862

3119863

11198613119862

lowast

2

1198633

0 1198611

0 0

119863lowast

1119862

lowast

3119861lowast

10 0 0

119861lowast

30 0 0 119860

lowast

2

1198622

0 0 11986020

]]]]]

]

minus 119903 (1198611) minus 119903 (119860

2)

(47)

Substituting (44)-(47) into (38) and (41) yields (31)ndash(34)respectively

Corollary 8 Let 1198601 119862

1 119861

1 119863

1 119860

2 119862

2 119860

3 119861

3 119862

3 119863

3

and 119864119894 (119894 = 1 2 5) be as in Theorem 7 and suppose

that the system of matrix equations (13) and (11) is consistentrespectively Denote the set of all solutions to (13) by 119878 and (11)by 119866 Then one has the following

(a) there exist 119883 isin 119866 and 119884 isin 119878 such that 1198603minus 119861

3119883 minus

(1198613119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

gt 0 if and only if

119894+(119864

1) minus 119903 (119860

1) minus 119903 (119860

2) ge 119901

119894+(119864

2) minus 119903 (119861

1) minus 119903 (119860

2) ge 119901

(48)

(b) there exist 119883 isin 119866 and 119884 isin 119878 such that 1198603minus 119861

3119883 minus

(1198613119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

lt 0 if and only if

119894minus(119864

1) minus 119903 (119860

1) minus 119903 (119860

2) ge 119901

119894minus(119864

2) minus 119903 (119861

1) minus 119903 (119860

2) ge 119901

(49)

(c) there exist 119883 isin 119866 and 119884 isin 119878 such that 1198603minus 119861

3119883 minus

(1198613119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

ge 0 if and only if

119903 (1198643) minus 119903 [

1198613

1198602

] + 119894minus(119864

1) minus 119903 (119864

4) le 0

119903 (1198643) minus 119903 [

1198613

1198602

] + 119894minus(119864

2) minus 119903 (119864

5) le 0

(50)

(d) there exist 119883 isin 119866 and 119884 isin 119878 such that 1198603minus 119861

3119883 minus

(1198613119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

le 0 if and only if

119903 (1198643) minus 119903 [

1198613

1198602

] + 119894+(119864

1) minus 119903 (119864

4) le 0

119903 (1198643) minus 119903 [

1198613

1198602

] + 119894+(119864

2) minus 119903 (119864

5) le 0

(51)

(e) 1198603minus 119861

3119883 minus (119861

3119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

gt 0 for all119883 isin 119866 and 119884 isin 119878 if and only if

119903 (1198643) minus 119903 [

1198613

1198602

] + 119894+(119864

1) minus 119903 (119864

4) = 119901

119900119903 119903 (1198643) minus 119903 [

1198613

1198602

] + 119894+(119864

2) minus 119903 (119864

5) = 119901

(52)

(f) 1198603minus 119861

3119883 minus (119861

3119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

lt 0 for all119883 isin 119866 and 119884 isin 119878 if and only if

119903 (1198643) minus 119903 [

1198613

1198602

] + 119894minus(119864

1) minus 119903 (119864

4) = 119901

119900119903 119903 (1198643) minus 119903 [

1198613

1198602

] + 119894minus(119864

2) minus 119903 (119864

5) = 119901

(53)

(g) 1198603minus 119861

3119883 minus (119861

3119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

ge 0 for all119883 isin 119866 and 119884 isin 119878 if and only if

119894minus(119864

1) minus 119903 (119860

1) minus 119903 (119860

2) le 0

119900119903 119894minus(119864

2) minus 119903 (119861

1) minus 119903 (119860

2) le 0

(54)

(h) 1198603minus 119861

3119883 minus (119861

3119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

le 0 for all119883 isin 119866 and 119884 isin 119878 if and only if

119894+(119864

1) minus 119903 (119860

1) minus 119903 (119860

2) le 0

119900119903 119894+(119864

2) minus 119903 (119861

1) minus 119903 (119860

2) le 0

(55)

(i) there exist 119883 isin 119866 and 119884 isin 119878 such that 1198603minus 119861

3119883 minus

(1198613119883)

lowast

minus1198623119884119863

3minus(119862

3119884119863

3)lowast is nonsingular if and only

if

119903 (1198641) minus 2119903 (119860

1) minus 2119903 (119860

2) ge 119901

119903 (1198642) minus 2119903 (119861

1) minus 2119903 (119860

2) ge 119901

119903 (1198643) minus 2119903 (119860

2) minus 119903 (119860

1) minus 119903 (119861

1) ge 119901

(56)

3 Relations between the Hermitian Part ofthe Solutions to (13) and (11)

Now we consider the extremal ranks and inertias of thedifference between the Hermitian part of the solutions to (13)and (11)

Theorem 9 Let 1198601isin C119898times119901 119862

1isin C119898times119901 119861

1isin C119901times119897 119863

1isin

C119901times119897 1198602isin C119905times119901 and 119862

2isin C119905times119901 be given Suppose that

the system of matrix equations (13) and (11) is consistent

Journal of Applied Mathematics 7

respectively Denote the set of all solutions to (13) by 119878 and (11)by 119866 Put

1198671=

[[[[[

[

0 119868 119862lowast

1minus119868 119862

lowast

2

119868 0 119860lowast

10 0

1198621119860

10 0 0

minus119868 0 0 0 119860lowast

2

11986220 0 119860

20

]]]]]

]

1198672= 119903

[[[[[

[

0 119868 1198631minus119868 119862

lowast

2

119868 0 1198611

0 0

119863lowast

1119861lowast

10 0 0

minus119868 0 0 0 119860lowast

2

11986220 0 119860

20

]]]]]

]

1198673=

[[[[[

[

0 minus119868 119868 119868 119862lowast

2

minus119868 0 0 0 119860lowast

2

119863lowast

10 0 119861

lowast

10

1198621

0 11986010 0

1198622119860

20 0 0

]]]]]

]

1198674=

[[[[[[[

[

0 minus119868 119868 119868 119862lowast

2119862

lowast

1

minus119868 0 0 0 119860lowast

20

119868 0 0 0 0 119860lowast

1

0 0 0 119861lowast

10 0

11986210 119860

10 0 0

1198622119860

20 0 0 0

]]]]]]]

]

1198675=

[[[[[[[

[

0 minus119868 119868 119868 119862lowast

2119863

1

minus119868 0 0 0 119860lowast

20

119868 0 0 0 0 1198611

119863lowast

10 0 119860

lowast

10 0

0 0 11986010 0 0

1198622119860

20 0 0 0

]]]]]]]

]

(57)

Then one has the following

max119883isin119866119884isin119878

119903 [(119883 + 119883lowast

) minus (119884 + 119884lowast

)]

= min 119901 119903 (1198671) minus 2119903 (119860

1) minus 2119903 (119860

2) 119903 (119867

2) minus 2119903 (119861

1)

minus2119903 (1198602) 119903 (119867

3) minus 2119903 (119860

2) minus 119903 (119860

1) minus 119903 (119861

1)

min119883isin119866119884isin119878

119903 [(119883 + 119883lowast

) minus (119884 + 119884lowast

)]

= 2119903 (1198673) minus 2119901

+max 119903 (1198671) minus 2119903 (119867

4) 119903 (119867

2) minus 2119903 (119867

5)

119894+(119867

1) + 119894

minus(119867

2) minus 119903 (119867

4) minus 119903 (119867

5)

119894minus(119867

1) + 119894

+(119867

2) minus 119903 (119867

4) minus 119903 (119867

5)

max119883isin119866119884isin119878

119894plusmn[(119883 + 119883

lowast

) minus (119884 + 119884lowast

)]

= min 119894plusmn(119867

1) minus 119903 (119860

1) minus 119903 (119860

2)

119894plusmn(119867

2) minus 119903 (119861

1) minus 119903 (119860

2)

min119883isin119866119884isin119878

119894plusmn[(119883 + 119883

lowast

) minus (119884 + 119884lowast

)]

= 119903 (1198643) minus 119901 +max 119894

plusmn(119867

1) minus 119903 (119867

4) 119894

plusmn(119867

2) minus 119903 (119867

5)

(58)

Proof By letting 1198603= 0 119861

3= minus119868 119862

3= 119868 and 119863

3= 119868 in

Theorem 7 we can get the results

Corollary 10 Let 1198601 119862

1 119861

1 119863

1 119860

2 119862

2 and 119867

119894 (119894 =

1 2 5) be as in Theorem 9 and suppose that the system ofmatrix equations (13) and (11) is consistent respectively Denotethe set of all solutions to (13) by 119878 and (11) by 119866 Then one hasthe following

(a) there exist 119883 isin 119866 and 119884 isin 119878 such that (119883 + 119883lowast

) gt

(119884 + 119884lowast

) if and only if

119894+(119867

1) minus 119903 (119860

1) minus 119903 (119860

2) ge 119901

119894+(119867

2) minus 119903 (119861

1) minus 119903 (119860

2) ge 119901

(59)

(b) there exist 119883 isin 119866 and 119884 isin 119878 such that (119883 + 119883lowast

) lt

(119884 + 119884lowast

) if and only if

119894minus(119867

1) minus 119903 (119860

1) minus 119903 (119860

2) ge 119901

119894minus(119867

2) minus 119903 (119861

1) minus 119903 (119860

2) ge 119901

(60)

(c) there exist 119883 isin 119866 and 119884 isin 119878 such that (119883 + 119883lowast

) ge

(119884 + 119884lowast

) if and only if

119903 (1198673) minus 119901 + 119894

minus(119867

1) minus 119903 (119867

4) le 0

119903 (1198673) minus 119901 + 119894

minus(119867

1) minus 119903 (119867

4) le 0

(61)

(d) there exist 119883 isin 119866 and 119884 isin 119878 such that (119883 + 119883lowast

) le

(119884 + 119884lowast

) if and only if

119903 (1198673) minus 119901 + 119894

+(119867

1) minus 119903 (119867

4) le 0

119903 (1198673) minus 119901 + 119894

+(119867

2) minus 119903 (119867

5) le 0

(62)

(e) (119883 + 119883lowast

) gt (119884 + 119884lowast

) for all 119883 isin 119866 and 119884 isin 119878 if andonly if

119903 (1198673) minus 119901 + 119894

+(119867

1) minus 119903 (119867

4) = 119901

119900119903 119903 (1198673) minus 119901 + 119894

+(119867

2) minus 119903 (119867

5) = 119901

(63)

(f) (119883 + 119883lowast

) lt (119884 + 119884lowast

) for all 119883 isin 119866 and 119884 isin 119878 if andonly if

119903 (1198673) minus 119901 + 119894

minus(119867

1) minus 119903 (119867

4) = 119901

119900119903 119903 (1198673) minus 119901 + 119894

minus(119867

2) minus 119903 (119867

5) = 119901

(64)

(g) (119883 + 119883lowast

) ge (119884 + 119884lowast

) for all 119883 isin 119866 and 119884 isin 119878 if andonly if

119894minus(119867

1) minus 119903 (119860

1) minus 119903 (119860

2) le 0

119900119903 119894minus(119867

2) minus 119903 (119861

1) minus 119903 (119860

2) le 0

(65)

8 Journal of Applied Mathematics

(h) (119883 + 119883lowast

) le (119884 + 119884lowast

) for all 119883 isin 119866 and 119884 isin 119878 if andonly if

119894+(119867

1) minus 119903 (119860

1) minus 119903 (119860

2) le 0

119900119903 119894+(119867

2) minus 119903 (119861

1) minus 119903 (119860

2) le 0

(66)

(i) there exist119883 isin 119866 and 119884 isin 119878 such that (119883 +119883lowast

) minus (119884 +

119884lowast

) is nonsingular if and only if119903 (119867

1) minus 2119903 (119860

1) minus 2119903 (119860

2) ge 119901

119903 (1198672) minus 2119903 (119861

1) minus 2119903 (119860

2) ge 119901

119903 (1198673) minus 2119903 (119860

2) minus 119903 (119860

1) minus 119903 (119861

1) ge 119901

(67)

4 The Solvability Conditions and the GeneralSolution to System (6)

We now turn our attention to (6) We in this section useTheorem 9 to give some necessary and sufficient conditionsfor the existence of a solution to (6) and present an expressionof the general solution to (6) We begin with a lemma whichis used in the latter part of this section

Lemma 11 (see [14]) Let 1198601isin C119898times1198991 119861

1isin C119898times1198992 119862

1isin

C119902times119898 and 1198641isin C119898times119898

ℎbe given Let 119860 = 119877

11986011198611 119861 = 119862

1119877

1198601

119864 = 1198771198601119864

1119877

1198601119872 = 119877

119860119861lowast119873 = 119860

lowast

119871119861 and 119878 = 119861lowast

119871119872 Then

the following statements are equivalent(1) equation (5) is consistent(2)119877

119872119877

119860119864 = 0 119877

119860119864119877

119860= 0 119871

119861119864119871

119861= 0 (68)

(3)

119903 [119864

11198611119862

lowast

1119860

1

119860lowast

10 0 0

] = 119903 [1198611119862

lowast

1119860

1] + 119903 (119860

1)

119903 [

[

11986411198611119860

1

119860lowast

10 0

119861lowast

10 0

]

]

= 2119903 [1198611119860

1]

119903 [

[

1198641119862

lowast

1119860

1

119860lowast

10 0

11986210 0

]

]

= 2119903 [119862lowast

1119860

1]

(69)

In this case the general solution of (5) can be expressed as

119884 =1

2[119860

dagger

119864119861dagger

minus 119860dagger

119861lowast

119872dagger

119864119861dagger

minus 119860dagger

119878(119861dagger

)lowast

119864119873dagger

119860lowast

119861dagger

+119860dagger

119864(119872dagger

)lowast

+ (119873dagger

)lowast

119864119861dagger

119878dagger

119878] + 1198711198601198811+ 119881

2119877

119861

+ 1198801119871

119878119871

119872+ 119877

119873119880

lowast

2119871

119872minus 119860

dagger

1198781198802119877

119873119860

lowast

119861dagger

119883 = 119860dagger

1[119864

1minus 119861

1119884119862

1minus (119861

1119884119862

1)lowast

]

minus1

2119860

dagger

1[119864

1minus 119861

1119884119862

1minus (119861

1119884119862

1)lowast

] 1198601119860

dagger

1

minus 119860dagger

1119882

1119860

lowast

1+119882

lowast

1119860

1119860

dagger

1+ 119871

1198601119882

2

(70)

where 1198801 119880

2 119881

1 119881

2119882

1 and119882

2are arbitrary matrices over

C with appropriate sizes

Now we give the main theorem of this section

Theorem 12 Let 119860119894 119862

119894 (119894 = 1 2 3) 119861

119895 and 119863

119895 (119895 = 1 3) be

given Set

119860 = 1198613119871

1198602 119861 = 119862

3119871

1198601

119862 = 1198771198611119863

3 119865 = 119877

119860119861

119866 = 119862119877119860 119872 = 119877

119865119866

lowast

119873 = 119865lowast

119871119866 119878 = 119866

lowast

119871119872

(71)

119863 = 1198603minus 119861

3119860

dagger

2119862

2minus (119861

3119860

dagger

2119862

2)lowast

minus 1198623(119860

dagger

1119862

1+ 119871

1198601119863

1119861dagger

1)119863

3

minus 119863lowast

3(119860

dagger

1119862

1+ 119871

1198601119863

1119861dagger

1)lowast

119862lowast

3

(72)

119864 = 119877119860119863119877

119860 (73)

Then the following statements are equivalent

(1) system (6) is consistent

(2) the equalities in (14) and (17) hold and

119877119872119877

119865119864 = 0 119877

119865119864119877

119865= 0 119871

119866119864119871

119866= 0 (74)

(3) the equalities in (15) and (18) hold and

119903

[[[[[

[

1198603

1198623119863

lowast

31198613119862

lowast

2

119861lowast

30 0 0 119860

lowast

2

1198621119863

3119860

10 0 0

119863lowast

1119862

lowast

30 119861

lowast

10 0

1198622

0 0 11986020

]]]]]

]

= 119903

[[[

[

1198623119863

lowast

31198613

11986010 0

0 119861lowast

10

0 0 1198602

]]]

]

+ 119903 [119860

2

1198613

]

119903

[[[[[

[

1198603

11986231198613119863

lowast

3119862

lowast

1119862

lowast

2

119862lowast

30 0 119860

lowast

10

119861lowast

30 0 0 119860

lowast

2

1198621119863

3119860

10 0 0

1198622

0 1198602

0 0

]]]]]

]

= 2119903[

[

11986231198613

11986010

0 1198602

]

]

119903

[[[[[

[

1198603

119863lowast

31198613119862

3119863

1119862

lowast

2

1198633

0 0 1198611

0

119861lowast

30 0 0 119860

lowast

2

119863lowast

1119862

lowast

3119861lowast

10 0 0

1198622

0 1198602

0 0

]]]]]

]

= 2119903[

[

119863lowast

31198613

119861lowast

10

0 1198602

]

]

(75)

In this case the general solution of system (6) can be expressedas

119883 = 119860dagger

2119862

2+ 119871

1198602119880

119884 = 119860dagger

1119862

1+ 119871

1198601119863

1119861dagger

1+ 119871

1198601119881119877

1198611

(76)

Journal of Applied Mathematics 9

where

119881 =1

2[119865

dagger

119864119866dagger

minus 119865dagger

119866lowast

119872dagger

119864119866dagger

minus 119865dagger

119878(119866dagger

)lowast

119864119873dagger

119865lowast

119866dagger

+119865dagger

119864(119872dagger

)lowast

+ (119873dagger

)lowast

119864119866dagger

119878dagger

119878] + 1198711198651198811

+ 1198812119877

119866+ 119880

1119871

119878119871

119872+ 119877

119873119880

lowast

2119871

119872minus 119865

dagger

1198781198802119877

119873119865

lowast

119866dagger

119880 = 119860dagger

[119863 minus 119861119881119862 minus (119861119881119862)lowast

]

minus1

2119860

dagger

[119863 minus 119861119881119862 minus (119861119881119862)lowast

] 119860119860dagger

minus 119860dagger

1198821119860

lowast

+119882lowast

1119860119860

dagger

+ 119871119860119882

2

(77)

where 1198801 119880

2 119881

1 119881

2119882

1 and119882

2are arbitrary matrices over

C with appropriate sizes

Proof (2)hArr (3) Applying Lemma 3 and Lemma 11 gives

119877119872119877

119865119864 = 0 lArrrArr 119903 (119877

119872119877

119865119864)

= 0 lArrrArr 119903[119863 119861 119862

lowast

119860

119860lowast

0 0 0]

= 119903 [119861 119862lowast

119860] + 119903 (119860)

lArrrArr 119903

[[[[[

[

119863 1198623119863

lowast

31198613

0

119861lowast

30 0 0 119860

lowast

2

0 11986010 0 0

0 0 119861lowast

10 0

0 0 0 11986020

]]]]]

]

= 119903

[[[

[

1198623119863

lowast

31198613

11986010 0

0 119861lowast

10

0 0 1198602

]]]

]

+ 119903 [119860

2

1198613

]

lArrrArr 119903

[[[[[

[

1198603

1198623119863

lowast

31198613119862

lowast

2

119861lowast

30 0 0 119860

lowast

2

1198621119863

3119860

10 0 0

119863lowast

1119862

lowast

30 119861

lowast

10 0

1198622

0 0 11986020

]]]]]

]

= 119903

[[[

[

1198623119863

lowast

31198613

11986010 0

0 119861lowast

10

0 0 1198602

]]]

]

+ 119903 [119860

2

1198613

]

(78)

By a similar approach we can obtain that

119877119865119864119877

119865= 0 lArrrArr 119903

[[[[[

[

1198603

11986231198613119863

lowast

3119862

lowast

1119862

lowast

2

119862lowast

30 0 119860

lowast

10

119861lowast

30 0 0 119860

lowast

2

1198621119863

3119860

10 0 0

1198622

0 1198602

0 0

]]]]]

]

= 2119903[

[

11986231198613

11986010

0 1198602

]

]

119871119866119864119871

119866= 0 lArrrArr 119903

[[[[[[[[

[

1198603

119863lowast

31198613119862

3119863

1119862

lowast

2

1198633

0 0 1198611

0

119861lowast

30 0 0 119860

lowast

2

119863lowast

1119862

lowast

3119861lowast

10 0 0

1198622

0 1198602

0 0

]]]]]]]]

]

= 2119903[[

[

119863lowast

31198613

119861lowast

10

0 1198602

]]

]

(79)

(1)hArr (2) We separate the four equations in system (6) intothree groups

1198601119884 = 119862

1 119884119861

1= 119863

1 (80)

1198602119883 = 119862

2 (81)

1198613119883 + (119861

3119883)

lowast

+ 1198623119884119863

3+ (119862

3119884119863

3)lowast

= 1198603 (82)

By Lemma 1 we obtain that system (80) is solvable if andonly if (14) (81) is consistent if and only if (17) The generalsolutions to system (80) and (81) can be expressed as (16) and(19) respectively Substituting (16) and (19) into (82) yields

119860119880 + (119860119880)lowast

+ 119861119881119862 + (119861119881119862)lowast

= 119863 (83)

Hence the system (5) is consistent if and only if (80) (81) and(83) are consistent respectively It follows fromLemma 11 that(83) is solvable if and only if

119877119872119877

119865119864 = 0 119877

119865119864119877

119865= 0 119871

119866119864119871

119866= 0 (84)

We know by Lemma 11 that the general solution of (83) canbe expressed as (77)

InTheorem 12 let 1198601and119863

1vanishThen we can obtain

the general solution to the following system

1198602119883 = 119862

2 119884119861

1= 119863

1

1198613119883 + (119861

3119883)

lowast

+ 1198623119884119863

3+ (119862

3119884119863

3)lowast

= 1198603

(85)

Corollary 13 Let 1198602 119862

2 119861

1 119863

1 119861

3 119862

3 119863

3 and 119860

3= 119860

lowast

3

be given Set

119860 = 1198613119871

1198602 119862 = 119877

1198611119863

3

119865 = 119877119860119862

3 119866 = 119862119877

119860

119872 = 119877119865119866

lowast

119873 = 119865lowast

119871119866

119878 = 119866lowast

119871119872

119863 = 1198603minus 119861

3119860

dagger

2119862

2minus (119861

3119860

dagger

2119862

2)lowast

minus 1198623119863

1119861dagger

1119863

3minus (119862

3119863

1119861dagger

1119863

3)lowast

119864 = 119877119860119863119877

119860

(86)

10 Journal of Applied Mathematics

Then the following statements are equivalent

(1) system (85) is consistent(2)

1198771198602119862

2= 0 119863

1119871

1198611= 0 119877

119872119877

119865119864 = 0 (87)

119877119865119864119877

119865= 0 119871

119866119864119871

119866= 0 (88)

(3)

119903 [1198602119862

2] = 119903 (119860

2) [

1198631

1198611

] = 119903 (1198611)

119903

[[[

[

1198603

1198623119863

lowast

31198613119862

lowast

2

119861lowast

30 0 0 119860

lowast

2

119863lowast

1119862

lowast

30 119861

lowast

10 0

1198622

0 0 11986020

]]]

]

= 119903[

[

1198623119863

lowast

31198613

0 119861lowast

10

0 0 1198602

]

]

+ 119903 [119860

2

1198613

]

119903

[[[

[

1198603119862

31198613119862

lowast

2

119862lowast

30 0 0

119861lowast

30 0 119860

lowast

2

11986220 119860

20

]]]

]

= 2119903 [119862

31198613

0 1198602

]

119903

[[[[[

[

1198603

119863lowast

31198613119862

3119863

1119862

lowast

2

1198633

0 0 1198611

0

119861lowast

30 0 0 119860

lowast

2

119863lowast

1119862

lowast

3119861lowast

10 0 0

1198622

0 1198602

0 0

]]]]]

]

= 2119903[

[

119863lowast

31198613

119861lowast

10

0 1198602

]

]

(89)

In this case the general solution of system (6) can be expressedas

119883 = 119860dagger

2119862

2+ 119871

1198602119880

119884 = 1198631119861dagger

1+ 119881119877

1198611

(90)

where

119881 =1

2[119865

dagger

119864119866dagger

minus 119865dagger

119866lowast

119872dagger

119864119866dagger

minus 119865dagger

119878(119866dagger

)lowast

119864119873dagger

119865lowast

119866dagger

+119865dagger

119864(119872dagger

)lowast

+ (119873dagger

)lowast

119864119866dagger

119878dagger

119878] + 1198711198651198811

+ 1198812119877

119866+ 119880

1119871

119878119871

119872+ 119877

119873119880

lowast

2119871

119872minus 119865

dagger

1198781198802119877

119873119865

lowast

119866dagger

119880 = 119860dagger

[119863 minus 1198623119881119862 minus (119862

3119881119862)

lowast

]

minus1

2119860

dagger

[119863 minus 1198623119881119862 minus (119862

3119881119862)

lowast

] 119860119860dagger

minus 119860dagger

1198821119860

lowast

+119882lowast

1119860119860

dagger

+ 119871119860119882

2

(91)

where 1198801 119880

2 119881

1 119881

2119882

1 and119882

2are arbitrary matrices over

C with appropriate sizes

Acknowledgments

The authors would like to thank Dr Mohamed Dr Sivaku-mar and a referee very much for their valuable suggestions

and comments which resulted in a great improvement of theoriginal paper This research was supported by the Grantsfrom the National Natural Science Foundation of China(NSFC (11161008)) and Doctoral Program Fund of Ministryof Education of PRChina (20115201110002)

References

[1] H Zhang ldquoRank equalities for Moore-Penrose inverse andDrazin inverse over quaternionrdquo Annals of Functional Analysisvol 3 no 2 pp 115ndash127 2012

[2] L Bao Y Lin and YWei ldquoA new projectionmethod for solvinglarge Sylvester equationsrdquo Applied Numerical Mathematics vol57 no 5ndash7 pp 521ndash532 2007

[3] MDehghan andMHajarian ldquoOn the generalized reflexive andanti-reflexive solutions to a system of matrix equationsrdquo LinearAlgebra and Its Applications vol 437 no 11 pp 2793ndash2812 2012

[4] F O Farid M S Moslehian Q-W Wang and Z-C Wu ldquoOnthe Hermitian solutions to a system of adjointable operatorequationsrdquo Linear Algebra and Its Applications vol 437 no 7pp 1854ndash1891 2012

[5] Z H He and Q W Wang ldquoA real quaternion matrix equationwith with applicationsrdquo Linear Multilinear Algebra vol 61 pp725ndash740 2013

[6] Q-W Wang and Z-C Wu ldquoCommon Hermitian solutionsto some operator equations on Hilbert 119862119909-modulesrdquo LinearAlgebra and Its Applications vol 432 no 12 pp 3159ndash3171 2010

[7] Q-W Wang and C-Z Dong ldquoPositive solutions to a systemof adjointable operator equations over Hilbert 119862119909-modulesrdquoLinear Algebra and Its Applications vol 433 no 7 pp 1481ndash14892010

[8] Q-W Wang J W van der Woude and H-X Chang ldquoA systemof real quaternion matrix equations with applicationsrdquo LinearAlgebra and Its Applications vol 431 no 12 pp 2291ndash2303 2009

[9] Q W Wang H-S Zhang and G-J Song ldquoA new solvable con-dition for a pair of generalized Sylvester equationsrdquo ElectronicJournal of Linear Algebra vol 18 pp 289ndash301 2009

[10] H W Braden ldquoThe equations 119860119879

119883 plusmn 119883119879

119860 = 119861rdquo SIAM Journalon Matrix Analysis and Applications vol 20 no 2 pp 295ndash3021999

[11] D S Djordjevic ldquoExplicit solution of the operator equation119860

lowast

119883 + 119883plusmn

119860 = 119861rdquo Journal of Computational and Applied Math-ematics vol 200 no 2 pp 701ndash704 2007

[12] W Cao ldquoSolvability of a quaternion matrix equationrdquo AppliedMathematics vol 17 no 4 pp 490ndash498 2002

[13] Q Xu L Sheng and Y Gu ldquoThe solutions to some operatorequationsrdquo Linear Algebra and Its Applications vol 429 no 8-9pp 1997ndash2024 2008

[14] Q-W Wang and Z-H He ldquoSome matrix equations with appli-cationsrdquo Linear and Multilinear Algebra vol 60 no 11-12 pp1327ndash1353 2012

[15] Y Tian ldquoMaximization and minimization of the rank andinertia of the Hermitian matrix expression 119860 minus 119861119883 minus (119861119883)

119909rdquoLinear Algebra and Its Applications vol 434 no 10 pp 2109ndash2139 2011

[16] Z-H He and Q-WWang ldquoSolutions to optimization problemson ranks and inertias of a matrix function with applicationsrdquoAppliedMathematics andComputation vol 219 no 6 pp 2989ndash3001 2012

[17] Y Liu andY Tian ldquoMax-min problems on the ranks and inertiasof the matrix expressions 119860 minus 119861119883119862 plusmn (119861119883119862)

lowastrdquo Journal of

Journal of Applied Mathematics 11

Optimization Theory and Applications vol 148 no 3 pp 593ndash622 2011

[18] DChu Y SHung andH JWoerdeman ldquoInertia and rank cha-racterizations of some matrix expressionsrdquo SIAM Journal onMatrix Analysis and Applications vol 31 no 3 pp 1187ndash12262009

[19] Y Liu and Y Tian ldquoA simultaneous decomposition of a mat-rix triplet with applicationsrdquoNumerical Linear Algebra with Ap-plications vol 18 no 1 pp 69ndash85 2011

[20] X Zhang Q-W Wang and X Liu ldquoInertias and ranks of someHermitian matrix functions with applicationsrdquo Central Euro-pean Journal of Mathematics vol 10 no 1 pp 329ndash351 2012

[21] Q-W Wang ldquoThe general solution to a system of real quater-nion matrix equationsrdquo Computers amp Mathematics with Appli-cations vol 49 no 5-6 pp 665ndash675 2005

[22] Y Tian ldquoEqualities and inequalities for inertias of Hermitianmatrices with applicationsrdquo Linear Algebra and Its Applicationsvol 433 no 1 pp 263ndash296 2010

[23] G Marsaglia and G P H Styan ldquoEqualities and inequalities forranks of matricesrdquo Linear and Multilinear Algebra vol 2 pp269ndash292 1974

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 230408 6 pageshttpdxdoiorg1011552013230408

Research ArticleLeft and Right Inverse Eigenpairs Problemfor 120581-Hermitian Matrices

Fan-Liang Li1 Xi-Yan Hu2 and Lei Zhang2

1 Institute of Mathematics and Physics School of Sciences Central South University of Forestry and TechnologyChangsha 410004 China

2 College of Mathematics and Econometrics Hunan University Changsha 410082 China

Correspondence should be addressed to Fan-Liang Li lfl302tomcom

Received 6 December 2012 Accepted 20 March 2013

Academic Editor Panayiotis J Psarrakos

Copyright copy 2013 Fan-Liang Li et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Left and right inverse eigenpairs problem for 120581-hermitian matrices and its optimal approximate problem are considered Based onthe special properties of 120581-hermitian matrices the equivalent problem is obtained Combining a new inner product of matricesthe necessary and sufficient conditions for the solvability of the problem and its general solutions are derived Furthermore theoptimal approximate solution and a calculation procedure to obtain the optimal approximate solution are provided

1 Introduction

Throughout this paper we use some notations as follows Let119862119899times119898 be the set of all 119899times119898 complex matrices119880119862

119899times119899119867119862119899times119899

119878119867119862119899times119899 denote the set of all 119899times119899 unitary matrices hermitian

matrices skew-hermitian matrices respectively Let 119860 119860119867and119860

+ be the conjugate conjugate transpose and theMoore-Penrose generalized inverse of 119860 respectively For 119860 119861 isin

119862119899times119898 ⟨119860 119861⟩ = re(tr(119861119867119860)) where re(tr(119861119867119860)) denotes the

real part of tr(119861119867119860) the inner product of matrices 119860 and 119861The induced matrix norm is called Frobenius norm That is119860 = ⟨119860 119860⟩

12

= (tr(119860119867119860))12

Left and right inverse eigenpairs problem is a specialinverse eigenvalue problem That is giving partial left andright eigenpairs (eigenvalue and corresponding eigenvector)(120582119894 119909119894) 119894 = 1 ℎ (120583

119895 119910119895) 119895 = 1 119897 a special matrix set

119878 finding a matrix 119860 isin 119878 such that

119860119909119894= 120582119894119909119894 119894 = 1 ℎ

119910119879

119895119860 = 120583

119895119910119879

119895 119895 = 1 119897

(1)

This problem which usually arises in perturbation analysisof matrix eigenvalues and in recursive matters has profoundapplication background [1ndash6] When the matrix set 119878 isdifferent it is easy to obtain different left and right inverse

eigenpairs problem For example we studied the left andright inverse eigenpairs problem of skew-centrosymmetricmatrices and generalized centrosymmetric matrices respec-tively [5 6] Based on the special properties of left and righteigenpairs of these matrices we derived the solvability condi-tions of the problem and its general solutions In this papercombining the special properties of 120581-hermitianmatrices anda new inner product ofmatrices we first obtain the equivalentproblem then derive the necessary and sufficient conditionsfor the solvability of the problem and its general solutions

Hill and Waters [7] introduced the following matrices

Definition 1 Let 120581 be a fixed product of disjoint transposi-tions and let119870 be the associated permutation matrix that is119870 = 119870

119867

= 119870 1198702 = 119868119899 a matrix 119860 isin 119862

119899times119899 is said to be 120581-hermitian matrices (skew 120581-hermitian matrices) if and onlyif 119886119894119895= 119886119896(119895)119896(119894)

(119886119894119895= minus119886119896(119895)119896(119894)

) 119894 119895 = 1 119899 We denote theset of 120581-hermitian matrices (skew 120581-hermitian matrices) by119870119867119862119899times119899

(119878119870119867119862119899times119899

)

FromDefinition 1 it is easy to see that hermitianmatricesand perhermitian matrices are special cases of 120581-hermitianmatrices with 119896(119894) = 119894 and 119896(119894) = 119899 minus 119894 + 1 respectivelyHermitian matrices and perhermitian matrices which areone of twelve symmetry patterns of matrices [8] are appliedin engineering statistics and so on [9 10]

2 Journal of Applied Mathematics

From Definition 1 it is also easy to prove the followingconclusions

(1) 119860 isin 119870119867119862119899times119899 if and only if 119860 = 119870119860

119867

119870(2) 119860 isin 119878119870119867119862

119899times119899 if and only if 119860 = minus119870119860119867

119870(3) If 119870 is a fixed permutation matrix then 119870119867119862

119899times119899 and119878119870119867119862

119899times119899 are the closed linear subspaces of 119862119899times119899 andsatisfy

119862119899times119899

= 119870119867119862119899times119899

⨁119878119870119867119862119899times119899

(2)

The notation 1198811oplus 1198812stands for the orthogonal direct sum of

linear subspace 1198811and 119881

2

(4) 119860 isin 119870119867119862119899times119899 if and only if there is a matrix isin

119867119862119899times119899 such that = 119870119860

(5) 119860 isin 119878119870119867119862119899times119899 if and only if there is a matrix isin

119878119867119862119899times119899 such that = 119870119860

Proof (1) From Definition 1 if 119860 = (119886119894119895) isin 119870119867119862

119899times119899 then119886119894119895= 119886119896(119895)119896(119894)

this implies119860 = 119870119860119867

119870 for119870119860119867

119870 = (119886119896(119895)119896(119894)

)(2)With the samemethod we can prove (2) So the proof

is omitted

(3) (a) For any 119860 isin 119862119899times119899 there exist 119860

1isin 119870119867119862

119899times1198991198602isin 119878119870119867119862

119899times119899 such that

119860 = 1198601+ 1198602 (3)

where 1198601= (12)(A + 119870119860

119867

119870) 1198602= (12)(119860 minus

119870119860119867

119870)(b) If there exist another 119860

1isin 119870119867119862

119899times119899 1198602

isin

119878119870119867119862119899times119899 such that

119860 = 1198601+ 1198602 (4)

(3)-(4) yields

1198601minus 1198601= minus (119860

2minus 1198602) (5)

Multiplying (5) on the left and on the right by119870 respectively and according to (1) and (2) weobtain

1198601minus 1198601= 1198602minus 1198602 (6)

Combining (5) and (6) gives1198601= 11986011198602= 1198602

(c) For any1198601isin 119870119867119862

119899times1198991198602isin 119878119870119867119862

119899times119899 we have

⟨1198601 1198602⟩ = re (tr (119860119867

21198601)) = re (tr (119870119860

119867

21198701198701198601119870))

= re (tr (minus11986011986721198601)) = minus ⟨119860

1 1198602⟩

(7)

This implies ⟨1198601 1198602⟩ = 0 Combining (a) (b)

and (c) gives (3)

(4) Let = 119870119860 if 119860 isin 119870119867119862119899times119899 then

119867

= isin 119867119862119899times119899

If 119867 = isin 119867119862119899timesn then 119860 = 119870 and 119870119860

119867

119870 = 119870119867

119870119870 =

119870 = 119860 isin 119870119867119862119899times119899

(5)With the samemethod we can prove (5) So the proofis omitted

In this paper we suppose that 119870 is a fixed permutationmatrix and assume (120582

119894 119909119894) 119894 = 1 ℎ be right eigenpairs

of 119860 (120583119895 119910119895) 119895 = 1 119897 be left eigenpairs of 119860 If we let

119883 = (1199091 119909

ℎ) isin 119862

119899timesℎ Λ = diag (1205821 120582

ℎ) isin 119862

ℎtimesℎ119884 = (119910

1 119910

119897) isin 119862119899times119897 Γ = diag(120583

1 120583

119897) isin 119862119897times119897 then the

problems studied in this paper can be described as follows

Problem 2 Giving 119883 isin 119862119899timesℎ Λ = diag(120582

1 120582

ℎ) isin 119862

ℎtimesℎ119884 isin 119862

119899times119897 Γ = diag(1205831 120583

119897) isin 119862119897times119897 find 119860 isin 119870119867119862

119899times119899 suchthat

119860119883 = 119883Λ

119884119879

119860 = Γ119884119879

(8)

Problem 3 Giving 119861 isin 119862119899times119899 find isin 119878

119864such that

10038171003817100381710038171003817119861 minus

10038171003817100381710038171003817= minforall119860isin119878119864

119861 minus 119860 (9)

where 119878119864is the solution set of Problem 2

This paper is organized as follows In Section 2 we firstobtain the equivalent problemwith the properties of119870119867119862

119899times119899

and then derive the solvability conditions of Problem 2 and itsgeneral solutionrsquos expression In Section 3 we first attest theexistence and uniqueness theorem of Problem 3 then presentthe unique approximation solution Finally we provide acalculation procedure to compute the unique approximationsolution and numerical experiment to illustrate the resultsobtained in this paper correction

2 Solvability Conditions of Problem 2

We first discuss the properties of119870119867119862119899times119899

Lemma 4 Denoting 119872 = 119870119864119870119866119864 and 119864 isin 119867119862119899times119899 one has

the following conclusions(1) If 119866 isin 119870119867119862

119899times119899 then119872 isin 119870119867119862119899times119899

(2) If 119866 isin 119878119870119867119862119899times119899 then119872 isin 119878119870119867119862

119899times119899(3) If 119866 = 119866

1+ 1198662 where 119866

1isin 119870119867119862

119899times119899 1198662isin 119878119870119867119862

119899times119899then 119872 isin 119870119867119862

119899times119899 if and only if 1198701198641198701198662119864 = 0 In

addition one has119872 = 1198701198641198701198661119864

Proof (1) 119870119872119867

119870 = 119870119864119866119867

119870119864119870119870 = 119870119864(119870119866119870)119870119864 =

119870119864119870119866119864 = 119872Hence we have119872 isin 119870119867119862

119899times119899(2) 119870119872

119867

119870 = 119870119864119866119867

119870119864119870119870 = 119870119864(minus119870119866119870)119870119864 =

minus119870119864119870119866119864 = minus119872Hence we have119872 isin 119878119870119867119862

119899times119899(3) 119872 = 119870119864119870(119866

1+ 1198662)119864 = 119870119864119870119866

1119864 + 119870119864119870119866

2119864 we

have 1198701198641198701198661119864 isin 119870119867119862

119899times119899 1198701198641198701198662119864 isin 119878119870119867119862

119899times119899 from (1)and (2) If 119872 isin 119870119867119862

119899times119899 then 119872 minus 1198701198641198701198661119864 isin 119870119867119862

119899times119899while119872minus119870119864119870119866

1119864 = 119870119864119870119866

2119864 isin 119878119870119867119862

119899times119899Therefore fromthe conclusion (3) of Definition 1 we have119870119864119870119866

2119864 = 0 that

is119872 = 1198701198641198701198661119864 On the contrary if119870119864119870119866

2119864 = 0 it is clear

that119872 = 1198701198641198701198661119864 isin 119870119867119862

119899times119899 The proof is completed

Lemma 5 Let 119860 isin 119870119867119862119899times119899 if (120582 119909) is a right eigenpair of 119860

then (120582 119870119909) is a left eigenpair of 119860

Journal of Applied Mathematics 3

Proof If (120582 119909) is a right eigenpair of 119860 then we have

119860119909 = 120582119909 (10)

From the conclusion (1) of Definition 1 it follows that

119870119860119867

119870119909 = 120582119909 (11)

This implies

(119870119909)119879

119860 = 120582(119870119909)119879

(12)

So (120582 119870119909) is a left eigenpair of 119860

From Lemma 5 without loss of the generality we mayassume that Problem 2 is as follows

119883 isin 119862119899timesℎ

Λ = diag (1205821 120582

ℎ) isin 119862ℎtimesℎ

119884 = 119870119883 isin 119862119899timesℎ

Γ = Λ isin 119862ℎtimesℎ

(13)

Combining (13) and the conclusion (4) of Definition 1 itis easy to derive the following lemma

Lemma 6 If 119883 Λ 119884 Γ are given by (13) then Problem 2 isequivalent to the following problem If 119883 Λ 119884 Γ are given by(13) find 119870119860 isin 119867119862

119899times119899 such that

119870119860119883 = 119870119883Λ (14)

Lemma 7 (see [11]) If giving119883 isin 119862119899timesℎ119861 isin 119862

119899timesℎ thenmatrixequation 119883 = 119861 has solution isin 119867119862

119899times119899 if and only if

119861 = 119861119883+

119883 119861119867

119883 = 119883119867

119861 (15)

Moreover the general solution can be expressed as

= 119861119883+

+ (119861119883+

)119867

(119868119899minus 119883119883

+

)

+ (119868119899minus 119883119883

+

) (119868119899minus 119883119883

+

) forall isin 119867119862119899times119899

(16)

Theorem 8 If119883Λ119884 Γ are given by (13) then Problem 2 hasa solution in 119870119867119862

119899times119899 if and only if

119883119867

119870119883Λ = Λ119883119867

119870119883 119883Λ = 119883Λ119883+

119883 (17)

Moreover the general solution can be expressed as

119860 = 1198600+ 119870119864119870119866119864 forall119866 isin 119870119867119862

119899times119899

(18)

where

1198600= 119883Λ119883

+

+ 119870(119883Λ119883+

)119867

119870119864 119864 = 119868119899minus 119883119883

+

(19)

Proof Necessity If there is a matrix 119860 isin 119870119867119862119899times119899 such that

(119860119883 = 119883Λ 119884119879119860 = Γ119884119879

) then from Lemma 6 there exists amatrix 119870119860 isin 119867119862

119899times119899 such that 119870119860119883 = 119870119883Λ and accordingto Lemma 7 we have

119870119883Λ = 119870119883Λ119883+

119883 (119870119883Λ)119867

119883 = 119883119867

(119870119883Λ) (20)

It is easy to see that (20) is equivalent to (17)

Sufficiency If (17) holds then (20) holds Hence matrixequation119870119860119883 = 119870119883Λ has solution119870119860 isin 119867119862

119899times119899 Moreoverthe general solution can be expressed as follows

119870119860 = 119870119883Λ119883+

+ (119870119883Λ119883+

)119867

(119868119899minus 119883119883

+

)

+ (119868119899minus 119883119883

+

) (119868119899minus 119883119883

+

) forall isin 119867119862119899times119899

(21)

Let

1198600= 119883Λ119883

+

+ 119870(119883Λ119883+

)119867

119870119864 119864 = 119868119899minus 119883119883

+

(22)

This implies 119860 = 1198600+ 119870119864119864 Combining the definition of

119870 119864 and the first equation of (17) we have

119870119860119867

0119870 = 119870(119883Λ119883

+

)119867

119870 + 119883Λ119883+

minus 119870(119883119883+

)119867

119870(119883Λ119883+

)

= 119883Λ119883+

+ 119870(119883Λ119883+

)119867

119870 minus 119870(119883Λ119883+

)119879

119870119883119883+

= 1198600

(23)

Hence1198600isin 119870119867119862

119899times119899 Combining the definition of119870 119864 (13)and (17) we have

1198600119883 = 119883Λ119883

+

119883 + 119870(119883Λ119883+

)119867

119870(119868119899minus 119883119883

+

)119883 = 119883Λ

119884119879

1198600= 119883119867

119870119883Λ119883+

+ 119883119867

119870119870(119883Λ119883+

)119867

119870119864

= Λ119883119867

119870119883119883+

+ (119870119883Λ119883+

119883)119867

(119868119899minus 119883119883

+

)

= Λ119883119867

119870119883119883+

+ Λ119883119867

119870(119868119899minus 119883119883

+

)

= Λ119883119867

119870 = Γ119884119879

(24)

Therefore 1198600is a special solution of Problem 2 Combining

the conclusion (4) of Definition 1 Lemma 4 and 119864 = 119868119899minus

119883119883+

isin 119867119862119899times119899 it is easy to prove that 119860 = 119860

0+ 119870119864119870119866119864 isin

119870119867119862119899times119899 if and only if 119866 isin 119870119867119862

119899times119899 Hence the solution setof Problem 2 can be expressed as (18)

3 An Expression of the Solution of Problem 3

From (18) it is easy to prove that the solution set 119878119864of

Problem 2 is a nonempty closed convex set if Problem 2 hasa solution in 119870119867119862

119899times119899 We claim that for any given 119861 isin 119877119899times119899

there exists a unique optimal approximation for Problem 3

Theorem9 Giving119861 isin 119862119899times119899 if the conditions of119883119884Λ Γ are

the same as those in Theorem 8 then Problem 3 has a uniquesolution isin 119878

119864 Moreover can be expressed as

= 1198600+ 119870119864119870119861

1119864 (25)

where 1198600 119864 are given by (19) and 119861

1= (12)(119861 + 119870119861

119867

119870)

Proof Denoting1198641= 119868119899minus119864 it is easy to prove thatmatrices119864

and1198641are orthogonal projectionmatrices satisfying119864119864

1= 0

It is clear that matrices 119870119864119870 and 1198701198641119870 are also orthogonal

4 Journal of Applied Mathematics

projection matrices satisfying (119870119864119870)(1198701198641119870) = 0 According

to the conclusion (3) of Definition 1 for any 119861 isin 119862119899times119899 there

exists unique

1198611isin 119870119867119862

119899times119899

1198612isin 119878119870119867119862

119899times119899 (26)

such that

119861 = 1198611+ 1198612 ⟨119861

1 1198612⟩ = 0 (27)

where

1198611=

1

2(119861 + 119870119861

119867

119870) 1198612=

1

2(119861 minus 119870119861

119867

119870) (28)

CombiningTheorem 8 for any 119860 isin 119878119864 we have

119861 minus 1198602

=1003817100381710038171003817119861 minus 119860

0minus 119870119864119870119866119864

1003817100381710038171003817

2

=10038171003817100381710038171198611 + 119861

2minus 1198600minus 119870119864119870119866119864

1003817100381710038171003817

2

=10038171003817100381710038171198611 minus 119860

0minus 119870119864119870119866119864

1003817100381710038171003817

2

+10038171003817100381710038171198612

1003817100381710038171003817

2

=1003817100381710038171003817(1198611 minus 119860

0) (119864 + 119864

1) minus 119870119864119870119866119864

1003817100381710038171003817

2

+10038171003817100381710038171198612

1003817100381710038171003817

2

=1003817100381710038171003817(1198611 minus 119860

0) 119864 minus 119870119864119870119866119864

1003817100381710038171003817

2

+1003817100381710038171003817(1198611 minus 119860

0) 1198641

1003817100381710038171003817

2

+10038171003817100381710038171198612

1003817100381710038171003817

2

=1003817100381710038171003817119870 (119864 + 119864

1)119870 (119861

1minus 1198600) 119864 minus 119870119864119870119866119864

1003817100381710038171003817

2

+1003817100381710038171003817(1198611 minus 119860

0) 1198641

1003817100381710038171003817

2

+10038171003817100381710038171198612

1003817100381710038171003817

2

=1003817100381710038171003817119870119864119870 (119861

1minus 1198600) 119864 minus 119870119864119870119866119864

1003817100381710038171003817

2

+10038171003817100381710038171198701198641119870(1198611minus 1198600) 119864

1003817100381710038171003817

2

+1003817100381710038171003817(1198611 minus 119860

0)1198641

1003817100381710038171003817

2

+10038171003817100381710038171198612

1003817100381710038171003817

2

(29)

It is easy to prove that 1198701198641198701198600119864 = 0 according to the

definitions of 1198600 119864 So we have

119861 minus 1198602

=1003817100381710038171003817119870119864119870119861

1119864 minus 119870119864119870119866119864

1003817100381710038171003817

2

+10038171003817100381710038171198701198641119870(1198611minus 1198600) 119864

1003817100381710038171003817

2

+1003817100381710038171003817(1198611 minus 119860

0)1198641

1003817100381710038171003817

2

+10038171003817100381710038171198612

1003817100381710038171003817

2

(30)

Obviously min119860isin119878119864

119861 minus 119860 is equivalent to

min119866isin119870119867119862

119899times119899

10038171003817100381710038171198701198641198701198611119864 minus 119870119864119870119866119864

1003817100381710038171003817 (31)

Since 1198641198641

= 0 (119870119864119870)(1198701198641119870) = 0 it is clear that 119866 =

1198611+ 11987011986411198701198641 for any isin 119870119867119862

119899times119899 is a solution of (31)Substituting this result to (18) we can obtain (25)

Algorithm 10 (1) Input 119883 Λ 119884 Γ according to (13) (2)Compute 119883

119867

119870119883Λ Λ119883119867

119870119883 119883Λ119883+

119883 119883Λ if (17) holdsthen continue otherwise stop (3) Compute 119860

0according to

(19) and compute 1198611according to (28) (4) According to (25)

calculate

Example 11 (119899 = 8 ℎ = 119897 = 4)

119883 =

((((

(

05661 minus02014 minus 01422119894 01446 + 02138119894 0524

minus02627 + 01875119894 05336 minus02110 minus 04370119894 minus00897 + 03467119894

minus04132 + 02409119894 00226 minus 00271119894 minus01095 + 02115119894 minus03531 minus 00642119894

minus00306 + 02109119894 minus03887 minus 00425119894 02531 + 02542119894 00094 + 02991119894

00842 minus 01778119894 minus00004 minus 03733119894 03228 minus 01113119894 01669 + 01952119894

00139 minus 03757119894 minus02363 + 03856119894 02583 + 00721119894 01841 minus 02202119894

00460 + 03276119894 minus01114 + 00654119894 minus00521 minus 02556119894 minus02351 + 03002119894

00085 minus 01079119894 00974 + 03610119894 05060 minus02901 minus 00268119894

))))

)

Λ = (

minus03967 minus 04050119894 0 0 0

0 minus03967 + 04050119894 0 0

0 0 00001 0

0 0 0 minus00001119894

)

119870 =

((((

(

0 0 0 0 0 0 0 1

0 0 0 0 0 0 1 0

0 0 0 0 0 1 0 0

0 0 0 1 0 0 0 0

0 0 0 0 1 0 0 0

0 0 1 0 0 0 0 0

0 1 0 0 0 0 0 0

1 0 0 0 0 0 0 0

))))

)

119884 = 119870119883 Γ = Λ

(32)

Journal of Applied Mathematics 5

119861 =

From the first column to the fourth column

((((

(

minus05218 + 00406119894 02267 minus 00560119894 minus01202 + 00820119894 minus00072 minus 03362119894

03909 minus 03288119894 02823 minus 02064119894 minus00438 minus 00403119894 02707 + 00547119894

02162 minus 01144119894 minus04307 + 02474119894 minus00010 minus 00412119894 02164 minus 01314119894

minus01872 minus 00599119894 minus00061 + 04698119894 03605 minus 00247119894 04251 + 01869119894

minus01227 minus 00194119894 02477 minus 00606119894 03918 + 06340119894 01226 + 00636119894

minus00893 + 04335119894 00662 + 00199119894 minus00177 minus 01412119894 04047 + 02288119894

01040 minus 02015119894 01840 + 02276119894 02681 minus 03526119894 minus05252 + 01022119894

01808 + 02669119894 02264 + 03860119894 minus01791 + 01976119894 minus00961 minus 00117119894

(33)

From the fifth column to the eighth column

minus02638 minus 04952119894 minus00863 minus 01664119894 02687 + 01958119894 minus02544 minus 01099119894

minus02741 minus 01656119894 minus00227 + 02684119894 01846 + 02456119894 minus00298 + 05163119894

minus01495 minus 03205119894 01391 + 02434119894 01942 minus 05211119894 minus03052 minus 01468119894

minus02554 + 02690119894 minus04222 minus 01080119894 02232 + 00774119894 00965 minus 00421119894

03856 minus 00619119894 01217 minus 00270119894 01106 minus 03090119894 minus01122 + 02379119894

minus01130 + 00766119894 07102 minus 00901119894 01017 + 01397119894 minus00445 + 00038119894

01216 + 00076119894 02343 minus 01772119894 05242 minus 00089119894 minus00613 + 00258119894

minus00750 minus 03581119894 00125 + 00964119894 00779 minus 01074119894 06735 minus 00266119894

))))

)

(34)

It is easy to see that matrices 119883 Λ 119884 Γ satisfy (17) Hencethere exists the unique solution for Problem 3 Using the

software ldquoMATLABrdquo we obtain the unique solution ofProblem 3

From the first column to the fourth column

((((

(

minus01983 + 00491119894 01648 + 00032119894 00002 + 01065119894 01308 + 02690119894

01071 minus 02992119894 02106 + 02381119894 minus00533 minus 03856119894 minus00946 + 00488119894

01935 minus 01724119894 minus00855 minus 00370119894 00200 minus 00665119894 minus02155 minus 00636119894

00085 minus 02373119894 minus00843 minus 01920119894 00136 + 00382119894 minus00328 minus 00000119894

minus00529 + 01703119894 01948 minus 00719119894 01266 + 01752119894 00232 minus 02351119894

00855 + 01065119894 00325 minus 02068119894 02624 + 00000119894 00136 minus 00382119894

01283 minus 01463119894 minus00467 + 00000119894 00325 + 02067119894 minus00843 + 01920119894

02498 + 00000119894 01283 + 01463119894 00855 minus 01065119894 00086 + 02373119894

(35)

From the fifth column to the eighth column

02399 minus 01019119894 01928 minus 01488119894 minus03480 minus 02574119894 01017 minus 00000119894

minus01955 + 00644119894 02925 + 01872119894 03869 + 00000119894 minus03481 + 02574119894

00074 + 00339119894 minus03132 minus 00000119894 02926 minus 01872119894 01928 + 01488119894

00232 + 02351119894 minus02154 + 00636119894 minus00946 minus 00489119894 01309 minus 02691119894

minus00545 minus 00000119894 00074 minus 00339119894 minus01955 minus 00643119894 02399 + 01019119894

01266 minus 01752119894 00200 + 00665119894 minus00533 + 03857119894 00002 minus 01065119894

01949 + 00719119894 minus00855 + 00370119894 02106 minus 02381119894 01648 minus 00032119894

minus00529 minus 01703119894 01935 + 01724119894 01071 + 02992119894 minus01983 minus 00491119894

))))

)

(36)

6 Journal of Applied Mathematics

Conflict of Interests

There is no conflict of interests between the authors

Acknowledgments

This research was supported by National Natural ScienceFoundation of China (31170532)The authors are very gratefulto the referees for their valuable comments and also thank theeditor for his helpful suggestions

References

[1] J H Wilkinson The Algebraic Eigenvalue Problem OxfordUniversity Press Oxford UK 1965

[2] A Berman and R J Plemmons Nonnegative Matrices in theMathematical Sciences vol 9 ofClassics in AppliedMathematicsSIAM Philadelphia Pa USA 1994

[3] MAravDHershkowitz VMehrmann andH Schneider ldquoTherecursive inverse eigenvalue problemrdquo SIAM Journal on MatrixAnalysis and Applications vol 22 no 2 pp 392ndash412 2000

[4] R Loewy and V Mehrmann ldquoA note on the symmetricrecursive inverse eigenvalue problemrdquo SIAM Journal on MatrixAnalysis and Applications vol 25 no 1 pp 180ndash187 2003

[5] F L Li X YHu and L Zhang ldquoLeft and right inverse eigenpairsproblem of skew-centrosymmetric matricesrdquo Applied Mathe-matics and Computation vol 177 no 1 pp 105ndash110 2006

[6] F L Li X Y Hu and L Zhang ldquoLeft and right inverseeigenpairs problem of generalized centrosymmetric matricesand its optimal approximation problemrdquo Applied Mathematicsand Computation vol 212 no 2 pp 481ndash487 2009

[7] R D Hill and S R Waters ldquoOn 120581-real and 120581-Hermitianmatricesrdquo Linear Algebra and its Applications vol 169 pp 17ndash29 1992

[8] S P Irwin ldquoMatrices with multiple symmetry propertiesapplications of centro-Hermitian and per-Hermitian matricesrdquoLinear Algebra and its Applications vol 284 no 1ndash3 pp 239ndash258 1998

[9] L C Biedenharn and J D Louck Angular Momentum inQuantum Physics vol 8 of Encyclopedia of Mathematics and itsApplications Addison-Wesley Reading Mass USA 1981

[10] WM Gibson and B R Pollard Symmetry Principles in Elemen-tary Particle Physics Cambridge University Press CambridgeUK 1976

[11] W F Trench ldquoHermitian Hermitian R-symmetric and Her-mitian R-skew symmetric Procrustes problemsrdquo Linear Algebraand its Applications vol 387 pp 83ndash98 2004

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 271978 5 pageshttpdxdoiorg1011552013271978

Research ArticleCompleting a 2 times 2 Block Matrix of Real Quaternions witha Partial Specified Inverse

Yong Lin12 and Qing-Wen Wang1

1 Department of Mathematics Shanghai University Shanghai 200444 China2 School of Mathematics and Statistics Suzhou University Suzhou 234000 China

Correspondence should be addressed to Qing-Wen Wang wqw858yahoocomcn

Received 4 December 2012 Revised 23 February 2013 Accepted 20 March 2013

Academic Editor K Sivakumar

Copyright copy 2013 Y Lin and Q-W Wang This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

This paper considers a completion problem of a nonsingular 2 times 2 block matrix over the real quaternion algebra H Let1198981 1198982 1198991 1198992be nonnegative integers119898

1+ 1198982= 1198991+ 1198992= 119899 gt 0 and 119860

12isin H1198981times1198992 119860

21isin H1198982times1198991 119860

22isin H1198982times1198992 119861

11isin H1198991times1198981

be given We determine necessary and sufficient conditions so that there exists a variant block entry matrix11986011isin H1198981times1198991 such that

119860 = (11986011 11986012

11986021 11986022) isin H119899times119899 is nonsingular and 119861

11is the upper left block of a partitioning of 119860minus1 The general expression for 119860

11is also

obtained Finally a numerical example is presented to verify the theoretical findings

1 Introduction

The problem of completing a block-partitioned matrix of aspecified type with some of its blocks given has been studiedby many authors Fiedler and Markham [1] considered thefollowing completion problem over the real number field RSuppose119898

1 1198982 1198991 1198992are nonnegative integers119898

1+1198982=

1198991+1198992= 119899 gt 0 119860

11isin R1198981times1198991 119860

12isin R1198981times1198992 119860

21isin R1198982times1198991

and 11986122isin R1198992times1198982 Determine a matrix 119860

22isin R1198982times1198992 such

that

119860 = (11986011

11986012

11986021

11986022

) (1)

is nonsingular and 11986122is the lower right block of a partition-

ing of 119860minus1 This problem has the form of

(11986011

11986012

11986021

)

minus1

= (

11986122

) (2)

and the solution and the expression for 11986022were obtained in

[1] Dai [2] considered this form of completion problemswithsymmetric and symmetric positive definite matrices over R

Some other particular forms for 2times2 block matrices overR have also been examined (see eg [3]) such as

(11986011

11986012

11986021

)

minus1

= (11986111

)

(11986011

)

minus1

= (

11986122

)

(11986011

11986022

)

minus1

= ( 11986112

11986121

)

(3)

The real quaternion matrices play a role in computerscience quantum physics and so on (eg [4ndash6]) Quaternionmatrices are receiving much attention as witnessed recently(eg [7ndash9]) Motivated by the work of [1 10] and keepingsuch applications of quaternionmatrices in view in this paperwe consider the following completion problem over the realquaternion algebra

H = 1198860+ 1198861119894 + 1198862119895 + 1198863119896 |

1198942

= 1198952

= 1198962

= 119894119895119896 = minus1 and 1198860 1198861 1198862 1198863isin R

(4)

Problem 1 Suppose 1198981 1198982 1198991 1198992are nonnegative inte-

gers 1198981+ 1198982= 1198991+ 1198992= 119899 gt 0 and 119860

12isin H1198981times1198992

2 Journal of Applied Mathematics

11986021isin H1198982times1198991 119860

22isin 1198771198982times1198992 119861

11isin H1198991times1198981 Find a matrix

11986011isin H1198981times1198991 such that

119860 = (11986011

11986012

11986021

11986022

) isin H119899times119899 (5)

is nonsingular and11986111is the upper left block of a partitioning

of 119860minus1 That is

( 119860

12

11986021

11986022

)

minus1

= (11986111

) (6)

where H119898times119899 denotes the set of all 119898 times 119899matrices over H and119860minus1 denotes the inverse matrix of 119860

Throughout over the real quaternion algebra H wedenote the identity matrix with the appropriate size by 119868 thetranspose of 119860 by 119860119879 the rank of 119860 by 119903(119860) the conjugatetranspose of119860 by119860lowast = (119860)119879 a reflexive inverse of a matrix119860over H by 119860+ which satisfies simultaneously 119860119860+119860 = 119860 and119860+

119860119860+

= 119860+Moreover119871

119860= 119868minus119860

+

119860 119877119860= 119868minus119860119860

+ where119860+ is an arbitrary but fixed reflexive inverse of 119860 Clearly 119871

119860

and119877119860are idempotent and each is a reflexive inverse of itself

R(119860) denotes the right column space of the matrix 119860The rest of this paper is organized as follows In Section 2

we establish some necessary and sufficient conditions to solveProblem 1 over H and the general expression for 119860

11is also

obtained In Section 3 we present a numerical example toillustrate the developed theory

2 Main Results

In this section we begin with the following lemmas

Lemma 1 (singular-value decomposition [9]) Let 119860 isin H119898times119899

be of rank 119903 Then there exist unitary quaternion matrices 119880 isin

H119898times119898 and 119881 isin H119899times119899 such that

119880119860119881 = (1198631199030

0 0) (7)

where 119863119903= diag(119889

1 119889

119903) and the 119889

119895rsquos are the positive

singular values of 119860

Let H119899119888denote the collection of column vectors with 119899

components of quaternions and 119860 be an 119898 times 119899 quaternionmatrix Then the solutions of 119860119909 = 0 form a subspace of H119899

119888

of dimension 119899(119860) We have the following lemma

Lemma 2 Let

(11986011

11986012

11986021

11986022

) (8)

be a partitioning of a nonsingular matrix 119860 isin H119899times119899 and let

(11986111

11986112

11986121

11986122

) (9)

be the corresponding (ie transpose) partitioning of 119860minus1 Then119899(11986011) = 119899(119861

22)

Proof It is readily seen that

(11986122

11986121

11986112

11986111

)

(11986022

11986021

11986012

11986011

)

(10)

are inverse to each other so we may suppose that 119899(11986011) lt

119899(11986122)

If 119899(11986122) = 0 necessarily 119899(119860

11) = 0 and we are finished

Let 119899(11986122) = 119888 gt 0 then there exists a matrix 119865 with 119888 right

linearly independent columns such that 11986122119865 = 0 Then

using

1198601111986112+ 1198601211986122= 0 (11)

we have

1198601111986112119865 = 0 (12)

From

1198602111986112+ 1198602211986122= 119868 (13)

we have

1198602111986112119865 = 119865 (14)

It follows that the rank 119903(11986112119865) ge 119888 In view of (12) this

implies

119899 (11986011) ge 119903 (119861

12119865) ge 119888 = 119899 (119861

22) (15)

Thus

119899 (11986011) = 119899 (119861

22) (16)

Lemma 3 (see [10]) Let 119860 isin H119898times119899 119861 isin H119901times119902 119863 isin H119898times119902 beknown and 119883 isin H119899times119901 unknown Then the matrix equation

119860119883119861 = 119863 (17)

is consistent if and only if

119860119860+

119863119861+

119861 = 119863 (18)

In that case the general solution is

119883 = 119860+

119863119861+

+ 1198711198601198841+ 1198842119877119861 (19)

where11988411198842are any matrices with compatible dimensions over

H

By Lemma 1 let the singular value decomposition of thematrix 119860

22and 119861

11in Problem 1 be

11986022= 119876(

Λ 0

0 0)119877lowast

(20)

11986111= 119880(

Σ 0

0 0)119881lowast

(21)

Journal of Applied Mathematics 3

where Λ = diag(1205821 1205822 120582

119904) is a positive diagonal matrix

120582119894= 0 (119894 = 1 119904) are the singular values of 119860

22 119904 =

119903(11986022) Σ = diag(120590

1 1205902 120590

119903) is a positive diagonal matrix

120590119894= 0 (119894 = 1 119903) are the singular values of 119861

11and 119903 =

119903(11986111)119876 = (119876

11198762) isin H1198982times1198982 119877 = (119877

11198772) isin H1198992times1198992

119880 = (11988011198802) isin H1198991times1198991 119881 = (119881

11198812) isin H1198981times1198981 are unitary

quaternion matrices where 1198761isin H1198982times119904 119877

1isin H1198992times119904 119880

1isin

H1198991times119903 and 1198811isin H1198981times119903

Theorem 4 Problem 1 has a solution if and only if thefollowing conditions are satisfied

(a) 119903 ( 1198601211986022

) = 1198992

(b) 1198992minus 119903(119860

22) = 119898

1minus 119903(11986111) that is 119899

2minus 119904 = 119898

1minus 119903

(c) R(1198602111986111) subR(119860

22)

(d) R(119860lowast

12119861lowast

11) subR(119860

lowast

22)

In that case the general solution has the form of

11986011= 119861+

11+ 11986012119877(

Λminus1

119876lowast

1119860211198801Σ 0

119867 minus(119881lowast

2119860121198772)minus1)

times 119881lowast

119861+

11+ 119884 minus 119884119861

11119861+

11

(22)

where 119867 is an arbitrary matrix in H(1198992minus119904)times119903 and 119884 is anarbitrary matrix in H1198981times1198991

Proof If there exists an 1198981times 1198991matrix 119860

11such that 119860 is

nonsingular and 11986111is the corresponding block of 119860minus1 then

(a) is satisfied From 119860119861 = 119861119860 = 119868 we have that

1198602111986111+ 1198602211986121= 0

1198611111986012+ 1198611211986022= 0

(23)

so that (c) and (d) are satisfiedBy (11) we have

119903 (11986022) + 119899 (119860

22) = 1198992 119903 (119861

11) + 119899 (119861

11) = 119898

1 (24)

From Lemma 2 Notice that ( 11986011 1198601211986021 11986022

) is the correspondingpartitioning of 119861minus1 we have

119899 (11986111) = 119899 (119860

22) (25)

implying that (b) is satisfiedConversely from (c) we know that there exists a matrix

119870 isin H1198992times1198981 such that

1198602111986111= 11986022119870 (26)

Let

11986121= minus119870 (27)

From (20) (21) and (26) we have

11986021119880(

Σ 0

0 0)119881lowast

= 119876(Λ 0

0 0)119877lowast

119870 (28)

It follows that

119876lowast

11986021119880(

Σ 0

0 0)119881lowast

119881 = 119876lowast

119876(Λ 0

0 0)119877lowast

119870119881 (29)

This implies that

(

119876lowast

1119860211198801119876lowast

1119860211198802

119876lowast

2119860211198801119876lowast

2119860211198802

)(Σ 0

0 0)

= (Λ 0

0 0)(

119877lowast

11198701198811119877lowast

11198701198812

119877lowast

21198701198811119877lowast

21198701198812

)

(30)

Comparing corresponding blocks in (30) we obtain

119876lowast

2119860211198801= 0 (31)

Let 119877lowast119870119881 = From (29) (30) we have

= (Λminus1

119876lowast

1119860211198801Σ 0

119867 11987022

)

119867 isin H(1198992minus119904)times119903 119870

22isin H(1198992minus119904)times(1198981minus119903)

(32)

In the same way from (d) we can obtain

119881lowast

1119860121198772= 0 (33)

Notice that ( 1198601211986022

) in (a) is a full column rank matrix By (20)(21) and (33) we have

(0 119876lowast

119881lowast

0)(

11986012

11986022

)119877 = (

Λ 0

0 0

119881lowast

1119860121198771119881lowast

1119860121198772

119881lowast

2119860121198771119881lowast

2119860121198772

) (34)

so that

1198992= 119903(

11986012

11986022

) = 119903((0 119876lowast

119881lowast

0)(

11986012

11986022

)119877)

= 119903(

Λ 0

0 0

119881lowast

1119860121198771119881lowast

1119860121198772

119881lowast

2119860121198771119881lowast

2119860121198772

)

= 119903 (Λ) + 119903 (119881lowast

2119860121198772)

= 119904 + 119903 (119881lowast

2119860121198772)

(35)

It follows from (b) and (35) that 1198811198792119860121198772is a full column

rank matrix so it is nonsingularFrom 119860119861 = 119868 we have the following matrix equation

1198601111986111+ 1198601211986121= 119868 (36)

that is

1198601111986111= 119868 minus 119860

1211986121 119868 isin H

1198981times1198981 (37)

4 Journal of Applied Mathematics

where 11986111 11986012

were given 11986121

= minus119870 (from (27)) ByLemma 3 the matrix equation (37) has a solution if and onlyif

(119868 minus 1198601211986121) 119861+

1111986111= 119868 minus 119860

1211986121 (38)

By (21) (27) (32) and (33) we have that (38) is equivalent to

(119868 + 11986012119870)119881(

Σminus1

0

0 0)119880lowast

119880(Σ 0

0 0)119881lowast

= 119868 + 11986012119870 (39)

We simplify the equation aboveThe left hand side reduces to(119868 + 119860

12119870)1198811119881lowast

1and so we have

119860121198701198811119881lowast

1minus 11986012119870 = 119868 minus 119881

1119881lowast

1 (40)

So

11986012119877119881lowast

1198811119881lowast

1minus 11986012119877119881lowast

= (11988111198812) (119881lowast

1

119881lowast

2

) minus 1198811119881lowast

1

(41)

This implies that

11986012119877(

119881lowast

11198811

119881lowast

21198811

)119881lowast

1minus 11986012119877(

119881lowast

1

119881lowast

2

) = 1198812119881lowast

2 (42)

so that

11986012119877(

119868

0)119881lowast

1minus 11986012119877(

119881lowast

1

119881lowast

2

) = 1198812119881lowast

2 (43)

So

minus11986012119877(

0

119881lowast

2

) = 1198812119881lowast

2 (44)

and hence

minus (119860121198771119860121198772) (Λminus1

119876lowast

1119860211198801Σ 0

119867 11987022

)(0

119881lowast

2

) = 1198812119881lowast

2

(45)

Finally we obtain

11986012119877211987022119881lowast

2= minus1198812119881lowast

2 (46)

Multiplying both sides of (46) by 119881lowast from the left consider-ing (33) and the fact that 119881lowast

2119860121198772is nonsingular we have

11987022= minus(119881

lowast

2119860121198772)minus1

(47)

From Lemma 3 (38) (47) Problem 1 has a solution and thegeneral solution is

11986011= 119861+

11+ 11986012119877(

Λminus1

119876lowast

1119860211198801Σ 0

119867 minus(119881lowast

2119860121198772)minus1)

times 119881lowast

119861+

11+ 119884 minus 119884119861

11119861+

11

(48)

where 119867 is an arbitrary matrix in H(1198992minus119904)times119903 and 119884 is anarbitrary matrix in H1198981times1198991

3 An Example

In this section we give a numerical example to illustrate thetheoretical results

Example 5 Consider Problem 1 with the parameter matricesas follows

11986012= (

2 + 1198951

2119896

minus119896 1 +1

2119895

)

11986021= (

3

2+1

2119894 minus

1

2119895 minus

1

2119896

1

2119895 +

1

2119896

3

2+1

2119894

)

11986022= (

2 119894

2119895 119896) 119861

11= (

1 119894

119895 119896)

(49)

It is easy to show that (c) (d) are satisfied and that

1198992= 119903(

11986012

11986022

) = 2

1198992minus 119903 (119860

22) = 119898

1minus 119903 (119861

11) = 0

(50)

so (a) (b) are satisfied too Therefore we have

119861+

11= (

1

2minus1

2119895

minus1

2119894 minus

1

2119896

)

11986022= 119876(

Λ 0

0 0)119877lowast

11986111= 119880(

Σ 0

0 0)119881lowast

(51)

where

119876 =1

radic2

(1 119894

119895 119896) Λ = (

2radic2 0

0 radic2)

119877 = (1 0

0 1) 119880 =

1

radic2

(1 119894

119895 119896)

Σ = (radic2 0

0 radic2) 119881 = (

1 0

0 1)

(52)

We also have

1198761=

1

radic2

(1 119894

119895 119896) 119877

1= (

1 0

0 1)

1198801=

1

radic2

(1 119894

119895 119896) 119881

1= (

1 0

0 1)

(53)

Journal of Applied Mathematics 5

By Theorem 4 for an arbitrary matrices 119884 isin H2times2 wehave

11986011= 119861+

11+ 11986012119877 (Λminus1

119876lowast

1119860211198801Σ)119881lowast

119861+

11+ 119884 minus 119884119861

11119861+

11

= (

3

2+1

4119895 +

1

4119896

3

4+1

4119894 minus

3

2119895

1

2minus 119894 +

1

4119895 minus

1

41198961

4minus3

4119894 minus

1

2119895 minus 119896

)

(54)

it follows that

119860 =((

(

3

2

+

1

4

119895 +

1

4

119896

3

4

+

1

4

119894 minus

3

2

119895 2 + 119895

1

2

119896

1

2

minus 119894 +

1

4

119895 minus

1

4

119896

1

4

minus

3

4

119894 minus

1

2

119895 minus 119896 minus119896 1 +

1

2

119895

3

2

+

1

2

119894 minus

1

2

119895 minus

1

2

119896 2 119894

1

2

119895 +

1

2

119896

3

2

+

1

2

119894 2119895 119896

))

)

119860minus1

=(

1 119894 minus1 minus1

119895 119896 0 minus1

minus1 03

4

1

2minus3

4119895

minus1 minus11

2minus 119894

1

2minus1

2119894 minus

1

2119895 minus 119896

)

(55)

The results verify the theoretical findings of Theorem 4

Acknowledgments

The authors would like to give many thanks to the refereesand Professor K C Sivakumar for their valuable suggestionsand comments which resulted in a great improvement ofthe paper This research was supported by Grants from theKey Project of Scientific Research Innovation Foundation ofShanghai Municipal Education Commission (13ZZ080) theNational Natural Science Foundation of China (11171205) theNatural Science Foundation of Shanghai (11ZR1412500) theDiscipline Project at the corresponding level of Shanghai (A13-0101-12-005) and Shanghai Leading Academic DisciplineProject (J50101)

References

[1] M Fiedler and T L Markham ldquoCompleting a matrix whencertain entries of its inverse are specifiedrdquo Linear Algebra andIts Applications vol 74 pp 225ndash237 1986

[2] H Dai ldquoCompleting a symmetric 2 times 2 block matrix and itsinverserdquo Linear Algebra and Its Applications vol 235 pp 235ndash245 1996

[3] W W Barrett M E Lundquist C R Johnson and H JWoerdeman ldquoCompleting a block diagonal matrix with apartially prescribed inverserdquoLinearAlgebra and Its Applicationsvol 223224 pp 73ndash87 1995

[4] S L Adler Quaternionic Quantum Mechanics and QuantumFields Oxford University Press New York NY USA 1994

[5] A Razon and L P Horwitz ldquoUniqueness of the scalar productin the tensor product of quaternion Hilbert modulesrdquo Journalof Mathematical Physics vol 33 no 9 pp 3098ndash3104 1992

[6] J W Shuai ldquoThe quaternion NN model the identification ofcolour imagesrdquo Acta Comput Sinica vol 18 no 5 pp 372ndash3791995 (Chinese)

[7] F Zhang ldquoOn numerical range of normal matrices of quater-nionsrdquo Journal of Mathematical and Physical Sciences vol 29no 6 pp 235ndash251 1995

[8] F Z Zhang Permanent Inequalities and Quaternion Matrices[PhD thesis] University of California Santa Barbara CalifUSA 1993

[9] F Zhang ldquoQuaternions and matrices of quaternionsrdquo LinearAlgebra and Its Applications vol 251 pp 21ndash57 1997

[10] Q-W Wang ldquoThe general solution to a system of real quater-nion matrix equationsrdquo Computers amp Mathematics with Appli-cations vol 49 no 5-6 pp 665ndash675 2005

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 178209 7 pageshttpdxdoiorg1011552013178209

Research ArticleFuzzy Approximate Solution of Positive Fully Fuzzy LinearMatrix Equations

Xiaobin Guo1 and Dequan Shang2

1 College of Mathematics and Statistics Northwest Normal University Lanzhou 730070 China2Department of Public Courses Gansu College of Traditional Chinese Medicine Lanzhou 730000 China

Correspondence should be addressed to Dequan Shang gxbglz163com

Received 22 November 2012 Accepted 19 January 2013

Academic Editor Panayiotis J Psarrakos

Copyright copy 2013 X Guo and D ShangThis is an open access article distributed under theCreativeCommonsAttribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

The fuzzy matrix equations otimesotimes = in which and are119898times119898 119899times119899 and119898times119899 nonnegative LR fuzzy numbers matricesrespectively are investigated The fuzzy matrix systems is extended into three crisp systems of linear matrix equations accordingto arithmetic operations of LR fuzzy numbers Based on pseudoinverse of matrix the fuzzy approximate solution of original fuzzysystems is obtained by solving the crisp linear matrix systems In addition the existence condition of nonnegative fuzzy solution isdiscussed Two examples are calculated to illustrate the proposed method

1 Introduction

Since many real-world engineering systems are too com-plex to be defined in precise terms imprecision is ofteninvolved in any engineering design process Fuzzy systemshave an essential role in this fuzzy modeling which canformulate uncertainty in actual environment Inmanymatrixequations some or all of the system parameters are vagueor imprecise and fuzzy mathematics is a better tool thancrisp mathematics for modeling these problems and hencesolving a fuzzy matrix equation is becomingmore importantThe concept of fuzzy numbers and arithmetic operationswith these numbers were first introduced and investigatedby Zadeh [1 2] Dubois and Prade [3] and Nahmias [4]A different approach to fuzzy numbers and the structureof fuzzy number spaces was given by Puri and Ralescu [5]Goetschel Jr and Voxman [6] and Wu and Ma [7 8]

In the past decades many researchers have studied thefuzzy linear equations such as fuzzy linear systems (FLS) dualfuzzy linear systems (DFLS) general fuzzy linear systems(GFLS) fully fuzzy linear systems (FFLS) dual fully fuzzylinear systems (DFFLS) and general dual fuzzy linear systems(GDFLS) These works were performed mainly by Friedmanet al [9 10] Allahviranloo et al [11ndash17] Abbasbandy et al

[18ndash21]Wang andZheng [22 23] andDehghan et al [24 25]The general method they applied is the fuzzy linear equationswere converted to a crisp function system of linear equationswith high order according to the embedding principlesand algebraic operations of fuzzy numbers Then the fuzzysolution of the original fuzzy linear systems was derived fromsolving the crisp function linear systemsHowever for a fuzzymatrix equationwhich always has awide use in control theoryand control engineering few works have been done in thepast In 2009 Allahviranloo et al [26] discussed the fuzzy lin-ear matrix equations (FLME) of the form 119860119861 = in whichthematrices119860 and119861 are known119898times119898 and 119899times119899 real matricesrespectively is a given 119898 times 119899 fuzzy matrix By using theparametric form of fuzzy number they derived necessaryand sufficient conditions for the existence condition of fuzzysolutions and designed a numerical procedure for calculatingthe solutions of the fuzzy matrix equations In 2011 Guoet al [27ndash29] investigated a class of fuzzy matrix equations119860 = by means of the block Gaussian elimination methodand studied the least squares solutions of the inconsistentfuzzy matrix equation 119860 = by using generalized inversesof the matrix and discussed fuzzy symmetric solutions offuzzy matrix equations 119860 = What is more there are two

2 Journal of Applied Mathematics

shortcomings in the above fuzzy systems The first is that inthese fuzzy linear systems and fuzzy matrix systems the fuzzyelements were denoted by triangular fuzzy numbers so theextended model equations always contain parameter 119903 0 le

119903 le 1 which makes their computation especially numericalimplementation inconvenient in some sense The other oneis that the weak fuzzy solution of fuzzy linear systems119860 =

does not exist sometimes see [30]To make the multiplication of fuzzy numbers easy and

handle the fully fuzzy systems Dubois and Prade [3] intro-duced the LR fuzzy number in 1978 We know that triangularfuzzy numbers and trapezoidal fuzzy numbers [31] are justspecious cases of LR fuzzy numbers In 2006 Dehghan etal [25] discussed firstly computational methods for fullyfuzzy linear systems = whose coefficient matrix andthe right-hand side vector are LR fuzzy numbers In 2012Allahviranloo et al studied LR fuzzy linear systems [32]by the linear programming with equality constraints Otadiand Mosleh considered the nonnegative fuzzy solution [33]of fully fuzzy matrix equations = by employinglinear programming with equality constraints at the sameyear Recently Guo and Shang [34] investigated the fuzzyapproximate solution of LR fuzzy Sylvester matrix equations119860 + 119861 = In this paper we propose a general model forsolving the fuzzy linearmatrix equation otimesotimes = where and are119898times119898 119899times119899 and119898times119899 nonnegative LR fuzzynumbersmatrices respectivelyThemodel is proposed in thisway that is we extend the fuzzy linear matrix system intoa system of linear matrix equations according to arithmeticoperations of LR fuzzy numbers The LR fuzzy solution ofthe original matrix equation is derived from solving crispsystems of linearmatrix equationsThe structure of this paperis organized as follows

In Section 2 we recall the LR fuzzy numbers andpresent the concept of fully fuzzy linear matrix equationThecomputing model to the positive fully fuzzy linear matrixequation is proposed in detail and the fuzzy approximatesolution of the fuzzy linear matrix equation is obtained byusing pseudo-inverse in Section 3 Some examples are givento illustrate our method in Section 4 and the conclusion isdrawn in Section 5

2 Preliminaries

21 The LR Fuzzy Number

Definition 1 (see [1]) A fuzzy number is a fuzzy set like 119906

119877 rarr 119868 = [0 1] which satisfies the following

(1) 119906 is upper semicontinuous(2) 119906 is fuzzy convex that is 119906(120582119909 + (1 minus 120582)119910) ge

min119906(119909) 119906(119910) for all 119909 119910 isin 119877 120582 isin [0 1](3) 119906 is normal that is there exists 119909

0isin 119877 such that

119906(1199090) = 1

(4) supp 119906 = 119909 isin 119877 | 119906(119909) gt 0 is the support of the 119906and its closure cl(supp 119906) is compact

Let 1198641 be the set of all fuzzy numbers on 119877

Definition 2 (see [3]) A fuzzy number is said to be an LRfuzzy number if

120583(119909) =

119871(119898 minus 119909

120572) 119909 le 119898 120572 gt 0

119877(119909 minus 119898

120573) 119909 ge 119898 120573 gt 0

(1)

where 119898 120572 and 120573 are called the mean value left and rightspreads of respectively The function 119871(sdot) which is calledleft shape function satisfies

(1) 119871(119909) = 119871(minus119909)(2) 119871(0) = 1 and 119871(1) = 0(3) 119871(119909) is nonincreasing on [0infin)

The definition of a right shape function 119877(sdot) is similar tothat of 119871(sdot)

Clearly = (119898 120572 120573)LR is positive (negative) if and onlyif119898 minus 120572 gt 0 (119898 + 120573 lt 0)

Also two LR fuzzy numbers = (119898 120572 120573)LR and =

(119899 120574 120575)LR are said to be equal if and only if 119898 = 119899 120572 = 120574and 120573 = 120575

Definition 3 (see [5]) For arbitrary LR fuzzy numbers =

(119898 120572 120573)LR and = (119899 120574 120575)LR we have the following

(1) Addition

oplus = (119898 120572 120573)LR oplus (119899 120574 120575)LR = (119898 + 119899 120572 + 120574 120573 + 120575)LR

(2)

(2) Multiplication(i) If gt 0 and gt 0 then

otimes = (119898 120572 120573)LR otimes (119899 120574 120575)LR

cong (119898119899119898120574 + 119899120572119898120575 + 119899120573)LR(3)

(ii) if lt 0 and gt 0 then

otimes = (119898 120572 120573)RL otimes (119899 120574 120575)LR

cong (119898119899 119899120572 minus 119898120575 119899120573 minus 119898120574)RL(4)

(iii) if lt 0 and lt 0 then

otimes = (119898 120572 120573)RL otimes (119899 120574 120575)LR

cong (119898119899 minus119898120575 minus 119899120573 minus119898120574 minus 119899120572)RL(5)

(3) Scalar multiplication

120582 times = 120582 times (119898 120572 120573)LR

= (120582119898 120582120572 120582120573)LR 120582 ge 0

(120582119898 minus120582120573 minus120582120572)RL 120582 lt 0

(6)

Journal of Applied Mathematics 3

22 The LR Fuzzy Matrix

Definition 4 Amatrix = (119894119895) is called an LR fuzzy matrix

if each element 119894119895of is an LR fuzzy number

will be positive (negative) and denoted by gt 0 ( lt

0) if each element 119894119895of is positive (negative) where

119894119895=

(119886119894119895 119886119897

119894119895 119886119903

119894119895)LR Up to the rest of this paper we use positive

LR fuzzy numbers and formulas given in Definition 3 Forexample we represent 119898 times 119899 LR fuzzy matrix = (

119894119895) that

119894119895= (119886119894119895 120572119894119895 120573119894119895)LR with new notation = (119860119872119873) where

119860 = (119886119894119895) 119872 = (120572

119894119895) and 119873 = (120573

119894119895) are three 119898 times 119899 crisp

matrices In particular an 119899 dimensions LR fuzzy numbersvector can be denoted by (119909 119909119897 119909119903) where 119909 = (119909

119894) 119909119897 =

(119909119897

119894) and 119909119903 = (119909

119903

119894) are three 119899 dimensions crisp vectors

Definition 5 Let = (119894119895) and = (

119894119895) be two 119898 times 119899 and

119899 times 119901 fuzzy matrices we define otimes = = (119894119895) which is an

119898 times 119901 fuzzy matrix where

119894119895=

oplus

sum

119896=12119899

119894119896otimes 119887119896119895 (7)

23 The Fully Fuzzy Linear Matrix Equation

Definition 6 Thematrix system

(

11

12

sdot sdot sdot 1119898

21

22

sdot sdot sdot 2119898

1198981

1198982

sdot sdot sdot 119898119898

)otimes(

11

12

sdot sdot sdot 1119899

21

22

sdot sdot sdot 2119899

1198981

1198982

sdot sdot sdot 119898119899

)

otimes(

11

11988712

sdot sdot sdot 1198871119899

21

11988722

sdot sdot sdot 1198872119899

1198991

1198992

sdot sdot sdot 119899119899

)

=(

11

12

sdot sdot sdot 1119899

21

22

sdot sdot sdot 2119899

1198981

1198982

sdot sdot sdot 119898119899

)

(8)

where 119894119895 1 le 119894 119895 le 119898

119894119895 1 le 119894 119895 le 119899 and

119894119895 1 le 119894 le 119898

1 le 119895 le 119899 are LR fuzzy numbers is called an LR fully fuzzylinear matrix equation (FFLME)

Using matrix notation we have

otimes otimes = (9)

A fuzzy numbers matrix

= (119909119894119895 119910119894119895 119911119894119895)LR 1 le 119894 le 119898 1 le 119895 le 119899 (10)

is called an LR fuzzy approximate solution of fully fuzzy linearmatrix equation (8) if satisfies (9)

Up to the rest of this paper we will discuss the nonnega-tive solution = (119883 119884 119885) ge 0 of FFLME otimes otimes =

where = (119860119872119873) ge 0 = (119861 119864 119865) ge 0 and =

(119862 119866119867) ge 0

3 Method for Solving FFLME

First we extend the fuzzy linear matrix system (9) into threesystems of linear matrix equations according to the LR fuzzynumber and its arithmetic operations

Theorem 7 The fuzzy linear matrix system (9) can beextended into the following model

119860119883119861 = 119862

119860119883119864 + 119860119884119861 +119872119883119861 = 119866

119860119883119865 + 119860119885119861 + 119873119883119861 = 119867

(11)

Proof We denote = (119860119872119873) = (119861 119864 119865) and =

(119862 119866119867) and assume = (119883 119884 119885) ge 0 then

= (119860119872119873) otimes (119883 119884 119885) otimes (119861 119864 119865)

= (119860119883119860119884 +119872119883119860119885 + 119873119883) otimes (119861 119864 119865)

= (119860119883119861119860119883119864 + 119860119884119861 +119872119883119861119860119883119865 + 119860119885119861 + 119873119883119861)

= (119862 119866119867)

(12)

according tomultiplication of nonnegative LR fuzzy numbersof Definition 2 Thus we obtain a model for solving FFLME(9) as follows

119860119883119861 = 119862

119860119883119864 + 119860119884119861 +119872119883119861 = 119866

119860119883119865 + 119860119885119861 + 119873119883119861 = 119867

(13)

Secondly in order to solve the fuzzy linear matrix equa-tion (9) we need to consider the crisp systems of linearmatrix equation (11) Supposing119860 and119861 are nonsingular crispmatrices we have

119883 = 119860minus1

119862119861minus1

119884 = 119860minus1

119866 minus 119883119864119861minus1

minus 119860minus1

119872119883

119885 = 119860minus1

119867 minus 119883119865119861minus1

minus 119860minus1

119873119883

(14)

that is

119883 = 119860minus1

119862119861minus1

119884 = 119860minus1

119866 minus 119860minus1

119862119861minus1

119864119861minus1

minus 119860minus1

119872119860minus1

119862119861minus1

119885 = 119860minus1

119867 minus 119860minus1

119862119861minus1

119865119861minus1

minus 119860minus1

119873119860minus1

119862119861minus1

(15)

Definition 8 Let = (119883 119884 119885) be an LR fuzzy matrix If(119883 119884 119885) is an exact solution of (11) such that 119883 ge 0 119884 ge 0119885 ge 0 and 119883 minus 119884 ge 0 we call = (119883 119884 119885) a nonnegativeLR fuzzy approximate solution of (9)

4 Journal of Applied Mathematics

Theorem 9 Let = (119860119872119873) = (119861 119864 119865) and =

(119862 119866119867) be three nonnegative fuzzymatrices respectively andlet 119860 and 119861 be the product of a permutation matrix by adiagonal matrix with positive diagonal entries Moreover let119866119861 ge 119862119861

minus1

119864 + 119872119860minus1

119862 119867119861 ge 119862119861minus1

119865 + 119873119860minus1

119862 and119862 + 119862119861

minus1

119864 +119872119860minus1

119862 ge 119866119861 Then the systems otimes otimes =

has nonnegative fuzzy solutions

Proof Our hypotheses on 119860 and 119861 imply that 119860minus1 and119861minus1 exist and they are nonnegative matrices Thus 119883 =

119860minus1

119862119861minus1

ge 0On the other hand because 119866119861 ge 119862119861

minus1

119864 + 119872119860minus1

119862 and119867119861 ge 119862119861

minus1

119865 + 119873119860minus1

119862 so with 119884 = 119860minus1

(119866119861 minus 119862119861minus1

119864 minus

119872119860minus1

119862)119861minus1 and 119885 = 119860

minus1

(119867119861 minus 119862119861minus1

119865 minus 119873119860minus1

119862)119861minus1 we

have 119884 ge 0 and 119885 ge 0 Thus = (119883 119884 119885) is a fuzzy matrixwhich satisfies otimes otimes = Since119883 minus 119884 = 119860

minus1

(119862 minus 119866119861 +

119862119861minus1

119864 + 119872119860minus1

119862)119861minus1 the positivity property of can be

obtained from the condition 119862+119862119861minus1119864+119872119860minus1

119862 ge 119866119861

When 119860 or 119861 is a singular crisp matrix the followingresult is obvious

Theorem 10 (see [35]) For linearmatrix equations119860119883119861 = 119862where 119860 isin 119877

119898times119898 119861 isin 119877119899times119899 and 119862 isin 119877

119898times119899 Then

119883 = 119860dagger

119862119861dagger (16)

is its minimal norm least squares solutionBy the pseudoinverse of matrices we solve model (11) and

obtain its minimal norm least squares solution as follows

119883 = 119860dagger

119862119861dagger

119884 = 119860dagger

119866 minus 119883119864119861dagger

minus 119860dagger

119872119883

119885 = 119860dagger

119867 minus 119883119865119861dagger

minus 119860dagger

119873119883

(17)

that is

119883 = 119860dagger

119862119861dagger

119884 = 119860dagger

119866 minus 119860dagger

119862119861dagger

119864119861dagger

minus 119860dagger

119872119860dagger

119862119861dagger

119885 = 119860dagger

119867 minus 119860dagger

119862119861dagger

119865119861dagger

minus 119860dagger

119873119860dagger

119862119861dagger

(18)

Definition 11 Let = (119883 119884 119885) be an LR fuzzy matrix If(119883 119884 119885) is a minimal norm least squares solution of (11) suchthat119883 ge 0119884 ge 0119885 ge 0 and119883minus119884 ge 0 we call = (119883 119884 119885)

a nonnegative LR fuzzy minimal norm least squares solutionof (9)

At last we give a sufficient condition for nonnegativefuzzy minimal norm least squares solution of FFLME (9) inthe same way

Theorem 12 Let 119860dagger and 119861dagger be nonnegative matrices More-over let 119866119861 ge 119862119861

dagger

119864 + 119872119860dagger

119862 119867119861 ge 119862119861dagger

119865 + 119873119860dagger

119862 and119862+119862119861

dagger

119864+119872119860dagger

119862 ge 119866119861 Then the systems otimes otimes = hasnonnegative fuzzy minimal norm least squares solutions

Proof Since 119860dagger and 119861dagger are nonnegative matrices we have

119883 = 119860dagger

119862119861dagger

ge 0Now that 119866119861 ge 119862119861

dagger

119864+119872119860dagger

119862 and119867119861 ge 119862119861dagger

119865+119873119860dagger

119862therefore with 119884 = 119860

dagger

(119866119861 minus 119862119861dagger

119864 minus 119872119860dagger

119862)119861dagger and 119885 =

119860dagger

(119867119861 minus 119862119861dagger

119865 minus 119873119860dagger

119862)119861dagger we have 119884 ge 0 and 119885 ge 0 Thus

= (119883 119884 119885) is a fuzzy matrix which satisfies otimes otimes =

Since 119883 minus 119884 = 119860dagger

(119862 minus 119866119861 + 119862119861dagger

119864 + 119872119860dagger

119862)119861dagger

the nonnegativity property of can be obtained from thecondition 119862 + 119862119861

dagger

119864 +119872119860dagger

119862 ge 119866119861

The following Theorems give some results for such 119878minus1

and 119878dagger to be nonnegative As usual (sdot)⊤ denotes the transposeof a matrix (sdot)

Theorem 13 (see [36]) The inverse of a nonnegative matrix119860 is nonnegative if and only if 119860 is a generalized permutationmatrix

Theorem 14 (see [37]) Let119860 be an119898 times119898 nonnegativematrixwith rank 119903 Then the following assertions are equivalent

(a) 119860dagger ge 0

(b) There exists a permutation matrix 119875 such that 119875119860 hasthe form

119875119860 =(

1198781

1198782

119878119903

119874

) (19)

where each 119878119894has rank 1 and the rows of 119878

119894are

orthogonal to the rows of 119878119895 whenever 119894 = 119895 the zero

matrix may be absent

(c) 119860dagger = (119870119875⊤119870119876⊤

119870119875⊤119870119875⊤ ) for some positive diagonal matrix 119870

In this case

(119875 + 119876)dagger

= 119870(119875 + 119876)⊤

(119875 minus 119876)dagger

= 119870(119875 minus 119876)⊤

(20)

4 Numerical Examples

Example 15 Consider the fully fuzzy linear matrix system

((2 1 1)LR (1 0 1)LR(1 0 0)LR (2 1 0)LR

) otimes (11

12

13

21

22

23

)

otimes (

(1 0 1)LR (2 1 1)LR (1 1 0)LR(2 1 0)LR (1 0 0)LR (2 1 1)LR(1 0 0)LR (2 1 1)LR (1 0 1)LR

)

= ((17 12 23)LR (22 22 29)LR (17 16 28)LR(19 15 13)LR (23 23 16)LR (19 16 17)LR

)

(21)

Journal of Applied Mathematics 5

ByTheorem 7 the model of the above fuzzy linear matrixsystem is made of the following three crisp systems of linearmatrix equations

(2 1

1 2)(

11990911

11990912

11990913

11990921

11990922

11990923

)(

1 2 1

2 1 2

1 2 1

) = (17 22 17

19 23 19)

(2 1

1 2)(

11990911

11990912

11990913

11990921

11990922

11990923

)(

0 1 1

1 0 1

0 1 0

)

+ (2 1

1 2)(

11991011

11991012

11991013

11991021

11991022

11991023

)(

1 2 1

2 1 2

1 2 1

)

+ (1 0

0 1)(

11990911

11990912

11990913

11990921

11990922

11990923

)(

1 2 1

2 1 2

1 2 1

) = (12 22 16

15 23 20)

(2 1

1 2)(

11990911

11990912

11990913

11990921

11990922

11990923

)(

1 1 0

0 0 1

0 1 1

)

+ (2 1

1 2)(

11991111

11991112

11991113

11991121

11991122

11991123

)(

1 2 1

2 1 2

1 2 1

)

+ (1 1

0 0)(

11990911

11990912

11990913

11990921

11990922

11990923

)(

1 2 1

2 1 2

1 2 1

) = (23 29 28

13 16 17)

(22)

Now that the matrix 119861 is singular according to formula(18) the solutions of the above three systems of linear matrixequations are as follows

119883 = 119860dagger

119862119861dagger

= (2 1

1 2)

dagger

(17 22 17

19 23 19)(

1 2 1

2 1 2

1 2 1

)

dagger

= (15000 10000 15000

15000 20000 15000)

119884 = 119860dagger

119866 minus 119883119864119861dagger

minus 119860dagger

119872119883 = (2 1

1 2)

dagger

(12 22 16

19 23 19)

minus 119883(

0 1 1

1 0 1

0 1 0

)(

1 2 1

2 1 2

1 2 1

)

dagger

minus (2 1

1 2)

dagger

(1 0

0 1)119883

= (10333 06667 08725

11250 16553 12578)

119885 = 119860dagger

119867 minus 119883119865119861dagger

minus 119860dagger

119873119883 = (2 1

1 2)

dagger

(23 29 28

13 16 17)

minus 119883(

1 1 0

0 0 1

0 1 1

)(

1 2 1

2 1 2

1 2 1

)

dagger

minus (2 1

1 2)

dagger

(1 1

0 0)119883

= (83333 116667 93333

14167 13333 24167)

(23)

By Definition 11 we know that the original fuzzy linearmatrix equations have a nonnegative LR fuzzy solution

= (11

12

13

21

22

23

) = ((15000 10333 83333)LR (10000 06667 116667)LR (15000 08725 93333)LR(15000 11250 14167)LR (20000 16553 13333)LR (15000 12578 24167)LR

) (24)

since119883 ge 0 119884 ge 0 119885 ge 0 and119883 minus 119884 ge 0

Example 16 Consider the following fuzzy matrix system

((1 1 0)LR (2 0 1)LR(2 0 1)LR (1 0 0)LR

)(11

12

21

22

)((1 0 1)LR (2 1 0)LR(2 1 1)LR (3 1 2)LR

)

= ((15 14 18)LR (25 24 24)LR(12 10 12)LR (20 17 15)LR

)

(25)

ByTheorem 7 the model of the above fuzzy linear matrixsystem is made of following three crisp systems of linear

matrix equations

(2 1

1 2)(

11990911

11990912

11990921

11990922

)(1 2

2 3) = (

15 25

10 20)

(2 1

1 2)(

11990911

11990912

11990921

11990922

)(0 1

1 1) + (

2 1

1 2)(

11991011

11991012

11991021

11991022

)(1 2

2 3)

+ (1 0

0 0)(

11990911

11990912

11990921

11990922

)(1 2

2 3) = (

14 24

10 17)

(2 1

1 2)(

11990911

11990912

11990921

11990922

)(1 0

1 2) + (

2 1

1 2)(

11991111

11991112

11991121

11991122

)(1 2

2 3)

+ (0 1

1 0)(

11990911

11990912

11990921

11990922

)(1 2

2 3) = (

18 24

12 15)

(26)

6 Journal of Applied Mathematics

By the same way we obtain that the solutions of the abovethree systems of linear matrix equations are as follows

119883 = 119860minus1

119862119861minus1

= (1 1

2 2)

119884 = 119860minus1

119866 minus 119883119864119861minus1

minus 119860minus1

119872119883 = (0 1

0 1)

119885 = 119860minus1

119867 minus 119883119865119861minus1

minus 119860minus1

119873119883 = (0 0

1 0)

(27)

Since 119883 ge 0 119884 ge 0 119885 ge 0 and 119883 minus 119884 ge 0 we know thatthe original fuzzy linear matrix equations have a nonnegativeLR fuzzy solution given by

= (11

12

21

22

)

= ((1000 1000 0000)LR (1000 1000 0000)LR(2000 0000 1000)LR (2000 1000 0000)LR

)

(28)

5 Conclusion

In this work we presented a model for solving fuzzy linearmatrix equations otimes otimes = in which and are119898 times 119898 and 119899 times 119899 fuzzy matrices respectively and isan 119898 times 119899 arbitrary LR fuzzy numbers matrix The modelwas made of three crisp systems of linear equations whichdetermined the mean value and the left and right spreads ofthe solution The LR fuzzy approximate solution of the fuzzylinear matrix equation was derived from solving the crispsystems of linear matrix equations In addition the existencecondition of strong LR fuzzy solution was studied Numericalexamples showed that ourmethod is feasible to solve this typeof fuzzy matrix equations Based on LR fuzzy numbers andtheir operations we can investigate all kinds of fully fuzzymatrix equations in future

Acknowledgments

The work is supported by the Natural Scientific Funds ofChina (no 71061013) and the Youth Research Ability Projectof Northwest Normal University (NWNU-LKQN-1120)

References

[1] L A Zadeh ldquoFuzzy setsrdquo Information and Computation vol 8pp 338ndash353 1965

[2] L A Zadeh ldquoThe concept of a linguistic variable and itsapplication to approximate reasoning Irdquo Information Sciencevol 8 pp 199ndash249 1975

[3] D Dubois and H Prade ldquoOperations on fuzzy numbersrdquoInternational Journal of Systems Science vol 9 no 6 pp 613ndash626 1978

[4] S Nahmias ldquoFuzzy variablesrdquo Fuzzy Sets and Systems AnInternational Journal in Information Science and Engineeringvol 1 no 2 pp 97ndash110 1978

[5] M L Puri and D A Ralescu ldquoDifferentials of fuzzy functionsrdquoJournal of Mathematical Analysis and Applications vol 91 no 2pp 552ndash558 1983

[6] R Goetschel Jr and W Voxman ldquoElementary fuzzy calculusrdquoFuzzy Sets and Systems vol 18 no 1 pp 31ndash43 1986

[7] C X Wu and M Ma ldquoEmbedding problem of fuzzy numberspace Irdquo Fuzzy Sets and Systems vol 44 no 1 pp 33ndash38 1991

[8] C X Wu and M Ma ldquoEmbedding problem of fuzzy numberspace IIIrdquo Fuzzy Sets and Systems vol 46 no 2 pp 281ndash2861992

[9] M Friedman M Ming and A Kandel ldquoFuzzy linear systemsrdquoFuzzy Sets and Systems vol 96 no 2 pp 201ndash209 1998

[10] M Ma M Friedman and A Kandel ldquoDuality in fuzzy linearsystemsrdquo Fuzzy Sets and Systems vol 109 no 1 pp 55ndash58 2000

[11] T Allahviranloo and M Ghanbari ldquoOn the algebraic solutionof fuzzy linear systems based on interval theoryrdquo AppliedMathematical Modelling vol 36 no 11 pp 5360ndash5379 2012

[12] T Allahviranloo and S Salahshour ldquoFuzzy symmetric solutionsof fuzzy linear systemsrdquo Journal of Computational and AppliedMathematics vol 235 no 16 pp 4545ndash4553 2011

[13] T Allahviranloo and M Ghanbari ldquoA new approach to obtainalgebraic solution of interval linear systemsrdquo Soft Computingvol 16 no 1 pp 121ndash133 2012

[14] T Allahviranloo ldquoNumericalmethods for fuzzy systemof linearequationsrdquo Applied Mathematics and Computation vol 155 no2 pp 493ndash502 2004

[15] T Allahviranloo ldquoSuccessive over relaxation iterative methodfor fuzzy system of linear equationsrdquo Applied Mathematics andComputation vol 162 no 1 pp 189ndash196 2005

[16] T Allahviranloo ldquoThe Adomian decomposition method forfuzzy system of linear equationsrdquo Applied Mathematics andComputation vol 163 no 2 pp 553ndash563 2005

[17] RNuraei T Allahviranloo andMGhanbari ldquoFinding an innerestimation of the so- lution set of a fuzzy linear systemrdquoAppliedMathematical Modelling vol 37 no 7 pp 5148ndash5161 2013

[18] S Abbasbandy R Ezzati and A Jafarian ldquoLU decompositionmethod for solving fuzzy system of linear equationsrdquo AppliedMathematics and Computation vol 172 no 1 pp 633ndash6432006

[19] S Abbasbandy A Jafarian and R Ezzati ldquoConjugate gradientmethod for fuzzy symmetric positive definite system of linearequationsrdquo Applied Mathematics and Computation vol 171 no2 pp 1184ndash1191 2005

[20] S Abbasbandy M Otadi andM Mosleh ldquoMinimal solution ofgeneral dual fuzzy linear systemsrdquo Chaos Solitons amp Fractalsvol 37 no 4 pp 1113ndash1124 2008

[21] B Asady S Abbasbandy and M Alavi ldquoFuzzy general linearsystemsrdquo Applied Mathematics and Computation vol 169 no 1pp 34ndash40 2005

[22] K Wang and B Zheng ldquoInconsistent fuzzy linear systemsrdquoApplied Mathematics and Computation vol 181 no 2 pp 973ndash981 2006

[23] B Zheng and K Wang ldquoGeneral fuzzy linear systemsrdquo AppliedMathematics and Computation vol 181 no 2 pp 1276ndash12862006

[24] M Dehghan and B Hashemi ldquoIterative solution of fuzzy linearsystemsrdquo Applied Mathematics and Computation vol 175 no 1pp 645ndash674 2006

[25] M Dehghan B Hashemi and M Ghatee ldquoSolution of the fullyfuzzy linear systems using iterative techniquesrdquo Chaos Solitonsamp Fractals vol 34 no 2 pp 316ndash336 2007

Journal of Applied Mathematics 7

[26] T Allahviranloo N Mikaeilvand and M Barkhordary ldquoFuzzylinear matrix equationrdquo Fuzzy Optimization and Decision Mak-ing vol 8 no 2 pp 165ndash177 2009

[27] X Guo and Z Gong ldquoBlock Gaussian elimination methodsfor fuzzy matrix equationsrdquo International Journal of Pure andApplied Mathematics vol 58 no 2 pp 157ndash168 2010

[28] Z Gong and X Guo ldquoInconsistent fuzzy matrix equationsand its fuzzy least squares solutionsrdquo Applied MathematicalModelling vol 35 no 3 pp 1456ndash1469 2011

[29] X Guo and D Shang ldquoFuzzy symmetric solutions of fuzzymatrix equationsrdquo Advances in Fuzzy Systems vol 2012 ArticleID 318069 9 pages 2012

[30] T Allahviranloo M Ghanbari A A Hosseinzadeh E Haghiand R Nuraei ldquoA note on lsquoFuzzy linear systemsrsquordquo Fuzzy Sets andSystems vol 177 pp 87ndash92 2011

[31] C-T Yeh ldquoWeighted semi-trapezoidal approximations of fuzzynumbersrdquo Fuzzy Sets and Systems vol 165 pp 61ndash80 2011

[32] T Allahviranloo F H Lotfi M K Kiasari and M KhezerlooldquoOn the fuzzy solution of LR fuzzy linear systemsrdquo AppliedMathematical Modelling vol 37 no 3 pp 1170ndash1176 2013

[33] MOtadi andMMosleh ldquoSolving fully fuzzymatrix equationsrdquoApplied Mathematical Modelling vol 36 no 12 pp 6114ndash61212012

[34] X B Guo and D Q Shang ldquoApproximate solutions of LR fuzzySylvester matrix equationsrdquo Journal of Applied Mathematics Inpress

[35] A Ben-Israel and T N E Greville Generalized Inverses Theoryand Applications Springer-Verlag New York NY USA 2ndedition 2003

[36] A Berman and R J Plemmons Nonnegative Matrices in theMathematical Sciences Academic Press New York NY USA1979

[37] R J Plemmons ldquoRegular nonnegative matricesrdquo Proceedings ofthe American Mathematical Society vol 39 pp 26ndash32 1973

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 514984 9 pageshttpdxdoiorg1011552013514984

Research ArticleRanks of a Constrained Hermitian MatrixExpression with Applications

Shao-Wen Yu

Department of Mathematics East China University of Science and Technology Shanghai 200237 China

Correspondence should be addressed to Shao-Wen Yu yushaowenecusteducn

Received 12 November 2012 Accepted 4 January 2013

Academic Editor Yang Zhang

Copyright copy 2013 Shao-Wen YuThis is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We establish the formulas of the maximal and minimal ranks of the quaternion Hermitian matrix expression 1198624minus 1198604119883119860lowast

4where

119883 is a Hermitian solution to quaternion matrix equations 1198601119883 = 119862

1 1198831198611= 1198622 and 119860

3119883119860lowast

3= 1198623 As applications we give a new

necessary and sufficient condition for the existence of Hermitian solution to the system of matrix equations 1198601119883 = 119862

11198831198611= 1198622

1198603119883119860lowast

3= 1198623 and 119860

4119883119860lowast

4= 1198624 which was investigated by Wang and Wu 2010 by rank equalities In addition extremal ranks of

the generalized Hermitian Schur complement 1198624minus 1198604119860sim

3119860lowast

4with respect to a Hermitian g-inverse 119860sim

3of 1198603 which is a common

solution to quaternion matrix equations 1198601119883 = 119862

1and119883119861

1= 1198622 are also considered

1 Introduction

Throughout this paper we denote the real number field byRthe complex number field by C the set of all 119898 times 119899 matricesover the quaternion algebra

H = 1198860+ 1198861119894 + 1198862119895 + 1198863119896 | 1198942

= 1198952

= 1198962

= 119894119895119896 = minus1 1198860 1198861 1198862 1198863isin R

(1)

by H119898times119899 the identity matrix with the appropriate size by 119868the column right space the row left space of a matrix 119860 overH by R(119860) N(119860) respectively the dimension of R(119860) bydimR(119860) a Hermitian g-inverse of a matrix 119860 by 119883 = 119860

∽

which satisfies 119860119860∽119860 = 119860 and 119883 = 119883lowast and the Moore-

Penrose inverse of matrix119860 overH by119860dagger which satisfies fourPenrose equations 119860119860dagger119860 = 119860 119860

dagger

119860119860dagger

= 119860dagger

(119860119860dagger

)lowast

=

119860119860dagger

and (119860dagger

119860)lowast

= 119860dagger

119860 In this case 119860dagger is unique and(119860dagger

)lowast

= (119860lowast

)dagger Moreover 119877

119860and 119871

119860stand for the two

projectors 119871119860

= 119868 minus 119860dagger

119860 119877119860

= 119868 minus 119860119860dagger induced by 119860

Clearly119877119860and 119871

119860are idempotent Hermitian and119877

119860= 119871119860lowast

By [1] for a quaternion matrix 119860 dimR(119860) = dimN(119860)dimR(119860) is called the rank of a quaternion matrix 119860 anddenoted by 119903(119860)

Mitra [2] investigated the system of matrix equations

1198601119883 = 119862

1 119883119861

1= 1198622 (2)

Khatri and Mitra [3] gave necessary and sufficient con-ditions for the existence of the common Hermitian solutionto (2) and presented an explicit expression for the generalHermitian solution to (2) by generalized inverses Using thesingular value decomposition (SVD) Yuan [4] investigatedthe general symmetric solution of (2) over the real numberfield R By the SVD Dai and Lancaster [5] considered thesymmetric solution of equation

119860119883119860lowast

= 119862 (3)

over R which was motivated and illustrated with an inverseproblem of vibration theory Groszlig [6] Tian and Liu [7]gave the solvability conditions for Hermitian solution and itsexpressions of (3) over C in terms of generalized inversesrespectively Liu Tian and Takane [8] investigated ranksof Hermitian and skew-Hermitian solutions to the matrixequation (3) By using the generalized SVD Chang andWang[9] examined the symmetric solution to the matrix equations

1198603119883119860lowast

3= 1198623 119860

4119883119860lowast

4= 1198624

(4)

2 Journal of Applied Mathematics

over R Note that all the matrix equations mentioned aboveare special cases of

1198601119883 = 119862

1 119883119861

1= 1198622

1198603119883119860lowast

3= 1198623 119860

4119883119860lowast

4= 1198624

(5)

Wang and Wu [10] gave some necessary and sufficientconditions for the existence of the common Hermitiansolution to (5) for operators between Hilbert Clowast-modulesby generalized inverses and range inclusion of matrices Inview of the complicated computations of the generalizedinverses of matrices we naturally hope to establish a morepractical necessary and sufficient condition for system (5)over quaternion algebra to have Hermitian solution by rankequalities

As is known to us solutions tomatrix equations and ranksof solutions to matrix equations have been considered previ-ously by many authors [10ndash34] and extremal ranks of matrixexpressions can be used to characterize their rank invariancenonsingularity range inclusion and solvability conditionsof matrix equations Tian and Cheng [35] investigated themaximal and minimal ranks of 119860 minus 119861119883119862 with respect to 119883with applications Tian [36] gave the maximal and minimalranks of 119860

1minus 11986111198831198621subject to a consistent matrix equation

11986121198831198622= 1198602 Tian and Liu [7] established the solvability

conditions for (4) to have a Hermitian solution over C bythe ranks of coefficientmatricesWang and Jiang [20] derivedextreme ranks of (skew)Hermitian solutions to a quaternionmatrix equation 119860119883119860

lowast

+ 119861119884119861lowast

= 119862 Wang Yu and Lin[31] derived the extremal ranks of 119862

4minus 11986041198831198614subject to a

consistent system of matrix equations

1198601119883 = 119862

1 119883119861

1= 1198622 119860

31198831198613= 1198623

(6)

over H and gave a new solvability condition to system

1198601119883 = 119862

1 119883119861

1= 1198622

11986031198831198613= 1198623 119860

41198831198614= 1198624

(7)

In matrix theory and its applications there are manymatrix expressions that have symmetric patterns or involveHermitian (skew-Hermitian) matrices For example

119860 minus 119861119883119861lowast

119860 minus 119861119883 plusmn 119883lowast

119861lowast

119860 minus 119861119883119861lowast

minus 119862119884119862lowast

119860 minus 119861119883119862 plusmn (119861119883119862)lowast

(8)

where 119860 = plusmn119860lowast

119861 and 119862 are given and 119883 and 119884 arevariable matrices In recent papers [7 8 37 38] Liu and Tianconsidered some maximization and minimization problemson the ranks of Hermitian matrix expressions (8)

Define a Hermitian matrix expression

119891 (119883) = 1198624minus 1198604119883119860lowast

4 (9)

where 1198624= 119862lowast

4 we have an observation that by investigating

extremal ranks of (9) where 119883 is a Hermitian solution to asystem of matrix equations

1198601119883 = 119862

1 119883119861

1= 1198622 119860

3119883119860lowast

3= 1198623 (10)

A new necessary and sufficient condition for system (5) tohave Hermitian solution can be given by rank equalitieswhich is more practical than one given by generalizedinverses and range inclusion of matrices

It is well known that Schur complement is one of themostimportant matrix expressions in matrix theory there havebeen many results in the literature on Schur complementsand their applications [39ndash41] Tian [36 42] has investigatedthe maximal and minimal ranks of Schur complements withapplications

Motivated by the workmentioned above we in this paperinvestigate the extremal ranks of the quaternion Hermitianmatrix expression (9) subject to the consistent system ofquaternion matrix equations (10) and its applications InSection 2 we derive the formulas of extremal ranks of (9)with respect to Hermitian solution of (10) As applications inSection 3 we give a new necessary and sufficient conditionfor the existence of Hermitian solution to system (5) byrank equalities In Section 4 we derive extremal ranks ofgeneralized Hermitian Schur complement subject to (2) Wealso consider the rank invariance problem in Section 5

2 Extremal Ranks of (9) Subject to System (10)Corollary 8 in [10] over Hilbert Clowast-modules can be changedinto the following lemma over H

Lemma 1 Let 1198601 1198621

isin H119898times119899 1198611 1198622

isin H119899times119904 1198603

isin

H119903times119899 1198623isin H119903times119903 be given and 119865 = 119861

lowast

11198711198601 119872 = 119878119871

119865 119878 =

11986031198711198601 119863 = 119862

lowast

2minus119861lowast

1119860dagger

11198621 119869 = 119860

dagger

11198621+119865dagger

119863 119866 = 1198623minus1198603(119869+

1198711198601119871lowast

119865119869lowast

)119860lowast

3 then the following statements are equivalent

(1) the system (10) have a Hermitian solution(2) 1198623= 119862lowast

3

11986011198622= 11986211198611 119860

1119862lowast

1= 1198621119860lowast

1 119861

lowast

11198622= 119862lowast

21198611 (11)

11987711986011198621= 0 119877

119865119863 = 0 119877

119872119866 = 0 (12)

(3) 1198623= 119862lowast

3 the equalities in (11) hold and

119903 [11986011198621] = 119903 (119860

1) 119903 [

11986011198621

119861lowast

1119862lowast

2

] = 119903 [1198601

119861lowast

1

]

119903[[

[

11986011198621119860lowast

3

119861lowast

1119862lowast

2119860lowast

3

1198603

1198623

]]

]

= 119903[

[

1198601

119861lowast

1

1198603

]

]

(13)

In that case the general Hermitian solution of (10) can beexpressed as

119883 = 119869 + 1198711198601119871119865119869lowast

+ 1198711198601119871119865119872dagger

119866(119872dagger

)lowast

1198711198651198711198601

+ 11987111986011198711198651198711198721198811198711198651198711198601+ 1198711198601119871119865119881lowast

1198711198721198711198651198711198601

(14)

where 119881 is Hermitian matrix over H with compatible size

Journal of Applied Mathematics 3

Lemma 2 (see Lemma 24 in [24]) Let 119860 isin H119898times119899 119861 isin

H119898times119896 119862 isin H119897times119899 119863 isin H119895times119896 and 119864 isin H119897times119894 Then the followingrank equalities hold

(a) 119903(119862119871119860) = 119903 [

119860

119862] minus 119903(119860)

(b) 119903 [ 119861 119860119871119862 ] = 119903 [119861 119860

0 119862] minus 119903(119862)

(c) 119903 [ 119862119877119861119860

] = 119903 [119862 0

119860 119861] minus 119903(119861)

(d) 119903 [ 119860 119861119871119863119877119864119862 0

] = 119903 [119860 119861 0

119862 0 119864

0 119863 0

] minus 119903(119863) minus 119903(119864)

Lemma 2 plays an important role in simplifying ranks ofvarious block matrices

Liu and Tian [38] has given the following lemma over afield The result can be generalized to H

Lemma 3 Let 119860 = plusmn119860lowast

isin H119898times119898 119861 isin H119898times119899 and 119862 isin H119901times119898

be given then

max119883isinH119899times119901

119903 [119860 minus 119861119883119862 ∓ (119861119883119862)lowast

]

= min119903 [119860 119861 119862lowast

] 119903 [119860 119861

119861lowast

0] 119903 [

119860 119862lowast

119862 0]

min119883isinH119899times119901

119903 [119860 minus 119861119883119862 ∓ (119861119883119862)lowast

]

= 2119903 [119860 119861 119862lowast

] +max 1199041 1199042

(15)

where

1199041= 119903 [

119860 119861

119861lowast

0] minus 2119903 [

119860 119861 119862lowast

119861lowast

0 0]

1199042= 119903 [

119860 119862lowast

119862 0] minus 2119903 [

119860 119861 119862lowast

119862 0 0]

(16)

IfR(119861) sube R(119862lowast

)

max119883

119903 [119860 minus 119861119883119862 minus (119861119883119862)lowast

] = min119903 [119860 119862lowast

] 119903 [119860 119861

119861lowast

0]

max119883

119903 [119860 minus 119861119883119862 minus (119861119883119862)lowast

] = min119903 [119860 119862lowast

] 119903 [119860 119861

119861lowast

0]

(17)

Now we consider the extremal ranks of the matrixexpression (9) subject to the consistent system (10)

Theorem 4 Let 1198601 1198621 1198611 1198622 1198603 and 119862

3be defined as

Lemma 11198624isin H119905times119905 and 119860

4isin H119905times119899Then the extremal ranks

of the quaternionmatrix expression119891(119883) defined as (9) subjectto system (10) are the following

max 119903 [119891 (119883)] = min 119886 119887 (18)

where

119886 = 119903

[[[[

[

1198624

1198604

119862lowast

2119860lowast

4119861lowast

1

1198621119860lowast

41198601

]]]]

]

minus 119903 [119861lowast

1

1198601

]

119887 = 119903

[[[[[[[[

[

0 119860lowast

4119860lowast

31198611

119860lowast

1

11986041198624

0 0 0

1198603

0 minus1198623

minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]]

]

minus 2119903[

[

1198603

119861lowast

1

1198601

]

]

(19)

min 119903 [119891 (119883)] = 2119903

[[[[

[

1198624

1198604

119862lowast

2119860lowast

4119861lowast

1

1198621119860lowast

41198601

]]]]

]

+ 119903

[[[[[[[[[

[

0 119860lowast

4119860lowast

31198611

119860lowast

1

11986041198624

0 0 0

1198603

0 minus1198623

minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]]]

]

minus 2119903

[[[[[[[[[

[

0 119860lowast

41198611

119860lowast

1

11986041198624

0 0

1198603

0 minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus11986211198611minus1198621119860lowast

1

]]]]]]]]]

]

(20)

Proof By Lemma 1 the general Hermitian solution of thesystem (10) can be expressed as

119883 = 119869 + 1198711198601119871119865119869lowast

+ 1198711198601119871119865119872dagger

119866(119872dagger

)lowast

1198711198651198711198601

+ 11987111986011198711198651198711198721198811198711198651198711198601+ 1198711198601119871119865119881lowast

1198711198721198711198651198711198601

(21)

where 119881 is Hermitian matrix over H with appropriate sizeSubstituting (21) into (9) yields

119891 (119883) = 1198624minus 1198604(119869 + 119871

1198601119871119865119869lowast

+ 1198711198601119871119865119872dagger

119866(119872dagger

)lowast

1198711198651198711198601)119860

lowast

4

minus 119860411987111986011198711198651198711198721198811198711198651198711198601119860lowast

4

minus 11986041198711198601119871119865119881lowast

1198711198721198711198651198711198601119860lowast

4

(22)

4 Journal of Applied Mathematics

Put

1198624minus 1198604(119869 + 119871

1198601119871119865119869lowast

+ 1198711198601119871119865119872dagger

119866(119872dagger

)lowast

1198711198651198711198601)119860lowast

4= 119860

119869 + 1198711198601119871119865119869lowast

+ 1198711198601119871119865119872dagger

119866(119872dagger

)lowast

1198711198651198711198601

= 1198691015840

11986041198711198601119871119865119871119872= 119873

1198711198651198711198601119860lowast

4= 119875

(23)

then

119891 (119883) = 119860 minus 119873119881119875 minus (119873119881119875)lowast

(24)

Note that119860 = 119860lowast andR(119873) sube R(119875

lowast

) Thus applying (17) to(24) we get the following

max 119903 [119891 (119883)] = max119881

119903 (119860 minus 119873119881119875 minus (119873119881119875)lowast

)

= min119903 [119860 119875lowast

] 119903 [119860 119873

119873lowast

0]

min 119903 [119891 (119883)] = min119881

119903 (119860 minus 119873119881119875 minus (119873119881119875)lowast

)

= 2119903 [119860 119875lowast

] + 119903 [119860 119873

119873lowast

0] minus 2119903 [

119860 119873

119875 0]

(25)

Now we simplify the ranks of block matrices in (25)In view of Lemma 2 block Gaussian elimination (11)

(12) and (23) we have the following

119903 (119865) = 119903 (119861lowast

11198711198601) = 119903 [

119861lowast

1

1198601

] minus 119903 (1198601)

119903 (119872) = 119903 (119878119871119865) = 119903 [

119878

119865] minus 119903 (119865)

= 119903 [

11986031198711198601

119861lowast

11198711198601

] minus 119903 (119865)

= 119903[

[

1198603

119861lowast

1

1198601

]

]

minus 119903 (1198601) minus 119903 (119865)

119903 [119860 119875lowast

] = 119903 [1198624minus 1198604119869119860lowast

4119875lowast

]

= 119903 [1198624minus 1198604119869119860lowast

411986041198711198601

0 119865] minus 119903 (119865)

= 119903[

[

1198624minus 1198604119869119860lowast

41198604

0 119861lowast

1

0 1198601

]

]

minus 119903 (119865) minus 119903 (1198601)

= 119903

[[[

[

1198624

1198604

119862lowast

2119860lowast

4119861lowast

1

1198621119860lowast

41198601

]]]

]

minus 119903 [119861lowast

1

1198601

]

119903 [119860 119873

119873lowast

0] = 119903 [

1198624minus 11986041198691015840

119860lowast

411986041198711198601119871119865119871119872

119877119872lowast119877119865lowast119877119860lowast

1

119860lowast

40

]

= 119903

[[[[[[

[

1198624minus 11986041198691015840

119860lowast

41198604

0 0 0

119860lowast

40 119860lowast

31198611119860lowast

1

0 1198603

0 0 0

0 119861lowast

10 0 0

0 1198601

0 0 0

]]]]]]

]

minus 2119903 (119872) minus 2119903 (119865) minus 2119903 (1198601)

= 119903

[[[[[[[[[[[[

[

1198624

1198604

0 0 0

119860lowast

40 119860lowast

31198611119860lowast

1

11986031198691015840

119860lowast

41198603

0 0 0

119861lowast

11198691015840

119860lowast

4119861lowast

10 0 0

11986011198691015840

119860lowast

41198601

0 0 0

]]]]]]]]]]]]

]

minus 2119903[

[

1198603

119861lowast

1

1198601

]

]

= 119903

[[[[[[[[[

[

11986241198604

0 0 0

119860lowast

40 119860

lowast

31198611

119860lowast

1

0 1198603

minus1198623

minus11986031198622minus1198603119862lowast

1

0 119861lowast

1minus119862lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

0 1198601minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]]]

]

minus 2119903[

[

1198603

119861lowast

1

1198601

]

]

= 119903

[[[[[[[[[

[

0 119860lowast

4119860lowast

31198611

119860lowast

1

11986041198624

0 0 0

1198603

0 minus1198623

minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]]]

]

minus 2119903[

[

1198603

119861lowast

1

1198601

]

]

119903 [119860 119873

119875 0] = 119903 [

1198624minus 11986041198691015840

119860lowast

411986041198711198601119871119865119871119872

119877119865lowast119877119860lowast

1

119860lowast

40

]

= 119903

[[[[[[[[[[

[

11986241198604

0 0

119860lowast

40 119861

1119860lowast

1

0 1198603minus11986031198622minus1198603119862lowast

1

0 119861lowast

1minus119862lowast

21198611minus119862lowast

2119860lowast

1

0 1198601minus11986211198611minus1198621119860lowast

1

]]]]]]]]]]

]

minus 119903[

[

1198603

119861lowast

1

1198601

]

]

minus 119903 [119861lowast

1

1198601

]

Journal of Applied Mathematics 5

= 119903

[[[[[[[[[[

[

0 119860lowast

41198611

119860lowast

1

11986041198624

0 0

1198603

0 minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus11986211198611minus1198621119860lowast

1

]]]]]]]]]]

]

minus 119903[[

[

1198603

119861lowast

1

1198601

]]

]

minus 119903 [

119861lowast

1

1198601

]

(26)

Substituting (26) into (25) yields (18) and (20)

In Theorem 4 letting 1198624vanish and 119860

4be 119868 with

appropriate size respectively we have the following

Corollary 5 Assume that 1198601 1198621

isin H119898times119899 11986111198622

isin

H119899times119904 1198603isin H119903times119899 and 119862

3isin H119903times119903 are given then the maximal

and minimal ranks of the Hermitian solution 119883 to the system(10) can be expressed as

max 119903 (119883) = min 119886 119887 (27)

where

119886 = 119899 + 119903 [119862lowast

2

1198621

] minus 119903 [119861lowast

1

1198601

]

119887 = 2119899 + 119903

[[[[

[

1198623

119860311986221198603119862lowast

1

119862lowast

2119860lowast

3119862lowast

21198611119862lowast

2119860lowast

1

1198621119860lowast

3119862111986111198621119860lowast

1

]]]]

]

minus 2119903[

[

1198603

119861lowast

1

1198601

]

]

min 119903 (119883) = 2119903 [

119862lowast

2

1198621

]

+ 119903

[[[

[

1198623

119860311986221198603119862lowast

1

119862lowast

2119860lowast

3119862lowast

21198611119862lowast

2119860lowast

1

1198621119860lowast

3119862111986111198621119860lowast

1

]]]

]

minus 2119903

[[[[

[

119860311986221198603119862lowast

1

119862lowast

21198611119862lowast

2119860lowast

1

119862111986111198621119860lowast

1

]]]]

]

(28)

InTheorem 4 assuming that 1198601 1198611 1198621 and 119862

2vanish

we have the following

Corollary 6 Suppose that the matrix equation 1198603119883119860lowast

3= 1198623

is consistent then the extremal ranks of the quaternion matrixexpression 119891(119883) defined as (9) subject to 119860

3119883119860lowast

3= 1198623are the

following

max 119903 [119891 (119883)]

= min

119903 [11986241198604] 119903 [

[

0 119860lowast

4119860lowast

3

11986041198624

0

1198603

0 minus1198623

]

]

minus 2119903 (1198603)

min 119903 [119891 (119883)] = 2119903 [11986241198604]

+ 119903[

[

0 119860lowast

4119860lowast

3

11986041198624

0

1198603

0 minus1198623

]

]

minus 2119903[

[

0 119860lowast

4

11986041198624

1198603

0

]

]

(29)

3 A Practical Solvability Condition forHermitian Solution to System (5)

In this section we use Theorem 4 to give a necessary andsufficient condition for the existence of Hermitian solutionto system (5) by rank equalities

Theorem 7 Let 1198601 1198621

isin H119898times119899 11986111198622

isin H119899times119904 1198603

isin

H119903times119899 1198623isin H119903times119903 119860

4isin H119905times119899 and 119862

4isin H119905times119905be given then

the system (5) have Hermitian solution if and only if 1198623= 119862lowast

3

(11) (13) hold and the following equalities are all satisfied

119903 [11986041198624] = 119903 (119860

4) (30)

119903

[[[

[

1198624

1198604

119862lowast

2119860lowast

4119861lowast

1

1198621119860lowast

41198601

]]]

]

= 119903[

[

1198604

119861lowast

1

1198601

]

]

(31)

119903

[[[[[[[[

[

0 119860lowast

4119860lowast

31198611

119860lowast

1

11986041198624

0 0 0

1198603

0 minus1198623

minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]]

]

= 2119903

[[[[

[

1198604

1198603

119861lowast

1

1198601

]]]]

]

(32)

Proof It is obvious that the system (5) have Hermitiansolution if and only if the system (10) haveHermitian solutionand

min 119903 [119891 (119883)] = 0 (33)

where 119891(119883) is defined as (9) subject to system (10) Let1198830be

aHermitian solution to the system (5) then1198830is aHermitian

solution to system (10) and1198830satisfies119860

41198830119860lowast

4= 1198624 Hence

Lemma 1 yields 1198623

= 119862lowast

3 (11) (13) and (30) It follows

6 Journal of Applied Mathematics

from

[[[[[[

[

119868 0 0 0 0

0 119868 0 0 0

119860311988300 119868 0 0

119861lowast

111988300 0 119868 0

119860111988300 0 0 119868

]]]]]]

]

times

[[[[[[[

[

0 119860lowast

4119860lowast

31198611

119860lowast

1

11986041198624

0 0 0

1198603

0 minus1198623

minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]

]

times

[[[[[

[

119868 minus1198830119860lowast

40 0 0

0 119868 0 0 0

0 0 119868 0 0

0 0 0 119868 0

0 0 0 0 119868

]]]]]

]

=

[[[[[

[

0 119860lowast

4119860lowast

31198611119860lowast

1

1198604

0 0 0 0

1198603

0 0 0 0

119861lowast

10 0 0 0

1198601

0 0 0 0

]]]]]

]

(34)

that (32) holds Similarly we can obtain (31)Conversely assume that 119862

3= 119862lowast

3 (11) (13) hold then by

Lemma 1 system (10) have Hermitian solution By (20) (31)-(32) and

119903

[[[[[[[

[

0 119860lowast

41198611

119860lowast

1

11986041198624

0 0

1198603

0 minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus11986211198611minus1198621119860lowast

1

]]]]]]]

]

ge 119903

[[[

[

1198604

1198603

119861lowast

1

1198601

]]]

]

+ 119903[

[

1198604

119861lowast

1

1198601

]

]

(35)

we can get

min 119903 [119891 (119883)] le 0 (36)

However

min 119903 [119891 (119883)] ge 0 (37)

Hence (33) holds implying that the system (5) have Hermi-tian solution

ByTheorem 7 we can also get the following

Corollary 8 Suppose that 1198603 1198623 1198604 and 119862

4are those in

Theorem 7 then the quaternion matrix equations 1198603119883119860lowast

3=

1198623and 119860

4119883119860lowast

4= 1198624have common Hermitian solution if and

only if (30) hold and the following equalities are satisfied

119903 [11986031198623] = 119903 (119860

3)

119903 [

[

0 119860lowast

4119860lowast

3

11986041198624

0

1198603

0 minus1198623

]

]

= 2119903 [1198603

1198604

]

(38)

Corollary 9 Suppose that11986011198621isin H119898times119899119861

11198622isin H119899times119904 and

119860 119861 isin H119899times119899 are Hermitian Then 119860 and 119861 have a common

Hermitian g-inverse which is a solution to the system (2) if andonly if (11) holds and the following equalities are all satisfied

119903[[

[

11986011198621119860

119861lowast

1119862lowast

2119860

119860 119860

]]

]

= 119903[

[

1198601

119861lowast

1

119860

]

]

119903[[

[

11986011198621119861

119861lowast

1119862lowast

2119861

119861 119861

]]

]

= 119903[

[

1198601

119861lowast

1

119861

]

]

(39)

119903

[[[[[[[

[

0 119861 119860 1198611

119860lowast

1

119861 119861 0 0 0

119860 0 minus119860 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

2119860 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

1119860 minus119862

11198611minus1198621119860lowast

1

]]]]]]]

]

= 2119903

[[[[

[

119861

119860

119861lowast

1

1198601

]]]]

]

(40)

4 Extremal Ranks of Schur ComplementSubject to (2)

As is well known for a given block matrix

119872 = [119860 119861

119861lowast

119863] (41)

where 119860 and 119863 are Hermitian quaternion matrices withappropriate sizes then the Hermitian Schur complement of119860 in119872 is defined as

119878119860= 119863 minus 119861

lowast

119860sim

119861 (42)

where 119860sim is a Hermitian g-inverse of 119860 that is 119860sim isin 119883 |

119860119883119860 = 119860119883 = 119883lowast

Now we use Theorem 4 to establish the extremal ranks

of 119878119860given by (42) with respect to 119860sim which is a solution to

system (2)

Theorem 10 Suppose 1198601 1198621isin H119898times119899 119861

1 1198622isin H119899times119904 119863 isin

H119905times119905 119861 isin H119899times119905 and 119860 isin H119899times119899 are given and system (2)is consistent then the extreme ranks of 119878

119860given by (42) with

respect to 119860sim which is a solution of (2) are the following

max1198601119860sim=1198621

119860sim1198611=1198622

119903 (119878119860) = min 119886 119887

(43)

Journal of Applied Mathematics 7

where

119886 = 119903[

[

119863 119861lowast

119862lowast

2119861 119861lowast

1

1198621119861 1198601

]

]

minus 119903 [119861lowast

1

1198601

]

119887 = 119903

[[[[[[

[

0 119861 119860 1198611

119860lowast

1

119861lowast

119863 0 0 0

119860 0 minus119860 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

2119860 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

1119860 minus119862

11198611minus1198621119860lowast

1

]]]]]]

]

minus 2119903[

[

119860

119861lowast

1

1198601

]

]

min1198601119860sim=1198621

119860sim1198611=1198622

119903 (119878119860) = 2119903[

[

119863 119861lowast

119862lowast

2119861 119861lowast

1

1198621119861 1198601

]

]

+ 119903

[[[[[[[[

[

0 119861 119860 1198611

119860lowast

1

119861lowast

119863 0 0 0

119860 0 minus119860 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

2119860 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

1119860 minus119862

11198611minus1198621119860lowast

1

]]]]]]]]

]

minus 2119903

[[[[[[[[

[

0 119861 1198611

119860lowast

1

119861lowast

119863 0 0

119860 0 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

11198611minus1198621119860lowast

1

]]]]]]]]

]

(44)

Proof It is obvious that

max1198601119860sim=1198621 119860

sim1198611=1198622

119903 (119863 minus 119861lowast

119860sim

119861)

= max1198601119883=1198621 1198831198611=1198622 119860119883119860=119860

119903 (119863 minus 119861lowast

119883119861)

min1198601119860sim=1198621 119860

sim1198611=1198622

119903 (119863 minus 119861lowast

119860sim

119861)

= min1198601119883=1198621 1198831198611=1198622 119860119883119860=119860

119903 (119863 minus 119861lowast

119883119861)

(45)

Thus in Theorem 4 and its proof letting 1198603= 119860lowast

3= 1198623= 119860

1198604= 119861lowast

and 1198624= 119863 we can easily get the proof

In Theorem 10 let 1198601 1198621 1198611 and 119862

2vanish Then we

can easily get the following

Corollary 11 The extreme ranks of 119878119860given by (42) with

respect to 119860sim are the following

max119860sim119903 (119878119860) = min

119903 [119863 119861lowast

] 119903 [

[

0 119861 119860

119861lowast

119863 0

119860 0 minus119860

]

]

minus 2119903 (119860)

min119860sim119903 (119878119860) = 2119903 [119863 119861

lowast

] + 119903[

[

0 119861 119860

119861lowast

119863 0

119860 0 minus119860

]

]

minus 2119903[

[

0 119861

119861lowast

119863

119860 0

]

]

(46)

5 The Rank Invariance of (9)As another application of Theorem 4 we in this sectionconsider the rank invariance of thematrix expression (9) withrespect to the Hermitian solution of system (10)

Theorem 12 Suppose that (10) have Hermitian solution thenthe rank of 119891(119883) defined by (9) with respect to the Hermitiansolution of (10) is invariant if and only if

119903

[[[[[[[[

[

0 119860lowast

41198611

119860lowast

1

11986041198624

0 0

1198603

0 minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus11986211198611minus1198621119860lowast

1

]]]]]]]]

]

= 119903[[

[

1198624

1198604

119862lowast

2119860lowast

4119861lowast

1

1198621119860lowast

41198601

]]

]

+ 119903[

[

1198603

119861lowast

1

1198601

]

]

119903

[[[[[[[[

[

0 119860lowast

4119860lowast

31198611

119860lowast

1

11986041198624

0 0 0

1198603

0 minus1198623

minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]]

]

+ 119903 [119861lowast

1

1198601

]

= 119903

[[[[[[

[

0 119860lowast

41198611

119860lowast

1

11986041198624

0 0

1198603

0 minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus11986211198611minus1198621119860lowast

1

]]]]]]

]

+ 119903[

[

1198603

119861lowast

1

1198601

]

]

(47)

or

119903

[[[[[[[

[

0 119860lowast

41198611

119860lowast

1

11986041198624

0 0

1198603

0 minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus11986211198611minus1198621119860lowast

1

]]]]]]]

]

= 119903[[

[

1198624

1198604

119862lowast

2119860lowast

4119861lowast

1

1198621119860lowast

41198601

]]

]

+ 119903[[

[

1198603

119861lowast

1

1198601

]]

]

(48)

Proof It is obvious that the rank of 119891(119883) with respect toHermitian solution of system (10) is invariant if and only if

max 119903 [119891 (119883)] minusmin 119903 [119891 (119883)] = 0 (49)

By (49) Theorem 4 and simplifications we can get (47)and (48)

8 Journal of Applied Mathematics

Corollary 13 The rank of 119878119860defined by (42) with respect to

119860sim which is a solution to system (2) is invariant if and only if

119903

[[[[[[

[

0 119861 1198611

119860lowast

1

119861lowast

119863 0 0

119860 0 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

11198611minus1198621119860lowast

1

]]]]]]

]

= 119903[

[

119863 119861lowast

119862lowast

2119861 119861lowast

1

1198621119861 1198601

]

]

+ 119903[

[

119860

119861lowast

1

1198601

]

]

119903

[[[[[[[[

[

0 119861 119860 1198611

119860lowast

1

119861lowast

119863 0 0 0

119860 0 minus119860 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

2119860 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

1119860 minus119862

11198611minus1198621119860lowast

1

]]]]]]]]

]

+ 119903 [119861lowast

1

1198601

]

= 119903

[[[[[[

[

0 119861 1198611

119860lowast

1

119861lowast

119863 0 0

119860 0 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

11198611minus1198621119860lowast

1

]]]]]]

]

+ 119903[

[

119860

119861lowast

1

1198601

]

]

(50)

or

119903

[[[[[[

[

0 119861 1198611

119860lowast

1

119861lowast

119863 0 0

119860 0 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

11198611minus1198621119860lowast

1

]]]]]]

]

= 119903[

[

119863 119861lowast

119862lowast

2119861 119861lowast

1

1198621119861 1198601

]

]

+ 119903[

[

119860

119861lowast

1

1198601

]

]

(51)

Acknowledgments

This research was supported by the National Natural ScienceFoundation of China Tian Yuan Foundation (11226067) theFundamental Research Funds for the Central Universities(WM1214063) and China Postdoctoral Science Foundation(2012M511014)

References

[1] T W Hungerford Algebra Springer New York NY USA 1980[2] S K Mitra ldquoThe matrix equations 119860119883 = 119862 119883119861 = 119863rdquo Linear

Algebra and its Applications vol 59 pp 171ndash181 1984[3] C G Khatri and S K Mitra ldquoHermitian and nonnegative

definite solutions of linear matrix equationsrdquo SIAM Journal onApplied Mathematics vol 31 no 4 pp 579ndash585 1976

[4] Y X Yuan ldquoOn the symmetric solutions of matrix equation(119860119883119883119862) = (119861119863)rdquo Journal of East China Shipbuilding Institutevol 15 no 4 pp 82ndash85 2001 (Chinese)

[5] H Dai and P Lancaster ldquoLinear matrix equations from aninverse problem of vibration theoryrdquo Linear Algebra and itsApplications vol 246 pp 31ndash47 1996

[6] J Groszlig ldquoA note on the general Hermitian solution to 119860119883119860lowast =119861rdquo Malaysian Mathematical Society vol 21 no 2 pp 57ndash621998

[7] Y G Tian and Y H Liu ldquoExtremal ranks of some symmetricmatrix expressions with applicationsrdquo SIAM Journal on MatrixAnalysis and Applications vol 28 no 3 pp 890ndash905 2006

[8] Y H Liu Y G Tian and Y Takane ldquoRanks of Hermitian andskew-Hermitian solutions to the matrix equation 119860119883119860lowast = 119861rdquoLinear Algebra and its Applications vol 431 no 12 pp 2359ndash2372 2009

[9] X W Chang and J S Wang ldquoThe symmetric solution of thematrix equations 119860119883 + 119884119860 = 119862 119860119883119860119879 + 119861119884119861

119879

= 119862 and(119860119879

119883119860 119861119879

119883119861) = (119862119863)rdquo Linear Algebra and its Applicationsvol 179 pp 171ndash189 1993

[10] Q-W Wang and Z-C Wu ldquoCommon Hermitian solutionsto some operator equations on Hilbert 119862lowast-modulesrdquo LinearAlgebra and its Applications vol 432 no 12 pp 3159ndash3171 2010

[11] F O Farid M S Moslehian Q-W Wang and Z-C Wu ldquoOnthe Hermitian solutions to a system of adjointable operatorequationsrdquo Linear Algebra and its Applications vol 437 no 7pp 1854ndash1891 2012

[12] Z-H He and Q-WWang ldquoSolutions to optimization problemson ranks and inertias of a matrix function with applicationsrdquoAppliedMathematics andComputation vol 219 no 6 pp 2989ndash3001 2012

[13] Q-W Wang ldquoThe general solution to a system of real quater-nion matrix equationsrdquo Computers amp Mathematics with Appli-cations vol 49 no 5-6 pp 665ndash675 2005

[14] Q-W Wang ldquoBisymmetric and centrosymmetric solutions tosystems of real quaternion matrix equationsrdquo Computers ampMathematics with Applications vol 49 no 5-6 pp 641ndash6502005

[15] Q W Wang ldquoA system of four matrix equations over vonNeumann regular rings and its applicationsrdquo Acta MathematicaSinica vol 21 no 2 pp 323ndash334 2005

[16] Q-W Wang ldquoA system of matrix equations and a linear matrixequation over arbitrary regular rings with identityrdquo LinearAlgebra and its Applications vol 384 pp 43ndash54 2004

[17] Q-WWang H-X Chang andQNing ldquoThe common solutionto six quaternion matrix equations with applicationsrdquo AppliedMathematics and Computation vol 198 no 1 pp 209ndash2262008

[18] Q-W Wang H-X Chang and C-Y Lin ldquoP-(skew)symmetriccommon solutions to a pair of quaternion matrix equationsrdquoApplied Mathematics and Computation vol 195 no 2 pp 721ndash732 2008

[19] Q W Wang and Z H He ldquoSome matrix equations withapplicationsrdquo Linear and Multilinear Algebra vol 60 no 11-12pp 1327ndash1353 2012

[20] Q W Wang and J Jiang ldquoExtreme ranks of (skew-)Hermitiansolutions to a quaternion matrix equationrdquo Electronic Journal ofLinear Algebra vol 20 pp 552ndash573 2010

[21] Q-W Wang and C-K Li ldquoRanks and the least-norm of thegeneral solution to a system of quaternion matrix equationsrdquoLinear Algebra and its Applications vol 430 no 5-6 pp 1626ndash1640 2009

[22] Q-W Wang X Liu and S-W Yu ldquoThe common bisymmetricnonnegative definite solutions with extreme ranks and inertiasto a pair of matrix equationsrdquo Applied Mathematics and Com-putation vol 218 no 6 pp 2761ndash2771 2011

Journal of Applied Mathematics 9

[23] Q-W Wang F Qin and C-Y Lin ldquoThe common solution tomatrix equations over a regular ring with applicationsrdquo IndianJournal of Pure andAppliedMathematics vol 36 no 12 pp 655ndash672 2005

[24] Q-W Wang G-J Song and C-Y Lin ldquoExtreme ranks of thesolution to a consistent system of linear quaternion matrixequations with an applicationrdquo Applied Mathematics and Com-putation vol 189 no 2 pp 1517ndash1532 2007

[25] Q-W Wang G-J Song and C-Y Lin ldquoRank equalities relatedto the generalized inverse 119860

(2)

119879119878with applicationsrdquo Applied

Mathematics and Computation vol 205 no 1 pp 370ndash3822008

[26] Q W Wang G J Song and X Liu ldquoMaximal and minimalranks of the common solution of some linear matrix equationsover an arbitrary division ring with applicationsrdquo AlgebraColloquium vol 16 no 2 pp 293ndash308 2009

[27] Q-W Wang Z-C Wu and C-Y Lin ldquoExtremal ranks ofa quaternion matrix expression subject to consistent systemsof quaternion matrix equations with applicationsrdquo AppliedMathematics and Computation vol 182 no 2 pp 1755ndash17642006

[28] Q W Wang and S W Yu ldquoRanks of the common solution tosome quaternion matrix equations with applicationsrdquo Bulletinof Iranian Mathematical Society vol 38 no 1 pp 131ndash157 2012

[29] Q W Wang S W Yu and W Xie ldquoExtreme ranks of realmatrices in solution of the quaternion matrix equation 119860119883119861 =

119862with applicationsrdquoAlgebra Colloquium vol 17 no 2 pp 345ndash360 2010

[30] Q-W Wang S-W Yu and Q Zhang ldquoThe real solutions toa system of quaternion matrix equations with applicationsrdquoCommunications in Algebra vol 37 no 6 pp 2060ndash2079 2009

[31] Q-W Wang S-W Yu and C-Y Lin ldquoExtreme ranks of alinear quaternionmatrix expression subject to triple quaternionmatrix equations with applicationsrdquo Applied Mathematics andComputation vol 195 no 2 pp 733ndash744 2008

[32] Q-W Wang and F Zhang ldquoThe reflexive re-nonnegativedefinite solution to a quaternion matrix equationrdquo ElectronicJournal of Linear Algebra vol 17 pp 88ndash101 2008

[33] Q W Wang X Zhang and Z H He ldquoOn the Hermitianstructures of the solution to a pair of matrix equationsrdquo Linearand Multilinear Algebra vol 61 no 1 pp 73ndash90 2013

[34] X Zhang Q-W Wang and X Liu ldquoInertias and ranks ofsome Hermitian matrix functions with applicationsrdquo CentralEuropean Journal of Mathematics vol 10 no 1 pp 329ndash3512012

[35] Y G Tian and S Z Cheng ldquoThemaximal andminimal ranks of119860 minus 119861119883119862 with applicationsrdquo New York Journal of Mathematicsvol 9 pp 345ndash362 2003

[36] Y G Tian ldquoUpper and lower bounds for ranks of matrixexpressions using generalized inversesrdquo Linear Algebra and itsApplications vol 355 pp 187ndash214 2002

[37] Y H Liu and Y G Tian ldquoMore on extremal ranks of thematrix expressions119860minus119861119883plusmn119883

lowast

119861lowast with statistical applicationsrdquo

Numerical Linear Algebra with Applications vol 15 no 4 pp307ndash325 2008

[38] Y H Liu and Y G Tian ldquoMax-min problems on the ranksand inertias of the matrix expressions 119860 minus 119861119883119862 plusmn (119861119883119862)

lowast withapplicationsrdquo Journal of Optimization Theory and Applicationsvol 148 no 3 pp 593ndash622 2011

[39] T Ando ldquoGeneralized Schur complementsrdquo Linear Algebra andits Applications vol 27 pp 173ndash186 1979

[40] D Carlson E Haynsworth and T Markham ldquoA generalizationof the Schur complement by means of the Moore-Penroseinverserdquo SIAM Journal onAppliedMathematics vol 26 pp 169ndash175 1974

[41] M Fiedler ldquoRemarks on the Schur complementrdquo Linear Algebraand its Applications vol 39 pp 189ndash195 1981

[42] Y G Tian ldquoMore on maximal and minimal ranks of Schurcomplements with applicationsrdquo Applied Mathematics andComputation vol 152 no 3 pp 675ndash692 2004

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 217540 14 pageshttpdxdoiorg1011552013217540

Research ArticleAn Efficient Algorithm for the Reflexive Solution of theQuaternion Matrix Equation 119860119883119861 + 119862119883119867119863 = 119865

Ning Li12 Qing-Wen Wang1 and Jing Jiang3

1 Department of Mathematics Shanghai University Shanghai 200444 China2 School of Mathematics and Quantitative Economics Shandong University of Finance and Economics Jinan 250002 China3Department of Mathematics Qilu Normal University Jinan 250013 China

Correspondence should be addressed to Qing-Wen Wang wqw858yahoocomcn

Received 3 October 2012 Accepted 4 December 2012

Academic Editor P N Shivakumar

Copyright copy 2013 Ning Li et al This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We propose an iterative algorithm for solving the reflexive solution of the quaternion matrix equation 119860119883119861 + 119862119883119867

119863 = 119865 Whenthe matrix equation is consistent over reflexive matrix X a reflexive solution can be obtained within finite iteration steps in theabsence of roundoff errors By the proposed iterative algorithm the least Frobenius norm reflexive solution of the matrix equationcan be derived when an appropriate initial iterative matrix is chosen Furthermore the optimal approximate reflexive solution to agiven reflexive matrix119883

0can be derived by finding the least Frobenius norm reflexive solution of a new corresponding quaternion

matrix equation Finally two numerical examples are given to illustrate the efficiency of the proposed methods

1 Introduction

Throughout the paper the notations R119898times119899 and H119898times119899 repre-sent the set of all 119898 times 119899 real matrices and the set of all 119898 times 119899

matrices over the quaternion algebraH = 1198861+ 119886

2119894+ 119886

3119895+ 119886

4119896 |

1198942

= 1198952

= 1198962

= 119894119895119896 = minus1 1198861 119886

2 119886

3 119886

4isin R We denote the

identity matrix with the appropriate size by 119868 We denote theconjugate transpose the transpose the conjugate the tracethe column space the real part the 119898119899 times 1 vector formedby the vertical concatenation of the respective columnsof a matrix 119860 by 119860

119867

119860119879

119860 tr(119860) 119877(119860)Re(119860) and vec(119860)respectivelyThe Frobenius norm of119860 is denoted by 119860 thatis 119860 = radictr(119860119867119860) Moreover119860otimes119861 and119860⊙119861 stand for theKronecker matrix product and Hadmard matrix product ofthe matrices 119860 and 119861

Let 119876 isin H119899times119899 be a generalized reflection matrix that is119876

2

= 119868 and 119876119867

= 119876 A matrix 119860 is called reflexive withrespect to the generalized reflection matrix 119876 if 119860 = 119876119860119876It is obvious that any matrix is reflexive with respect to 119868Let RH119899times119899

(119876) denote the set of order 119899 reflexive matriceswith respect to 119876 The reflexive matrices with respect to ageneralized reflection matrix 119876 have been widely used inengineering and scientific computations [1 2]

In the field of matrix algebra quaternion matrix equa-tions have received much attention Wang et al [3] gavenecessary and sufficient conditions for the existence andthe representations of P-symmetric and P-skew-symmetricsolutions of quaternion matrix equations 119860

119886119883 = 119862

119886and

119860119887119883119861

119887= 119862

119887 Yuan and Wang [4] derived the expressions of

the least squares 120578-Hermitian solution with the least normand the expressions of the least squares anti-120578-Hermitiansolution with the least norm for the quaternion matrixequation 119860119883119861 + 119862119883119863 = 119864 Jiang and Wei [5] derivedthe explicit solution of the quaternion matrix equation 119883 minus

119860119883119861 = 119862 Li and Wu [6] gave the expressions of symmetricand skew-antisymmetric solutions of the quaternion matrixequations 119860

1119883 = 119862

1and 119883119861

3= 119862

3 Feng and Cheng [7]

gave a clear description of the solution set of the quaternionmatrix equation 119860119883 minus 119883119861 = 0

The iterative method is a very important method tosolve matrix equations Peng [8] constructed a finite iterationmethod to solve the least squares symmetric solutions oflinear matrix equation 119860119883119861 = 119862 Also Peng [9ndash11] presentedseveral efficient iteration methods to solve the constrainedleast squares solutions of linear matrix equations 119860119883119861 =

119862 and 119860119883119861 + 119862119884119863 = 119864 by using Paigersquos algorithm [12]

2 Journal of Applied Mathematics

as the frame method Duan et al [13ndash17] proposed iterativealgorithms for the (Hermitian) positive definite solutionsof some nonlinear matrix equations Ding et al proposedthe hierarchical gradient-based iterative algorithms [18] andhierarchical least squares iterative algorithms [19] for solvinggeneral (coupled) matrix equations based on the hierarchi-cal identification principle [20] Wang et al [21] proposedan iterative method for the least squares minimum-normsymmetric solution of 119860X119861 = 119864 Dehghan and Hajarianconstructed finite iterative algorithms to solve several linearmatrix equations over (anti)reflexive [22ndash24] generalizedcentrosymmetric [25 26] and generalized bisymmetric [2728] matrices Recently Wu et al [29ndash31] proposed iterativealgorithms for solving various complex matrix equations

However to the best of our knowledge there has beenlittle information on iterative methods for finding a solutionof a quaternionmatrix equation Due to the noncommutativemultiplication of quaternions the study of quaternionmatrixequations is more complex than that of real and complexequations Motivated by the work mentioned above andkeeping the interests and wide applications of quaternionmatrices in view (eg [32ndash45]) we in this paper consideran iterative algorithm for the following two problems

Problem 1 For given matrices 119860119862 isin H119898times119899

119861 119863 isin

H119899times119901

119865 isin H119898times119901 and the generalized reflectionmatrix119876 find119883 isin RH119899times119899

(119876) such that

119860119883119861 + 119862119883119867

119863 = 119865 (1)

Problem 2 When Problem 1 is consistent let its solutionset be denoted by 119878

119867 For a given reflexive matrix 119883

0isin

RH119899times119899

(119876) find119883 isin RH119899times119899

(119876) such that10038171003817100381710038171003817119883 minus 119883

0

10038171003817100381710038171003817= min

119883isin119878119867

1003817100381710038171003817119883 minus 1198830

1003817100381710038171003817 (2)

The remainder of this paper is organized as followsIn Section 2 we give some preliminaries In Section 3 weintroduce an iterative algorithm for solving Problem 1 Thenwe prove that the given algorithm can be used to obtain areflexive solution for any initial matrix within finite stepsin the absence of roundoff errors Also we prove that theleast Frobenius norm reflexive solution can be obtained bychoosing a special kind of initial matrix In addition theoptimal reflexive solution of Problem 2 by finding the leastFrobenius norm reflexive solution of a new matrix equationis given In Section 4 we give two numerical examples toillustrate our results In Section 5 we give some conclusionsto end this paper

2 Preliminary

In this section we provide some results which will playimportant roles in this paper First we give a real innerproduct for the space H119898times119899 over the real field R

Theorem 3 In the space H119898times119899 over the field R a real innerproduct can be defined as

⟨119860 119861⟩ = Re [tr (119861119867

119860)] (3)

for 119860 119861 isin H119898times119899 This real inner product space is denoted as(H119898times119899

R ⟨sdot sdot⟩)

Proof (1) For 119860 isin H119898times119899 let 119860 = 1198601+ 119860

2119894 + 119860

3119895 + 119860

4119896 then

⟨119860 119860⟩ = Re [tr (119860119867

119860)]

= tr (119860119879

1119860

1+ 119860

119879

2119860

2+ 119860

119879

3119860

3+ 119860

119879

4119860

4)

(4)

It is obvious that ⟨119860 119860⟩ gt 0 and ⟨119860 119860⟩ = 0 hArr 119860 = 0(2) For 119860 119861 isin H119898times119899 let 119860 = 119860

1+ 119860

2119894 + 119860

3119895 + 119860

4119896 and

119861 = 1198611+ 119861

2119894 + 119861

3119895 + 119861

4119896 then we have

⟨119860 119861⟩ = Re [tr (119861119867

119860)]

= tr (119861119879

1119860

1+ 119861

119879

2119860

2+ 119861

119879

3119860

3+ 119861

119879

4119860

4)

= tr (119860119879

1119861

1+ 119860

119879

2119861

2+ 119860

119879

3119861

3+ 119860

119879

4119861

4)

= Re [tr (119860119867

119861)] = ⟨119861 119860⟩

(5)

(3) For 119860 119861 119862 isin H119898times119899

⟨119860 + 119861 119862⟩ = Re tr [119862119867

(119860 + 119861)]

= Re [tr (119862119867

119860 + 119862119867

119861)]

= Re [tr (119862119867

119860)] + Re [tr (119862119867

119861)]

= ⟨119860 119862⟩ + ⟨119861 119862⟩

(6)

(4) For 119860 119861 isin H119898times119899 and 119886 isin R

⟨119886119860 119861⟩ = Re tr [119861119867

(119886119860)] = Re [tr (119886119861119867

119860)]

= 119886Re [tr (119861119867

119860)] = 119886 ⟨119860 119861⟩

(7)

All the above arguments reveal that the space H119898times119899 overfield R with the inner product defined in (3) is an innerproduct space

Let sdot Δrepresent the matrix norm induced by the inner

product ⟨sdot sdot⟩ For an arbitrary quaternion matrix 119860 isin H119898times119899it is obvious that the following equalities hold

119860Δ= radic⟨119860119860⟩ = radicRe [tr (119860119867119860)] = radictr (119860119867119860) = 119860

(8)

which reveals that the induced matrix norm is exactly theFrobenius norm For convenience we still use sdot to denotethe induced matrix norm

Let 119864119894119895

denote the 119898 times 119899 quaternion matrix whose(119894 119895) entry is 1 and the other elements are zeros In innerproduct space (H119898times119899

R ⟨sdot sdot⟩) it is easy to verify that 119864119894119895 119864

119894119895119894

119864119894119895119895 119864

119894119895119896 119894 = 1 2 119898 119895 = 1 2 119899 is an orthonormal

basis which reveals that the dimension of the inner productspace (H119898times119899

R ⟨sdot sdot⟩) is 4119898119899Next we introduce a real representation of a quaternion

matrix

Journal of Applied Mathematics 3

For an arbitrary quaternion matrix 119872 = 1198721+ 119872

2119894 +

1198723119895+ 119872

4119896 a map 120601(sdot) fromH119898times119899 toR4119898times4119899 can be defined

as

120601 (119872) =

[[[

[

1198721

minus1198722

minus1198723

minus1198724

1198722

1198721

minus1198724

1198723

1198723

1198724

1198721

minus1198722

1198724

minus1198723

1198722

1198721

]]]

]

(9)

Lemma4 (see [41]) Let119872 and119873 be two arbitrary quaternionmatrices with appropriate size The map 120601(sdot) defined by (9)satisfies the following properties

(a) 119872 = 119873 hArr 120601(119872) = 120601(119873)

(b) 120601(119872 + 119873) = 120601(119872) + 120601(119873) 120601(119872119873) = 120601(119872)120601(119873)

120601(119896119872) = 119896120601(119872) 119896 isin R

(c) 120601(119872119867

) = 120601119879

(119872)

(d) 120601(119872) = 119879minus1

119898120601(119872)119879

119899= 119877

minus1

119898120601(119872)119877

119899= 119878

minus1

119898120601(119872)119878

119899

where

119879119905=

[[[

[

0 minus119868119905

0 0

119868119905

0 0 0

0 0 0 119868119905

0 0 minus119868119905

0

]]]

]

119877119905= [

0 minus1198682119905

1198682119905

0]

119878119905=

[[[

[

0 0 0 minus119868119905

0 0 119868119905

0

0 minus119868119905

0 0

119868119905

0 0 0

]]]

]

119905 = 119898 119899

(10)

By (9) it is easy to verify that

1003817100381710038171003817120601 (119872)1003817100381710038171003817 = 2 119872 (11)

Finally we introduce the commutation matrixA commutation matrix 119875(119898 119899) is a 119898119899 times 119898119899 matrix

which has the following explicit form

119875 (119898 119899) =

119898

sum

119894=1

119899

sum

119895=1

119864119894119895otimes 119864

119879

119894119895= [119864

119879

119894119895] 119894=1119898

119895=1119899

(12)

Moreover 119875(119898 119899) is a permutation matrix and 119875(119898 119899) =

119875119879

(119899119898) = 119875minus1

(119899119898) We have the following lemmas on thecommutation matrix

Lemma 5 (see [46]) Let 119860 be a 119898 times 119899 matrix There is acommutation matrix 119875(119898 119899) such that

vec (119860119879

) = 119875 (119898 119899) vec (119860) (13)

Lemma 6 (see [46]) Let 119860 be a 119898 times 119899 matrix and 119861 a 119901 times

119902 matrix There exist two commutation matrices 119875(119898 119901) and119875(119899 119902) such that

119861 otimes 119860 = 119875119879

(119898 119901) (119860 otimes 119861) 119875 (119899 119902) (14)

3 Main Results

31 The Solution of Problem 1 In this subsection we willconstruct an algorithm for solving Problem 1 Then somelemmaswill be given to analyse the properties of the proposedalgorithm Using these lemmas we prove that the proposedalgorithm is convergent

Algorithm 7 (Iterative algorithm for Problem 1)

(1) Choose an initial matrix119883(1) isin RH119899times119899

(119876)(2) Calculate

119877 (1) = 119865 minus 119860119883 (1) 119861 minus 119862119883119867

(1)119863

119875 (1) =1

2(119860

119867

119877 (1) 119861119867

+ 119863119877119867

(1) 119862

+ 119876119860119867

119877 (1) 119861119867

119876 + 119876119863119877119867

(1) 119862119876)

119896 = 1

(15)

(3) If 119877(119896) = 0 then stop and 119883(119896) is the solution ofProblem 1 else if 119877(119896) =0 and 119875(119896) = 0 then stopand Problem 1 is not consistent else 119896 = 119896 + 1

(4) Calculate

119883 (119896) = 119883 (119896 minus 1) +119877 (119896 minus 1)

2

119875 (119896 minus 1)2119875 (119896 minus 1)

119877 (119896) = 119877 (119896 minus 1) minus119877 (119896 minus 1)

2

119875 (119896 minus 1)2

times (119860119875 (119896 minus 1) 119861 + 119862119875119867

(119896 minus 1)119863)

119875 (119896) =1

2(119860

119867

119877 (119896) 119861119867

+ 119863119877119867

(119896) 119862

+ 119876119860119867

119877 (119896) 119861119867

119876 +119876119863119877119867

(119896) 119862119876)

+119877 (119896)

2

119877 (119896 minus 1)2119875 (119896 minus 1)

(16)

(5) Go to Step (3)

Lemma 8 Assume that the sequences 119877(119894) and 119875(119894)

are generated by Algorithm 7 then ⟨119877(119894) 119877(119895)⟩ =

0 and ⟨119875(119894) 119875(119895)⟩ = 0 for 119894 119895 = 1 2

119894 =119895

Proof Since ⟨119877(119894) 119877(119895)⟩ = ⟨119877(119895) 119877(119894)⟩ and⟨119875(119894) 119875(119895)⟩ = ⟨119875(119895) 119875(119894)⟩ for 119894 119895 = 1

2 we only need to prove that ⟨119877(119894)

119877(119895)⟩ = 0 and ⟨119875(119894) 119875(119895)⟩ = 0 for 1 le 119894 lt 119895Now we prove this conclusion by induction

Step 1 We show that

⟨119877 (119894) 119877 (119894 + 1)⟩ = 0

⟨119875 (119894) 119875 (119894 + 1)⟩ = 0 for 119894 = 1 2

(17)

4 Journal of Applied Mathematics

We also prove (17) by induction When 119894 = 1 we have

⟨119877 (1) 119877 (2)⟩

= Re tr [119877119867

(2) 119877 (1)]

= Retr[(119877 (1) minus119877 (1)

2

119875 (1)2(119860119875 (1) 119861 + 119862119875

119867

(1)119863))

119867

times 119877 (1) ]

= 119877 (1)2

minus119877 (1)

2

119875 (1)2

times Re tr [119875119867

(1) (119860119867

119877 (1) 119861119867

+ 119863119877119867

(1) 119862)]

= 119877(1)2

minus119877(1)

2

119875(1)2

timesRetr [119875119867

(1) ((119860119867

119877 (1) 119861119867

+119863119877119867

(1) 119862+119876119860119867

119877 (1)

times 119861119867

119876 + 119876119863119877119867

(1) 119862119876) times (2)minus1

)]

= 119877 (1)2

minus119877 (1)

2

119875 (1)2119875 (1)

2

= 0

(18)

Also we can write

⟨119875 (1) 119875 (2)⟩

= Re tr [119875119867

(2) 119875 (1)]

= Retr[(1

2(119860

119867

119877 (2) 119861119867

+ 119863119877119867

(2) 119862

+ 119876119860119867

119877 (2) 119861119867

119876 + 119876119863119877119867

(2) 119862119876)

+119877 (2)

2

119877 (1)2119875 (1))

119867

119875 (1)]

=119877 (2)

2

119877 (1)2119875 (1)

2

+ Re tr [119875119867

(1) times ((119860119867

119877 (2) 119861119867

+ 119863119877119867

(2) 119862

+ 119876119860119867

119877 (2) 119861119867

119876

+119876119863119877119867

(2) 119862119876) times (2)minus1

)]

=119877 (2)

2

119877 (1)2119875 (1)

2

+ Re tr [119875119867

(1) (119860119867

119877 (2) 119861119867

+ 119863119877119867

(2) 119862)]

=119877 (2)

2

119877 (1)2119875 (1)

2

+ Re tr [119877119867

(2) (119860119875 (1) 119861 + 119862119875119867

(1)119863)]

=119877 (2)

2

119877 (1)2119875 (1)

2

+119875 (1)

2

119877 (1)2Re tr [119877119867

(2) (119877 (1) minus 119877 (2))]

=119877 (2)

2

119877 (1)2119875 (1)

2

minus119875 (1)

2

119877 (1)2119877 (2)

2

= 0

(19)

Now assume that conclusion (17) holds for 1 le 119894 le 119904 minus 1 then

⟨119877 (119904) 119877 (119904 + 1)⟩

= Re tr [119877119867

(119904 + 1) 119877 (119904)]

= Retr[(119877 (119904) minus119877 (119904)

2

119875 (119904)2(119860119875 (119904) 119861 + 119862119875

119867

(119904)119863))

119867

times 119877 (119904) ]

= 119877 (s)2 minus119877 (119904)

2

119875 (119904)2

times Re tr [119875119867

(119904) (119860119867

119877 (119904) 119861119867

+ 119863119877119867

(119904) 119862)]

= 119877 (119904)2

minus119877 (119904)

2

119875 (119904)2

times Re tr [119875119867

(119904) times ((119860119867

119877 (119904) 119861119867

+ 119863119877119867

(119904) 119862

+ 119876119860119867

119877 (119904) 119861119867

119876

+ 119876119863119877119867

(119904) 119862119876) times (2)minus1

)]

= 119877 (119904)2

minus119877 (119904)

2

119875 (119904)2

times Retr[119875119867

(119904) (119875 (119904) minus119877 (119904)

2

119877 (119904 minus 1)2119875 (119904 minus 1))]

= 0

(20)

Journal of Applied Mathematics 5

And it can also be obtained that

⟨119875 (119904) 119875 (119904 + 1)⟩

= Re tr [119875119867

(119904 + 1) 119875 (119904)]

= Retr[(((119860119867

119877 (119904 + 1) 119861119867

+ 119863119877119867

(119904 + 1) 119862

+ 119876119860119867

119877 (119904 + 1) 119861119867

119876

+ 119876119863119877119867

(119904 + 1) 119862119876) times (2)minus1

)

+119877 (119904 + 1)

2

119877 (119904)2

119875(119904))

119867

119875 (119904)]

=119877 (119904 + 1)

2

119877 (119904)2

119875 (119904)2

+ Re tr [119875119867

(119904) (119860119867

119877 (119904 + 1) 119861119867

+ 119863119877119867

(119904 + 1) 119862)]

=119877 (119904 + 1)

2

119877 (119904)2

119875 (119904)2

+ Re tr [119877119867

(119904 + 1) (119860119875 (119904) 119861 + 119862119875119867

(119904)119863)]

=119877 (119904 + 1)

2

119877 (119904)2

119875 (119904)2

+119875 (119904)

2

119877 (119904)2Re tr [119877119867

(119904 + 1) (119877 (119904) minus 119877 (119904 + 1))]

= 0

(21)

Therefore the conclusion (17) holds for 119894 = 1 2 Step 2 Assume that ⟨119877(119894) 119877(119894+119903)⟩ = 0 and ⟨119875(119894) 119875(119894+119903)⟩ = 0

for 119894 ge 1 and 119903 ge 1 We will show that

⟨119877 (119894) 119877 (119894 + 119903 + 1)⟩ = 0 ⟨119875 (119894) 119875 (119894 + 119903 + 1)⟩ = 0

(22)

We prove the conclusion (22) in two substepsSubstep 21 In this substep we show that

⟨119877 (1) 119877 (119903 + 2)⟩ = 0 ⟨119875 (1) 119875 (119903 + 2)⟩ = 0 (23)

It follows from Algorithm 7 that

⟨119877 (1) 119877 (119903 + 2)⟩

= Re tr [119877119867

(119903 + 2) 119877 (1)]

= Retr[(119877 (119903 + 1) minus119877 (119903 + 1)

2

119875 (119903 + 1)2

times (119860119875 (119903 + 1) 119861+119862119875119867

(119903 + 1)119863))

119867

119877 (1)]

= Re tr [119877119867

(119903 + 1) 119877 (1)] minus119877 (119903 + 1)

2

119875 (119903 + 1)2

times Re tr [119875119867

(119903 + 1) (119860119867

119877 (1) 119861119867

+ 119863119877119867

(1) 119862)]

= Re tr [119877119867

(119903 + 1) 119877 (1)] minus119877 (119903 + 1)

2

119875 (119903 + 1)2

times Re tr [119875119867

(119903 + 1) ((119860119867

119877 (1) 119861119867

+ 119863119877119867

(1) 119862

+ 119876119860119867

119877 (1) 119861119867

119876

+ 119876119863119877119867

(1) 119862119876) times (2)minus1

)]

= Re tr [119877119867

(119903 + 1) 119877 (1)]

minus119877 (119903 + 1)

2

119875 (119903 + 1)2Re tr [119875119867

(119903 + 1) 119875 (1)]

= 0

(24)

Also we can write

⟨119875 (1) 119875 (119903 + 2)⟩

= Re tr [119875119867

(119903 + 2) 119875 (1)]

= Retr[( ((119860119867

119877 (119903 + 2) 119861119867

+ 119863119877119867

(119903 + 2) 119862

+119876119860119867

119877 (119903 + 2) 119861119867

119876 + 119876119863119877119867

(119903 + 2) 119862119876)

times (2)minus1

) +119877(119903 + 2)

2

119877(119903 + 1)2119875(119903 + 1))

119867

119875 (1)]

= Re tr [(119860119867

119877 (119903 + 2) 119861119867

+ 119863119877119867

(119903 + 2) 119862)119867

119875 (1)]

+119877 (119903 + 2)

2

119877 (119903 + 1)2Re tr [119875119867

(119903 + 1) 119875 (1)]

= Re tr [119877119867

(119903 + 2) (119860119875 (1) 119861 + 119862119875119867

(1)119863)]

+119877 (119903 + 2)

2

119877 (119903 + 1)2Re tr [119875119867

(119903 + 1) 119875 (1)]

6 Journal of Applied Mathematics

=119875 (1)

2

119877 (1)2Re tr [119877119867

(119903 + 2) (119877 (1) minus 119877 (2))]

+119877 (119903 + 2)

2

119877 (119903 + 1)2Re tr [119875119867

(119903 + 1) 119875 (1)]

= 0

(25)

Substep 22 In this substep we prove the conclusion (22) inStep 2 It follows from Algorithm 7 that

⟨119877 (119894) 119877 (119894 + 119903 + 1)⟩

= Re tr [119877119867

(119894 + 119903 + 1) 119877 (119894)]

= Retr[(119877 (119894 + 119903) minus119877 (119894 + 119903)

2

119875 (119894 + 119903)2

times (119860119875 (119894 + 119903) 119861 + 119862119875119867

(119894 + 119903)119863))

119867

119877 (119894) ]

= Re tr [119877119867

(119894 + 119903) 119877 (119894)] minus119877 (119894 + 119903)

2

119875 (119894 + 119903)2

times Re tr [119875119867

(119894 + 119903) (119860119867

119877 (119894) 119861119867

+ 119863119877119867

(119894) 119862)]

= minus119877 (119894 + 119903)

2

119875 (119894 + 119903)2

times Re tr [119875119867

(119894 + 119903) ((119860119867

119877 (119894) 119861119867

+ 119863119877119867

(119894) 119862

+ 119876119860119867

119877 (119894) 119861119867

119876

+ 119876119863119877119867

(119894) 119862119876) times (2)minus1

)]

= minus119877 (119894 + 119903)

2

119875 (119894 + 119903)2

times Retr[119875119867

(119894 + 119903) (119875 (119894) minus119877 (119894)

2

119877 (119894 minus 1)2119875 (119894 minus 1))]

=119877 (119894 + 119903)

2

119877 (119894)2

119875 (119894 + 119903)2

119877 (119894 minus 1)2Re tr [119875119867

(119894 + 119903) 119875 (119894 minus 1)]

⟨119875 (119894) 119875 (119894 + 119903 + 1)⟩

= Re tr [119875119867

(119894 + 119903 + 1) 119875 (119894)]

= Retr[(1

2(119860

119867

119877 (119894 + 119903 + 1) 119861119867

+ 119863119877119867

(119894 + 119903 + 1) 119862

+ 119876119860119867

119877 (119894 + 119903 + 1) 119861119867

119876

+ 119876119863119877119867

(119894 + 119903 + 1) 119862119876)

+119877(119894 + 119903 + 1)

2

119877(119894 + 119903)2

119875(119894 + 119903))

119867

119875 (119894)]

= Re tr [(119860119867

119877 (119894 + 119903 + 1) 119861119867

+119863119877119867

(119894+119903+1) 119862)119867

119875 (119894)]

= Re tr [119877119867

(119894 + 119903 + 1) (119860119875 (119894)B + 119862119875119867

(119894) 119863)]

=119875 (119894)

2

119877 (119894)2Re tr [119877119867

(119894 + 119903 + 1) (119877 (119894) minus 119877 (119894 + 1))]

=119875 (119894)

2

119877 (119894)2Re tr [119877119867

(119894 + 119903 + 1) 119877 (119894)]

minus119875 (119894)

2

119877 (119894)2Re tr [119877119867

(119894 + 119903 + 1) 119877 (119894 + 1)]

=119875 (119894)

2

119877 (119894 + 119903)2

119877 (119894)2

119877 (119894)2

119875 (119894 + 119903)2

119877 (119894 minus 1)2

times Re tr [119875119867

(119894 + 119903) 119875 (119894 minus 1)]

(26)

Repeating the above process (26) we can obtain

⟨119877 (119894) 119877 (119894 + 119903 + 1)⟩ = sdot sdot sdot = 120572Re tr [119875119867

(119903 + 2) 119875 (1)]

⟨119875 (119894) 119875 (119894 + 119903 + 1)⟩ = sdot sdot sdot = 120573Re tr [119875119867

(119903 + 2) 119875 (1)]

(27)

Combining these two relations with (24) and (25) it impliesthat (22) holds So by the principle of induction we knowthat Lemma 8 holds

Lemma 9 Assume that Problem 1 is consistent and let 119883 isin

RH119899times119899

(119876) be its solution Then for any initial matrix 119883(1) isin

RH119899times119899

(119876) the sequences 119877(119894) 119875(119894) and 119883(119894) generatedby Algorithm 7 satisfy

⟨119875 (119894) 119883 minus 119883 (119894)⟩ = 119877 (119894)2

119894 = 1 2 (28)

Proof We also prove this conclusion by inductionWhen 119894 = 1 it follows from Algorithm 7 that

⟨119875 (1) 119883 minus 119883 (1)⟩

= Re tr [(119883 minus 119883 (1))119867

119875 (1)]

= Re tr [(119883 minus 119883 (1))119867

times ((119860119867

119877 (1) 119861119867

+ 119863119877119867

(1) 119862+119876119860119867

119877 (1) 119861119867

119876

+ 119876119863119877119867

(1) 119862119876) times (2)minus1

) ]

Journal of Applied Mathematics 7

= Re tr [(119883 minus 119883 (1))119867

(119860119867

119877 (1) 119861119867

+ 119863119877119867

(1) 119862)]

= Retr[119877119867

(1) (119860 (119883 minus 119883 (1)) 119861+119862 (119883 minus 119883 (1))119867

119863)]

= Re tr [119877119867

(1) 119877 (1)]

= 119877 (1)2

(29)

This implies that (28) holds for 119894 = 1Now it is assumed that (28) holds for 119894 = 119904 that is

⟨119875 (119904) 119883 minus 119883 (119904)⟩ = 119877 (119904)2

(30)

Then when 119894 = 119904 + 1

⟨119875 (119904 + 1) 119883 minus 119883 (119904 + 1)⟩

= Re tr [(119883 minus 119883 (119904 + 1))119867

119875 (119904 + 1)]

= Retr[(119883 minus 119883 (119904 + 1))119867

times (1

2(119860

119867

119877 (119904 + 1) 119861119867

+ 119863119877119867

(119904 + 1) 119862

+ 119876119860119867

119877 (119904 + 1) 119861119867

119876

+ 119876119863119877119867

(119904 + 1) 119862119876)

+119877(119904 + 1)

2

119877(119904)2

119875 (119904))]

= Re tr [(119883 minus 119883 (119904 + 1))119867

times (119860119867

119877 (119904 + 1) 119861119867

+ 119863119877119867

(119904 + 1) 119862) ]

+119877 (119904 + 1)

2

119877 (119904)2

Re tr [(119883 minus 119883 (119904 + 1))119867

119875 (119904)]

= Re tr [119877119867

(119904 + 1)

times (119860 (119883 minus 119883 (119904 + 1)) 119861

+ 119862 (119883 minus 119883 (119904 + 1))119867

119863)]

+119877 (119904 + 1)

2

119877 (119904)2

Re tr [(119883 minus 119883 (119904))119867

119875 (119904)]

minus Re tr [(119883 (119904 + 1) minus 119883 (119904))119867

119875 (119904)]

= 119877 (119904 + 1)2

+119877 (119904 + 1)

2

119877 (119904)2

times 119877 (119904)2

minus119877 (119904)

2

119875 (119904)2Re tr [119875119867

(119904) 119875 (119904)]

= 119877 (119904 + 1)2

(31)

Therefore Lemma 9 holds by the principle of induction

From the above two lemmas we have the followingconclusions

Remark 10 If there exists a positive number 119894 such that119877(119894) =0 and 119875(119894) = 0 then we can get from Lemma 9 thatProblem 1 is not consistent Hence the solvability of Problem1 can be determined by Algorithm 7 automatically in theabsence of roundoff errors

Theorem 11 Suppose that Problem 1 is consistentThen for anyinitial matrix119883(1) isin RH119899times119899

(119876) a solution of Problem 1 can beobtained within finite iteration steps in the absence of roundofferrors

Proof In Section 2 it is known that the inner productspace (H119898times119901

119877 ⟨sdot sdot⟩) is 4119898119901-dimensional According toLemma 9 if 119877(119894) =0 119894 = 1 2 4119898119901 then we have119875(119894) =0 119894 = 1 2 4119898119901 Hence 119877(4119898119901+1) and 119875(4119898119901+1)

can be computed From Lemma 8 it is not difficult to get

⟨119877 (119894) 119877 (119895)⟩ = 0 for 119894 119895 = 1 2 4119898119901 119894 =119895 (32)

Then 119877(1) 119877(2) 119877(4119898119901) is an orthogonal basis of theinner product space (H119898times119901

119877 ⟨sdot sdot⟩) In addition we can getfrom Lemma 8 that

⟨119877 (119894) 119877 (4119898119901 + 1)⟩ = 0 for 119894 = 1 2 4119898119901 (33)

It follows that 119877(4119898119901+1) = 0 which implies that119883(4119898119901+1)

is a solution of Problem 1

32 The Solution of Problem 2 In this subsection firstly weintroduce some lemmas Then we will prove that the leastFrobenius norm reflexive solution of (1) can be derived bychoosing a suitable initial iterative matrix Finally we solveProblem 2 by finding the least Frobenius norm reflexivesolution of a new-constructed quaternion matrix equation

Lemma 12 (see [47]) Assume that the consistent system oflinear equations119872119910 = 119887 has a solution 119910

0isin 119877(119872

119879

) then 1199100is

the unique least Frobenius norm solution of the system of linearequations

Lemma 13 Problem 1 is consistent if and only if the system ofquaternion matrix equations

119860119883119861 + 119862119883119867

119863 = 119865

119860119876119883119876119861 + 119862119876119883119867

119876119863 = 119865

(34)

8 Journal of Applied Mathematics

is consistent Furthermore if the solution sets of Problem 1 and(34) are denoted by 119878

119867and 119878

1

119867 respectively then we have 119878

119867sube

1198781

119867

Proof First we assume that Problem 1 has a solution 119883 By119860119883119861 + 119862119883

119867

119863 = 119865 and 119876119883119876 = 119883 we can obtain 119860119883119861 +

119862119883119867

119863 = 119865 and 119860119876119883119876119861 + 119862119876119883119867

119876119863 = 119865 which impliesthat 119883 is a solution of quaternion matrix equations (34) and119878119867

sube 1198781

119867

Conversely suppose (34) is consistent Let119883 be a solutionof (34) Set 119883

119886= (119883 + 119876119883119876)2 It is obvious that 119883

119886isin

RH119899times119899

(119876) Now we can write

119860119883119886119861 + 119862119883

119867

119886119863

=1

2[119860 (119883 + 119876119883119876)119861 + 119862(119883 + 119876119883119876)

119867

119863]

=1

2[119860119883119861 + 119862119883

119867

119863 + 119860119876119883119876119861 + 119862119876119883119867

119876119863]

=1

2[119865 + 119865]

= 119865

(35)

Hence 119883119886is a solution of Problem 1 The proof is completed

Lemma 14 The system of quaternion matrix equations (34) isconsistent if and only if the system of real matrix equations

120601 (119860) [119883119894119895]4 times 4

120601 (119861) + 120601 (119862) [119883119894119895]119879

4 times 4

120601 (119863) = 120601 (119865)

120601 (119860) 120601 (119876) [119883119894119895]4 times 4

120601 (119876) 120601 (119861)

+ 120601 (119862) 120601 (119876) [119883119894119895]119879

4 times 4

120601 (119876) 120601 (119863) = 120601 (119865)

(36)

is consistent where 119883119894119895

isin H119899times119899

119894 119895 = 1 2 3 4 are submatri-ces of the unknown matrix Furthermore if the solution sets of(34) and (36) are denoted by 119878

1

119867and 119878

2

119877 respectively then we

have 120601(1198781

119867) sube 119878

2

119877

Proof Suppose that (34) has a solution

119883 = 1198831+ 119883

2119894 + 119883

3119895 + 119883

4119896 (37)

Applying (119886) (119887) and (119888) in Lemma 4 to (34) yields

120601 (119860) 120601 (119883) 120601 (119861) + 120601 (119862) 120601119879

(119883) 120601 (119863) = 120601 (119865)

120601 (119860) 120601 (119876) 120601 (119883) 120601 (119876) 120601 (119861)

+ 120601 (119862) 120601 (119876) 120601119879

(119883) 120601 (119876) 120601 (119863) = 120601 (119865)

(38)

which implies that 120601(119883) is a solution of (36) and 120601(1198781

119867) sube 119878

2

119877

Conversely suppose that (36) has a solution

119883 = [119883119894119895]4 times 4

(39)

By (119889) in Lemma 4 we have that

119879minus1

119898120601 (119860) 119879

119899119883119879

minus1

119899120601 (119861) 119879

119901

+ 119879minus1

119898120601 (119862) 119879

119899119883

119879

119879minus1

119899120601 (119863)119879

119901= 119879

minus1

119898120601 (119865) 119879

119901

119877minus1

119898120601 (119860) 119877

119899119883119877

minus1

119899120601 (119861) 119877

119901

+ 119877minus1

119898120601 (119862) 119877

119899119883

119879

119877minus1

119899120601 (119863) 119877

119901= 119877

minus1

119898120601 (119865) 119877

119901

119878minus1

119898120601 (119860) 119878

119899119883119878

minus1

119899120601 (119861) 119878

119901

+ Sminus1

119898120601 (119862) 119878

119899119883

119879

119878minus1

119899120601 (119863) 119878

119901= 119878

minus1

119898120601 (119865) 119878

119901

119879minus1

119898120601 (119860) 120601 (119876) 119879

119899119883119879

minus1

119899120601 (119876) 120601 (119861) 119879

119901

+ 119879minus1

119898120601 (119862) 120601 (119876) 119879

119899119883

119879

119879minus1

119899120601 (119876) 120601 (119863) 119879

119901= 119879

minus1

119898120601 (119865) 119879

119901

119877minus1

119898120601 (119860) 120601 (119876) 119877

119899119883119877

minus1

119899120601 (119876) 120601 (119861) 119877

119901

+ 119877minus1

119898120601 (119862) 120601 (119876) 119877

119899119883

119879

119877minus1

119899120601 (119876) 120601 (119863) 119877

119901= 119877

minus1

119898120601 (119865) 119877

119901

119878minus1

119898120601 (119860) 120601 (119876) 119878

119899119883119878

minus1

119899120601 (119876) 120601 (119861) 119878

119901

+ 119878minus1

119898120601 (119862) 120601 (119876) 119878

119899119883

119879

119878minus1

119899120601 (119876) 120601 (119863) 119878

119901= 119878

minus1

119898120601 (119865) 119878

119901

(40)

Hence

120601 (119860) 119879119899119883119879

minus1

119899120601 (119861) + 120601 (119862) (119879

119899119883119879

minus1

119899)119879

120601 (119863) = 120601 (119865)

120601 (119860) 119877119899119883119877

minus1

119899120601 (119861) + 120601 (119862) (R

119899119883119877

minus1

119899)119879

120601 (119863) = 120601 (119865)

120601 (119860) 119878119899119883119878

minus1

119899120601 (119861) + 120601 (119862) (119878

119899119883119878

minus1

119899)119879

120601 (119863) = 120601 (119865)

120601 (119860) 120601 (119876) 119879119899119883119879

minus1

119899120601 (119876) 120601 (119861)

+ 120601 (119862) 120601 (119876) (119879119899119883119879

minus1

119899)119879

120601 (119876) 120601 (119863) = 120601 (119865)

120601 (119860) 120601 (119876) 119877119899119883119877

minus1

119899120601 (119876) 120601 (119861)

+ 120601 (119862) 120601 (119876) (119877119899119883119877

minus1

119899)119879

120601 (119876) 120601 (119863) = 120601 (119865)

120601 (119860) 120601 (119876) 119878119899119883119878

minus1

119899120601 (119876) 120601 (119861)

+ 120601 (119862) 120601 (119876) (119878119899119883119878

minus1

119899)119879

120601 (119876) 120601 (119863) = 120601 (119865)

(41)

which implies that 119879119899119883119879

minus1

119899 119877

119899119883119877

minus1

119899 and 119878

119899119883119878

minus1

119899are also

solutions of (36) Thus

1

4(119883 + 119879

119899119883119879

minus1

119899+ 119877

119899119883119877

minus1

119899+ 119878

119899119883119878

minus1

119899) (42)

Journal of Applied Mathematics 9

is also a solution of (36) where

119883 + 119879119899119883119879

minus1

119899+ 119877

119899119883119877

minus1

119899+ 119878

119899119883119878

minus1

119899

= [119883119894119895]4 times 4

119894 119895 = 1 2 3 4

11988311

= 11988311

+ 11988322

+ 11988333

+ 11988344

11988312

= 11988312

minus 11988321

+ 11988334

minus 11988343

11988313

= 11988313

minus 11988324

minus 11988331

+ 11988342

11988314

= 11988314

+ 11988323

minus 11988332

minus 11988341

11988321

= minus11988312

+ 11988321

minus 11988334

+ 11988343

11988322

= 11988311

+ 11988322

+ 11988333

+ 11988344

11988323

= 11988314

+ 11988323

minus 11988332

minus 11988341

11988324

= minus11988313

+ 11988324

+ 11988331

minus 11988342

11988331

= minus11988313

+ 11988324

+ 11988331

minus 11988342

11988332

= minus11988314

minus 11988323

+ 11988332

+ 11988341

11988333

= 11988311

+ 11988322

+ 11988333

+ 11988344

11988334

= 11988312

minus 11988321

+ 11988334

minus 11988343

11988341

= minus11988314

minus 11988323

+ 11988332

+ 11988341

11988342

= 11988313

minus 11988324

minus 11988331

+ 11988342

11988343

= minus11988312

+ 11988321

minus 11988334

+ 11988343

11988344

= 11988311

+ 11988322

+ 11988333

+ 11988344

(43)

Let

119883 =1

4(119883

11+ 119883

22+ 119883

33+ 119883

44)

+1

4(minus119883

12+ 119883

21minus 119883

34+ 119883

43) 119894

+1

4(minus119883

13+ 119883

24+ 119883

31minus 119883

42) 119895

+1

4(minus119883

14minus 119883

23+ 119883

32+ 119883

41) 119896

(44)

Then it is not difficult to verify that

120601 (119883) =1

4(119883 + 119879

119899119883119879

minus1

119899+ 119877

119899119883119877

minus1

119899+ 119878

119899119883119878

minus1

119899) (45)

We have that 119883 is a solution of (34) by (119886) in Lemma 4 Theproof is completed

0 5 10 15 20

0

2

4

Iteration number k

minus12

minus10

minus8

minus6

minus4

minus2

log10r (k)

Figure 1 The convergence curve for the Frobenius norm of theresiduals from Example 18

0 5 10 15 20

0

2

4

Iteration number k

minus12

minus10

minus8

minus6

minus4

minus2

log10r (k)

Figure 2 The convergence curve for the Frobenius norm of theresiduals from Example 19

Lemma 15 There exists a permutation matrix P(4n4n) suchthat (36) is equivalent to

[120601

119879

(119861) otimes 120601 (119860) + (120601119879

(119863) otimes 120601 (119862)) 119875 (4119899 4119899)

120601119879

(119861) 120601 (119876) otimes 120601 (119860) 120601 (119876) + (120601119879

(119863) 120601 (119876) otimes 120601 (119862) 120601 (119876)) 119875 (4119899 4119899)

]

times vec ([119883119894119895]4 times 4

) = [vec (120601 (119865))

vec (120601 (119865))]

(46)

Lemma 15 is easily proven through Lemma 5 So we omitit here

10 Journal of Applied Mathematics

Theorem 16 When Problem 1 is consistent let its solution setbe denoted by 119878

119867 If

∘

119883 isin 119878119867 and

∘

119883 can be expressed as

∘

119883 = 119860119867

119866119861119867

+ 119863119866119867

119862 + 119876119860119867

119866119861119867

119876

+ 119876119863119866119867

119862119876 119866 isin H119898times119901

(47)

then∘

119883 is the least Frobenius norm solution of Problem 1

Proof By (119886) (119887) and (119888) in Lemmas 4 5 and 6 we have that

vec(120601(

∘

119883))

= vec (120601119879

(119860) 120601 (119866) 120601119879

(119861) + 120601 (119863) 120601119879

(119866) 120601 (119862)

+ 120601 (119876) 120601119879

(119860) 120601 (119866) 120601119879

(119861) 120601 (119876)

+ 120601 (119876) 120601 (119863) 120601119879

(119866) 120601 (119862) 120601 (119876))

= [120601 (119861) otimes 120601119879

(119860) + (120601119879

(119862) otimes 120601 (119863)) 119875 (4119898 4119901)

120601 (119876) 120601 (119861) otimes 120601 (119876) 120601119879

(119860)

+ (120601 (119876) 120601119879

(119862) otimes 120601 (119876) 120601 (119863)) 119875 (4119898 4119901)]

times [vec (120601 (119866))

vec (120601 (119866))]

= [120601

119879

(119861) otimes 120601(119860) + (120601119879

(119863) otimes 120601(119862))119875(4119899 4119899)

120601119879

(119861)120601(119876) otimes 120601(119860)120601(119876) + (120601119879

(119863)120601(119876) otimes 120601(119862)120601(119876))119875(4119899 4119899)

]

119879

times [vec (120601 (119866))

vec (120601 (119866))]

isin 119877 [120601

119879

(119861) otimes 120601 (119860)

120601119879

(119861) 120601 (119876) otimes 120601 (119860) 120601 (119876)

+ (120601119879

(119863) otimes 120601 (119862)) 119875 (4119899 4119899)

+ (120601119879

(119863) 120601 (119876) otimes 120601 (119862) 120601 (119876)) 119875 (4119899 4119899)]

119879

(48)

By Lemma 12 120601(∘

119883) is the least Frobenius norm solution ofmatrix equations (46)

Noting (11) we derive from Lemmas 13 14 and 15 that∘

119883

is the least Frobenius norm solution of Problem 1

From Algorithm 7 it is obvious that if we consider

119883(1) = 119860119867

119866119861119867

+ 119863119866119867

119862 + 119876119860119867

119866119861119867

119876

+ 119876119863119866119867

119862119876 119866 isin H119898times119901

(49)

then all119883(119896) generated by Algorithm 7 can be expressed as

119883 (119896) = 119860119867

119866119896119861

119867

+ 119863119866119867

119896119862 + 119876119860

119867

119866119896119861

119867

119876

+ 119876119863119866119867

119896119862119876 119866

119896isin H

119898times119901

(50)

Using the above conclusion and consideringTheorem 16we propose the following theorem

Theorem 17 Suppose that Problem 1 is consistent Let theinitial iteration matrix be

119883 (1) = 119860119867

119866119861H+ 119863119866

119867

119862 + 119876119860119867

119866119861119867

119876 + 119876119863119866119867

119862119876 (51)

where 119866 is an arbitrary quaternion matrix or especially119883(1) = 0 then the solution 119883

lowast generated by Algorithm 7 isthe least Frobenius norm solution of Problem 1

Now we study Problem 2 When Problem 1 is consistentthe solution set of Problem 1 denoted by 119878

119867is not empty

Then For a given reflexive matrix1198830isin RH119899times119899

(119876)

119860119883119861 + 119862119883119867

119863 = 119865 lArrrArr 119860(119883 minus 1198830) 119861 + 119862(119883 minus 119883

0)119867

119863

= 119865 minus 1198601198830119861 minus 119862119883

119867

0119863

(52)

Let 119883 = 119883 minus 1198830and 119865 = 119865 minus 119860119883

0119861 minus 119862119883

119867

0119863 then Problem

2 is equivalent to finding the least Frobenius norm reflexivesolution of the quaternion matrix equation

119860119883119861 + 119862119883119867

119863 = 119865 (53)

By using Algorithm 7 let the initial iteration matrix 119883(1) =

119860119867

119866119861119867

+ 119863119866119867

119862 + 119876119860119867

119866119861119867

119876 + 119876119863119866119867

119862Q where 119866 is anarbitrary quaternion matrix in H119898times119901 or especially 119883(1) = 0we can obtain the least Frobenius norm reflexive solution119883

lowast

of (53) Then we can obtain the solution of Problem 2 whichis

119883 = 119883lowast

+ 1198830 (54)

Journal of Applied Mathematics 11

4 Examples

In this section we give two examples to illustrate the effi-ciency of the theoretical results

Example 18 Consider the quaternion matrix equation

119860119883119861 + 119862119883119867

119863 = 119865 (55)with

119860 = [1 + 119894 + 119895 + 119896 2 + 2119895 + 2119896 119894 + 3119895 + 3119896 2 + 119894 + 4119895 + 4119896

3 + 2 119894 + 2 119895 + 119896 3 + 3119894 + 119895 1 + 7119894 + 3119895 + 119896 6 minus 7119894 + 2119895 minus 4119896]

119861 =

[[[

[

minus2 + 119894 + 3119895 + 119896 3 + 3119894 + 4119895 + 119896

5 minus 4119894 minus 6119895 minus 5119896 minus1 minus 5119894 + 4119895 + 3119896

minus1 + 2119894 + 119895 + 119896 2 + 2119895 + 6119896

3 minus 2119894 + 119895 + 119896 4 + 119894 minus 2119895 minus 4119896

]]]

]

119862 = [1 + 119894 + 119895 + 119896 2 + 2119894 + 2119895 119896 2 + 2119894 + 2119895 + 119896

minus1 + 119894 + 3119895 + 3119896 minus2 + 3119894 + 4119895 + 2119896 6 minus 2119894 + 6119895 + 4119896 3 + 119894 + 119895 + 5119896]

119863 =

[[[

[

minus1 + 4119895 + 2119896 1 + 2119894 + 3119895 + 3119896

7 + 6119894 + 5119895 + 6119896 3 + 7119894 minus 119895 + 9119896

4 + 119894 + 6119895 + 119896 7 + 2119894 + 9119895 + 119896

1 + 119894 + 3119895 minus 3119896 1 + 3119894 + 2119895 + 2119896

]]]

]

119865 = [1 + 3119894 + 2119895 + 119896 2 minus 2119894 + 3119895 + 3119896

minus1 + 4119894 + 2119895 + 119896 minus2 + 2119894 minus 1119895]

(56)

We apply Algorithm 7 to find the reflexive solution with res-pect to the generalized reflection matrix

119876 =

[[[

[

028 0 096119896 0

0 minus1 00 0

minus096119896 0 minus028 0

0 0 0 minus1

]]]

]

(57)

For the initial matrix119883(1) = 119876 we obtain a solution that is

119883lowast

= 119883 (20)

=[[

[

027257 minus 023500119894 + 0034789119895 + 0054677119896 0085841 minus 0064349119894 + 010387119895 minus 026871119896

0028151 minus 0013246119894 minus 0091097119895 + 0073137119896 minus046775 + 0048363119894 minus 0015746119895 + 021642119896

0043349 + 0079178119894 + 0085167119895 minus 084221119896 035828 minus 013849119894 minus 0085799119895 + 011446119896

013454 minus 0028417119894 minus 0063010119895 minus 010574119896 010491 minus 0039417119894 + 018066119895 minus 0066963119896

minus0043349 + 0079178119894 + 0085167119895 + 084221119896 minus0014337 minus 014868119894 minus 0011146119895 minus 000036397119896

0097516 + 012146119894 minus 0017662119895 minus 0037534119896 minus0064483 + 0070885119894 minus 0089331119895 + 0074969119896

minus021872 + 018532119894 + 0011399119895 + 0029390119896 000048529 + 0014861119894 minus 019824119895 minus 0019117119896

minus014099 + 0084013119894 minus 0037890119895 minus 017938119896 minus063928 minus 0067488119894 + 0042030119895 + 010106119896

]]

]

(58)

with corresponding residual 119877(20) = 22775 times 10minus11 The

convergence curve for the Frobenius norm of the residuals119877(119896) is given in Figure 1 where 119903(119896) = 119877(119896)

Example 19 In this example we choose the matrices119860 119861 119862 119863 119865 and 119876 as same as in Example 18 Let

1198830= [

[

1 minus036119894 minus 075119896 0 minus05625119894 minus 048119896

036119894 128 048119895 minus096119895

0 1 minus 048119895 1 064 minus 075119895

048119896 096119895 064 072

]

]

isin RH119899times119899

(119876)

(59)

In order to find the optimal approximation reflexive solutionto the given matrix 119883

0 let 119883 = 119883 minus 119883

0and 119865 = 119865 minus

1198601198830119861 minus 119862119883

119867

0119863 Now we can obtain the least Frobenius

norm reflexive solution119883lowast of the quaternionmatrix equation

119860119883119861 + 119862 119883119867

119863 = 119865 by choosing the initial matrix 119883(1) = 0that is

12 Journal of Applied Mathematics

119883lowast

= 119883 (20)

=

[[[

[

minus047933 + 0021111119894 + 011703119895 minus 0066934119896 minus052020 minus 019377119894 + 017420119895 minus 059403119896

minus031132 minus 011705119894 minus 033980119895 + 011991119896 minus086390 minus 021715119894 minus 0037089119895 + 040609119896

minus024134 minus 0025941119894 minus 031930119895 minus 026961119896 minus044552 + 013065119894 + 014533119895 + 039015119896

minus011693 minus 027043119894 + 022718119895 minus 00000012004119896 minus044790 minus 031260119894 minus 10271119895 + 027275119896

024134 minus 0025941119894 minus 031930119895 + 026961119896 0089010 minus 067960119894 + 043044119895 minus 0045772119896

minus0089933 minus 025485119894 + 0087788119895 minus 023349119896 minus017004 minus 037655119894 + 080481119895 + 037636119896

minus063660 + 016515119894 minus 013216119895 + 0073845119896 minus0034329 + 032283119894 + 050970119895 minus 0066757119896

000000090029 + 017038119894 + 020282119895 minus 0087697119896 minus044735 + 021846119894 minus 021469119895 minus 015347119896

]]]

]

(60)

with corresponding residual 119877(20) = 43455 times 10minus11 The

convergence curve for the Frobenius norm of the residuals119877(119896) is given in Figure 2 where 119903(119896) = 119877(119896)

Therefore the optimal approximation reflexive solutionto the given matrix119883

0is

119883 = 119883lowast

+ 1198830

=

[[[

[

052067 + 0021111119894 + 011703119895 minus 0066934119896 minus052020 minus 055377119894 + 017420119895 minus 13440119896

minus031132 + 024295119894 minus 033980119895 + 011991119896 041610 minus 021715119894 minus 0037089119895 + 040609119896

minus024134 minus 0025941119894 minus 031930119895 minus 026961119896 055448 + 013065119894 minus 033467119895 + 039015119896

minus011693 minus 027043119894 + 022718119895 + 048000119896 minus044790 minus 031260119894 minus 0067117119895 + 027275119896

024134 minus 0025941119894 minus 031930119895 + 026961119896 0089010 minus 12421119894 + 043044119895 minus 052577119896

minus0089933 minus 025485119894 + 056779119895 minus 023349119896 minus017004 minus 037655119894 minus 015519119895 + 037636119896

036340 + 016515119894 minus 013216119895 + 0073845119896 060567 + 032283119894 minus 024030119895 minus 0066757119896

064000 + 017038119894 + 020282119895 minus 0087697119896 027265 + 021846119894 minus 021469119895 minus 015347119896

]]]

]

(61)

The results show that Algorithm 7 is quite efficient

5 Conclusions

In this paper an algorithm has been presented for solving thereflexive solution of the quaternion matrix equation 119860119883119861 +

119862119883119867

119863 = 119865 By this algorithm the solvability of the problemcan be determined automatically Also when the quaternionmatrix equation 119860119883119861 + 119862119883

119867

119863 = 119865 is consistent overreflexive matrix 119883 for any reflexive initial iterative matrixa reflexive solution can be obtained within finite iterationsteps in the absence of roundoff errors It has been proventhat by choosing a suitable initial iterative matrix we canderive the least Frobenius norm reflexive solution of thequaternion matrix equation 119860119883119861 + 119862119883

119867

119863 = 119865 throughAlgorithm 7 Furthermore by using Algorithm 7 we solvedProblem 2 Finally two numerical examples were given toshow the efficiency of the presented algorithm

Acknowledgments

This research was supported by Grants from InnovationProgram of Shanghai Municipal Education Commission(13ZZ080) the National Natural Science Foundation ofChina (11171205) theNatural Science Foundation of Shanghai

(11ZR1412500) the Discipline Project at the correspondinglevel of Shanghai (A 13-0101-12-005) and Shanghai LeadingAcademic Discipline Project (J50101)

References

[1] H C Chen and A H Sameh ldquoAmatrix decompositionmethodfor orthotropic elasticity problemsrdquo SIAM Journal on MatrixAnalysis and Applications vol 10 no 1 pp 39ndash64 1989

[2] H C Chen ldquoGeneralized reflexive matrices special propertiesand applicationsrdquo SIAM Journal on Matrix Analysis and Appli-cations vol 19 no 1 pp 140ndash153 1998

[3] Q W Wang H X Chang and C Y Lin ldquo119875-(skew)symmetriccommon solutions to a pair of quaternion matrix equationsrdquoApplied Mathematics and Computation vol 195 no 2 pp 721ndash732 2008

[4] S F Yuan and Q W Wang ldquoTwo special kinds of least squaressolutions for the quaternionmatrix equation119860119883119861+119862119883119863 = 119864rdquoElectronic Journal of Linear Algebra vol 23 pp 257ndash274 2012

[5] T S Jiang and M S Wei ldquoOn a solution of the quaternionmatrix equation 119883 minus 119860119883119861 = 119862 and its applicationrdquo ActaMathematica Sinica (English Series) vol 21 no 3 pp 483ndash4902005

[6] Y T Li and W J Wu ldquoSymmetric and skew-antisymmetricsolutions to systems of real quaternion matrix equationsrdquoComputers amp Mathematics with Applications vol 55 no 6 pp1142ndash1147 2008

Journal of Applied Mathematics 13

[7] L G Feng and W Cheng ldquoThe solution set to the quaternionmatrix equation119860119883minus119883119861 = 0rdquo Algebra Colloquium vol 19 no1 pp 175ndash180 2012

[8] Z Y Peng ldquoAn iterative method for the least squares symmetricsolution of the linear matrix equation 119860119883119861 = 119862rdquo AppliedMathematics and Computation vol 170 no 1 pp 711ndash723 2005

[9] Z Y Peng ldquoA matrix LSQR iterative method to solve matrixequation 119860119883119861 = 119862rdquo International Journal of ComputerMathematics vol 87 no 8 pp 1820ndash1830 2010

[10] Z Y Peng ldquoNew matrix iterative methods for constraintsolutions of the matrix equation 119860119883119861 = 119862rdquo Journal ofComputational and Applied Mathematics vol 235 no 3 pp726ndash735 2010

[11] Z Y Peng ldquoSolutions of symmetry-constrained least-squaresproblemsrdquo Numerical Linear Algebra with Applications vol 15no 4 pp 373ndash389 2008

[12] C C Paige ldquoBidiagonalization of matrices and solutions of thelinear equationsrdquo SIAM Journal on Numerical Analysis vol 11pp 197ndash209 1974

[13] X F Duan A P Liao and B Tang ldquoOn the nonlinear matrixequation 119883 minus sum

119898

119894=1119860

119909

2119886119894119883

120575119894119860119894= 119876rdquo Linear Algebra and Its

Applications vol 429 no 1 pp 110ndash121 2008[14] X F Duan and A P Liao ldquoOn the existence of Hermitian

positive definite solutions of thematrix equation119883119904

+119860lowast

119883minus119905

119860 =

119876rdquo Linear Algebra and Its Applications vol 429 no 4 pp 673ndash687 2008

[15] X FDuan andA P Liao ldquoOn the nonlinearmatrix equation119883+

119860lowast

119883minus119902

119860 = 119876(119902 ge 1)rdquo Mathematical and Computer Modellingvol 49 no 5-6 pp 936ndash945 2009

[16] X F Duan and A P Liao ldquoOn Hermitian positive definitesolution of the matrix equation119883minussum

119898

119894=1119860

lowast

119894119883

119903

119860119894= 119876rdquo Journal

of Computational and Applied Mathematics vol 229 no 1 pp27ndash36 2009

[17] X F Duan CM Li and A P Liao ldquoSolutions and perturbationanalysis for the nonlinear matrix equation119883+sum

119898

119894=1119860

lowast

119894119883

minus1

119860119894=

119868rdquo Applied Mathematics and Computation vol 218 no 8 pp4458ndash4466 2011

[18] F Ding P X Liu and J Ding ldquoIterative solutions of thegeneralized Sylvester matrix equations by using the hierarchicalidentification principlerdquo Applied Mathematics and Computa-tion vol 197 no 1 pp 41ndash50 2008

[19] F Ding and T Chen ldquoIterative least-squares solutions ofcoupled Sylvester matrix equationsrdquo Systems amp Control Lettersvol 54 no 2 pp 95ndash107 2005

[20] FDing andTChen ldquoHierarchical gradient-based identificationof multivariable discrete-time systemsrdquo Automatica vol 41 no2 pp 315ndash325 2005

[21] M H Wang M S Wei and S H Hu ldquoAn iterative method forthe least-squares minimum-norm symmetric solutionrdquo CMESComputer Modeling in Engineering amp Sciences vol 77 no 3-4pp 173ndash182 2011

[22] M Dehghan and M Hajarian ldquoAn iterative algorithm for thereflexive solutions of the generalized coupled Sylvester matrixequations and its optimal approximationrdquoAppliedMathematicsand Computation vol 202 no 2 pp 571ndash588 2008

[23] M Dehghan and M Hajarian ldquoFinite iterative algorithmsfor the reflexive and anti-reflexive solutions of the matrixequation119860

1119883

1119861

1+119860

2119883

2119861

2= 119862rdquoMathematical and Computer

Modelling vol 49 no 9-10 pp 1937ndash1959 2009[24] M Hajarian and M Dehghan ldquoSolving the generalized

Sylvester matrix equation119901

sum

119894=1

119860119894119883119861

119894+

119902

sum

119895=1

119862119895119884119863

119895= 119864 over

reflexive and anti-reflexive matricesrdquo International Journal ofControl and Automation vol 9 no 1 pp 118ndash124 2011

[25] M Dehghan and M Hajarian ldquoAn iterative algorithm forsolving a pair of matrix equations 119860119884119861 = 119864 119862119884119863 = 119865

over generalized centro-symmetric matricesrdquo Computers ampMathematics with Applications vol 56 no 12 pp 3246ndash32602008

[26] M Dehghan and M Hajarian ldquoAnalysis of an iterative algo-rithm to solve the generalized coupled Sylvester matrix equa-tionsrdquo Applied Mathematical Modelling vol 35 no 7 pp 3285ndash3300 2011

[27] M Dehghan and M Hajarian ldquoThe general coupled matrixequations over generalized bisymmetric matricesrdquo Linear Alge-bra and Its Applications vol 432 no 6 pp 1531ndash1552 2010

[28] M Dehghan and M Hajarian ldquoAn iterative method for solvingthe generalized coupled Sylvester matrix equations over gener-alized bisymmetric matricesrdquo Applied Mathematical Modellingvol 34 no 3 pp 639ndash654 2010

[29] A G Wu B Li Y Zhang and G R Duan ldquoFinite iterativesolutions to coupled Sylvester-conjugate matrix equationsrdquoApplied Mathematical Modelling vol 35 no 3 pp 1065ndash10802011

[30] A G Wu L L Lv and G R Duan ldquoIterative algorithmsfor solving a class of complex conjugate and transpose matrixequationsrdquo Applied Mathematics and Computation vol 217 no21 pp 8343ndash8353 2011

[31] AGWu L L Lv andMZHou ldquoFinite iterative algorithms forextended Sylvester-conjugate matrix equationsrdquo Mathematicaland Computer Modelling vol 54 no 9-10 pp 2363ndash2384 2011

[32] N L Bihan and J Mars ldquoSingular value decomposition ofmatrices of Quaternions matrices a new tool for vector-sensorsignal processingrdquo Signal Process vol 84 no 7 pp 1177ndash11992004

[33] N L Bihan and S J Sangwine ldquoJacobi method for quaternionmatrix singular value decompositionrdquoAppliedMathematics andComputation vol 187 no 2 pp 1265ndash1271 2007

[34] F O Farid Q W Wang and F Zhang ldquoOn the eigenvalues ofquaternion matricesrdquo Linear and Multilinear Algebra vol 59no 4 pp 451ndash473 2011

[35] S de Leo and G Scolarici ldquoRight eigenvalue equation inquaternionic quantummechanicsrdquo Journal of Physics A vol 33no 15 pp 2971ndash2995 2000

[36] S J Sangwine and N L Bihan ldquoQuaternion singular valuedecomposition based on bidiagonalization to a real or com-plex matrix using quaternion Householder transformationsrdquoApplied Mathematics and Computation vol 182 no 1 pp 727ndash738 2006

[37] C C Took D P Mandic and F Zhang ldquoOn the unitarydiagonalisation of a special class of quaternion matricesrdquoApplied Mathematics Letters vol 24 no 11 pp 1806ndash1809 2011

[38] QWWang H X Chang and Q Ning ldquoThe common solutionto six quaternion matrix equations with applicationsrdquo AppliedMathematics and Computation vol 198 no 1 pp 209ndash2262008

[39] Q W Wang J W van der Woude and H X Chang ldquoA systemof real quaternion matrix equations with applicationsrdquo LinearAlgebra and Its Applications vol 431 no 12 pp 2291ndash2303 2009

[40] Q W Wang S W Yu and W Xie ldquoExtreme ranks of realmatrices in solution of the quaternion matrix equation 119860119883119861 =

119862with applicationsrdquoAlgebra Colloquium vol 17 no 2 pp 345ndash360 2010

14 Journal of Applied Mathematics

[41] Q W Wang and J Jiang ldquoExtreme ranks of (skew-)Hermitiansolutions to a quaternion matrix equationrdquo Electronic Journal ofLinear Algebra vol 20 pp 552ndash573 2010

[42] Q W Wang X Liu and S W Yu ldquoThe common bisymmetricnonnegative definite solutions with extreme ranks and inertiasto a pair of matrix equationsrdquo Applied Mathematics and Com-putation vol 218 no 6 pp 2761ndash2771 2011

[43] Q W Wang Y Zhou and Q Zhang ldquoRanks of the commonsolution to six quaternion matrix equationsrdquo Acta Mathemati-cae Applicatae Sinica (English Series) vol 27 no 3 pp 443ndash4622011

[44] F Zhang ldquoGersgorin type theorems for quaternionic matricesrdquoLinear Algebra and Its Applications vol 424 no 1 pp 139ndash1532007

[45] F Zhang ldquoQuaternions and matrices of quaternionsrdquo LinearAlgebra and Its Applications vol 251 pp 21ndash57 1997

[46] R A Horn and C R Johnson Topics in Matrix AnalysisCambridge University Press Cambridge UK 1991

[47] Y X Peng X Y Hu and L Zhang ldquoAn iterationmethod for thesymmetric solutions and the optimal approximation solutionof the matrix equation 119860119883119861 = 119862rdquo Applied Mathematics andComputation vol 160 no 3 pp 763ndash777 2005

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 961568 6 pageshttpdxdoiorg1011552013961568

Research ArticleThe Optimization on Ranks and Inertias of a QuadraticHermitian Matrix Function and Its Applications

Yirong Yao

Department of Mathematics Shanghai University Shanghai 200444 China

Correspondence should be addressed to Yirong Yao yryaostaffshueducn

Received 3 December 2012 Accepted 9 January 2013

Academic Editor Yang Zhang

Copyright copy 2013 Yirong Yao This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We solve optimization problems on the ranks and inertias of the quadratic Hermitian matrix function 119876 minus 119883119875119883lowast subject to a

consistent system of matrix equations 119860119883 = 119862 and 119883119861 = 119863 As applications we derive necessary and sufficient conditions for thesolvability to the systems of matrix equations and matrix inequalities 119860119883 = 119862119883119861 = 119863 and 119883119875119883lowast = (gt lt ge le)119876 in the Lownerpartial ordering to be feasible respectively The findings of this paper widely extend the known results in the literature

1 Introduction

Throughout this paper we denote the complex number fieldby C The notations C119898times119899 and C119898times119898

ℎstand for the sets

of all 119898 times 119899 complex matrices and all 119898 times 119898 complexHermitian matrices respectivelyThe identity matrix with anappropriate size is denoted by 119868 For a complex matrix 119860the symbols 119860lowast and 119903(119860) stand for the conjugate transposeand the rank of119860 respectivelyTheMoore-Penrose inverse of119860 isin C119898times119899 denoted by119860dagger is defined to be the unique solution119883 to the following four matrix equations

(1) 119860119883119860 = 119860 (2) 119883119860119883 = 119883

(3) (119860119883)lowast

= 119860119883 (4) (119883119860)lowast

= 119883119860

(1)

Furthermore 119871119860and 119877

119860stand for the two projectors 119871

119860=

119868 minus 119860dagger

119860 and 119877119860= 119868 minus 119860119860

dagger induced by 119860 respectively It isknown that 119871

119860= 119871lowast

119860and 119877

119860= 119877lowast

119860 For 119860 isin C119898times119898

ℎ its inertia

I119899(119860) = (119894

+(119860) 119894minus(119860) 1198940(119860)) (2)

is the triple consisting of the numbers of the positive nega-tive and zero eigenvalues of 119860 counted with multiplicitiesrespectively It is easy to see that 119894

+(119860) + 119894

minus(119860) = 119903(119860) For

two Hermitian matrices 119860 and 119861 of the same sizes we say119860 gt 119861 (119860 ge 119861) in the Lowner partial ordering if 119860 minus 119861 ispositive (nonnegative) definite

The investigation on maximal and minimal ranks andinertias of linear and quadratic matrix function is activein recent years (see eg [1ndash24]) Tian [21] considered themaximal and minimal ranks and inertias of the Hermitianquadratic matrix function

ℎ (119883) = 119860119883119861119883lowast

119860lowast

+ 119860119883119862 + 119862lowast

119883lowast

119860lowast

+ 119863 (3)

where 119861 and 119863 are Hermitian matrices Moreover Tian [22]investigated the maximal and minimal ranks and inertias ofthe quadratic Hermitian matrix function

119891 (119883) = 119876 minus 119883119875119883lowast (4)

such that 119860119883 = 119862The goal of this paper is to give the maximal andminimal

ranks and inertias of the matrix function (4) subject to theconsistent system of matrix equations

119860119883 = 119862 119883119861 = 119863 (5)

where 119876 isin C119899times119899ℎ

119875 isin C119901times119901

ℎare given complex matrices

As applications we consider the necessary and sufficient

2 Journal of Applied Mathematics

conditions for the solvability to the systems of matrix equa-tions and inequality

119860119883 = 119862 119883119861 = 119863 119883119875119883lowast

= 119876

119860119883 = 119862 119883119861 = 119863 119883119875119883lowast

gt 119876

119860119883 = 119862 119883119861 = 119863 119883119875119883lowast

lt 119876

119860119883 = 119862 119883119861 = 119863 119883119875119883lowast

ge 119876

119860119883 = 119862 119883119861 = 119863 119883119875119883lowast

le 119876

(6)

in the Lowner partial ordering to be feasible respectively

2 The Optimization on Ranks and Inertiasof (4) Subject to (5)

In this section we consider the maximal and minimal ranksand inertias of the quadratic Hermitian matrix function (4)subject to (5) We begin with the following lemmas

Lemma 1 (see [3]) Let 119860 isin C119898times119898ℎ

119861 isin C119898times119901 and 119862 isin C119902times119898

be given and denote

1198751= [

119860 119861

119861lowast

0] 119875

2= [

119860 119862lowast

119862 0]

1198753= [

119860 119861 119862lowast

119861lowast

0 0] 119875

4= [

119860 119861 119862lowast

119862 0 0]

(7)

Then

max119884isinC119901times119902

119903 [119860 minus 119861119884119862 minus (119861119884119862)lowast

]

= min 119903 [119860 119861 119862lowast

] 119903 (1198751) 119903 (119875

2)

min119884isinC119901times119902

119903 [119860 minus 119861119884119862 minus (119861119884119862)lowast

]

= 2119903 [119860 119861 119862lowast

]

+max 119908++ 119908minus 119892++ 119892minus 119908++ 119892minus 119908minus+ 119892+

max119884isinC119901times119902

119894plusmn[119860 minus 119861119884119862 minus (119861119884119862)

lowast

] = min 119894plusmn(1198751) 119894plusmn(1198752)

min119884isinC119901times119902

119894plusmn[119860 minus 119861119884119862 minus (119861119884119862)

lowast

]

= 119903 [119860 119861 119862lowast

] +max 119894plusmn(1198751) minus 119903 (119875

3) 119894plusmn(1198752) minus 119903 (119875

4)

(8)

where

119908plusmn= 119894plusmn(1198751) minus 119903 (119875

3) 119892

plusmn= 119894plusmn(1198752) minus 119903 (119875

4) (9)

Lemma 2 (see [4]) Let 119860 isin C119898times119899 119861 isin C119898times119896 119862 isin C119897times119899 119863 isin

C119898times119901 119864 isin C119902times119899 119876 isin C1198981times119896 and 119875 isin C119897times1198991 be given Then

(1) 119903 (119860) + 119903 (119877119860119861) = 119903 (119861) + 119903 (119877

119861119860) = 119903 [119860 119861]

(2) 119903 (119860) + 119903 (119862119871119860) = 119903 (119862) + 119903 (119860119871

119862) = 119903 [

119860

119862]

(3) 119903 (119861) + 119903 (119862) + 119903 (119877119861119860119871119862) = 119903 [

119860 119861

119862 0]

(4) 119903 (119875) + 119903 (119876) + 119903 [119860 119861119871

119876

119877119875119862 0

] = 119903[

[

119860 119861 0

119862 0 119875

0 119876 0

]

]

(5) 119903 [119877119861119860119871119862119877119861119863

119864119871119862

0] + 119903 (119861) + 119903 (119862) = 119903[

[

119860 119863 119861

119864 0 0

119862 0 0

]

]

(10)

Lemma 3 (see [23]) Let 119860 isin C119898times119898ℎ

119861 isin C119898times119899 119862 isin C119899times119899ℎ

119876 isin C119898times119899 and 119875 isin C119901times119899 be given and 119879 isin C119898times119898 benonsingular Then

(1) 119894plusmn(119879119860119879

lowast

) = 119894plusmn(119860)

(2) 119894plusmn[119860 0

0 119862] = 119894plusmn(119860) + 119894

plusmn(119862)

(3) 119894plusmn[0 119876

119876lowast

0] = 119903 (119876)

(4) 119894plusmn[

119860 119861119871119875

119871119875119861lowast

0] + 119903 (119875) = 119894

plusmn

[

[

119860 119861 0

119861lowast

0 119875lowast

0 119875 0

]

]

(11)

Lemma 4 Let 119860 119862 119861 and 119863 be given Then the followingstatements are equivalent

(1) System (5) is consistent

(2) Let

119903 [119860 119862] = 119903 (119860) [119863

119861] = 119903 (119861) 119860119863 = 119862119861 (12)

In this case the general solution can be written as

119883 = 119860dagger

119862 + 119871119860119863119861dagger

+ 119871119860119881119877119861 (13)

where 119881 is an arbitrary matrix over C with appropriate size

Now we give the fundamental theorem of this paper

Journal of Applied Mathematics 3

Theorem5 Let119891(119883) be as given in (4) and assume that119860119883 =

119862 and 119883119861 = 119863 in (5) is consistent Then

max119860119883=119862119883119861=119863

119903 (119876 minus 119883119875119883lowast

)

= min

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861)

minus 119903 (119875) 2119899 + 119903 (119860119876119860lowast

minus 119862119875119862lowast

)

minus2119903 (119860) 119903 [

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 2119903 (119861)

min119860119883=119862119883119861=119863

119903 (119876 minus 119883119875119883lowast

)

= 2119899 + 2119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 2119903 (119860) minus 2119903 (119861) minus 119903 (119875)

+max 119904++ 119904minus 119905++ 119905minus 119904++ 119905minus 119904minus+ 119905+

max119860119883=119862119883119861=119863

119894plusmn(119876 minus 119883119875119883

lowast

)

= min

119899 + 119894plusmn(119860119876119860

lowast

minus 119862119875119862lowast

)

minus119903 (119860) 119894plusmn

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 119903 (119861)

(14)

min119860119883=119862119883119861=119863

119894plusmn(119876 minus 119883119875119883

lowast

)

= 119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861)

minus 119894plusmn(119875) +max 119904

plusmn 119905plusmn

(15)

where

119904plusmn=minus119899+119903 (119860)minus119894

∓(119875)+119894

plusmn(119860119876119860

lowast

minus119862119875119862lowast

)minus119903 [119862119875 119860119876119860

lowast

119861lowast

119863lowast

119860lowast]

119905plusmn=minus119899+119903 (119860)minus119894

∓(119875)+119894

plusmn

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus[

[

0 119875 119861

119860119876 119862119875 0

119863lowast

119861lowast

0

]

]

(16)

Proof It follows from Lemma 4 that the general solution of(4) can be expressed as

119883 = 1198830+ 119871119860119881119877119861 (17)

where 119881 is an arbitrary matrix over C and 1198830is a special

solution of (5) Then

119876 minus 119883119875119883lowast

= 119876 minus (1198830+ 119871119860119881119877119861) 119875(119883

0+ 119871119860119881119877119861)lowast

(18)

Note that

119903 [119876 minus (1198830+ 119871119860119881119877119861) 119875(119883

0+ 119871119860119881119877119861)lowast

]

= 119903 [119876 (119883

0+ 119871119860119881119877119861) 119875

119875(1198830+ 119871119860119881119877119861)lowast

119875] minus 119903 (119875)

= 119903 [[119876 119883

0119875

119875119883lowast

0119875] + [

119871119860

0]119881 [0 119877

119861119875]

+([119871119860

0]119881 [0 119877

119861119875])

lowast

] minus 119903 (119875)

(19)

119894plusmn[119876 minus (119883

0+ 119871119860119881119877119861) 119875(119883

0+ 119871119860119881119877119861)lowast

]

= 119894plusmn[

119876 (1198830+ 119871119860119881119877119861) 119875

119875(1198830+ 119871119860119881119877119861)lowast

119875] minus 119894plusmn(119875)

= 119894plusmn[[

119876 1198830119875

119875119883lowast

0119875] + [

119871119860

0]119881 [0 119877

119861119875]

+([119871119860

0]119881 [0 119877

119861119875])

lowast

] minus 119894plusmn(119875)

(20)

Let

119902 (119881) = [119876 119883

0119875

119875119883lowast

0119875] + [

119871119860

0]119881 [0 119877

119861119875]

+ ([119871119860

0]119881 [0 119877

119861119875])

lowast

(21)

Applying Lemma 1 to (19) and (20) yields

max119881

119903 [119902 (119881)] = min 119903 (119872) 119903 (1198721) 119903 (119872

2)

min119881

119903 [119902 (119881)]

= 2119903 (119872) +max 119904++ 119904minus 119905++ 119905minus 119904++ 119905minus 119904minus+ 119905+

max119881

119894plusmn[119902 (119881)] = min 119894

plusmn(1198721) 119894plusmn(1198722)

min119881

119894plusmn[119902 (119881)] = 119903 (119872) +max 119904

plusmn 119905plusmn

(22)

where

119872=[119876 119883

0119875 119871119860

0

119875119883lowast

0119875 0 119875119877

119861

] 1198721=[

[

119876 1198830119875 119871119860

119875119883lowast

0119875 0

119871119860

0 0

]

]

1198722=[

[

119876 1198830119875 0

119875119883lowast

0119875 119875119877

119861

0 119877119861119875 0

]

]

1198723=[

[

119876 1198830119875 119871119860

0

119875119883lowast

0119875 0 119875119877

119861

119871119860

0 0 0

]

]

1198724= [

[

119876 1198830119875 119871119860

0

119875119883lowast

0119875 0 119875119877

119861

0 119877119861119875 0 0

]

]

119904plusmn= 119894plusmn(1198721) minus 119903 (119872

3) 119905

plusmn= 119894plusmn(1198722) minus 119903 (119872

4)

(23)

4 Journal of Applied Mathematics

Applying Lemmas 2 and 3 elementary matrix operations andcongruence matrix operations we obtain

119903 (119872) = 119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861)

119903 (1198721) = 2119899 + 119903 (119860119876119860

lowast

minus 119862119875119862lowast

) minus 2119903 (119860) + 119903 (119875)

119894plusmn(1198721) = 119899 + 119894

plusmn(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 (119860) + 119894plusmn(119875)

119903 (1198722) = 119903[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 2119903 (119861) + 119903 (119875)

119894plusmn(1198722) = 119894plusmn

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 119903 (119861) + 119894plusmn(119875)

119903 (1198723) = 2119899 + 119903 (119875) minus 2119903 (119860) minus 119903 (119861) + 119903 [

119862119875 119860119876119860lowast

119861lowast

119863lowast

119860lowast]

119903 (1198724) = 119899 + 119903 (119875) + 119903[

[

0 119875 119861

119860119876 119862119875 0

119863lowast

119861lowast

0

]

]

minus 2119903 (119861) minus 119903 (119860)

(24)

Substituting (24) into (22) we obtain the results

Using immediately Theorem 5 we can easily get thefollowing

Theorem 6 Let 119891(119883) be as given in (4) 119904plusmnand let 119905

plusmnbe as

given in Theorem 5 and assume that 119860119883 = 119862 and 119883119861 = 119863 in(5) are consistent Then we have the following

(a) 119860119883 = 119862 and 119883119861 = 119863 have a common solution suchthat 119876 minus 119883119875119883

lowast

ge 0 if and only if

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894minus(119875) + 119904

minusle 0

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894minus(119875) + 119905

minusle 0

(25)

(b) 119860119883 = 119862 and 119883119861 = 119863 have a common solution suchthat 119876 minus 119883119875119883

lowast

le 0 if and only if

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894+(119875) + 119904

+le 0

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894+(119875) + 119905

+le 0

(26)

(c) 119860119883 = 119862 and 119883119861 = 119863 have a common solution suchthat 119876 minus 119883119875119883

lowast

gt 0 if and only if

119894+(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 (119860) ge 0

119894+

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 119903 (119861) ge 119899

(27)

(d) 119860119883 = 119862 and 119883119861 = 119863 have a common solution suchthat 119876 minus 119883119875119883

lowast

lt 0 if and only if

119894minus(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 (119860) ge 0

119894minus

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 119903 (119861) ge 119899

(28)

(e) All common solutions of 119860119883 = 119862 and 119883119861 = 119863 satisfy119876 minus 119883119875119883

lowast

ge 0 if and only if

119899 + 119894minus(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 (119860) = 0

or 119894minus

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 119903 (119861) = 0

(29)

(f) All common solutions of 119860119883 = 119862 and 119883119861 = 119863 satisfy119876 minus 119883119875119883

lowast

le 0 if and only if

119899 + 119894+(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 (119860) = 0

or 119894+

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 119903 (119861) = 0

(30)

(g) All common solutions of 119860119883 = 119862 and 119883119861 = 119863 satisfy119876 minus 119883119875119883

lowast

gt 0 if and only if

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894+(119875) + 119904

+= 119899

(31)or

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894+(119875) + 119905

+= 119899

(32)

(h) All common solutions of 119860119883 = 119862 and 119883119861 = 119863 satisfy119876 minus 119883119875119883

lowast

lt 0 if and only if

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894minus(119875) + 119904

minus= 119899

(33)or

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894minus(119875) + 119905

minus= 119899

(34)

Journal of Applied Mathematics 5

(i) 119860119883 = 119862 119883119861 = 119863 and 119876 = 119883119875119883lowast have a common

solution if and only if

2119899+2119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus2119903 (119860)minus2119903 (119861)minus119903 (119875)+119904++119904minusle0

2119899+2119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus2119903 (119860)minus2119903 (119861)minus119903 (119875)+119905++119905minusle0

2119899+2119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus2119903 (119860)minus2119903 (119861)minus119903 (119875) + 119904++119905minusle0

2119899+2119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus2119903 (119860)minus2119903 (119861)minus119903 (119875) + 119904minus+119905+le0

(35)

Let 119875 = 119868 in Theorem 5 we get the following corollary

Corollary 7 Let 119876 isin C119899times119899 119860 119861 119862 and119863 be given Assumethat (5) is consistent Denote

1198791= [

119862 119860119876

119861lowast

119863lowast] 119879

2= 119860119876119860

lowast

minus 119862119862lowast

1198793= [

119876 119863

119863lowast

119861lowast

119861] 119879

4= [

119862 119860119876119860lowast

119861lowast

119863lowast

119860lowast]

1198795= [

119862119861 119860119876

119861lowast

119861 119863lowast]

(36)

Then

max119860119883=119862119883119861=119863

119903 (119876 minus 119883119883lowast

)

= min 119899 + 119903 (1198791) minus 119903 (119860) minus 119903 (119861) 2119899 + 119903 (119879

2)

minus2119903 (119860) 119899 + 119903 (1198793) minus 2119903 (119861)

min119860119883=119862119883119861=119863

119903 (119876 minus 119883119883lowast

)

= 2119903 (1198791) +max 119903 (119879

2) minus 2119903 (119879

4) minus119899 + 119903 (119879

3)

minus 2119903 (1198795) 119894+(1198792) + 119894minus(1198793)

minus 119903 (1198794) minus 119903 (119879

5) minus119899 + 119894

minus(1198792)

+119894+(1198793) minus 119903 (119879

4) minus 119903 (119879

5)

max119860119883=119862119883119861=119863

119894+(119876 minus 119883119883

lowast

)

= min 119899 + 119894+(1198792) minus 119903 (119860) 119894

+(1198793) minus 119903 (119861)

max119860119883=119862119883119861=119863

119894minus(119876 minus 119883119883

lowast

)

= min 119899 + 119894minus(1198792) minus 119903 (119860) 119899 + 119894

minus(1198793) minus 119903 (119861)

min119860119883=119862119883119861=119863

119894+(119876 minus 119883119883

lowast

)

= 119903 (1198791) +max 119894

+(1198792) minus 119903 (119879

4) 119894+(1198793) minus 119899 minus 119903 (119879

5)

min119860119883=119862119883119861=119863

119894minus(119876 minus 119883119883

lowast

)

= 119903 (1198791) +max 119894

minus(1198792) minus 119903 (119879

4) 119894minus(1198793) minus 119903 (119879

5)

(37)

Remark 8 Corollary 7 is one of the results in [24]

Let 119861 and 119863 vanish in Theorem 5 then we can obtainthe maximal and minimal ranks and inertias of (4) subjectto 119860119883 = 119862

Corollary 9 Let 119891(119883) be as given in (4) and assume that119860119883 = 119862 is consistent Then

max119860119883=119862

119903 (119876 minus 119883119875119883lowast

)

= min 119899 + 119903 [119860119876 119862119875] minus 119903 (119860) minus 119903 (119861)

2119899 + 119903 (119860119876119860lowast

minus 119862119875119862lowast

) minus 2119903 (119860) 119903 (119876) + 119903 (119875)

min119860119883=119862

119903 (119876 minus 119883119875119883lowast

)

= 2119899 + 2119903 [119860119876 119862119875] minus 2119903 (119860)

+max 119904++ 119904minus 119905++ 119905minus 119904++ 119905minus 119904minus+ 119905+

max119860119883=119862

119894plusmn(119876 minus 119883119875119883

lowast

)

= min 119899 + 119894plusmn(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 (119860) 119894plusmn(119876) + 119894

∓(119875)

min119860119883=119862

119894plusmn(119876 minus 119883119875119883

lowast

)

= 119899 + 119903 [119860119876 119862119875] minus 119903 (119860) + 119894∓(119875) +max 119904

plusmn 119905plusmn

(38)

where

119904plusmn= minus 119899 + 119903 (119860) minus 119894

∓(119875)

+ 119894plusmn(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 [119862119875 119860119876119860lowast

]

119905plusmn= minus119899 + 119903 (119860) + 119894

plusmn(119876) minus 119903 (119875) minus [119860119876 119862119875]

(39)

Remark 10 Corollary 9 is one of the results in [22]

References

[1] D L Chu Y S Hung and H J Woerdeman ldquoInertia and rankcharacterizations of somematrix expressionsrdquo SIAM Journal onMatrix Analysis and Applications vol 31 no 3 pp 1187ndash12262009

[2] Z H He and Q W Wang ldquoSolutions to optimization problemson ranks and inertias of a matrix function with applicationsrdquo

6 Journal of Applied Mathematics

AppliedMathematics andComputation vol 219 no 6 pp 2989ndash3001 2012

[3] Y Liu andY Tian ldquoMax-min problems on the ranks and inertiasof the matrix expressions119860minus119861119883119862plusmn(119861119883119862)lowast with applicationsrdquoJournal of OptimizationTheory and Applications vol 148 no 3pp 593ndash622 2011

[4] G Marsaglia and G P H Styan ldquoEqualities and inequalities forranks of matricesrdquo Linear and Multilinear Algebra vol 2 pp269ndash292 197475

[5] X Zhang Q W Wang and X Liu ldquoInertias and ranks ofsome Hermitian matrix functions with applicationsrdquo CentralEuropean Journal of Mathematics vol 10 no 1 pp 329ndash3512012

[6] Q W Wang Y Zhou and Q Zhang ldquoRanks of the commonsolution to six quaternion matrix equationsrdquo Acta Mathemati-cae Applicatae Sinica vol 27 no 3 pp 443ndash462 2011

[7] Q Zhang and Q W Wang ldquoThe (119875 119876)-(skew)symmetricextremal rank solutions to a system of quaternion matrixequationsrdquo Applied Mathematics and Computation vol 217 no22 pp 9286ndash9296 2011

[8] D Z Lian QWWang and Y Tang ldquoExtreme ranks of a partialbanded block quaternion matrix expression subject to somematrix equationswith applicationsrdquoAlgebraColloquium vol 18no 2 pp 333ndash346 2011

[9] H S Zhang andQWWang ldquoRanks of submatrices in a generalsolution to a quaternion system with applicationsrdquo Bulletin ofthe Korean Mathematical Society vol 48 no 5 pp 969ndash9902011

[10] Q W Wang and J Jiang ldquoExtreme ranks of (skew-)Hermitiansolutions to a quaternion matrix equationrdquo Electronic Journal ofLinear Algebra vol 20 pp 552ndash573 2010

[11] I A Khan Q W Wang and G J Song ldquoMinimal ranks ofsome quaternionmatrix expressions with applicationsrdquoAppliedMathematics and Computation vol 217 no 5 pp 2031ndash20402010

[12] Q Wang S Yu and W Xie ldquoExtreme ranks of real matricesin solution of the quaternion matrix equation 119860119883119861 = 119862 withapplicationsrdquo Algebra Colloquium vol 17 no 2 pp 345ndash3602010

[13] Q W Wang H S Zhang and G J Song ldquoA new solvable con-dition for a pair of generalized Sylvester equationsrdquo ElectronicJournal of Linear Algebra vol 18 pp 289ndash301 2009

[14] Q W Wang S W Yu and Q Zhang ldquoThe real solutions toa system of quaternion matrix equations with applicationsrdquoCommunications in Algebra vol 37 no 6 pp 2060ndash2079 2009

[15] Q W Wang and C K Li ldquoRanks and the least-norm of thegeneral solution to a system of quaternion matrix equationsrdquoLinear Algebra and its Applications vol 430 no 5-6 pp 1626ndash1640 2009

[16] Q W Wang H S Zhang and S W Yu ldquoOn solutions to thequaternionmatrix equation119860119883119861+119862119884119863 = 119864rdquoElectronic Journalof Linear Algebra vol 17 pp 343ndash358 2008

[17] Q W Wang G J Song and X Liu ldquoMaximal and minimalranks of the common solution of some linear matrix equationsover an arbitrary division ring with applicationsrdquo AlgebraColloquium vol 16 no 2 pp 293ndash308 2009

[18] Q W Wang G J Song and C Y Lin ldquoRank equalities relatedto the generalized inverse 119860

(2)

119879119878with applicationsrdquo Applied

Mathematics and Computation vol 205 no 1 pp 370ndash3822008

[19] Q W Wang S W Yu and C Y Lin ldquoExtreme ranks of alinear quaternionmatrix expression subject to triple quaternionmatrix equations with applicationsrdquo Applied Mathematics andComputation vol 195 no 2 pp 733ndash744 2008

[20] Q W Wang G J Song and C Y Lin ldquoExtreme ranks ofthe solution to a consistent system of linear quaternion matrixequations with an applicationrdquo Applied Mathematics and Com-putation vol 189 no 2 pp 1517ndash1532 2007

[21] Y Tian ldquoFormulas for calculating the extremum ranks andinertias of a four-term quadratic matrix-valued function andtheir applicationsrdquo Linear Algebra and its Applications vol 437no 3 pp 835ndash859 2012

[22] Y Tian ldquoSolving optimization problems on ranks and inertiasof some constrained nonlinearmatrix functions via an algebraiclinearization methodrdquo Nonlinear Analysis Theory Methods ampApplications vol 75 no 2 pp 717ndash734 2012

[23] Y Tian ldquoMaximization and minimization of the rank andinertia of the Hermitianmatrix expression119860minus119861119883minus(119861119883)lowast withapplicationsrdquo Linear Algebra and its Applications vol 434 no10 pp 2109ndash2139 2011

[24] Q W Wang X Zhang and Z H He ldquoOn the Hermitianstructures of the solution to a pair of matrix equationsrdquo Linearand Multilinear Algebra vol 61 no 1 pp 73ndash90 2013

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2012 Article ID 471573 19 pagesdoi1011552012471573

Research ArticleConstrained Solutions of a System ofMatrix Equations

Qing-Wen Wang1 and Juan Yu1 2

1 Department of Mathematics Shanghai University 99 Shangda Road Shanghai 200444 China2 Department of Basic Mathematics China University of Petroleum Qingdao 266580 China

Correspondence should be addressed to Qing-Wen Wang wqw858yahoocomcn

Received 26 September 2012 Accepted 7 December 2012

Academic Editor Panayiotis J Psarrakos

Copyright q 2012 Q-W Wang and J Yu This is an open access article distributed under theCreative Commons Attribution License which permits unrestricted use distribution andreproduction in any medium provided the original work is properly cited

We derive the necessary and sufficient conditions of and the expressions for the orthogonalsolutions the symmetric orthogonal solutions and the skew-symmetric orthogonal solutions ofthe system of matrix equations AX = B and XC = D respectively When the matrix equationsare not consistent the least squares symmetric orthogonal solutions and the least squares skew-symmetric orthogonal solutions are respectively given As an auxiliary an algorithm is providedto compute the least squares symmetric orthogonal solutions and meanwhile an example ispresented to show that it is reasonable

1 Introduction

Throughout this paper the following notations will be used Rmtimesn OR

ntimesn SRntimesn and ASR

ntimesn

denote the set of all m times n real matrices the set of all n times n orthogonal matrices the set of alln times n symmetric matrices and the set of all n times n skew-symmetric matrices respectively In isthe identity matrix of order n (middot)T and tr(middot) represent the transpose and the trace of the realmatrix respectively middot stands for the Frobenius norm induced by the inner product Thefollowing two definitions will also be used

Definition 11 (see [1]) A real matrix X isin Rntimesn is said to be a symmetric orthogonal matrix if

XT = X and XTX = In

Definition 12 (see [2]) A real matrix X isin R2mtimes2m is called a skew-symmetric orthogonal

matrix if XT = minusX and XTX = In

The set of all n times n symmetric orthogonal matrices and the set of all 2m times 2m skew-symmetric orthogonal matrices are respectively denoted by SOR

ntimesn and SSOR2mtimes2m Since

2 Journal of Applied Mathematics

the linear matrix equation(s) and its optimal approximation problem have great applicationsin structural design biology control theory and linear optimal control and so forth seefor example [3ndash5] there has been much attention paid to the linear matrix equation(s) Thewell-known system of matrix equations

AX = B XC = D (11)

as one kind of linear matrix equations has been investigated by many authors and a seriesof important and useful results have been obtained For instance the system (11) withunknown matrix X being bisymmetric centrosymmetric bisymmetric nonnegative definiteHermitian and nonnegative definite and (PQ)-(skew) symmetric has been respectivelyinvestigated by Wang et al [6 7] Khatri and Mitra [8] and Zhang and Wang [9] Of courseif the solvability conditions of system (11) are not satisfied we may consider its least squaressolution For example Li et al [10] presented the least squares mirrorsymmetric solutionYuan [11] got the least-squares solution Some results concerning the system (11) can also befound in [12ndash18]

Symmetric orthogonal matrices and skew-symmetric orthogonal matrices play impor-tant roles in numerical analysis and numerical solutions of partial differential equationsPapers [1 2] respectively derived the symmetric orthogonal solution X of the matrixequation XC = D and the skew-symmetric orthogonal solution X of the matrix equationAX = B Motivated by the work mentioned above we in this paper will respectively studythe orthogonal solutions symmetric orthogonal solutions and skew-symmetric orthogonalsolutions of the system (11) Furthermore if the solvability conditions are not satisfiedthe least squares skew-symmetric orthogonal solutions and the least squares symmetricorthogonal solutions of the system (11) will be also given

The remainder of this paper is arranged as follows In Section 2 some lemmas areprovided to give the main results of this paper In Sections 3 4 and 5 the necessary andsufficient conditions of and the expression for the orthogonal the symmetric orthogonaland the skew-symmetric orthogonal solutions of the system (11) are respectively obtainedIn Section 6 the least squares skew-symmetric orthogonal solutions and the least squaressymmetric orthogonal solutions of the system (11) are presented respectively In additionan algorithm is provided to compute the least squares symmetric orthogonal solutions andmeanwhile an example is presented to show that it is reasonable Finally in Section 7 someconcluding remarks are given

2 Preliminaries

In this section we will recall some lemmas and the special C-S decomposition which will beused to get the main results of this paper

Lemma 21 (see [1 Lemmas 1 and 2]) Given C isin R2mtimesn D isin R

2mtimesn The matrix equation YC =D has a solution Y isin OR

2mtimes2m if and only if DTD = CTC Let the singular value decompositions ofC and D be respectively

C = U

(Π 00 0

)V T D = W

(Π 00 0

)V T (21)

Journal of Applied Mathematics 3

where

Π = diag(σ1 σk) gt 0 k = rank(C) = rank(D)

U =(U1 U2

) isin OR2mtimes2m U1 isin R

2mtimesk

W =(W1 W2

) isin OR2mtimes2m W1 isin R

2mtimesk V =(V1 V2

) isin ORntimesn V1 isin R

ntimesk

(22)

Then the orthogonal solutions of YC = D can be described as

Y = W

(Ik 00 P

)UT (23)

where P isin OR(2mminusk)times(2mminusk) is arbitrary

Lemma 22 (see [2 Lemmas 1 and 2]) GivenA isin Rntimesm B isin R

ntimesm The matrix equationAX = Bhas a solution X isin OR

mtimesm if and only if AAT = BBT Let the singular value decompositions of Aand B be respectively

A = U

(Σ 00 0

)V T B = U

(Σ 00 0

)QT (24)

where

Σ = diag(δ1 δl) gt 0 l = rank(A) = rank(B) U =(U1 U2

) isin ORntimesn U1 isin R

ntimesl

V =(V1 V2

) isin ORmtimesm V1 isin R

mtimesl Q =(Q1 Q2

) isin ORmtimesm Q1 isin R

mtimesl(25)

Then the orthogonal solutions of AX = B can be described as

X = V

(Il 00 W

)QT (26)

whereW isin OR(mminusl)times(mminusl) is arbitrary

Lemma 23 (see [2 Theorem 1]) If

X =(X11 X12

X21 X22

)isin OR

2mtimes2m X11 isin ASRktimesk (27)

then the C-S decomposition of X can be expressed as

(D1 00 D2

)T(X11 X12

X21 X22

)(D1 00 R2

)=(Σ11 Σ12

Σ21 Σ22

) (28)

4 Journal of Applied Mathematics

where D1 isin ORktimesk D2 R2 isin OR

(2mminusk)times(2mminusk)

Σ11 =

⎛⎜⎝

I 0 00 C 00 0 0

⎞⎟⎠ Σ12 =

⎛⎝0 0 0

0 S 00 0 I

⎞⎠

Σ21 =

⎛⎝0 0 0

0 S 00 0 I

⎞⎠ Σ22 =

⎛⎜⎝

I 0 00 minusCT 00 0 0

⎞⎟⎠

I = diag(I1 Ir1

) I1 = middot middot middot = Ir1 =

(0 1minus1 0

) C = diag(C1 Cl1)

Ci =(

0 ciminusci 0

) i = 1 l1 S = diag(S1 Sl1)

Si =(si 00 si

) ST

i Si + CTi Ci = I2 i = 1 l1

2r1 + 2l1 = rank(X11) Σ21 = ΣT12 k minus 2r1 = rank(X21)

(29)

Lemma 24 (see [1 Theorem 1]) If

K =(K11 K12

K21 K22

)isin OR

mtimesm K11 isin SRltimesl (210)

then the C-S decomposition of K can be described as

(D1 00 D2

)T(K11 K12

K21 K22

)(D1 00 R2

)=(Π11 Π12

Π21 Π22

) (211)

where D1 isin ORltimesl D2 R2 isin OR

(mminusl)times(mminusl)

Π11 =

⎛⎜⎝

I 0 00 C 00 0 0

⎞⎟⎠ Π12 =

⎛⎝0 0 0

0 S 00 0 I

⎞⎠

Π21 =

⎛⎝0 0 0

0 S 00 0 I

⎞⎠ Π22 =

⎛⎜⎝

I 0 00 minusC 00 0 0

⎞⎟⎠

I = diag(i1 ir) ij = plusmn1 j = 1 r

S = diag(s1prime slprime) C = diag(c1prime clprime)

s2iprime =radic

1 minus c2iprime i = 1prime lprime r + lprime = rank(K11) Π21 = ΠT

12

(212)

Journal of Applied Mathematics 5

Remarks 25 In order to know the C-S decomposition of an orthogonal matrix with ak times k leading (skew-) symmetric submatrix for details one can deeply study the proof ofTheorem 1 in [1] and [2]

Lemma 26 Given A isin Rntimesm B isin R

ntimesm Then the matrix equation AX = B has a solution X isinSOR

mtimesm if and only ifAAT = BBT andABT = BAT When these conditions are satisfied the generalsymmetric orthogonal solutions can be expressed as

X = V

(I2lminusr 0

0 G

)QT (213)

where

Q = JQdiag(ID2) isin ORmtimesm V = JV diag(I R2) isin OR

mtimesm

J =

⎛⎝I 0 0

0 0 I0 I 0

⎞⎠

(214)

and G isin SOR(mminus2l+r)times(mminus2l+r) is arbitrary

Proof The Necessity Assume X isin SORmtimesm is a solution of the matrix equation AX = B then

we have

BBT = AXXTAT = AAT

BAT = AXAT = AXTAT = ABT(215)

The Sufficiency Since the equality AAT = BBT holds then by Lemma 22 the singularvalue decompositions of A and B can be respectively expressed as (24) Moreover thecondition ABT = BAT means

U

(Σ 00 0

)V TQ

(Σ 00 0

)UT = U

(Σ 00 0

)QTV

(Σ 00 0

)UT (216)

which can be written as

V T1 Q1 = QT

1V1 (217)

From Lemma 22 the orthogonal solutions of the matrix equation AX = B can be describedas (26) Now we aim to find that X in (26) is also symmetric Suppose that X is symmetricthen we have

(Il 00 WT

)V TQ = QTV

(Il 00 W

) (218)

6 Journal of Applied Mathematics

together with the partitions of the matrices Q and V in Lemma 22 we get

V T1 Q1 = QT

1V1 (219)

QT1V2W = V T

1 Q2 (220)

QT2 V2W = WTV T

2 Q2 (221)

By (217) we can get (219) Now we aim to find the orthogonal solutions of the systemof matrix equations (220) and (221) Firstly we obtain from (220) that QT

1V2(QT1V2)

T =V T

1 Q2(V T1 Q2)

T then by Lemma 22 (220) has an orthogonal solution W By (217) the l times lleading principal submatrix of the orthogonal matrix V TQ is symmetric Then we have fromLemma 24

V T1 Q2 = D1Π12R

T2 (222)

QT1V2 = D1Π12D

T2 (223)

V T2 Q2 = D2Π22R

T2 (224)

From (220) (222) and (223) the orthogonal solution W of (220) is

W = D2

⎛⎝G 0 0

0 I 00 0 I

⎞⎠RT

2 (225)

where G isin OR(mminus2l+r)times(mminus2l+r) is arbitrary Combining (221) (224) and (225) yields GT =

G that is G is a symmetric orthogonal matrix Denote

V = V diag(ID2) Q = Qdiag(I R2) (226)

then the symmetric orthogonal solutions of the matrix equation AX = B can be expressed as

X = V

⎛⎝I 0 0

0 G 00 0 I

⎞⎠QT (227)

Let the partition matrix J be

J =

⎛⎝I 0 0

0 0 I0 I 0

⎞⎠ (228)

Journal of Applied Mathematics 7

compatible with the block matrix

⎛⎝I 0 0

0 G 00 0 I

⎞⎠ (229)

Put

V = JV Q = JQ (230)

then the symmetric orthogonal solutions of the matrix equation AX = B can be described as(213)

Setting A = CT B = DT and X = YT in [2 Theorem 2] and then by Lemmas 21 and23 we can have the following result

Lemma 27 Given C isin R2mtimesn D isin R

2mtimesn Then the equation has a solution Y isin SSOR2mtimes2m if

and only ifDTD = CTC andDTC = minusCTD When these conditions are satisfied the skew-symmetricorthogonal solutions of the matrix equation YC = D can be described as

Y = W

(I 00 H

)UT (231)

where

W = J primeW diag(IminusD2) isin OR2mtimes2m U = J primeUdiag(I R2) isin OR

2mtimes2m

J prime =

⎛⎝I 0 0

0 0 I0 I 0

⎞⎠

(232)

and H isin SSOR2rtimes2r is arbitrary

3 The Orthogonal Solutions of the System (11)

The following theorems give the orthogonal solutions of the system (11)

Theorem 31 Given A B isin Rntimesm and C D isin R

mtimesn suppose the singular value decompositions ofA and B are respectively as (24) Denote

QTC =(C1

C2

) V TD =

(D1

D2

) (31)

8 Journal of Applied Mathematics

where C1 D1 isin Rltimesn and C2 D2 isin R

(mminusl)timesn Let the singular value decompositions of C2 and D2 berespectively

C2 = U

(Π 00 0

)V T D2 = W

(Π 00 0

)V T (32)

where U W isin OR(mminusl)times(mminusl) V isin OR

ntimesn Π isin Rktimesk is diagonal whose diagonal elements are

nonzero singular values of C2 or D2 Then the system (11) has orthogonal solutions if and only if

AAT = BBT C1 = D1 DT2 D2 = CT

2 C2 (33)

In which case the orthogonal solutions can be expressed as

X = V

(Ik+l 00 Gprime

)QT (34)

where

V = V

(Il 00 W

)isin OR

mtimesm Q = Q

(Il 00 U

)isin OR

mtimesm (35)

and Gprime isin OR(mminuskminusl)times(mminuskminusl) is arbitrary

Proof Let the singular value decompositions of A and B be respectively as (24) Since thematrix equation AX = B has orthogonal solutions if and only if

AAT = BBT (36)

then by Lemma 22 its orthogonal solutions can be expressed as (26) Substituting (26) and(31) into the matrix equation XC = D we have C1 = D1 and WC2 = D2 By Lemma 21 thematrix equation WC2 = D2 has orthogonal solution W if and only if

DT2 D2 = CT

2C2 (37)

Let the singular value decompositions of C2 and D2 be respectively

C2 = U

(Π 00 0

)V T D2 = W

(Π 00 0

)V T (38)

where U W isin OR(mminusl)times(mminusl) V isin OR

ntimesn Π isin Rktimesk is diagonal whose diagonal elements are

nonzero singular values of C2 or D2 Then the orthogonal solutions can be described as

W = W

(Ik 00 Gprime

)UT (39)

Journal of Applied Mathematics 9

where Gprime isin OR(mminuskminusl)times(mminuskminusl) is arbitrary Therefore the common orthogonal solutions of the

system (11) can be expressed as

X = V

(Il 00 W

)QT = V

(Il 00 W

)⎛⎝Il 0 0

0 Ik 00 0 Gprime

⎞⎠(Il 00 UT

)QT = V

(Ik+l 00 Gprime

)QT (310)

where

V = V

(Il 00 W

)isin OR

mtimesm Q = Q

(Il 00 U

)isin OR

mtimesm (311)

and Gprime isin OR(mminuskminusl)times(mminuskminusl) is arbitrary

The following theorem can be shown similarly

Theorem 32 GivenA B isin Rntimesm and C D isin R

mtimesn let the singular value decompositions of C andD be respectively as (21) Partition

AW =(A1 A2

) BU =

(B1 B2

) (312)

where A1 B1 isin Rntimesk A2 B2 isin R

ntimes(mminusk) Assume the singular value decompositions of A2 and B2

are respectively

A2 = U

(Σ 00 0

)V T B2 = U

(Σ 00 0

)QT (313)

where V Q isin OR(mminusk)times(mminusk) U isin OR

ntimesn Σ isin Rlprimetimeslprime is diagonal whose diagonal elements are

nonzero singular values of A2 or B2 Then the system (11) has orthogonal solutions if and onlyif

DTD = CTC A1 = B1 A2AT2 = B2B

T2 (314)

In which case the orthogonal solutions can be expressed as

X = W

(Ik+lprime 0

0 H prime

)UT (315)

where

W = W

(Ik 00 W

)isin OR

mtimesm U = U

(Ik 00 U

)isin OR

mtimesm (316)

and H prime isin OR(mminuskminuslprime)times(mminuskminuslprime) is arbitrary

10 Journal of Applied Mathematics

4 The Symmetric Orthogonal Solutions of the System (11)

We now present the symmetric orthogonal solutions of the system (11)

Theorem 41 Given AB isin Rntimesm CD isin R

mtimesn Let the symmetric orthogonal solutions of thematrix equation AX = B be described as in Lemma 26 Partition

QTC =(Cprime

1Cprime

2

) V TD =

(Dprime

1Dprime

2

) (41)

where Cprime1 D

prime1 isin R

(2lminusr)timesn Cprime2 D

prime2 isin R

(mminus2l+r)timesn Then the system (11) has symmetric orthogonalsolutions if and only if

AAT = BBT ABT = BAT Cprime1 = Dprime

1 DprimeT2 Dprime

2 = CprimeT2 Cprime

2 DprimeT2 Cprime

2 = CprimeT2 Dprime

2 (42)

In which case the solutions can be expressed as

X = V

(I 00 Gprimeprime

)QT (43)

where

V = V

(I2lminusr 0

0 W

)isin OR

mtimesm Q = Q

(I2lminusr 0

0 U

)isin OR

mtimesm (44)

and Gprimeprime isin SOR(mminus2l+rminus2lprime+r prime)times(mminus2l+rminus2lprime+r prime) is arbitrary

Proof From Lemma 26 we obtain that the matrix equation AX = B has symmetricorthogonal solutions if and only if AAT = BBT and ABT = BAT When these conditionsare satisfied the general symmetric orthogonal solutions can be expressed as

X = V

(I2lminusr 0

0 G

)QT (45)

where G isin SOR(mminus2l+r)times(mminus2l+r) is arbitrary Q isin OR

mtimesm V isin ORmtimesm Inserting (41) and (45)

into the matrix equation XC = D we get Cprime1 = Dprime

1 and GCprime2 = Dprime

2 By [1 Theorem 2] thematrix equation GCprime

2 = Dprime2 has a symmetric orthogonal solution if and only if

DprimeT2 Dprime

2 = CprimeT2 Cprime

2 DprimeT2 Cprime

2 = CprimeT2 Dprime

2 (46)

In which case the solutions can be described as

G = W

(I2lprimeminusr prime 0

0 Gprimeprime

)UT (47)

Journal of Applied Mathematics 11

where Gprimeprime isin SOR(mminus2l+rminus2lprime+r prime)times(mminus2l+rminus2lprime+r prime) is arbitrary W isin OR

(mminus2l+r)times(mminus2l+r) and U isinOR

(mminus2l+r)times(mminus2l+r) Hence the system (11) has symmetric orthogonal solutions if and onlyif all equalities in (42) hold In which case the solutions can be expressed as

X = V

(I2lminusr 0

0 G

)QT = V

(I2lminusr 0

0 W

)⎛⎝I2lminusr 0

0(I 00 Gprimeprime

)⎞⎠(I2lminusr 0

0 UT

)QT (48)

that is the expression in (43)

The following theorem can also be obtained by the method used in the proof ofTheorem 41

Theorem 42 Given AB isin Rntimesm CD isin R

mtimesn Let the symmetric orthogonal solutions of thematrix equation XC = D be described as

X = M

(I2kminusr 0

0 G

)NT (49)

where M N isin ORmtimesm G isin SOR

(mminus2k+r)times(mminus2k+r) Partition

AM =(M1 M2

) BN =

(N1 N2

) M1N1 isin R

ntimes(2kminusr) M2N2 isin Rntimes(mminus2k+r) (410)

Then the system (11) has symmetric orthogonal solutions if and only if

DTD = CTC DTC = CTD M1 = N1 M2MT2 = N2N

T2 M2N

T2 = N2M

T2

(411)

In which case the solutions can be expressed as

X = M

(I 00 H primeprime

)NT (412)

where

M = M

(I2kminusr 0

0 W1

)isin OR

mtimesm N = N

(I2kminusr 0

0 U1

)isin OR

mtimesm (413)

and H primeprime isin SOR(mminus2k+rminus2kprime+r prime)times(mminus2k+rminus2kprime+r prime) is arbitrary

12 Journal of Applied Mathematics

5 The Skew-Symmetric Orthogonal Solutions of the System (11)

In this section we show the skew-symmetric orthogonal solutions of the system (11)

Theorem 51 Given AB isin Rntimes2m CD isin R

2mtimesn Suppose the matrix equation AX = B has skew-symmetric orthogonal solutions with the form

X = V1

(I 00 H

)QT

1 (51)

whereH isin SSOR2rtimes2r is arbitrary V1 Q1 isin OR

2mtimes2m Partition

QT1C =

(Q1

Q2

) V T

1 D =(V1

V2

) (52)

where Q1 V1 isin R(2mminus2r)timesn Q2 V2 isin R

2rtimesn Then the system (11) has skew-symmetric orthogonalsolutions if and only if

AAT = BBT ABT = minusBAT Q1 = V1 QT2 Q2 = V T

2 V2 QT2V2 = minusV T

2 Q2 (53)

In which case the solutions can be expressed as

X = V1

(I 00 J prime

)QT

1 (54)

where

V = V1

(I2mminus2r 0

0 W

)isin OR

2mtimes2m Q = Q1

(I2mminus2r 0

0 U

)isin OR

2mtimes2m (55)

and J prime isin SSOR2kprimetimes2kprime

is arbitrary

Proof By [2 Theorem 2] the matrix equation AX = B has the skew-symmetric orthogonalsolutions if and only if AAT = BBT and ABT = minusBAT When these conditions are satisfiedthe general skew-symmetric orthogonal solutions can be expressed as (51) Substituting (51)and (52) into the matrix equation XC = D we get Q1 = V1 and HQ2 = V2 From Lemma 27equation HQ2 = V2 has a skew-symmetric orthogonal solution H if and only if

QT2 Q2 = V T

2 V2 QT2V2 = minusV T

2 Q2 (56)

When these conditions are satisfied the solution can be described as

H = W

(I 00 J prime

)UT (57)

Journal of Applied Mathematics 13

where J prime isin SSOR2kprimetimes2kprime

is arbitrary W U isin OR2rtimes2r Inserting (57) into (51) yields that the

system (11) has skew-symmetric orthogonal solutions if and only if all equalities in (53)hold In which case the solutions can be expressed as (54)

Similarly the following theorem holds

Theorem 52 Given AB isin Rntimes2m CD isin R

2mtimesn Suppose the matrix equation XC = D hasskew-symmetric orthogonal solutions with the form

X = W

(I 00 K

)UT (58)

where K isin SSOR2ptimes2p is arbitrary W U isin OR

2mtimes2m Partition

AW =(W1 W2

) BU =

(U1 U2

) (59)

where W1 U1 isin Rntimes(2mminus2p) W2 U2 isin R

ntimes2p Then the system (11) has skew-symmetric orthogonalsolutions if and only if

DTD = CTC DTC = minusCTD W1 = U1 W2WT2 = U2U

T2 U2W

T2 = minusW2U

T2

(510)

In which case the solutions can be expressed as

X = W1

(I 00 J primeprime

)UT

1 (511)

where

W1 = W

(I2mminus2p 0

0 W1

)isin OR

2mtimes2m U1 =

(I2mminus2p 0

0 UT1

)U isin OR

2mtimes2m (512)

and J primeprime isin SSOR2qtimes2q is arbitrary

6 The Least Squares (Skew-) Symmetric Orthogonal Solutions ofthe System (11)

If the solvability conditions of a system of matrix equations are not satisfied it is natural toconsider its least squares solution In this section we get the least squares (skew-) symmetricorthogonal solutions of the system (11) that is seek X isin SSOR

2mtimes2m(SORntimesn) such that

minXisinSSOR2mtimes2m(SORntimesn)

AX minus B2 + XC minusD2 (61)

14 Journal of Applied Mathematics

With the help of the definition of the Frobenius norm and the properties of the skew-symmetric orthogonal matrix we get that

AX minus B2 + XC minusD2 = A2 + B2 + C2 + D2 minus 2 tr(XT(ATB +DCT

)) (62)

Let

ATB +DCT = K

KT minusK

2= T

(63)

Then it follows from the skew-symmetric matrix X that

tr(XT(ATB +DCT

))= tr(XTK

)= tr(minusXK) = tr

(X

(KT minusK

2

))= tr(XT) (64)

Therefore (61) holds if and only if (64) reaches its maximum Now we pay our attention tofind the maximum value of (64) Assume the eigenvalue decomposition of T is

T = E

(Λ 00 0

)ET (65)

with

Λ = diag(Λ1 Λl) Λi =(

0 αi

minusαi 0

) αi gt 0 i = 1 l 2l = rank(T) (66)

Denote

ETXE =(X11 X12

X21 X22

) (67)

partitioned according to

(Λ 00 0

) (68)

then (64) has the following form

tr(XT) = tr(ETXE

(Λ 00 0

))= tr(X11Λ) (69)

Journal of Applied Mathematics 15

Thus by

X11 = I = diag(I1 Il

) (610)

where

Ii =(

0 minus11 0

) i = 1 l (611)

Equation (69) gets its maximum Since ETXE is skew-symmetric it follows from

X = E

(I 00 G

)ET (612)

where G isin SSOR(2mminus2l)times(2mminus2l) is arbitrary that (61) obtains its minimum Hence we have the

following theorem

Theorem 61 Given AB isin Rntimes2m and CD isin R

2mtimesn denote

ATB +DCT = KKT minusK

2= T (613)

and let the spectral decomposition of T be (65) Then the least squares skew-symmetric orthogonalsolutions of the system (11) can be expressed as (612)

If X in (61) is a symmetric orthogonal matrix then by the definition of the Frobeniusnorm and the properties of the symmetric orthogonal matrix (62) holds Let

ATB +DCT = HHT +H

2= N (614)

Then we get that

minXisinSORntimesn

AX minus B2 + XC minusD2 = minXisinSORntimesn

[A2 + B2 + C2 + D2 minus 2 tr(XN)

] (615)

Thus (615) reaches its minimum if and only if tr(XN) obtains its maximum Now we focuson finding the maximum value of tr(XN) Let the spectral decomposition of the symmetricmatrix N be

N = M

(Σ 00 0

)MT (616)

16 Journal of Applied Mathematics

where

Σ =(Σ+ 00 Σminus

) Σ+ = diag(λ1 λs)

λ1 ge λ2 ge middot middot middot ge λs gt 0 Σminus = diag(λs+1 λt)

λt le λtminus1 le middot middot middot le λs+1 lt 0 t = rank(N)

(617)

Denote

MTXM =

(X11 X12

X21 X22

)(618)

being compatible with

(Σ 00 0

) (619)

Then

tr(XN) = tr(MTXM

(Σ 00 0

))= tr

((X11 X12

X21 X22

)(Σ 00 0

))= tr(X11Σ

) (620)

Therefore it follows from

X11 = I =(Is 00 minusItminuss

)(621)

that (620) reaches its maximum Since MTXM is a symmetric orthogonal matrix then whenX has the form

X = M

(I 00 L

)MT (622)

where L isin SOR(nminust)times(nminust) is arbitrary (615) gets its minimum Thus we obtain the following

theorem

Theorem 62 Given AB isin Rntimesm CD isin R

mtimesn denote

ATB +DCT = HHT +H

2= N (623)

and let the eigenvalue decomposition of N be (616) Then the least squares symmetric orthogonalsolutions of the system (11) can be described as (622)

Journal of Applied Mathematics 17

Algorithm 63 Consider the following

Step 1 Input AB isin Rntimesm and CD isin R

mtimesn

Step 2 Compute

ATB +DCT = H

N =HT +H

2

(624)

Step 3 Compute the spectral decomposition of N with the form (616)

Step 4 Compute the least squares symmetric orthogonal solutions of (11) according to(622)

Example 64 Assume

A =

⎛⎜⎜⎜⎜⎜⎝

122 84 minus56 63 94 107118 29 85 69 96 minus78106 23 115 78 67 8936 78 49 119 94 5945 67 78 31 56 116

⎞⎟⎟⎟⎟⎟⎠

B =

⎛⎜⎜⎜⎜⎜⎝

85 94 36 78 63 4727 36 79 94 56 7837 67 86 98 34 29minus43 62 57 74 54 9529 39 minus52 63 78 46

⎞⎟⎟⎟⎟⎟⎠

C =

⎛⎜⎜⎜⎜⎜⎜⎜⎝

54 64 37 56 9736 42 78 minus63 7867 35 minus46 29 28minus27 72 108 37 3819 39 82 56 11289 78 94 79 56

⎞⎟⎟⎟⎟⎟⎟⎟⎠

D =

⎛⎜⎜⎜⎜⎜⎜⎜⎝

79 95 54 28 8687 26 67 84 8157 39 minus29 52 1948 58 18 minus72 5828 79 45 67 9695 41 34 98 39

⎞⎟⎟⎟⎟⎟⎟⎟⎠

(625)

It can be verified that the given matrices ABC and D do not satisfy the solvabilityconditions in Theorem 41 or Theorem 42 So we intend to derive the least squares symmetricorthogonal solutions of the system (11) By Algorithm 63 we have the following results

(1) the least squares symmetric orthogonal solution

X =

⎛⎜⎜⎜⎜⎜⎜⎜⎝

001200 006621 minus024978 027047 091302 minus016218006621 065601 022702 minus055769 024587 037715minus024978 022702 080661 040254 004066 minus026784027047 minus055769 040254 006853 023799 062645091302 024587 004066 023799 minus012142 minus018135minus016218 037715 minus026784 062645 minus0181354 057825

⎞⎟⎟⎟⎟⎟⎟⎟⎠

(626)

(2)

minXisinSORmtimesm

AX minus B2 + XC minusD2 = 198366

∥∥∥XTX minus I6

∥∥∥ = 284882 times 10minus15∥∥∥XT minusX

∥∥∥ = 000000(627)

18 Journal of Applied Mathematics

Remark 65 (1) There exists a unique symmetric orthogonal solution such that (61) holds ifand only if the matrix

N =HT +H

2 (628)

where

H = ATB +DCT (629)

is invertible Example 64 just illustrates it(2) The algorithm about computing the least squares skew-symmetric orthogonal

solutions of the system (11) can be shown similarly we omit it here

7 Conclusions

This paper is devoted to giving the solvability conditions of and the expressions ofthe orthogonal solutions the symmetric orthogonal solutions and the skew-symmetricorthogonal solutions to the system (11) respectively and meanwhile obtaining the leastsquares symmetric orthogonal and skew-symmetric orthogonal solutions of the system (11)In addition an algorithm and an example have been provided to compute its least squaressymmetric orthogonal solutions

Acknowledgments

This research was supported by the grants from Innovation Program of Shanghai MunicipalEducation Commission (13ZZ080) the National Natural Science Foundation of China(11171205) the Natural Science Foundation of Shanghai (11ZR1412500) the DisciplineProject at the corresponding level of Shanghai (A 13-0101-12-005) and Shanghai LeadingAcademic Discipline Project (J50101)

References

[1] C J Meng and X Y Hu ldquoThe inverse problem of symmetric orthogonal matrices and its optimalapproximationrdquo Mathematica Numerica Sinica vol 28 pp 269ndash280 2006 (Chinese)

[2] C J Meng X Y Hu and L Zhang ldquoThe skew-symmetric orthogonal solutions of the matrix equationAX = Brdquo Linear Algebra and Its Applications vol 402 no 1ndash3 pp 303ndash318 2005

[3] D L Chu H C Chan and D W C Ho ldquoRegularization of singular systems by derivative andproportional output feedbackrdquo SIAM Journal on Matrix Analysis and Applications vol 19 no 1 pp21ndash38 1998

[4] K T Joseph ldquoInverse eigenvalue problem in structural designrdquo AIAA Journal vol 30 no 12 pp2890ndash2896 1992

[5] A Jameson and E Kreinder ldquoInverse problem of linear optimal controlrdquo SIAM Journal on Controlvol 11 no 1 pp 1ndash19 1973

[6] Q W Wang ldquoBisymmetric and centrosymmetric solutions to systems of real quaternion matrixequationsrdquo Computers and Mathematics with Applications vol 49 no 5-6 pp 641ndash650 2005

[7] Q W Wang X Liu and S W Yu ldquoThe common bisymmetric nonnegative definite solutions withextreme ranks and inertias to a pair of matrix equationsrdquo Applied Mathematics and Computation vol218 pp 2761ndash2771 2011

Journal of Applied Mathematics 19

[8] C G Khatri and S K Mitra ldquoHermitian and nonnegative definite solutions of linear matrixequationsrdquo SIAM Journal on Applied Mathematics vol 31 pp 578ndash585 1976

[9] Q Zhang and Q W Wang ldquoThe (PQ)-(skew)symmetric extremal rank solutions to a system ofquaternion matrix equationsrdquo Applied Mathematics and Computation vol 217 no 22 pp 9286ndash92962011

[10] F L Li X Y Hu and L Zhang ldquoLeast-squares mirrorsymmetric solution for matrix equations (AX =BXC = D)rdquo Numerical Mathematics vol 15 pp 217ndash226 2006

[11] Y X Yuan ldquoLeast-squares solutions to the matrix equations AX = B and XC = Drdquo AppliedMathematics and Computation vol 216 no 10 pp 3120ndash3125 2010

[12] A Dajic and J J Koliha ldquoPositive solutions to the equations AX = C and XB = D for Hilbert spaceoperatorsrdquo Journal of Mathematical Analysis and Applications vol 333 no 2 pp 567ndash576 2007

[13] F L Li X Y Hu and L Zhang ldquoThe generalized reflexive solution for a class of matrix equations(AX = BXC = D)rdquo Acta Mathematica Scientia vol 28 no 1 pp 185ndash193 2008

[14] S Kumar Mitra ldquoThe matrix equations AX = CXB = Drdquo Linear Algebra and Its Applications vol 59pp 171ndash181 1984

[15] Y Qiu Z Zhang and J Lu ldquoThe matrix equations AX = BXC = D with PX = sXP constraintrdquoApplied Mathematics and Computation vol 189 no 2 pp 1428ndash1434 2007

[16] Y Y Qiu and A D Wang ldquoLeast squares solutions to the equations AX = BXC = B with someconstraintsrdquo Applied Mathematics and Computation vol 204 no 2 pp 872ndash880 2008

[17] Q W Wang X Zhang and Z H He ldquoOn the Hermitian structures of the solution to a pair of matrixequationsrdquo Linear Multilinear Algebra vol 61 no 1 pp 73ndash90 2013

[18] Q X Xu ldquoCommon Hermitian and positive solutions to the adjointable operator equations AX =CXB = Drdquo Linear Algebra and Its Applications vol 429 no 1 pp 1ndash11 2008

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2012 Article ID 398085 14 pagesdoi1011552012398085

Research ArticleOn the Hermitian R-Conjugate Solution ofa System of Matrix Equations

Chang-Zhou Dong1 Qing-Wen Wang2 and Yu-Ping Zhang3

1 School of Mathematics and Science Shijiazhuang University of Economics ShijiazhuangHebei 050031 China

2 Department of Mathematics Shanghai University Shanghai Shanghai 200444 China3 Department of Mathematics Ordnance Engineering College Shijiazhuang Hebei 050003 China

Correspondence should be addressed to Qing-Wen Wang wqw858yahoocomcn

Received 3 October 2012 Accepted 26 November 2012

Academic Editor Yang Zhang

Copyright q 2012 Chang-Zhou Dong et al This is an open access article distributed under theCreative Commons Attribution License which permits unrestricted use distribution andreproduction in any medium provided the original work is properly cited

Let R be an n by n nontrivial real symmetric involution matrix that is R = Rminus1 = RT = In An n times n

complex matrix A is termed R-conjugate if A = RAR where A denotes the conjugate of A Wegive necessary and sufficient conditions for the existence of the Hermitian R-conjugate solutionto the system of complex matrix equations AX = C andXB = D and present an expression ofthe Hermitian R-conjugate solution to this system when the solvability conditions are satisfiedIn addition the solution to an optimal approximation problem is obtained Furthermore the leastsquares Hermitian R-conjugate solution with the least norm to this system mentioned above isconsidered The representation of such solution is also derived Finally an algorithm and numericalexamples are given

1 Introduction

Throughout we denote the complex m times n matrix space by Cmtimesn the real m times n matrix space

by Rmtimesn and the set of all matrices in R

mtimesn with rank r by Rmtimesnr The symbols IAAT Alowast Adagger

and A stand for the identity matrix with the appropriate size the conjugate the transposethe conjugate transpose the Moore-Penrose generalized inverse and the Frobenius norm ofA isin C

mtimesn respectively We use Vn to denote the ntimes n backward matrix having the elements 1along the southwest diagonal and with the remaining elements being zeros

Recall that an n times n complex matrix A is centrohermitian if A = VnAVn Centro-hermitian matrices and related matrices such as k-Hermitian matrices Hermitian Toeplitzmatrices and generalized centrohermitian matrices appear in digital signal processing andothers areas (see [1ndash4]) As a generalization of a centrohermitian matrix and related matrices

2 Journal of Applied Mathematics

Trench [5] gave the definition of R-conjugate matrix A matrix A isin Cntimesn is R-conjugate if

A = RAR where R is a nontrivial real symmetric involution matrix that is R = Rminus1 = RT andR= In At the same time Trench studied the linear equation Az = w for R-conjugate matricesin [5] where zw are known column vectors

Investigating the matrix equation

AX = B (11)

with the unknown matrix X being symmetric reflexive Hermitian-generalized Hamiltonianand repositive definite is a very active research topic [6ndash14] As a generalization of (11) theclassical system of matrix equations

AX = C XB = D (12)

has attracted many authorrsquos attention For instance [15] gave the necessary and sufficientconditions for the consistency of (12) [16 17] derived an expression for the general solutionby using singular value decomposition of a matrix and generalized inverses of matricesrespectively Moreover many results have been obtained about the system (12) with variousconstraints such as bisymmetric Hermitian positive semidefinite reflexive and generalizedreflexive solutions (see [18ndash28]) To our knowledge so far there has been little investigationof the Hermitian R-conjugate solution to (12)

Motivated by the work mentioned above we investigate Hermitian R-conjugatesolutions to (12) We also consider the optimal approximation problem

∥∥∥X minus E∥∥∥ = min

XisinSX

X minus E (13)

where E is a given matrix in Cntimesn and SX the set of all Hermitian R-conjugate solutions to

(12) In many cases the system (12) has not Hermitian R-conjugate solution Hence we needto further study its least squares solution which can be described as follows Let RHC

ntimesn

denote the set of all Hermitian R-conjugate matrices in Cntimesn

SL =X | min

XisinRHCntimesn

(AX minus C2 + XB minusD2

) (14)

Find X isin Cntimesn such that

∥∥∥X∥∥∥ = minXisinSL

X (15)

In Section 2 we present necessary and sufficient conditions for the existence of theHermitian R-conjugate solution to (12) and give an expression of this solution when thesolvability conditions are met In Section 3 we derive an optimal approximation solution to(13) In Section 4 we provide the least squares Hermitian R-conjugate solution to (15) InSection 5 we give an algorithm and a numerical example to illustrate our results

Journal of Applied Mathematics 3

2 R-Conjugate Hermitian Solution to (12)

In this section we establish the solvability conditions and the general expression for theHermitian R-conjugate solution to (12)

We denote RCntimesn and RHC

ntimesn the set of all R-conjugate matrices and Hermitian R-conjugate matrices respectively that is

RCntimesn =

A | A = RAR

HRCntimesn =

A | A = RARA = Alowast

(21)

where R is n times n nontrivial real symmetric involution matrixChang et al in [29] mentioned that for nontrivial symmetric involution matrix R isin

Rntimesn there exist positive integer r and n times n real orthogonal matrix [P Q] such that

R =[P Q

]⎡⎣Ir 0

0 minusInminusr

⎤⎦⎡⎣P

T

QT

⎤⎦ (22)

where P isin Rntimesr Q isin R

ntimes(nminusr) By (22)

RP = P RQ = minusQ PTP = Ir QTQ = Inminusr PTQ = 0 QTP = 0 (23)

Throughout this paper we always assume that the nontrivial symmetric involutionmatrix R is fixed which is given by (22) and (23) Now we give a criterion of judging amatrix is R-conjugate Hermitian matrix

Theorem 21 A matrix K isin HRCntimesn if and only if there exists a symmetric matrix H isin R

ntimesn suchthat K = ΓHΓlowast where

Γ =[P iQ

] (24)

with PQ being the same as (22)

Proof If K isin HRCntimesn then K = RKR By (22)

K = RKR =[P Q

]⎡⎣Ir 0

0 minusInminusr

⎤⎦⎡⎣P

T

QT

⎤⎦K[P Q

]⎡⎣Ir 0

0 minusInminusr

⎤⎦⎡⎣P

T

QT

⎤⎦ (25)

4 Journal of Applied Mathematics

which is equivalent to

⎡⎣P

T

QT

⎤⎦K[P Q

]

=

⎡⎣Ir 0

0 minusInminusr

⎤⎦⎡⎣P

T

QT

⎤⎦K[P Q

]⎡⎣Ir 0

0 minusInminusr

⎤⎦

(26)

Suppose that

⎡⎣P

T

QT

⎤⎦K[P Q

]=

⎡⎣K11 K12

K21 K22

⎤⎦ (27)

Substituting (27) into (26) we obtain

⎡⎣K11 K12

K21 K22

⎤⎦ =

⎡⎣Ir 0

0 minusInminusr

⎤⎦⎡⎣K11 K12

K21 K22

⎤⎦⎡⎣Ir 0

0 minusInminusr

⎤⎦ =

⎡⎣ K11 minusK12

minusK21 K22

⎤⎦ (28)

Hence K11 = K11 K12 = minusK12 K21 = minusK21 K22 = K22 that is K11 iK12 iK21 K22 are realmatrices If we denote M = iK12 N = minusiK21 then by (27)

K =[P Q

]⎡⎣K11 K12

K21 K22

⎤⎦⎡⎣P

T

QT

⎤⎦ =

[P iQ

]⎡⎣K11 M

N K22

⎤⎦⎡⎣ PT

minusiQT

⎤⎦ (29)

Let Γ = [P iQ] and

H =

⎡⎣K11 M

N K22

⎤⎦ (210)

Then K can be expressed as ΓHΓlowast where Γ is unitary matrix and H is a real matrix ByK = Klowast

ΓHTΓlowast = Klowast = K = ΓHΓlowast (211)

we obtain H = HT

Journal of Applied Mathematics 5

Conversely if there exists a symmetric matrix H isin Rntimesn such that K = ΓHΓlowast then it

follows from (23) that

RKR = RΓHΓlowastR = R[P iQ

]H

⎡⎣ PT

minusiQT

⎤⎦R =

[P minusiQ]H

⎡⎣ PT

iQT

⎤⎦ = ΓHΓlowast = K

Klowast = ΓHTΓlowast = ΓHΓlowast = K

(212)

that is K isin HRCntimesn

Theorem 21 implies that an arbitrary complex Hermitian R-conjugate matrix is equiv-alent to a real symmetric matrix

Lemma 22 For any matrix A isin Cmtimesn A = A1 + iA2 where

A1 =A +A

2 A2 =

A minusA

2i (213)

Proof For any matrix A isin Cmtimesn it is obvious that A = A1 + iA2 where A1 A2 are defined

as (213) Now we prove that the decomposition A = A1 + iA2 is unique If there exist B1 B2

such that A = B1 + iB2 then

A1 minus B1 + i(B2 minusA2) = 0 (214)

It follows from A1 A2 B1 andB2 are real matrix that

A1 = B1 A2 = B2 (215)

Hence A = A1 + iA2 holds where A1 A2 are defined as (213)

By Theorem 21 for X isin HRCntimesn we may assume that

X = ΓYΓlowast (216)

where Γ is defined as (24) and Y isin Rntimesn is a symmetric matrix

Suppose that AΓ = A1 + iA2 isin Cmtimesn CΓ = C1 + iC2 isin C

mtimesn ΓlowastB = B1 + iB2 isin Cntimesl and

ΓlowastD = D1 + iD2 isin Cntimesl where

A1 =AΓ +AΓ

2 A2 =

AΓ minusAΓ2i

C1 =CΓ + CΓ

2 C2 =

CΓ minus CΓ2i

B1 =ΓlowastB + ΓlowastB

2 B2 =

ΓlowastB minus ΓlowastB2i

D1 =ΓlowastD + ΓlowastD

2 D2 =

ΓlowastD minus ΓlowastD2i

(217)

6 Journal of Applied Mathematics

Then system (12) can be reduced into

(A1 + iA2)Y = C1 + iC2 Y (B1 + iB2) = D1 + iD2 (218)

which implies that

⎡⎣A1

A2

⎤⎦Y =

⎡⎣C1

C2

⎤⎦ Y

[B1 B2

]=[D1 D2

] (219)

Let

F =

⎡⎣A1

A2

⎤⎦ G =

⎡⎣C1

C2

⎤⎦ K =

[B1 B2

]

L =[D1 D2

] M =

⎡⎣ F

KT

⎤⎦ N =

⎡⎣G

LT

⎤⎦

(220)

Then system (12) has a solution X in HRCntimesn if and only if the real system

MY = N (221)

has a symmetric solution Y in Rntimesn

Lemma 23 (Theorem 1 in [7]) Let A isin Rmtimesn The SVD of matrix A is as follows

A = U

⎡⎣Σ 0

0 0

⎤⎦V T (222)

where U = [U1 U2] isin Rmtimesm and V = [V1 V2] isin R

ntimesn are orthogonal matrices Σ = diag(σ1 σr) σi gt 0 (i = 1 r) r = rank(A) U1 isin R

mtimesr V1 isin Rntimesr Then (11) has a symmetric

solution if and only if

ABT = BAT UT2 B = 0 (223)

In that case it has the general solution

X = V1Σminus1UT1 B + V2V

T2 B

TU1Σminus1V T1 + V2GVT

2 (224)

where G is an arbitrary (n minus r) times (n minus r) symmetric matrix

By Lemma 23 we have the following theorem

Journal of Applied Mathematics 7

Theorem 24 Given A isin Cmtimesn C isin C

mtimesn B isin Cntimesl and D isin C

ntimesl Let A1 A2 C1C2B1 B2 D1 D2 F G K L M andN be defined in (217) (220) respectively Assume thatthe SVD of M isin R

(2m+2l)timesn is as follows

M = U

⎡⎣M1 0

0 0

⎤⎦V T (225)

where U = [U1 U2] isin R(2m+2l)times(2m+2l) and V = [V1 V2] isin R

ntimesn are orthogonal matrices M1 =diag(σ1 σr) σi gt 0 (i = 1 r) r = rank(M) U1 isin R

(2m+2l)timesr V1 isin Rntimesr Then system

(12) has a solution in HRCntimesn if and only if

MNT = NMT UT2 N = 0 (226)

In that case it has the general solution

X = Γ(V1M

minus11 UT

1 N + V2VT2 N

TU1Mminus11 V T

1 + V2GVT2

)Γlowast (227)

where G is an arbitrary (n minus r) times (n minus r) symmetric matrix

3 The Solution of Optimal Approximation Problem (13)

When the set SX of all Hermitian R-conjugate solution to (12) is nonempty it is easy to verifySX is a closed set Therefore the optimal approximation problem (13) has a unique solutionby [30]

Theorem 31 GivenA isin Cmtimesn C isin C

mtimesn B isin Cntimesl D isin C

ntimesl E isin Cntimesn and E1 = (12)(ΓlowastEΓ+

ΓlowastEΓ) Assume SX is nonempty then the optimal approximation problem (13) has a unique solutionX and

X = Γ(V1M

minus11 UT

1 N + V2VT2 N

TU1Mminus11 V T

1 + V2VT2 E1V2V

T2

)Γlowast (31)

Proof Since SX is nonempty X isin SX has the form of (227) By Lemma 22 ΓlowastEΓ can be writtenas

ΓlowastEΓ = E1 + iE2 (32)

where

E1 =12

(ΓlowastEΓ + ΓlowastEΓ

) E2 =

12i

(ΓlowastEΓ minus ΓlowastEΓ

) (33)

8 Journal of Applied Mathematics

According to (32) and the unitary invariance of Frobenius norm

X minus E =∥∥∥Γ(V1M

minus1UT1N + V2V

T2 N

TU1Mminus1V T

1 + V2GVT2

)Γlowast minus E

∥∥∥=∥∥∥(E1 minus V1M

minus1UT1N minus V2V

T2 N

TU1Mminus1V T

1 minus V2GVT2

)+ iE2

∥∥∥(34)

We get

X minus E2 =∥∥∥E1 minus V1M

minus1UT1 N minus V2V

T2 N

TU1Mminus1V T

1 minus V2GVT2

∥∥∥2+ E22 (35)

Then minXisinSXX minus E is consistent if and only if there exists G isin R(nminusr)times(nminusr) such that

min∥∥∥E1 minus V1M

minus1UT1 N minus V2V

T2 N

TU1Mminus1V T

1 minus V2GVT2

∥∥∥ (36)

For the orthogonal matrix V

∥∥∥E1 minus V1Mminus1UT

1N minus V2VT2 N

TU1Mminus1V T

1 minus V2GVT2

∥∥∥2

=∥∥∥V T (E1 minus V1M

minus1UT1 N minus V2V

T2 N

TU1Mminus1V T

1 minus V2GVT2 )V

∥∥∥2

=∥∥∥V T

1

(E1 minus V1M

minus1UT1 N)V1

∥∥∥2+∥∥∥V T

1 (E1 minus V1Mminus1UT

1 N)V2

∥∥∥2

+∥∥∥V T

2

(E1 minus V2V

T2 N

TU1Mminus1V T

1

)V1

∥∥∥2+∥∥∥V T

2

(E1 minus V2GVT

2

)V2

∥∥∥

(37)

Therefore

min∥∥∥E1 minus V1M

minus1UT1 N minus V2V

T2 N

TU1Mminus1V T

1 minus V2GVT2

∥∥∥ (38)

is equivalent to

G = V T2 E1V2 (39)

Substituting (39) into (227) we obtain (31)

4 The Solution of Problem (15)

In this section we give the explicit expression of the solution to (15)

Theorem 41 Given A isin Cmtimesn C isin C

mtimesn B isin Cntimesl and D isin C

ntimesl Let A1 A2 C1 C2B1 B2 D1 D2 F G K L M and N be defined in (217) (220) respectively Assume that the

Journal of Applied Mathematics 9

SVD of M isin R(2m+2l)timesn is as (225) and system (12) has not a solution in HRC

ntimesn Then X isin SL

can be expressed as

X = ΓV

⎡⎣Mminus1

1 UT1NV1 Mminus1

1 UT1NV2

V T2 N

TU1Mminus11 Y22

⎤⎦V TΓlowast (41)

where Y22 isin R(nminusr)times(nminusr) is an arbitrary symmetric matrix

Proof It yields from (217)ndash(221) and (225) that

AX minus C2 + XB minusD2 = MY minusN2

=

∥∥∥∥∥∥U⎡⎣M1 0

0 0

⎤⎦V TY minusN

∥∥∥∥∥∥2

=

∥∥∥∥∥∥⎡⎣M1 0

0 0

⎤⎦V TYV minusUTNV

∥∥∥∥∥∥2

(42)

Assume that

V TYV =

⎡⎣Y11 Y12

Y21 Y22

⎤⎦ Y11 isin R

rtimesr Y22 isin R(nminusr)times(nminusr) (43)

Then we have

AX minus C2 + XB minusD2

=∥∥∥M1Y11 minusUT

1 NV1

∥∥∥2+∥∥∥M1Y12 minusUT

1NV2

∥∥∥2

+∥∥∥UT

2NV1

∥∥∥2+∥∥∥UT

2 NV2

∥∥∥2

(44)

Hence

min(AX minus C2 + XB minusD2

)(45)

is solvable if and only if there exist Y11 Y12 such that

∥∥∥M1Y11 minusUT1 NV1

∥∥∥2= min

∥∥∥M1Y12 minusUT1NV2

∥∥∥2= min

(46)

10 Journal of Applied Mathematics

It follows from (46) that

Y11 = Mminus11 UT

1 NV1

Y12 = Mminus11 UT

1 NV2(47)

Substituting (47) into (43) and then into (216) we can get that the form of elements in SL is(41)

Theorem 42 Assume that the notations and conditions are the same as Theorem 41 Then

∥∥∥X∥∥∥ = minXisinSL

X (48)

if and only if

X = ΓV

⎡⎣Mminus1

1 UT1NV1 Mminus1

1 UT1NV2

V T2 N

TU1Mminus11 0

⎤⎦V TΓlowast (49)

Proof In Theorem 41 it implies from (41) that minXisinSLX is equivalent to X has theexpression (41) with Y22 = 0 Hence (49) holds

5 An Algorithm and Numerical Example

Base on the main results of this paper we in this section propose an algorithm for finding thesolution of the approximation problem (13) and the least squares problem with least norm(15) All the tests are performed by MATLAB 65 which has a machine precision of around10minus16

Algorithm 51 (1) Input A isin Cmtimesn C isin C

mtimesn B isin Cntimesl D isin C

ntimesl(2) Compute A1 A2 C1 C2 B1 B2 D1 D2 F G K L M andN by (217) and

(220)(3) Compute the singular value decomposition of M with the form of (225)(4) If (226) holds then input E isin C

ntimesn and compute the solution X of problem (13)according (31) else compute the solution X to problem (15) by (49)

To show our algorithm is feasible we give two numerical example Let an nontrivialsymmetric involution be

R =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 0 0 0

0 minus1 0 0

0 0 1 0

0 0 0 minus1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦ (51)

Journal of Applied Mathematics 11

We obtain [P Q] in (22) by using the spectral decomposition of R then by (24)

Γ =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 0 0 0

0 0 i 0

0 1 0 0

0 0 0 i

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦ (52)

Example 52 Suppose A isin C2times4 C isin C

2times4 B isin C4times3 D isin C

4times3 and

A =

⎡⎣333 minus 5987i 45i 721 minusi

0 minus066i 7694 1123i

⎤⎦

C =

⎡⎣02679 minus 00934i 00012 + 40762i minus00777 minus 01718i minus12801i

02207 minus01197i 00877 07058i

⎤⎦

B =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

4 + 12i 2369i 4256 minus 5111i

4i 466i 821 minus 5i

0 483i 56 + i

222i minus4666 7i

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

D =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

00616 + 01872i minus00009 + 01756i 16746 minus 00494i

00024 + 02704i 01775 + 04194i 07359 minus 06189i

minus00548 + 03444i 00093 minus 03075i minus04731 minus 01636i

00337i 01209 minus 01864i minus02484 minus 38817i

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

(53)

We can verify that (226) holds Hence system (12) has an Hermitian R-conjugate solutionGiven

E =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

735i 8389i 99256 minus 651i minus46i

155 456i 771 minus 75i i

5i 0 minus4556i minus799

422i 0 51i 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦ (54)

12 Journal of Applied Mathematics

Applying Algorithm 51 we obtain the following

X =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

15597 00194i 28705 00002i

minus00194i 90001 02005i minus39997

28705 minus02005i minus00452 79993i

minus00002i minus39997 minus79993i 56846

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦ (55)

Example 52 illustrates that we can solve the optimal approximation problem withAlgorithm 51 when system (12) have Hermitian R-conjugate solutions

Example 53 Let A B andC be the same as Example 52 and let D in Example 52 be changedinto

D =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

00616 + 01872i minus00009 + 01756i 16746 + 00494i

00024 + 02704i 01775 + 04194i 07359 minus 06189i

minus00548 + 03444i 00093 minus 03075i minus04731 minus 01636i

00337i 01209 minus 01864i minus02484 minus 38817i

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦ (56)

We can verify that (226) does not hold By Algorithm 51 we get

X =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

052 22417i 04914 03991i

minus22417i 86634 01921i minus28232

04914 minus01921i 01406 13154i

minus03991i minus28232 minus13154i 63974

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦ (57)

Example 53 demonstrates that we can get the least squares solution with Algo-rithm 51 when system (12) has not Hermitian R-conjugate solutions

Acknowledgments

This research was supported by the Grants from the Key Project of Scientific ResearchInnovation Foundation of Shanghai Municipal Education Commission (13ZZ080) theNational Natural Science Foundation of China (11171205) the Natural Science Foundationof Shanghai (11ZR1412500) the PhD Programs Foundation of Ministry of Education ofChina (20093108110001) the Discipline Project at the corresponding level of Shanghai (A13-0101-12-005) Shanghai Leading Academic Discipline Project (J50101) the Natural ScienceFoundation of Hebei province (A2012403013) and the Natural Science Foundation of Hebeiprovince (A2012205028) The authors are grateful to the anonymous referees for their helpfulcomments and constructive suggestions

Journal of Applied Mathematics 13

References

[1] R D Hill R G Bates and S R Waters ldquoOn centro-Hermitian matricesrdquo SIAM Journal on MatrixAnalysis and Applications vol 11 no 1 pp 128ndash133 1990

[2] R Kouassi P Gouton and M Paindavoine ldquoApproximation of the Karhunen-Loeve tranformationand Its application to colour imagesrdquo Signal Processing Image Communication vol 16 pp 541ndash5512001

[3] A Lee ldquoCentro-Hermitian and skew-centro-Hermitian matricesrdquo Linear Algebra and Its Applicationsvol 29 pp 205ndash210 1980

[4] Z-Y Liu H-D Cao and H-J Chen ldquoA note on computing matrix-vector products with generalizedcentrosymmetric (centrohermitian) matricesrdquo Applied Mathematics and Computation vol 169 no 2pp 1332ndash1345 2005

[5] W F Trench ldquoCharacterization and properties of matrices with generalized symmetry or skewsymmetryrdquo Linear Algebra and Its Applications vol 377 pp 207ndash218 2004

[6] K-W E Chu ldquoSymmetric solutions of linear matrix equations by matrix decompositionsrdquo LinearAlgebra and Its Applications vol 119 pp 35ndash50 1989

[7] H Dai ldquoOn the symmetric solutions of linear matrix equationsrdquo Linear Algebra and Its Applicationsvol 131 pp 1ndash7 1990

[8] F J H Don ldquoOn the symmetric solutions of a linear matrix equationrdquo Linear Algebra and ItsApplications vol 93 pp 1ndash7 1987

[9] I Kyrchei ldquoExplicit representation formulas for the minimum norm least squares solutions of somequaternion matrix equationsrdquo Linear Algebra and Its Applications vol 438 no 1 pp 136ndash152 2013

[10] Y Li Y Gao and W Guo ldquoA Hermitian least squares solution of the matrix equation AXB = Csubject to inequality restrictionsrdquo Computers amp Mathematics with Applications vol 64 no 6 pp 1752ndash1760 2012

[11] Z-Y Peng and X-Y Hu ldquoThe reflexive and anti-reflexive solutions of the matrix equation AX = BrdquoLinear Algebra and Its Applications vol 375 pp 147ndash155 2003

[12] L Wu ldquoThe re-positive definite solutions to the matrix inverse problem AX = Brdquo Linear Algebra andIts Applications vol 174 pp 145ndash151 1992

[13] S F Yuan Q W Wang and X Zhang ldquoLeast-squares problem for the quaternion matrix equationAXB + CYD = E over different constrained matricesrdquo Article ID 722626 International Journal ofComputer Mathematics In press

[14] Z-Z Zhang X-Y Hu and L Zhang ldquoOn the Hermitian-generalized Hamiltonian solutions of linearmatrix equationsrdquo SIAM Journal on Matrix Analysis and Applications vol 27 no 1 pp 294ndash303 2005

[15] F Cecioni ldquoSopra operazioni algebricherdquo Annali della Scuola Normale Superiore di Pisa vol 11 pp17ndash20 1910

[16] K-W E Chu ldquoSingular value and generalized singular value decompositions and the solution oflinear matrix equationsrdquo Linear Algebra and Its Applications vol 88-89 pp 83ndash98 1987

[17] S K Mitra ldquoA pair of simultaneous linear matrix equations A1XB1 = C1 A2XB2 = C2 and a matrixprogramming problemrdquo Linear Algebra and Its Applications vol 131 pp 107ndash123 1990

[18] H-X Chang and Q-W Wang ldquoReflexive solution to a system of matrix equationsrdquo Journal of ShanghaiUniversity vol 11 no 4 pp 355ndash358 2007

[19] A Dajic and J J Koliha ldquoEquations ax = c and xb = d in rings and rings with involution withapplications to Hilbert space operatorsrdquo Linear Algebra and Its Applications vol 429 no 7 pp 1779ndash1809 2008

[20] A Dajic and J J Koliha ldquoPositive solutions to the equations AX = C and XB = D for Hilbert spaceoperatorsrdquo Journal of Mathematical Analysis and Applications vol 333 no 2 pp 567ndash576 2007

[21] C-Z Dong Q-W Wang and Y-P Zhang ldquoThe common positive solution to adjointable operatorequations with an applicationrdquo Journal of Mathematical Analysis and Applications vol 396 no 2 pp670ndash679 2012

[22] X Fang J Yu and H Yao ldquoSolutions to operator equations on Hilbert Clowast-modulesrdquo Linear Algebraand Its Applications vol 431 no 11 pp 2142ndash2153 2009

[23] C G Khatri and S K Mitra ldquoHermitian and nonnegative definite solutions of linear matrixequationsrdquo SIAM Journal on Applied Mathematics vol 31 no 4 pp 579ndash585 1976

[24] I Kyrchei ldquoAnalogs of Cramerrsquos rule for the minimum norm least squares solutions of some matrixequationsrdquo Applied Mathematics and Computation vol 218 no 11 pp 6375ndash6384 2012

[25] F L Li X Y Hu and L Zhang ldquoThe generalized reflexive solution for a class of matrix equations(AX = CXB = D)rdquo Acta Mathematica Scientia vol 28B no 1 pp 185ndash193 2008

14 Journal of Applied Mathematics

[26] Q-W Wang and C-Z Dong ldquoPositive solutions to a system of adjointable operator equations overHilbert Clowast-modulesrdquo Linear Algebra and Its Applications vol 433 no 7 pp 1481ndash1489 2010

[27] Q Xu ldquoCommon Hermitian and positive solutions to the adjointable operator equations AX = CXB = Drdquo Linear Algebra and Its Applications vol 429 no 1 pp 1ndash11 2008

[28] F Zhang Y Li and J Zhao ldquoCommon Hermitian least squares solutions of matrix equations A1XAlowast1 =

B1 and A2XAlowast2 = B2 subject to inequality restrictionsrdquo Computers ampMathematics with Applications vol

62 no 6 pp 2424ndash2433 2011[29] H-X Chang Q-W Wang and G-J Song ldquo(R S)-conjugate solution to a pair of linear matrix

equationsrdquo Applied Mathematics and Computation vol 217 no 1 pp 73ndash82 2010[30] E W Cheney Introduction to Approximation Theory McGraw-Hill New York NY USA 1966

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2012 Article ID 784620 13 pagesdoi1011552012784620

Research ArticlePerturbation Analysis for the Matrix EquationX minussumm

i=1 Alowasti XAi +

sumnj=1 B

lowastj XBj = I

Xue-Feng Duan1 2 and Qing-Wen Wang3

1 School of Mathematics and Computational Science Guilin University of Electronic TechnologyGuilin 541004 China

2 Research Center on Data Science and Social Computing Guilin University of Electronic TechnologyGuilin 541004 China

3 Department of Mathematics Shanghai University Shanghai 200444 China

Correspondence should be addressed to Qing-Wen Wang wqw858yahoocomcn

Received 18 September 2012 Accepted 19 November 2012

Academic Editor Yang Zhang

Copyright q 2012 X-F Duan and Q-W Wang This is an open access article distributed underthe Creative Commons Attribution License which permits unrestricted use distribution andreproduction in any medium provided the original work is properly cited

We consider the perturbation analysis of the matrix equation X minussummi=1 A

lowasti XAi +

sumnj=1 B

lowastj XBj = I

Based on the matrix differentiation we first give a precise perturbation bound for the positivedefinite solution A numerical example is presented to illustrate the sharpness of the perturbationbound

1 Introduction

In this paper we consider the matrix equation

X minusmsumi=1

Alowasti XAi +

nsumj=1

Blowastj XBj = I (11)

where A1 A2 Am B1 B2 Bn are n times n complex matrices I is an n times n identity matrixm n are nonnegative integers and the positive definite solution X is practical interestHere Alowast

i and Blowasti denote the conjugate transpose of the matrices Ai and Bi respectively

Equation (11) arises in solving some nonlinear matrix equations with Newton method Seefor example the nonlinear matrix equation which appears in Sakhnovich [1] Solving these

2 Journal of Applied Mathematics

nonlinear matrix equations gives rise to (11) On the other hand (11) is the general case ofthe generalized Lyapunov equation

MYSlowast + SYMlowast +tsum

k=1

NkYNlowastk + CClowast = 0 (12)

whose positive definite solution is the controllability Gramian of the bilinear control system(see [2 3] for more details)

Mx(t) = Sx(t) +tsum

k=1

Nkx(t)uk(t) + Cu(t) (13)

Set

E =1radic2(M minus S + I) F =

1radic2(M + S + I) G = S minus I Q = CClowast

X = Qminus12YQminus12 Ai = Qminus12NiQ12 i = 1 2 t

At+1 = Qminus12FQ12 At+2 = Qminus12GQ12

B1 = Qminus12EQ12 B2 = Qminus12CQ12

(14)

then (12) can be equivalently written as (11) with m = t + 2 and n = 2Some special cases of (11) have been studied Based on the kronecker product and

fixed point theorem in partially ordered sets Reurings [4] and Ran and Reurings [5 6] gavesome sufficient conditions for the existence of a unique positive definite solution of the linearmatrix equations X minussumm

i=1 Alowasti XAi = I and X +

sumnj=1 B

lowasti XBi = I And the expressions for these

unique positive definite solutions were also derived under some constraint conditions Forthe general linear matrix equation (11) Reurings [[4] Page 61] pointed out that it is hard tofind sufficient conditions for the existence of a positive definite solution because the map

G(X) = I +msumi=1

Alowasti XAi minus

nsumj=1

Blowastj XBj (15)

is not monotone and does not map the set of n times n positive definite matrices into itselfRecently Berzig [7] overcame these difficulties by making use of Bhaskar-Lakshmikanthamcoupled fixed point theorem and gave a sufficient condition for (11) existing a uniquepositive definite solution An iterative method was constructed to compute the uniquepositive definite solution and the error estimation was given too

Recently the matrix equations of the form (11) have been studied by many authors(see [8ndash14]) Some numerical methods for solving the well-known Lyapunov equationX + ATXA = Q such as Bartels-Stewart method and Hessenberg-Schur method have beenproposed in [12] Based on the fixed point theorem the sufficient and necessary conditions forthe existence of a positive definite solution of the matrix equation Xs plusmnAlowastXminustA = Q s t isin Nhave been given in [8 9] The fixed point iterative method and inversion-free iterative method

Journal of Applied Mathematics 3

were developed for solving the matrix equations X plusmn AlowastXminusαA = Q α gt 0 in [13 14] Bymaking use of the fixed point theorem of mixed monotone operator the matrix equationX minussumm

i=1 Alowasti X

δiAi = Q 0 lt |δi| lt 1 was studied in [10] and derived a sufficient condition forthe existence of a positive definite solution Assume that F maps positive definite matriceseither into positive definite matrices or into negative definite matrices the general nonlinearmatrix equation X +AlowastF(X)A = Q was studied in [11] and the fixed point iterative methodwas constructed to compute the positive definite solution under some additional conditions

Motivated by the works and applications in [2ndash6] we continue to study the matrixequation (11) Based on a new mathematical tool (ie the matrix differentiation) we firstlygive a differential bound for the unique positive definite solution of (11) and then use it toderive a precise perturbation bound for the unique positive definite solution A numericalexample is used to show that the perturbation bound is very sharp

Throughout this paper we write B gt 0 (B ge 0) if the matrix B is positive definite(semidefinite) If B minus C is positive definite (semidefinite) then we write B gt C (B ge C) Ifa positive definite matrix X satisfies B le X le C we denote that X isin [BC] The symbolsλ1(B) and λn(B) denote the maximal and minimal eigenvalues of an n times n Hermitian matrixB respectively The symbol Hntimesn stands for the set of n times n Hermitian matrices The symbolB denotes the spectral norm of the matrix B

2 Perturbation Analysis for the Matrix Equation (11)

Based on the matrix differentiation we firstly give a differential bound for the unique positivedefinite solution XU of (11) and then use it to derive a precise perturbation bound for XU inthis section

Definition 21 ([15] Definition 36) Let F = (fij)mtimesn then the matrix differentiation of F isdF = (dfij)mtimesn For example let

F =(

s + t s2 minus 2t2s + t3 t2

) (21)

Then

dF =(

ds + dt 2sds minus 2dt2ds + 3t2dt 2tdt

) (22)

Lemma 22 ([15] Theorem 32) The matrix differentiation has the following properties

(1) d(F1 plusmn F2) = dF1 plusmn dF2

(2) d(kF) = k(dF) where k is a complex number

(3) d(Flowast) = (dF)lowast

(4) d(F1F2F3) = (dF1)F2F3 + F1(dF2)F3 + F1F2(dF3)

(5) dFminus1 = minusFminus1(dF)Fminus1

(6) dF = 0 where F is a constant matrix

4 Journal of Applied Mathematics

Lemma 23 ([7] Theorem 31) If

msumi=1

Alowasti Ai lt

12I

nsumj=1

Blowastj Bj lt

12I (23)

then (11) has a unique positive definite solution XU and

XU isin⎡⎣I minus 2

nsumj=1

Blowastj Bj I + 2

msumi=1

Alowasti Ai

⎤⎦ (24)

Theorem 24 If

msumi=1

Ai2 lt12

nsumj=1

∥∥Bj

∥∥2lt

12 (25)

then (11) has a unique positive definite solution XU and it satisfies

dXU le2(

1 + 2summ

i=1 Ai2)[summ

i=1(AidAi) +sumn

j=1(∥∥Bj

∥∥∥∥dBj

∥∥)]

1 minussummi=1 Ai2 minussumn

j=1

∥∥Bj

∥∥2 (26)

Proof Since

λ1(Alowast

i Ai

) le ∥∥Alowasti Ai

∥∥ le Ai2 i = 1 2 m

λ1

(Blowastj Bj

)le∥∥∥Blowast

j Bj

∥∥∥ le ∥∥Bj

∥∥2 j = 1 2 n

(27)

then

Alowasti Ai le λ1

(Alowast

i Ai

)I le ∥∥Alowast

i Ai

∥∥I le Ai2I i = 1 2 m

Blowastj Bj le λ1

(Blowastj Bj

)I le∥∥∥Blowast

j Bj

∥∥∥ le ∥∥Bj

∥∥2I j = 1 2 n

(28)

consequently

msumi=1

Alowasti Ai le

msumi=1

Ai2I

nsumj=1

Blowastj Bj le

nsumj=1

∥∥Bj

∥∥2I

(29)

Journal of Applied Mathematics 5

Combining (25)ndash(29) we have

msumi=1

Alowasti Ai le

msumi=1

Ai2I lt12I

nsumj=1

Blowastj Bj le

nsumj=1

∥∥Bj

∥∥2I lt

12I

(210)

Then by Lemma 23 we obtain that (11) has a unique positive definite solution XU whichsatisfies

XU isin⎡⎣I minus 2

nsumj=1

Blowastj Bj I + 2

msumi=1

Alowasti Ai

⎤⎦ (211)

Noting that XU is the unique positive definite solution of (11) then

XU minusmsumi=1

Alowasti XUAi +

nsumj=1

Blowastj XUBj = I (212)

It is known that the elements of XU are differentiable functions of the elements of Ai and BiDifferentiating (212) and by Lemma 22 we have

dXU minusmsumi=1

[(dAlowast

i

)XUAi +Alowast

i (dXU)Ai +Alowasti XU(dAi)

]

+nsumj=1

[(dBlowast

j

)XUBj + Blowast

j (dXU)Bj + Blowastj XU

(dBj

)]= 0

(213)

which implies that

dXU minusmsumi=1

Alowasti (dXU)Ai +

nsumj=1

Blowastj (dXU)Bj

=msumi=1

(dAlowast

i

)XUAi +

msumi=1

Alowasti XU(dAi) minus

nsumj=1

(dBlowast

j

)XUBj minus

nsumj=1

Blowastj XU

(dBj

)

(214)

6 Journal of Applied Mathematics

By taking spectral norm for both sides of (214) we obtain that

∥∥∥∥∥∥dXU minusmsumi=1

Alowasti (dXU)Ai +

nsumj=1

Blowastj (dXU)Bj

∥∥∥∥∥∥

=

∥∥∥∥∥∥msumi=1

(dAlowast

i

)XUAi +

msumi=1

Alowasti XU(dAi) minus

nsumj=1

(dBlowast

j

)XUBj minus

nsumj=1

Blowastj XU

(dBj

)∥∥∥∥∥∥

le∥∥∥∥∥

msumi=1

(dAlowast

i

)XUAi

∥∥∥∥∥ +∥∥∥∥∥

msumi=1

Alowasti XU(dAi)

∥∥∥∥∥ +∥∥∥∥∥∥

nsumj=1

(dBlowast

j

)XUBj

∥∥∥∥∥∥ +∥∥∥∥∥∥

nsumj=1

Blowastj XU

(dBj

)∥∥∥∥∥∥

lemsumi=1

∥∥(dAlowasti

)XUAi

∥∥ + msumi=1

∥∥Alowasti XU(dAi)

∥∥ + nsumj=1

∥∥∥(dBlowastj

)XUBj

∥∥∥ +nsumj=1

∥∥∥Blowastj XU

(dBj

)∥∥∥

lemsumi=1

∥∥dAlowasti

∥∥XUAi +msumi=1

∥∥Alowasti

∥∥XUdAi +nsumj=1

∥∥∥dBlowastj

∥∥∥XU∥∥Bj

∥∥ + nsumj=1

∥∥∥Blowastj

∥∥∥XU∥∥dBj

∥∥

= 2msumi=1

(AiXUdAi) + 2nsumj=1

(∥∥Bj

∥∥XU∥∥dBj

∥∥)

= 2

⎡⎣ msum

i=1

(AidAi) +nsumj=1

(∥∥Bj

∥∥∥∥dBj

∥∥)⎤⎦XU

(215)

and noting (211) we obtain that

XU le∥∥∥∥∥I + 2

msumi=1

Alowasti Ai

∥∥∥∥∥ le 1 + 2msumi=1

Ai2 (216)

Then

∥∥∥∥∥∥dXU minusmsumi=1

Alowasti (dXU)Ai +

nsumj=1

Blowastj (dXU)Bj

∥∥∥∥∥∥

le 2

⎡⎣ msum

i=1

(AidAi) +nsumj=1

(∥∥Bj

∥∥∥∥dBj

∥∥)⎤⎦XU

le 2

(1 + 2

msumi=1

Ai2

)⎡⎣ msum

i=1

(AidAi) +nsumj=1

(∥∥Bj

∥∥∥∥dBj

∥∥)⎤⎦

(217)

Journal of Applied Mathematics 7∥∥∥∥∥∥dXU minus

msumi=1

Alowasti (dXU)Ai +

nsumj=1

Blowastj (dXU)Bj

∥∥∥∥∥∥

ge dXU minus∥∥∥∥∥

msumi=1

Alowasti (dXU)Ai

∥∥∥∥∥ minus∥∥∥∥∥∥

nsumj=1

Blowastj (dXU)Bj

∥∥∥∥∥∥

ge dXU minusmsumi=1

∥∥Alowasti (dXU)Ai

∥∥ minus nsumj=1

∥∥∥Blowastj (dXU)Bj

∥∥∥

ge dXU minusmsumi=1

∥∥Alowasti

∥∥dXUAi minusnsumj=1

∥∥∥Blowastj

∥∥∥dXU∥∥Bj

∥∥

=

⎛⎝1 minus

msumi=1

Ai2 minusnsumj=1

∥∥Bj

∥∥2

⎞⎠dXU

(218)

Due to (25) we have

1 minusmsumi=1

Ai2 minusnsumj=1

∥∥Bj

∥∥2gt 0 (219)

Combining (217) (218) and noting (219) we have

⎛⎝1 minus

msumi=1

Ai2 minusnsumj=1

∥∥Bj

∥∥2

⎞⎠dXU

le∥∥∥∥∥∥dXU minus

msumi=1

Alowasti (dXU)Ai +

nsumj=1

Blowastj (dXU)Bj

∥∥∥∥∥∥

le 2

(1 + 2

msumi=1

Ai2

)⎡⎣ msum

i=1

(AidAi) +nsumj=1

(∥∥Bj

∥∥∥∥dBj

∥∥)⎤⎦

(220)

which implies that

dXU le2(

1 + 2summ

i=1 Ai2)[summ

i=1(AidAi) +sumn

j=1(∥∥Bj

∥∥∥∥dBj

∥∥)]

1 minussummi=1 Ai2 minussumn

j=1

∥∥Bj

∥∥2 (221)

8 Journal of Applied Mathematics

Theorem 25 Let A1 A2 Am B1 B2 Bn be perturbed matrices of A1 A2 AmB1 B2 Bn in (11) and Δi = Ai minusAi i = 1 2 m Δj = Bj minus Bj j = 1 2 n If

msumi=1

Ai2 lt12

nsumj=1

∥∥Bj

∥∥2lt

12 (222)

2msumi=1

(AiΔi) +msumi=1

Ai2 lt12minus

msumi=1

Ai2 (223)

2nsumj=1

(∥∥Bj

∥∥∥∥Δj

∥∥) + nsumj=1

∥∥Δj

∥∥2lt

12minus

nsumj=1

∥∥Bj

∥∥2 (224)

then (11) and its perturbed equation

X minusmsumi=1

Alowasti XAi +

nsumj=1

Blowastj XBj = I (225)

have unique positive definite solutions XU and XU respectively which satisfy

∥∥∥XU minusXU

∥∥∥ le Serr (226)

where

Serr =2[1 + 2

summi=1 (Ai + Δi)2

][summi=1 (Ai + Δi)2Δi +

sumnj=1(∥∥Bj

∥∥ + ∥∥Δj

∥∥)2∥∥Δj

∥∥]

1 minussummi=1 (Ai + Δi)2 minussumn

j=1(∥∥Bj

∥∥ + ∥∥Δj

∥∥)2

(227)

Proof By (222) and Theorem 24 we know that (11) has a unique positive definite solutionXU And by (223) we have

msumi=1

∥∥∥Ai

∥∥∥2=

msumi=1

Ai + Δi2 lemsumi=1

(Ai + Δi)2

=msumi=1

(Ai2 + 2AiΔi + Δi2

)

=msumi=1

Ai2 + 2msumi=1

(AiΔi) +msumi=1

Δi2

ltmsumi=1

Ai2 +12minus

msumi=1

Ai2 =12

(228)

Journal of Applied Mathematics 9

similarly by (224) we have

nsumj=1

∥∥∥Bj

∥∥∥2lt

12 (229)

By (228) (229) and Theorem 24 we obtain that the perturbed equation (225) has a uniquepositive definite solution XU

Set

Ai(t) = Ai + tΔi Bj(t) = Bj + tΔj t isin [0 1] (230)

then by (223) we have

msumi=1

Ai(t)2 =msumi=1

Ai + tΔi2 lemsumi=1

(Ai + tΔi)2

=msumi=1

(Ai2 + 2tAiΔi + t2Δi2

)

lemsumi=1

(Ai2 + 2AiΔi + Δi2

)

=msumi=1

Ai2 + 2msumi=1

(AiΔi) +msumi=1

Δi2

ltmsumi=1

Ai2 +12minus

msumi=1

Ai2 =12

(231)

similarly by (224) we have

nsumj=1

∥∥Bj(t)∥∥2 =

msumi=1

∥∥Bj + tΔj

∥∥2lt

12 (232)

Therefore by (231) (232) and Theorem 24 we derive that for arbitrary t isin [0 1] thematrix equation

X minusmsumi=1

Alowasti (t)XAi(t) +

nsumj=1

Blowastj (t)XBj(t) = I (233)

has a unique positive definite solution XU(t) especially

XU(0) = XU XU(1) = XU (234)

10 Journal of Applied Mathematics

From Theorem 24 it follows that

∥∥∥XU minusXU

∥∥∥ = XU(1) minusXU(0) =

∥∥∥∥∥int1

0dXU(t)

∥∥∥∥∥ leint1

0dXU(t)

leint1

0

2(

1 + 2summ

i=1 Ai(t)2)[summ

i=1(Ai(t)dAi(t)) +sumn

j=1(∥∥Bj(t)

∥∥∥∥dBj(t)∥∥)]

1 minussummi=1 Ai(t)2 minussumn

j=1

∥∥Bj(t)∥∥2

leint1

0

2(

1 + 2summ

i=1 Ai(t)2)[summ

i=1(Ai(t)Δidt) +sumn

j=1(∥∥Bj(t)

∥∥∥∥Δj

∥∥dt)]

1 minussummi=1 Ai(t)2 minussumn

j=1

∥∥Bj(t)∥∥2

=int1

0

2(

1 + 2summ

i=1 Ai(t)2)[summ

i=1(Ai(t)Δi) +sumn

j=1(∥∥Bj(t)

∥∥∥∥Δj

∥∥)]

1 minussummi=1 Ai(t)2 minussumn

j=1

∥∥Bj(t)∥∥2

dt

(235)

Noting that

Ai(t) = Ai + tΔi le Ai + tΔi i = 1 2 m∥∥Bj(t)

∥∥ =∥∥Bj + tΔi

∥∥ le ∥∥Bj

∥∥ + tΔi j = 1 2 n(236)

and combining Mean Value Theorem of Integration we have

∥∥∥XU minusXU

∥∥∥

leint1

0

2(

1 + 2summ

i=1 Ai(t)2)[summ

i=1(Ai(t)Δi) +sumn

j=1(∥∥Bj(t)

∥∥∥∥Δj

∥∥)]

1 minussummi=1 Ai(t)2 minussumn

j=1

∥∥Bj(t)∥∥2

dt

leint1

0

2[1 + 2

summi=1 (Ai + tΔi)2

][summi=1 (Ai + tΔi)2Δi +

sumnj=1(∥∥Bj

∥∥ + t∥∥Δj

∥∥)2∥∥Δj

∥∥]

1 minussummi=1 (Ai + tΔi)2 minussumn

j=1(∥∥Bj

∥∥ + t∥∥Δj

∥∥)2dt

=2[1 + 2

summi=1 (Ai + ξΔi)2

][summi=1 (Ai + ξΔi)2Δi +

sumnj=1(∥∥Bj

∥∥ + ξ∥∥Δj

∥∥)2∥∥Δj

∥∥]

1 minussummi=1 (Ai + ξΔi)2 minussumn

j=1(∥∥Bj

∥∥ + ξ∥∥Δj

∥∥)2

times (1 minus 0) ξ isin [0 1]

le2[1 + 2

summi=1 (Ai + Δi)2

][summi=1 (Ai + Δi)2Δi +

sumnj=1(∥∥Bj

∥∥ + ∥∥Δj

∥∥)2∥∥Δj

∥∥]

1 minussummi=1 (Ai + Δi)2 minussumn

j=1(∥∥Bj

∥∥ + ∥∥Δj

∥∥)2= Serr

(237)

Journal of Applied Mathematics 11

3 Numerical Experiments

In this section we use a numerical example to confirm the correctness of Theorem 25 and theprecision of the perturbation bound for the unique positive definite solution XU of (11)

Example 31 Consider the symmetric linear matrix equation

X minusAlowast1XA1 minusAlowast

2XA2 + Blowast1XB1 + Blowast

2XB2 = I (31)

and its perturbed equation

X minus Alowast1XA1 minus Alowast

2XA2 + Blowast1XB1 + Blowast

2XB2 = I (32)

where

A1 =

⎛⎝ 002 minus010 minus002

008 minus010 002minus006 minus012 014

⎞⎠ A2 =

⎛⎝ 008 minus010 minus002

008 minus010 002minus006 minus012 014

⎞⎠

B1 =

⎛⎝ 047 002 004

minus010 036 minus002minus004 001 047

⎞⎠ B2 =

⎛⎝010 010 005

015 0275 0075005 005 0175

⎞⎠

A1 = A1 +

⎛⎝ 05 01 minus02

minus04 02 06minus02 01 minus01

⎞⎠ times 10minusj A2 = A2 +

⎛⎝minus04 01 minus02

05 07 minus1311 09 06

⎞⎠ times 10minusj

B1 = B1 +

⎛⎝ 08 02 005

minus02 012 014minus025 minus02 026

⎞⎠ times 10minusj B2 = B2 +

⎛⎝ 02 02 01

minus03 015 minus01501 minus01 025

⎞⎠ times 10minusj j isin N

(33)

It is easy to verify that the conditions (222)ndash(224) are satisfied then (31) and itsperturbed equation (32) have unique positive definite solutions XU and XU respectivelyFrom Berzig [7] it follows that the sequences Xk and Yk generated by the iterative method

X0 = 0 Y0 = 2I

Xk+1 = I +Alowast1XkA1 +Alowast

2XkA2 minus Blowast1YkB1 minus Blowast

2YkB2

Yk+1 = I +Alowast1YkA1 +Alowast

2YkA2 minus Blowast1XkB1 minus Blowast

2XkB2 k = 0 1 2

(34)

both converge to XU Choose τ = 10 times 10minus15 as the termination scalar that is

R(X) =∥∥X minusAlowast

1XA1 minusAlowast2XA2 + Blowast

1XB1 + Blowast2XB2 minus I

∥∥ le τ = 10 times 10minus15 (35)

By using the iterative method (34) we can get the computed solution Xk of (31) SinceR(Xk) lt 10 times 10minus15 then the computed solution Xk has a very high precision For simplicity

12 Journal of Applied Mathematics

Table 1 Numerical results for the different values of j

j 2 3 4 5 6

XU minusXUXU 213 times 10minus4 616 times 10minus6 523 times 10minus8 437 times 10minus10 845 times 10minus12

SerrXU 342 times 10minus4 713 times 10minus6 663 times 10minus8 512 times 10minus10 977 times 10minus12

we write the computed solution as the unique positive definite solution XU Similarly we canalso get the unique positive definite solution XU of the perturbed equation (32)

Some numerical results on the perturbation bounds for the unique positive definitesolution XU are listed in Table 1

From Table 1 we see that Theorem 25 gives a precise perturbation bound for theunique positive definite solution of (31)

4 Conclusion

In this paper we study the matrix equation (11) which arises in solving some nonlinearmatrix equations and the bilinear control system A new method of perturbation analysisis developed for the matrix equation (11) By making use of the matrix differentiationand its elegant properties we derive a precise perturbation bound for the unique positivedefinite solution of (11) A numerical example is presented to illustrate the sharpness of theperturbation bound

Acknowledgments

This research was supported by the National Natural Science Foundation of China(11101100 11171205 11261014 11226323) the Natural Science Foundation of GuangxiProvince (2012GXNSFBA053006 2011GXNSFA018138) the Key Project of Scientific ResearchInnovation Foundation of Shanghai Municipal Education Commission (13ZZ080) theNatural Science Foundation of Shanghai (11ZR1412500) the PhD Programs Foundation ofMinistry of Education of China (20093108110001) the Discipline Project at the correspondinglevel of Shanghai (A 13-0101-12-005) and Shanghai Leading Academic Discipline Project(J50101) The authors wish to thank Professor Y Zhang and the anonymous referees forproviding very useful suggestions for improving this paper The authors also thank DrXiaoshan Chen for discussing the properties of matrix differentiation

References

[1] L A Sakhnovich Interpolation Theory and Its Applications Mathematics and Its Applications KluwerAcademic Publishers Dordrecht The Netherlands 1997

[2] T Damm ldquoDirect methods and ADI-preconditioned Krylov subspace methods for generalizedLyapunov equationsrdquo Numerical Linear Algebra with Applications vol 15 no 9 pp 853ndash871 2008

[3] L Zhang and J Lam ldquoOn H2 model reduction of bilinear systemsrdquo Automatica vol 38 no 2 pp205ndash216 2002

[4] M C B Reurings Symmetric matrix equations [PhD thesis] Vrije Universiteit Amsterdam TheNetherlands 2003

[5] A C M Ran and M C B Reurings ldquoThe symmetric linear matrix equationrdquo Electronic Journal ofLinear Algebra vol 9 pp 93ndash107 2002

Journal of Applied Mathematics 13

[6] A C M Ran and M C B Reurings ldquoA fixed point theorem in partially ordered sets and someapplications to matrix equationsrdquo Proceedings of the American Mathematical Society vol 132 no 5 pp1435ndash1443 2004

[7] M Berzig ldquoSolving a class of matrix equation via Bhaskar-Lakshmikantham coupled fixed pointtheoremrdquo Applied Mathematics Letters vol 25 no 11 pp 1638ndash1643 2012

[8] J Cai and G Chen ldquoSome investigation on Hermitian positive definite solutions of the matrixequation Xs +AlowastXminustA = Qrdquo Linear Algebra and Its Applications vol 430 no 8-9 pp 2448ndash2456 2009

[9] X Duan and A Liao ldquoOn the existence of Hermitian positive definite solutions of the matrix equationXs +AlowastXminustA = Qrdquo Linear Algebra and its Applications vol 429 no 4 pp 673ndash687 2008

[10] X Duan A Liao and B Tang ldquoOn the nonlinear matrix equation X minus summi=1 A

lowasti X

δiAi = Qrdquo LinearAlgebra and its Applications vol 429 no 1 pp 110ndash121 2008

[11] S M El-Sayed and A C M Ran ldquoOn an iteration method for solving a class of nonlinear matrixequationsrdquo SIAM Journal on Matrix Analysis and Applications vol 23 no 3 pp 632ndash645 2001

[12] Z Gajic and M T J Qureshi Lyapunov Matrix Equation in System Stability and Control Academic PressLondon UK 1995

[13] V I Hasanov ldquoNotes on two perturbation estimates of the extreme solutions to the equations X plusmnAlowastXminus1A = Qrdquo Applied Mathematics and Computation vol 216 no 5 pp 1355ndash1362 2010

[14] Z-Y Peng S M El-Sayed and X-l Zhang ldquoIterative methods for the extremal positive definitesolution of the matrix equation X + AlowastXminusαA = Qrdquo Journal of Computational and Applied Mathematicsvol 200 no 2 pp 520ndash527 2007

[15] B R Fang J D Zhou and Y M Li Matrix Theory Tsinghua University Press Beijing China 2006

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2012 Article ID 483624 11 pagesdoi1011552012483624

Research ArticleRefinements of Kantorovich Inequality forHermitian Matrices

Feixiang Chen

School of Mathematics and Statistics Chongqing Three Gorges University WanzhouChongqing 404000 China

Correspondence should be addressed to Feixiang Chen cfx2002126com

Received 28 August 2012 Accepted 5 November 2012

Academic Editor K C Sivakumar

Copyright q 2012 Feixiang Chen This is an open access article distributed under the CreativeCommons Attribution License which permits unrestricted use distribution and reproduction inany medium provided the original work is properly cited

Some new Kantorovich-type inequalities for Hermitian matrix are proposed in this paper Weconsider what happens to these inequalities when the positive definite matrix is allowed to beinvertible and provides refinements of the classical results

1 Introduction and Preliminaries

We first state the well-known Kantorovich inequality for a positive definite Hermite matrix(see [1 2]) let A isin Mn be a positive definite Hermitian matrix with real eigenvalues λ1 leλ2 le middot middot middot le λn Then

1 le xlowastAxxlowastAminus1x le (λ1 + λn)2

4λ1λn (11)

for any x isin Cn x = 1 where Alowast denotes the conjugate transpose of matrix A A matrix

A isin Mn is Hermitian if A = Alowast An equivalent form of this result is incorporated in

0 le xlowastAxxlowastAminus1x minus 1 le (λn minus λ1)2

4λ1λn (12)

for any x isin Cn x = 1

Attributed to Kantorovich the inequality has built up a considerable literatureThis typically comprises generalizations Examples are [3ndash5] for matrix versions Operator

2 Journal of Applied Mathematics

versions are developed in [6 7] Multivariate versions have been useful in statistics to assessthe robustness of least squares see [8 9] and the references therein

Due to the important applications of the original Kantorovich inequality for matrices[10] in Statistics [8 11 12] and Numerical Analysis [13 14] any new inequality of this typewill have a flow of consequences in the areas of applications

Motivated by the interest in both pure and applied mathematics outlined abovewe establish in this paper some improvements of Kantorovich inequalities The classicalKantorovich-type inequalities are modified to apply not only to positive definite but also toinvertible Hermitian matrices As natural tools in deriving the new results the recent Gruss-type inequalities for vectors in inner product in [6 15ndash19] are utilized

To simplify the proof we first introduce some lemmas

2 Lemmas

Let B be a Hermitian matrix with real eigenvalues μ1 le μ2 le middot middot middot le μn if A minus B is positivesemidefinite we write

A ge B (21)

that is λi ge μi i = 1 2 n On Cn we have the standard inner product defined by 〈x y〉 =sumn

i=1 xiyi where x = (x1 xn)lowast isin C

n and y = (y1 yn)lowast isin C

n

Lemma 21 Let a b c and d be real numbers then one has the following inequality

(a2 minus b2

)(c2 minus d2

)le (ac minus bd)2 (22)

Lemma 22 Let A and B be Hermitian matrices if AB = BA then

AB le (A + B)2

4 (23)

Lemma 23 Let A ge 0 B ge 0 if AB = BA then

AB ge 0 (24)

3 Some Results

The following lemmas can be obtained from [16ndash19] by replacing Hilbert space (H 〈middot middot〉) withinner product spaces C

n so we omit the details

Lemma 31 Let u v and e be vectors in Cn and e = 1 If α β δ and γ are real or complex

numbers such that

Relangβe minus u u minus αe

rang ge 0 Relangδe minus v v minus γe

rang ge 0 (31)

Journal of Applied Mathematics 3

then

|〈u v〉 minus 〈u e〉〈e v〉| le 14(β minus α

)(δ minus γ

) minus [Re

langβe minus u u minus αe

rangRe

langδe minus v v minus γe

rang]12 (32)

Lemma 32 With the assumptions in Lemma 31 one has

|〈u v〉 minus 〈u e〉〈e v〉| le 14(β minus α

)(γ minus δ

) minus∣∣∣∣〈u e〉 minus α + β

2

∣∣∣∣∣∣∣∣〈v e〉 minus γ + δ

2

∣∣∣∣ (33)

Lemma 33 With the assumptions in Lemma 31 if Re(βα) ge 0 Re(δγ) ge 0 one has

|〈u v〉 minus 〈u e〉〈e v〉| le∣∣β minus α

∣∣∣∣δ minus γ∣∣

4[Re

(βα

)Re

(δγ

)]12|〈u e〉〈e v〉| (34)

Lemma 34 With the assumptions in Lemma 33 one has

|〈u v〉 minus 〈u e〉〈e v〉|

le(∣∣α + β

∣∣ minus 2[Re

(βα

)]12)(∣∣δ + γ

∣∣ minus 2[Re

(δγ

)]12)12

[|〈u e〉〈e v〉|]12(35)

4 New Kantorovich Inequalities for Hermitian Matrices

For a Hermitian matrix A as in [6] we define the following transform

C(A) = (λnI minusA)(A minus λ1I) (41)

When A is invertible if λ1λn gt 0 then

C(Aminus1

)=(

1λ1

I minusAminus1)(

Aminus1 minus 1λn

I

) (42)

Otherwise λ1λn lt 0 then

C(Aminus1

)=(

1λk+1

I minusAminus1)(

Aminus1 minus 1λk

I

) (43)

where

λ1 le middot middot middot le λk lt 0 lt λk+1 le middot middot middot le λn (44)

From Lemma 23 we can conclude that C(A) ge 0 and C(Aminus1) ge 0

4 Journal of Applied Mathematics

For two Hermitian matrices A and B and x isin Cn x = 1 we define the following

functional

G(ABx) = 〈AxBx〉 minus 〈Ax x〉〈x Bx〉 (45)

When A = B we denote

G(Ax) = Ax2 minus 〈Ax x〉2 (46)

for x isin Cn x = 1

Lemma 41 With notations above and for x isin Cn x = 1 then

0 le 〈C(A)x x〉 le (λn minus λ1)2

4 (47)

if λnλ1 gt 0

0 lelangC(Aminus1

)x x

rangle (λn minus λ1)

2

4(λnλ1)2 (48)

if λnλ1 lt 0

0 lelangC(Aminus1

)x x

rangle (λk+1 minus λk)

2

4(λk+1λk)2 (49)

Proof From C(A) ge 0 then

〈C(A)x x〉 ge 0 (410)

While from Lemma 22 we can get

C(A) = (λnI minusA)(A minus λ1I) le (λn minus λ1)2

4I (411)

Then 〈C(A)x x〉 le (λn minus λ1)24 is straightforward The proof for C(Aminus1) is similar

Lemma 42 With notations above and for x isin Cn x = 1 then

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le G(Ax)G

(Aminus1x

) (412)

Journal of Applied Mathematics 5

Proof Thus

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2

=∣∣∣xlowast

((xlowastAminus1x

)I minusAminus1

)((xlowastAx)I minusA)x

∣∣∣2

le∥∥∥((xlowastAminus1x

)I minusAminus1

)x∥∥∥2

((xlowastAx)I minusA)x2

(413)

while

((xlowastAx)I minusA)x2 = xlowast((xlowastAx)2I minus 2(xlowastAx)A +A2

)x

= xlowastA2x minus (xlowastAx)2

= Ax2 minus 〈Ax x〉2

= G(Ax)

(414)

Similarly we can get ((xlowastAminus1x)I minusAminus1)x2 = G(Aminus1x) then we complete the proof

Theorem 43 Let A B be two Hermitian matrices and C(A) ge 0 C(B) ge 0 are defined as abovethen

|G(ABx)| le 14(λn minus λ1)

(μn minus μ1

) minus [〈C(A)x x〉〈C(B)x x〉]12

|G(ABx)| le 14(λn minus λ1)

(μn minus μ1

) minus∣∣∣∣lang(

A minus λ1 + λn2

)x x

rang∣∣∣∣∣∣∣∣lang(

B minus μ1 + μn

2

)x x

rang∣∣∣∣(415)

for any x isin Cn x = 1

If λ1λn gt 0 μ1μn gt 0 then

|G(ABx)| le (λn minus λ1)(μn minus μ1

)4[(λnλ1)

(μnμ1

)]12|〈Ax x〉〈Bx x〉|

|G(ABx)| le(

|λ1 + λn| minus 2(λ1λn)12

)(∣∣μ1 + μn

∣∣ minus 2(μ1μn

)12)12

[|〈Ax x〉〈Bx x〉|]12

(416)

for any x isin Cn x = 1

Proof The proof follows by Lemmas 31 32 33 and 34 on choosing u = Ax v = Bx ande = x β = λn α = λ1 δ = μn and γ = μ1 x isin C

n x = 1 respectively

6 Journal of Applied Mathematics

Corollary 44 Let A be a Hermitian matrices and C(A) ge 0 is defined as above then

|G(Ax)| le 14(λn minus λ1)

2 minus 〈C(A)x x〉 (417)

|G(Ax)| le 14(λn minus λ1)

2 minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣2

(418)

for any x isin Cn x = 1

If λ1λn gt 0 then

|G(Ax)| le (λn minus λ1)2

4(λnλ1)|〈Ax x〉|2 (419)

|G(Ax)| le(|λ1 + λn| minus 2

radicλ1λn

)|〈Ax x〉| (420)

for any x isin Cn x = 1

Proof The proof follows by Theorem 43 on choosing A = B respectively

Corollary 45 Let A be a Hermitian matrices and C(Aminus1) ge 0 is defined as above then one has thefollowing

If λ1λn gt 0 then

∣∣∣G(Aminus1x

)∣∣∣ le (λn minus λ1)2

4(λnλ1)2

minuslangC(Aminus1

)x x

rang (421)

∣∣∣G(Aminus1x

)∣∣∣ le (λn minus λ1)2

4(λnλ1)2

minus∣∣∣∣lang(

Aminus1 minus λ1 + λn2λnλ1

I

)x x

rang∣∣∣∣2

(422)

∣∣∣G(Aminus1x

)∣∣∣ le (λn minus λ1)2

4λnλ1

∣∣∣langAminus1x xrang∣∣∣2

(423)

∣∣∣G(Aminus1x

)∣∣∣ le(

|λ1 + λn|λ1λn

minus 21radicλ1λn

)∣∣∣langAminus1x xrang∣∣∣ (424)

for any x isin Cn x = 1

If λ1λn lt 0 then

∣∣∣G(Aminus1x

)∣∣∣ le (λk minus λk+1)2

4(λkλk+1)2

minuslangC(Aminus1

)x x

rang (425)

Journal of Applied Mathematics 7

where C(Aminus1) = ((1λk+1)I minusAminus1)(Aminus1 minus (1λk)I) and

∣∣∣G(Aminus1x

)∣∣∣ le (λk minus λk+1)2

4(λkλk+1)2

minus∣∣∣∣lang(

Aminus1 minus λk+1 + λk2λkλk+1

I

)x x

rang∣∣∣∣2

(426)

for any x isin Cn x = 1

Proof The proof follows by Corollary 44 by replacing A with Aminus1 respectively

Theorem 46 LetA be an ntimesn invertible Hermitian matrix with real eigenvalues λ1 le λ2 le middot middot middot le λnthen one has the following

If λ1λn gt 0 then

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le (λn minus λ1)

2

4(λnλ1)minusradic〈C(A)x x〉langC(Aminus1

)x x

rang (427)

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le (λn minus λ1)

2

4(λnλ1)minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣∣∣∣∣lang(

Aminus1 minus λ1 + λn2λnλ1

I

)x x

rang∣∣∣∣(428)

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le (λn minus λ1)

2

4(λnλ1)2 〈Ax x〉

langAminus1x x

rang (429)

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le

(radic|λn| minus

radic|λ1|

)2

radicλnλ1

radic〈Ax x〉〈Aminus1x x〉 (430)

for any x isin Cn x = 1

If λ1λn lt 0 then

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le (λn minus λ1)(λk+1 minus λk)

4|λkλk+1| minusradic〈C(A)x x〉langC(Aminus1

)x x

rang (431)

where C(Aminus1) = ((1λk+1)I minusAminus1)(Aminus1 minus (1λk)I)

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le (λn minus λ1)(λk+1 minus λk)

4|λkλk+1|

minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣∣∣∣∣lang(

Aminus1 minus λk + λk+1

2λ1λk+1I

)x x

rang∣∣∣∣(432)

Proof Considering

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le G(Ax)G

(Aminus1x

) (433)

8 Journal of Applied Mathematics

When λ1λn gt 0 from (417) and (421) we get

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le

14(λn minus λ1)

2 minus 〈C(A)x x〉

(λn minus λ1)2

4(λnλ1)2

minuslangC(Aminus1

)x x

rang (434)

From (a2 minus b2)(c2 minus d2) le (ac minus bd)2 we have

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le

(λn minus λ1)

2

4(λnλ1)minusradic[〈C(A)x x〉langC(Aminus1

)x x

rang]2

(435)

then the conclusion (427) holdsSimilarly from (418) and (422) we get

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le

14(λn minus λ1)

2 minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣2

times(λn minus λ1)

2

4(λnλ1)2

minus∣∣∣∣lang(

Aminus1 minus λ1 + λn2λnλ1

I

)x x

rang∣∣∣∣2

le(λn minus λ1)

2

4(λnλ1)minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣∣∣∣∣lang(

Aminus1 minus λ1 + λn2λnλ1

I

)x x

rang∣∣∣∣2

(436)

then the conclusion (428) holdsFrom (419) and (423) we get

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le (λn minus λ1)

2

4(λnλ1)|〈Ax x〉|2 (λn minus λ1)

2

4(λnλ1)

∣∣∣langAminus1x xrang∣∣∣2

(437)

then the conclusion (429) holdsFrom (420) and (424) we get

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le

(|λ1 + λn| minus 2

radicλ1λn

)|〈Ax x〉|

(|λ1 + λn|λ1λn

minus 21radicλ1λn

)∣∣∣langAminus1x xrang∣∣∣(438)

then the conclusion (430) holdsWhen λ1λn lt 0 from (417) and (425) we get

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2le

14(λn minus λ1)

2 minus 〈C(A)x x〉

(λk minus λk+1)2

4(λkλk+1)2

minuslangC(Aminus1

)x x

rang (439)

then the conclusion (431) holds

Journal of Applied Mathematics 9

From (418) and (426) we get

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le

14(λn minus λ1)

2 minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣2

times(λk minus λk+1)

2

4(λkλk+1)2

minus∣∣∣∣lang(

Aminus1 minus λk+1 + λk2λkλk+1

I

)x x

rang∣∣∣∣2

le(λn minus λ1)(λk+1 minus λk)

4|λkλk+1|

minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣∣∣∣∣lang(

Aminus1 minus λk + λk+1

2λ1λk+1I

)x x

rang∣∣∣∣2

(440)

then the conclusion (432) holds

Corollary 47 With the notations above for any x isin Cn x = 1 one lets

|G1(Ax)| = 14(λn minus λ1)

2 minus 〈C(A)x x〉

|G2(Ax)| = 14(λn minus λ1)

2 minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣2

(441)

If λ1λn gt 0 one lets

|G3(Ax)| = (λn minus λ1)2

4(λnλ1)|〈Ax x〉|2

|G4(Ax)| =(|λ1 + λn| minus 2

radicλ1λn

)|〈Ax x〉|

(442)

If λ1λn gt 0

∣∣∣G1(Aminus1x

)∣∣∣ = (λn minus λ1)2

4(λnλ1)2

minuslangC(Aminus1

)x x

rang

∣∣∣G2(Aminus1x

)∣∣∣ = (λn minus λ1)2

4(λnλ1)2

minus∣∣∣∣lang(

Aminus1 minus λ1 + λn2λnλ1

I

)x x

rang∣∣∣∣2

∣∣∣G3(Aminus1x

)∣∣∣ = (λn minus λ1)2

4λnλ1

∣∣∣langAminus1x xrang∣∣∣2

∣∣∣G4(Aminus1x

)∣∣∣ =(

|λ1 + λn|λ1λn

minus 21radicλ1λn

)∣∣∣langAminus1x xrang∣∣∣

(443)

10 Journal of Applied Mathematics

If λ1λn lt 0 then

∣∣∣G5(Aminus1x

)∣∣∣ = (λk minus λk+1)2

4(λkλk+1)2

minuslangC(Aminus1

)x x

rang (444)

where C(Aminus1) = ((1λk+1)I minusAminus1)(Aminus1 minus (1λk)I) and

∣∣∣G6(Aminus1x

)∣∣∣ = (λk minus λk+1)2

4(λkλk+1)2

minus∣∣∣∣lang(

Aminus1 minus λk+1 + λk2λkλk+1

I

)x x

rang∣∣∣∣2

(445)

Then one has the followingIf λ1λn gt 0

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le

radicGlowast(Ax)Glowast(Aminus1x

) (446)

where

Glowast(Ax) = minG1(Ax) G2(Ax) G3(Ax) G4(Ax)

Glowast(Ax) = minG1

(Aminus11x

) G2

(Aminus1x

) G3

(Aminus1x

) G4

(Aminus1x

)

(447)

If λ1λn lt 0

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le

radicG(Ax)Gbull(Aminus1x

) (448)

where

G(Ax) = minG1(Ax) G2(Ax)

Gbull(Ax) = minG5

(Aminus1x

) G6

(Aminus1x

)

(449)

Proof The proof follows from that the conclusions in Corollaries 44 and 45 are independent

Remark 48 It is easy to see that if λ1 gt 0 λn gt 0 our result coincides with the inequalityof operator versions in [6] So we conclude that our results give an improvement of theKantorovich inequality [6] that applies to all invertible Hermite matrices

5 Conclusion

In this paper we introduce some new Kantorovich-type inequalities for the invertibleHermitian matrices Inequalities (427) and (431) are the same as [4] but our proof is simpleIn Theorem 46 if λ1 gt 0 λn gt 0 the results are similar to the well-known Kantorovich-typeinequalities for operators in [6] Moreover for any invertible Hermitian matrix there exists asimilar inequality

Journal of Applied Mathematics 11

Acknowledgments

The authors are grateful to the associate editor and two anonymous referees whose commentshave helped to improve the final version of the present work This work was supported by theNatural Science Foundation of Chongqing Municipal Education Commission (no KJ091104)

References

[1] R A Horn and C R Johnson Matrix Analysis Cambridge University Press Cambridge UK 1985[2] A S Householder The Theory of Matrices in Numerical Analysis Blaisdell New York NY USA 1964[3] S Liu and H Neudecker ldquoSeveral matrix Kantorovich-type inequalitiesrdquo Journal of Mathematical

Analysis and Applications vol 197 no 1 pp 23ndash26 1996[4] A W Marshall and I Olkin ldquoMatrix versions of the Cauchy and Kantorovich inequalitiesrdquo

Aequationes Mathematicae vol 40 no 1 pp 89ndash93 1990[5] J K Baksalary and S Puntanen ldquoGeneralized matrix versions of the Cauchy-Schwarz and

Kantorovich inequalitiesrdquo Aequationes Mathematicae vol 41 no 1 pp 103ndash110 1991[6] S S Dragomir ldquoNew inequalities of the Kantorovich type for bounded linear operators in Hilbert

spacesrdquo Linear Algebra and its Applications vol 428 no 11-12 pp 2750ndash2760 2008[7] P G Spain ldquoOperator versions of the Kantorovich inequalityrdquo Proceedings of the American

Mathematical Society vol 124 no 9 pp 2813ndash2819 1996[8] S Liu and H Neudecker ldquoKantorovich inequalities and efficiency comparisons for several classes of

estimators in linear modelsrdquo Statistica Neerlandica vol 51 no 3 pp 345ndash355 1997[9] P Bloomfield and G S Watson ldquoThe inefficiency of least squaresrdquo Biometrika vol 62 pp 121ndash128

1975[10] L V Kantorovic ldquoFunctional analysis and applied mathematicsrdquo Uspekhi Matematicheskikh Nauk vol

3 no 6(28) pp 89ndash185 1948 (Russian)[11] C G Khatri and C R Rao ldquoSome extensions of the Kantorovich inequality and statistical

applicationsrdquo Journal of Multivariate Analysis vol 11 no 4 pp 498ndash505 1981[12] C T Lin ldquoExtrema of quadratic forms and statistical applicationsrdquo Communications in Statistics A

vol 13 no 12 pp 1517ndash1520 1984[13] A Galantai ldquoA study of Auchmutyrsquos error estimaterdquo Computers amp Mathematics with Applications vol

42 no 8-9 pp 1093ndash1102 2001[14] P D Robinson and A J Wathen ldquoVariational bounds on the entries of the inverse of a matrixrdquo IMA

Journal of Numerical Analysis vol 12 no 4 pp 463ndash486 1992[15] S S Dragomir ldquoA generalization of Grussrsquos inequality in inner product spaces and applicationsrdquo

Journal of Mathematical Analysis and Applications vol 237 no 1 pp 74ndash82 1999[16] S S Dragomir ldquoSome Gruss type inequalities in inner product spacesrdquo Journal of Inequalities in Pure

and Applied Mathematics vol 4 no 2 article 42 2003[17] S S Dragomir ldquoOn Bessel and Gruss inequalities for orthonormal families in inner product spacesrdquo

Bulletin of the Australian Mathematical Society vol 69 no 2 pp 327ndash340 2004[18] S S Dragomir ldquoReverses of Schwarz triangle and Bessel inequalities in inner product spacesrdquo Journal

of Inequalities in Pure and Applied Mathematics vol 5 no 3 article 76 2004[19] S S Dragomir ldquoReverses of the Schwarz inequality generalising a Klamkin-McLenaghan resultrdquo

Bulletin of the Australian Mathematical Society vol 73 no 1 pp 69ndash78 2006[20] Z Liu L Lu and K Wang ldquoOn several matrix Kantorovich-type inequalitiesrdquo Journal of Inequalities

and Applications vol 2010 Article ID 571629 5 pages 2010

Journal of Applied Mathematics

Advances in Matrices Finite and Infinitewith Applications

Guest Editors P N Shivakumar Panayiotis PsarrakosK C Sivakumar and Yang Zhang

Copyright copy 2013 Hindawi Publishing Corporation All rights reserved

This is a special issue published in ldquoJournal of Applied Mathematicsrdquo All articles are open access articles distributed under the CreativeCommons Attribution License which permits unrestricted use distribution and reproduction in any medium provided the originalwork is properly cited

Editorial Board

Saeid Abbasbandy IranMina B Abd-El-Malek EgyptMohamed A Abdou EgyptSubhas Abel IndiaMostafa Adimy FranceCarlos J S Alves PortugalMohamad Alwash USAIgor Andrianov GermanySabri Arik TurkeyFrancis TK Au Hong KongOlivier Bahn CanadaRoberto Barrio SpainAlfredo Bellen ItalyJafar Biazar IranHester Bijl The NetherlandsAnjan Biswas Saudi ArabiaStephane PA Bordas USAJames Robert Buchanan USAAlberto Cabada SpainXiao Chuan Cai USAJinde Cao ChinaAlexandre Carvalho BrazilSong Cen ChinaQianshun S Chang ChinaTai-Ping Chang TaiwanShih-sen Chang ChinaRushan Chen ChinaXinfu Chen USAKe Chen UKEric Cheng Hong KongFrancisco Chiclana UKJen-Tzung Chien TaiwanC S Chien TaiwanHan H Choi Republic of KoreaTin-Tai Chow ChinaM S H Chowdhury MalaysiaCarlos Conca ChileVitor Costa PortugalLivija Cveticanin SerbiaEric de Sturler USAOrazio Descalzi ChileKai Diethelm GermanyVit Dolejsi Czech RepublicBo-Qing Dong ChinaMagdy A Ezzat Egypt

Meng Fan ChinaYa Ping Fang ChinaAntonio J M Ferreira PortugalMichel Fliess FranceM A Fontelos SpainHuijun Gao ChinaBernard J Geurts The NetherlandsJamshid Ghaboussi USAPablo Gonzalez-Vera SpainLaurent Gosse ItalyK S Govinder South AfricaJose L Gracia SpainYuantong Gu AustraliaZhihong GUAN ChinaNicola Guglielmi ItalyFrederico G Guimaraes BrazilVijay Gupta IndiaBo Han ChinaMaoan Han ChinaPierre Hansen CanadaFerenc Hartung HungaryXiaoqiao He Hong KongLuis Javier Herrera SpainJ Hoenderkamp The NetherlandsYing Hu FranceNing Hu JapanZhilong L Huang ChinaKazufumi Ito USATakeshi Iwamoto JapanGeorge Jaiani GeorgiaZhongxiao Jia ChinaTarun Kant IndiaIdo Kanter IsraelAbdul Hamid Kara South AfricaHamid Reza Karimi NorwayJae-Wook Kim UKJong Hae Kim Republic of KoreaKazutake Komori JapanFanrong Kong USAVadim Krysko RussiaJin L Kuang SingaporeMiroslaw Lachowicz PolandHak-Keung Lam UKTak-Wah Lam Hong KongPGL Leach Cyprus

Yongkun Li ChinaWan-Tong Li ChinaJ Liang ChinaChing-Jong Liao TaiwanChong Lin ChinaMingzhu Liu ChinaChein-Shan Liu TaiwanKang Liu USAYansheng Liu ChinaFawang Liu AustraliaShutian Liu ChinaZhijun Liu ChinaJulian Lopez-Gomez SpainShiping Lu ChinaGert Lube GermanyNazim Idrisoglu Mahmudov TurkeyOluwole Daniel Makinde South AfricaFrancisco J Marcellan SpainGuiomar Martın-Herran SpainNicola Mastronardi ItalyMichael McAleer The NetherlandsStephane Metens FranceMichael Meylan AustraliaAlain Miranville FranceRam N Mohapatra USAJaime E Munoz Rivera BrazilJavier Murillo SpainRoberto Natalini ItalySrinivasan Natesan IndiaJiri Nedoma Czech RepublicJianlei Niu Hong KongRoger Ohayon FranceJavier Oliver SpainDonal OrsquoRegan IrelandMartin Ostoja-Starzewski USATurgut Ozis TurkeyClaudio Padra ArgentinaReinaldo Martinez Palhares BrazilFrancesco Pellicano ItalyJuan Manuel Pena SpainRicardo Perera SpainMalgorzata Peszynska USAJames F Peters CanadaMark A Petersen South AfricaMiodrag Petkovic Serbia

Vu Ngoc Phat VietnamAndrew Pickering SpainHector Pomares SpainMaurizio Porfiri USAMario Primicerio ItalyMorteza Rafei The NetherlandsRoberto Reno ItalyJacek Rokicki PolandDirk Roose BelgiumCarla Roque PortugalDebasish Roy IndiaSamir H Saker EgyptMarcelo A Savi BrazilWolfgang Schmidt GermanyEckart Schnack GermanyMehmet Sezer TurkeyNaseer Shahzad Saudi ArabiaFatemeh Shakeri IranJian Hua Shen ChinaHui-Shen Shen ChinaFernando Simoes PortugalTheodore E Simos Greece

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Qing-Wen Wang ChinaGuangchen Wang ChinaJunjie Wei ChinaLi Weili ChinaMartin Weiser GermanyFrank Werner GermanyShanhe Wu ChinaDongmei Xiao ChinaGongnan Xie ChinaYuesheng Xu USASuh-Yuh Yang TaiwanBo Yu ChinaJinyun Yuan BrazilAlejandro Zarzo SpainGuisheng Zhai JapanJianming Zhan ChinaZhihua Zhang ChinaJingxin Zhang AustraliaShan Zhao USAChongbin Zhao AustraliaRenat Zhdanov USAHongping Zhu China

Advances in Matrices Finite and Infinite with Applications P N Shivakumar Panayiotis PsarrakosK C Sivakumar and Yang ZhangVolume 2013 Article ID 156136 3 pages

On Nonnegative Moore-Penrose Inverses of Perturbed Matrices Shani Jose and K C SivakumarVolume 2013 Article ID 680975 7 pages

A New Construction of Multisender Authentication Codes from Polynomials over Finite FieldsXiuli WangVolume 2013 Article ID 320392 7 pages

Geometrical and Spectral Properties of the Orthogonal Projections of the Identity Luis GonzalezAntonio Suarez and Dolores GarcıaVolume 2013 Article ID 435730 8 pages

A Parameterized Splitting Preconditioner for Generalized Saddle Point ProblemsWei-Hua Luo and Ting-Zhu HuangVolume 2013 Article ID 489295 6 pages

Solving Optimization Problems on Hermitian Matrix Functions with ApplicationsXiang Zhang and Shu-Wen XiangVolume 2013 Article ID 593549 11 pages

Left and Right Inverse Eigenpairs Problem for κ-Hermitian Matrices Fan-Liang Li Xi-Yan Huand Lei ZhangVolume 2013 Article ID 230408 6 pages

Completing a 2times 2 Block Matrix of Real Quaternions with a Partial Specified InverseYong Lin and Qing-Wen WangVolume 2013 Article ID 271978 5 pages

Fuzzy Approximate Solution of Positive Fully Fuzzy Linear Matrix EquationsXiaobin Guo and Dequan ShangVolume 2013 Article ID 178209 7 pages

Ranks of a Constrained Hermitian Matrix Expression with Applications Shao-Wen YuVolume 2013 Article ID 514984 9 pages

An Efficient Algorithm for the Reflexive Solution of the Quaternion Matrix Equation AXB + CXHD = FNing Li Qing-Wen Wang and Jing JiangVolume 2013 Article ID 217540 14 pages

The Optimization on Ranks and Inertias of a Quadratic Hermitian Matrix Function and Its ApplicationsYirong YaoVolume 2013 Article ID 961568 6 pages

Constrained Solutions of a System of Matrix Equations Qing-Wen Wang and Juan YuVolume 2012 Article ID 471573 19 pages

Contents

On the Hermitian R-Conjugate Solution of a System of Matrix Equations Chang-Zhou DongQing-Wen Wang and Yu-Ping ZhangVolume 2012 Article ID 398085 14 pages

Perturbation Analysis for the Matrix Equation X minussummi=1 A

lowasti XAi +

sumnj=1 B

lowastj XBj = I

Xue-Feng Duan and Qing-Wen WangVolume 2012 Article ID 784620 13 pages

Refinements of Kantorovich Inequality for Hermitian Matrices Feixiang ChenVolume 2012 Article ID 483624 11 pages

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 156136 3 pageshttpdxdoiorg1011552013156136

EditorialAdvances in Matrices Finite and Infinite with Applications

P N Shivakumar1 Panayiotis Psarrakos2 K C Sivakumar3 and Yang Zhang1

1 Department of Mathematics University of Manitoba Winnipeg MB Canada R3T 2N22Department of Mathematics School of Applied Mathematical and Physical SciencesNational Technical Univeristy of Athens Zografou Campus 15780 Athens Greece

3 Department of Mathematics Indian Institute of Technology Madras Chennai 600 036 India

Correspondence should be addressed to P N Shivakumar shivakuccumanitobaca

Received 13 June 2013 Accepted 13 June 2013

Copyright copy 2013 P N Shivakumar et al This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

In the mathematical modeling of many problems includingbut not limited to physical sciences biological sciencesengineering image processing computer graphics medicalimaging and even social sciences lately solution methodsmost frequently involve matrix equations In conjunctionwith powerful computing machines complex numericalcalculations employing matrix methods could be performedsometimes even in real time It is well known that infinitematrices arise more naturally than finite matrices and havea colorful history in development from sequences seriesand quadratic forms Modern viewpoint considers infinitematrices more as operators defined between certain spe-cific infinite-dimensional normed spaces Banach spacesor Hilbert spaces Giant and rapid advancements in thetheory and applications of finite matrices have been madeover the past several decades These developments havebeen well documented in many books monographs andresearch journals The present topics of research includediagonally dominantmatrices theirmany extensions inverseof matrices their recursive computation singular matricesgeneralized inverses inverse positive matrices with specificemphasis on their applications to economics like the Leon-tief input-output models and use of finite difference andfinite element methods in the solution of partial differentialequations perturbationmethods and eigenvalue problems ofinterest and importance to numerical analysts statisticiansphysical scientists and engineers to name a few

The aim of this special issue is to announce certainrecent advancements in matrices finite and infinite andtheir applications For a review of infinite matrices andapplications see P N Shivakumar and K C Sivakumar

(Linear Algebra Appl 2009) Specifically an example of anapplication can be found in the classical problem ldquoshapeof a drumrdquo in P N Shivakumar W Yan and Y Zhang(WSES transactions in Mathematics 2011) The focal themesof this special issue are inverse positivematrices including119872-matrices applications of operator theorymatrix perturbationtheory and pseudospectra matrix functions and generalizedeigenvalue problems and inverse problems including scat-tering and matrices over quaternions In the following wepresent a brief overview of a few of the central topics of thisspecial issue

For119872-matrices let us start by mentioning the most citedbooks byR Berman andR J Plemmons (SIAM 1994) and byR A Horn and C R Johnson (Cambridge 1994) as excellentsources One of the most important properties of invertible119872-matrices is that their inverses are nonnegative There areseveral works that have considered generalizations of someof the important properties of 119872-matrices We only pointout a few of them here The article by D Mishra and K CSivakumar (Oper Matrices 2012) considers generalizationsof inverse nonnegativity to the Moore-Penrose inverse whilemore general classes of matrices are the objects of discussionin D Mishra and K C Sivakumar (Linear Algebra Appl2012) and A N Sushama K Premakumari and K CSivakumar (Electron J Linear Algebra 2012)

A brief description of the papers in the issue is as followsIn the work of Z Zhang the problem of solving cer-

tain optimization problems on Hermitian matrix functionsis studied Specifically the author considers the extremalinertias and ranks of a given function 119891(119883 119884) C119898times119899 rarr

C119898times119899 (which is defined in terms of matrices 119860 119861 119862 and 119863

2 Journal of Applied Mathematics

inC119898times119899) subject to119883 119884 satisfying certain matrix equationsAs applications the author presents necessary and sufficientconditions for 119891 to be positive definite positive semidefiniteand so forth As another application the author presents acharacterization for 119891(119883 119884) = 0 again subject to 119883 119884satisfying certain matrix equations

The task of determining parametrized splitting precon-ditioners for certain generalized saddle point problems isundertaken byW-H Luo and T-Z Huang in their workThewell-known and very widely applicable Sherman-Morrison-Woodbury formula for the sum of matrices is used indetermining a preconditioner for linear equations wherethe coefficient matrix may be singular and nonsymmetricIllustrations of the procedure are provided for Stokes andOseen problems

The work reported by A Gonzalez A Suarez and DGarcia deals with the problem of estimating the Frobeniuscondition of the matrix 119860119873 given a matrix 119860 isin R119899times119899 where119873 satisfies the condition that 119873 = argmin

119883isin119878119868 minus 119860119883

119865

Here 119878 is a certain subspace of R119899 and sdot 119865denotes the

Frobenius norm The authors derive certain spectral andgeometrical properties of the preconditioner119873 as above

Let 119861 119862 119863 and 119864 be (not necessarily square) matriceswith quaternion entries so that the matrix 119872 = (

119860 119861

119862 119863)

is square Y Lin and Q-W Wang present necessary andsufficient conditions so that the top left block 119860 of119872 existssatisfying the conditions that119872 is nonsingular and thematrix119864 is the top left block of119872minus1 A general expression for suchan 119860 when it exists is obtained A numerical illustration ispresented

For a complex hermitian 119860 of order 119899 with eigenvalues1205821le 1205822le sdot sdot sdot le 120582

119899 the celebrated Kantorovich inequality is

1 le 119909lowast

119860119909119909lowast

119860minus1

119909 le (120582119899+1205821)2

41205821120582119899 for all unit vectors 119909 isin

C119899 Equivalently 0 le 119909lowast119860119909119909lowast119860minus1119909minus1 le (120582119899minus1205821)2

41205821120582119899 for

all unit vectors 119909 isin C119899 F Chen in his work on refinements ofKantorovich inequality for hermitian matrices obtains somerefinements of the second inequality where the proofs useonly elementary ideas (httpwwwmathntuagrsimppsarr)

It is pertinent to point out that among other things aninteresting generalization of 119872-matrices is obtained by SJose and K C Sivakumar The authors consider the prob-lem of nonnegativity of the Moore-Penrose inverse of aperturbation of the form 119860minus119883119866119884

119879 given that 119860dagger ge 0 Usinga generalized version of the Sherman-Morrison-Woodburyformula conditions for (119860 minus 119883119866119884

119879

)dagger to be nonnegative

are derived Applications of the results are presented brieflyIterative versions of the results are also studied

It is well known that the concept of quaternions wasintroduced by Hamilton in 1843 and has played an importantrole in contemporary mathematics such as noncommuta-tive algebra analysis and topology Nowadays quaternionmatrices are widely and heavily used in computer sciencequantum physics altitude control robotics signal and colourimage processing and so on Many questions can be reducedto quaternion matrices and solving quaternion matrix equa-tionsThis special issue has provided an excellent opportunityfor researchers to report their recent results

Let H119899times119899 denote the space of all 119899 times 119899 matrices whoseentries are quaternions Let 119876 isin H119899times119899 be Hermitian and self-invertible 119883 isin H119899times119899 is called reflexive (relative to 119876) if 119883 =

119876119883119876 The authors F-L Li X-Y Hu and L Zhang presentan efficient algorithm for finding the reflexive solution of thequaternion matrix equation 119860119883119861 + 119862119883

lowast

119863 = 119865 They alsoshow that given an appropriate initialmatrix their procedureyields the least reflexive solutionwith respect to the Frobeniusnorm

In the work on the ranks of a constrained hermitianmatrix expression S W Yu gives formulas for the maximaland minimal ranks of the quaternion Hermitian matrixexpression 119862

4minus 1198604119883119860lowast

4 where 119883 is a Hermitian solution

to quaternion matrix equations 1198601119883 = 119862

1 1198831198611= 1198622 and

1198603119883119860lowast

3= 1198623 and also applies this result to some special

system of quaternion matrix equationsY Yao reports his findings on the optimization on ranks

and inertias of a quadratic hermitian matrix function Hepresents one solution for optimization problems on the ranksand inertias of the quadratic Hermitian matrix function 119876 minus119883119875119883lowast subject to a consistent system of matrix equations

119860119883 = 119862 and 119883119861 = 119863 As applications he derives necessaryand sufficient conditions for the solvability to the systems ofmatrix equations and matrix inequalities 119860119883 = 119862 119883119861 = 119863and119883119875119883lowast = (gt lt ge le)119876

Let 119877 be a nontrivial real symmetric involution matrixA complex matrix 119860 is called 119877-conjugate if 119860 = 119877119860119877In the work on the hermitian 119877-conjugate solution of asystem of matrix equations C-Z Dong Q-W Wang andY-P Zhang present necessary and sufficient conditions forthe existence of the hermitian 119877-conjugate solution to thesystem of complex matrix equations 119860119883 = 119862 and 119883119861 = 119863An expression of the Hermitian 119877-conjugate solution is alsopresented

A perturbation analysis for the matrix equation 119883 minus

sum119898

119894=1119860lowast

119894119883119860119894+ sum119899

119895=1119861lowast

119895119883119861119895= 119868 is undertaken by X-F Duan

and Q-W Wang They give a precise perturbation bound fora positive definite solution A numerical example is presentedto illustrate the sharpness of the perturbation bound

Linear matrix equations and their optimal approxima-tion problems have great applications in structural designbiology statistics control theory and linear optimal controland as a consequence they have attracted the attention ofseveral researchers for many decades In the work of Q-WWang and J Yu necessary and sufficient conditions of and theexpressions for orthogonal solutions symmetric orthogonalsolutions and skew-symmetric orthogonal solutions of thesystem of matrix equations 119860119883 = 119861 and 119883119862 = 119863 arederived When the solvability conditions are not satisfied theleast squares symmetric orthogonal solutions and the leastsquares skew-symmetric orthogonal solutions are obtainedA methodology is provided to compute the least squaressymmetric orthogonal solutions and an illustrative exampleis given to illustrate the proposed algorithm

Many real-world engineering systems are too complex tobe defined in precise terms and in many matrix equationssome or all of the system parameters are vague or impreciseThus solving a fuzzy matrix equation becomes important In

Journal of Applied Mathematics 3

the work reported by X Guo and D Shang the fuzzy matrixequation 119860 otimes 119883 otimes 119861 = 119862 in which 119860 119861 and 119862 are 119898 times 119898119899 times 119899 and 119898 times 119899 nonnegative LR fuzzy numbers matricesrespectively is investigated This fuzzy matrix system whichhas a wide use in control theory and control engineering isextended into three crisp systems of linear matrix equationsaccording to the arithmetic operations of LR fuzzy numbersBased on the pseudoinversematrix a computingmodel to thepositive fully fuzzy linear matrix equation is provided andthe fuzzy approximate solution of original fuzzy systems isobtained by solving the crisp linear matrix systems In addi-tion the existence condition of nonnegative fuzzy solutionis discussed and numerical examples are given to illustratethe proposed method Overall the technique of LR fuzzynumbers and their operations appears to be a strong tool ininvestigating fully fuzzy matrix equations

Left and right inverse eigenpairs problem arises in anatural way when studying spectral perturbations ofmatricesand relative recursive problems F-L Li X-Y Hu and LZhang consider in their work left and right inverse eigenpairsproblem for 119896-hermitian matrices and its optimal approx-imate problem The class of 119896-hermitian matrices containshermitian and perhermitian matrices and has numerousapplications in engineering and statistics Based on thespecial properties of 119896-hermitian matrices and their left andright eigenpairs an equivalent problem is obtained Thencombining a new inner product of matrices necessary andsufficient conditions for the solvability of the problem aregiven and its general solutions are derived Furthermorethe optimal approximate solution a calculation procedure tocompute the unique optimal approximation and numericalexperiments are provided

Finally let us point out a work on algebraic coding theoryRecall that a communication system in which a receivercan verify the authenticity of a message sent by a group ofsenders is referred to as a multisender authentication codeIn the work reported by X Wang the author constructsmulti-sender authentication codes using polynomials overfinite fields Here certain important parameters and theprobabilities of deceptions of these codes are also computed

Acknowledgments

We wish to thank all the authors for their contributions andthe reviewers for their valuable comments in bringing thespecial issue to fruition

P N ShivakumarPanayiotis Psarrakos

K C SivakumarYang Zhang

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 680975 7 pageshttpdxdoiorg1011552013680975

Research ArticleOn Nonnegative Moore-Penrose Inverses of Perturbed Matrices

Shani Jose and K C Sivakumar

Department of Mathematics Indian Institute of Technology Madras Chennai Tamil Nadu 600036 India

Correspondence should be addressed to K C Sivakumar kcskumariitmacin

Received 22 April 2013 Accepted 7 May 2013

Academic Editor Yang Zhang

Copyright copy 2013 S Jose and K C Sivakumar This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

Nonnegativity of theMoore-Penrose inverse of a perturbation of the form119860minus119883119866119884119879 is consideredwhen119860dagger ge 0 Using a generalizedversion of the Sherman-Morrison-Woodbury formula conditions for (119860 minus 119883119866119884119879)dagger to be nonnegative are derived Applications ofthe results are presented briefly Iterative versions of the results are also studied

1 Introduction

We consider the problem of characterizing nonnegativityof the Moore-Penrose inverse for matrix perturbations ofthe type 119860 minus 119883119866119884

119879 when the Moore-Penrose inverse of119860 is nonnegative Here we say that a matrix 119861 = (119887

119894119895) is

nonnegative and denote it by 119861 ge 0 if 119887119894119895ge 0 forall119894 119895 This

problemwasmotivated by the results in [1] where the authorsconsider an 119872-matrix 119860 and find sufficient conditions forthe perturbed matrix (119860 minus 119883119884119879) to be an119872-matrix Let usrecall that a matrix 119861 = (119887

119894119895) is said to 119885-matrix if 119887

119894119895le 0

for all 119894 119895 119894 = 119895 An119872-matrix is a nonsingular 119885-matrix withnonnegative inverse The authors in [1] use the well-knownSherman-Morrison-Woodbury (SMW) formula as one of theimportant tools to prove their main resultThe SMW formulagives an expression for the inverse of (119860minus119883119884119879) in terms of theinverse of119860 when it exists When119860 is nonsingular119860minus119883119884119879

is nonsingular if and only 119868 minus 119884119879119860minus1119883 is nonsingular In thatcase

(119860 minus 119883119884119879

)minus1

= 119860minus1

minus 119860minus1

119883(119868 minus 119884119879

119860minus1

119883)minus1

119884119879

119860minus1

(1)

Themain objective of the present work is to study certainstructured perturbations 119860 minus 119883119884119879 of matrices 119860 such thattheMoore-Penrose inverse of the perturbation is nonnegativewhenever the Moore-Penrose inverse of 119860 is nonnegativeClearly this class of matrices includes the class of matricesthat have nonnegative inverses especially119872-matrices In ourapproach extensions of SMW formula for singular matrices

play a crucial role Let us mention that this problem has beenstudied in the literature (See for instance [2] formatrices and[3] for operators over Hilbert spaces) We refer the reader tothe references in the latter for other recent extensions

In this paper first we present alternative proofs ofgeneralizations of the SMW formula for the cases of theMoore-Penrose inverse (Theorem 5) and the group inverse(Theorem 6) in Section 3 In Section 4 we characterize thenonnegativity of (119860 minus 119883119866119884119879)dagger This is done in Theorem 9and is one of the main results of the present work As aconsequence we present a result for119872-matrices which seemsnew We present a couple of applications of the main resultin Theorems 13 and 15 In the concluding section we studyiterative versions of the results of the second section Weprove two characterizations for (119860minus119883119884119879)dagger to be nonnegativein Theorems 18 and 21

Before concluding this introductory section let us givea motivation for the work that we have undertaken hereIt is a well-documented fact that 119872-matrices arise quiteoften in solving sparse systems of linear equations Anextensive theory of119872-matrices has been developed relativeto their role in numerical analysis involving the notion ofsplitting in iterativemethods and discretization of differentialequations in the mathematical modeling of an economyoptimization and Markov chains [4 5] Specifically theinspiration for the present study comes from the work of [1]where the authors consider a system of linear inequalitiesarising out of a problem in third generation wireless commu-nication systemsThematrix defining the inequalities there is

2 Journal of Applied Mathematics

an 119872-matrix In the likelihood that the matrix of thisproblem is singular (due to truncation or round-off errors)the earlier method becomes inapplicable Our endeavour isto extend the applicability of these results to more generalmatrices for instance matrices with nonnegative Moore-Penrose inverses Finally as mentioned earlier sincematriceswith nonnegative generalized inverses include in particular119872-matrices it is apparent that our results are expected toenlarge the applicability of themethods presently available for119872-matrices even in a very general framework including thespecific problem mentioned above

2 Preliminaries

Let R R119899 and R119898times119899 denote the set of all real numbers the119899-dimensional real Euclidean space and the set of all 119898 times 119899matrices over R For 119860 isin R119898times119899 let 119877(119860) 119873(119860) 119877(119860)perp and119860119879 denote the range space the null space the orthogonal

complement of the range space and the transpose of thematrix 119860 respectively For x = (119909

1 1199092 119909

119899)119879

isin R119899 wesay that x is nonnegative that is x ge 0 if and only if x

119894ge 0

for all 119894 = 1 2 119899 As mentioned earlier for a matrix 119861 weuse 119861 ge 0 to denote that all the entries of 119861 are nonnegativeAlso we write 119860 le 119861 if 119861 minus 119860 ge 0

Let 120588(119860) denote the spectral radius of the matrix 119860 If120588(119860) lt 1 for 119860 isin R119899times119899 then 119868 minus 119860 is invertible Thenext result gives a necessary and sufficient condition for thenonnegativity of (119868 minus 119860)minus1 This will be one of the results thatwill be used in proving the first main result

Lemma 1 (see [5 Lemma 21 Chapter 6]) Let 119860 isin R119899times119899 benonnegative Then 120588(119860) lt 1 if and only if (119868 minus 119860)minus1 exists and

(119868 minus 119860)minus1

=

infin

sum

119896=0

119860119896

ge 0 (2)

More generally matrices having nonnegative inversesare characterized using a property called monotonicity Thenotion of monotonicity was introduced by Collatz [6] A real119899 times 119899matrix 119860 is called monotone if 119860x ge 0 rArr x ge 0 It wasproved by Collatz [6] that 119860 is monotone if and only if 119860minus1exists and 119860minus1 ge 0

One of the frequently used tools in studying monotonematrices is the notion of a regular splitting We only refer thereader to the book [5] for more details on the relationshipbetween these concepts

The notion of monotonicity has been extended in a vari-ety of ways to singular matrices using generalized inversesFirst let us briefly review the notion of two importantgeneralized inverses

For 119860 isin R119898times119899 the Moore-Penrose inverse is the unique119885 isin R119899times119898 satisfying the Penrose equations 119860119885119860 = 119860119885119860119885 = 119885 (119860119885)119879 = 119860119885 and (119885119860)119879 = 119885119860 The uniqueMoore-Penrose inverse of119860 is denoted by119860dagger and it coincideswith 119860minus1 when 119860 is invertible

The following theorem by Desoer and Whalen whichis used in the sequel gives an equivalent definition for theMoore-Penrose inverse Let us mention that this result wasproved for operators between Hilbert spaces

Theorem 2 (see [7]) Let 119860 isin R119898times119899 Then 119860dagger isin R119899times119898 is theunique matrix119883 isin R119899times119898 satisfying

(i) 119885119860119909 = 119909 forall119909 isin 119877(119860119879)(ii) 119885119910 = 0 forall119910 isin 119873(119860119879)

Now for119860 isin R119899times119899 any119885 isin R119899times119899 satisfying the equations119860119885119860 = 119860 119885119860119885 = 119885 and 119860119885 = 119885119860 is called the groupinverse of 119860 The group inverse does not exist for everymatrix But whenever it exists it is unique A necessary andsufficient condition for the existence of the group inverse of119860 is that the index of119860 is 1 where the index of a matrix is thesmallest positive integer 119896 such that rank (119860119896+1) = rank (119860119896)

Some of the well-known properties of theMoore-Penroseinverse and the group inverse are given as follows 119877(119860119879) =119877(119860dagger

)119873(119860119879) = 119873(119860dagger) 119860dagger119860 = 119875119877(119860119879) and 119860119860dagger = 119875

119877(119860) In

particular x isin 119877(119860119879) if and only if x = 119860dagger119860x Also 119877(119860) =119877(119860

) 119873(119860) = 119873(119860

) and 119860119860 = 119860119860

= 119875119877(119860)119873(119860)

Herefor complementary subspaces 119871 and119872 of R119896 119875

119871119872denotes

the projection ofR119896 onto 119871 along119872 119875119871denotes 119875

119871119872if119872 =

119871perp For details we refer the reader to the book [8]Inmatrix analysis a decomposition (splitting) of amatrix

is considered in order to study the convergence of iterativeschemes that are used in the solution of linear systems of alge-braic equations As mentioned earlier regular splittings areuseful in characterizing matrices with nonnegative inverseswhereas proper splittings are used for studying singularsystems of linear equations Let us next recall this notionFor a matrix 119860 isin R119898times119899 a decomposition 119860 = 119880 minus 119881 iscalled a proper splitting [9] if 119877(119860) = 119877(119880) and 119873(119860) =119873(119880) It is rather well-known that a proper splitting existsfor every matrix and that it can be obtained using a full-rank factorization of the matrix For details we refer to [10]Certain properties of a proper splitting are collected in thenext result

Theorem 3 (see [9 Theorem 1]) Let 119860 = 119880 minus 119881 be a propersplitting of 119860 isin R119898times119899 Then

(a) 119860 = 119880(119868 minus 119880dagger119881)(b) 119868 minus 119880dagger119881 is nonsingular and(c) 119860dagger = (119868 minus 119880dagger119881)minus1119880dagger

The following result by Berman and Plemmons [9] givesa characterization for 119860dagger to be nonnegative when 119860 has aproper splitting This result will be used in proving our firstmain result

Theorem 4 (see [9 Corollary 4]) Let 119860 = 119880 minus 119881 be a propersplitting of 119860 isin R119898times119899 where 119880dagger ge 0 and 119880dagger119881 ge 0 Then119860dagger

ge 0 if and only if 120588(119880dagger119881) lt 1

3 Extensions of the SMW Formula forGeneralized Inverses

The primary objects of consideration in this paper are gen-eralized inverses of perturbations of certain types of a matrix

Journal of Applied Mathematics 3

119860 Naturally extensions of the SMW formula for generalizedinverses are relevant in the proofs Inwhat follows we presenttwo generalizations of the SMW formula for matrices Wewould like to emphasize that our proofs also carry overto infinite dimensional spaces (the proof of the first resultis verbatim and the proof of the second result with slightmodifications applicable to range spaces instead of ranks ofthe operators concerned) However we are confining ourattention to the case of matrices Let us also add that theseresults have been proved in [3] for operators over infinitedimensional spaces We have chosen to include these resultshere as our proofs are different from the ones in [3] and thatour intention is to provide a self-contained treatment

Theorem 5 (see [3 Theorem 21]) Let 119860 isin R119898times119899 119883 isin R119898times119896and 119884 isin R119899times119896 be such that

119877 (119883) sube 119877 (119860) 119877 (119884) sube 119877 (119860119879

) (3)

Let 119866 isin R119896times119896 and Ω = 119860 minus 119883119866119884119879 If

119877 (119883119879

) sube 119877 ((119866dagger

minus 119884119879

119860dagger

119883)119879

)

119877 (119884119879

) sube 119877 (119866dagger

minus 119884119879

119860dagger

119883) 119877 (119883119879

) sube 119877 (119866)

119877 (119884119879

) sube 119877 (119866119879

)

(4)

then

Ωdagger

= 119860dagger

+ 119860dagger

119883(119866dagger

minus 119884119879

119860dagger

119883)dagger

119884119879

119860dagger

(5)

Proof Set 119885 = 119860dagger + 119860dagger119883119867dagger119884119879119860dagger where119867 = 119866dagger

minus 119884119879

119860dagger

119883From the conditions (3) and (4) it follows that 119860119860dagger119883 = 119883119860dagger

119860119884 = 119884119867dagger119867119883119879 = 119883119879119867119867dagger119884119879 = 119884119879119866119866dagger119883119879 = 119883119879 and119866dagger

119866119884119879

= 119884119879

Now

119860dagger

119883119867dagger

119884119879

minus 119860dagger

119883119867dagger

119884119879

119860dagger

119883119866119884119879

= 119860dagger

119883119867dagger

(119866dagger

119866119884119879

minus 119884119879

119860dagger

119883119866119884119879

)

= 119860dagger

119883119867dagger

119867119866119884119879

= 119860dagger

119883119866119884119879

(6)

Thus 119885Ω = 119860dagger

119860 Since 119877(Ω119879) sube 119877(119860119879

) it follows that119885Ω(x) = 119909 forallx isin 119877(Ω119879)

Let y isin 119873(Ω119879

) Then 119860119879y minus 119884119866119879119883119879y = 0 so that119883119879

119860dagger119879

(119860119879

minus 119884119866119879

119883119879

)y = 0 Substituting 119883119879 = 119866119866dagger119883119879 andsimplifying it we get119867119879119866119879119883119879y = 0 Also119860119879y = 119884119866119879119883119879y =119884119867119867dagger

119866119879

119883119879y = 0 and so 119860daggery = 0 Thus 119885y = 0 for

y isin 119873(Ω119879) Hence by Theorem 2 119885 = Ωdagger

The result for the group inverse follows

Theorem 6 Let 119860 isin R119899times119899 be such that 119860 exists Let 119883119884 isin R119899times119896 and 119866 be nonsingular Assume that 119877(119883) sube 119877(119860)119877(119884) sube 119877(119860

119879

) and 119866minus1 minus 119884119879119860119883 is nonsingular Suppose that

rank (119860 minus 119883119866119884119879) = rank (119860) Then (119860 minus 119883119866119884119879) exists andthe following formula holds

(119860 minus 119883119866119884119879

)= 119860

+ 119860

119883(119866minus1

minus 119884119879

119860119883)minus1

119884119879

119860 (7)

Conversely if (119860minus119883119866119884119879) exists then the formula above holdsand we have rank (119860 minus 119883119866119884119879) = rank (119860)

Proof Since 119860 exists 119877(119860) and 119873(119860) are complementarysubspaces of R119899times119899

Suppose that rank (119860 minus 119883119866119884119879) = rank (119860) As 119877(119883) sube119877(119860) it follows that 119877(119860 minus 119883119866119884119879) sube 119877(119860) Thus 119877(119860 minus119883119866119884119879

) = 119877(119860) By the rank-nullity theorem the nullity ofboth 119860 minus 119883119866119884119879 and 119860 are the same Again since 119877(119884) sube119877(119860119879

) it follows that 119873(119860 minus 119883119866119884119879

) = 119873(119860) Thus119877(119860 minus 119883119866119884

119879

) and 119873(119860 minus 119883119866119884119879) are complementary sub-spaces This guarantees the existence of the group inverse of119860 minus 119883119866119884

119879Conversely suppose that (119860 minus 119883119866119884119879) exists It can be

verified by direct computation that 119885 = 119860+ 119860

119883(119866minus1

minus

119884119879

119860119883)minus1

119884119879

119860 is the group inverse of 119860 minus 119883119866119884119879 Also we

have (119860minus119883119866119884119879)(119860minus119883119866119884119879) = 119860119860 so that119877(119860minus119883119866119884119879) =119877(119860) and hence the rank (119860 minus 119883119866119884119879) = rank (119860)

We conclude this section with a fairly old result [2] as aconsequence of Theorem 5

Theorem 7 (see [2 Theorem 15]) Let 119860 isin R119898times119899 of rank 119903119883 isin R119898times119903 and 119884 isin R119899times119903 Let119866 be an 119903times119903 nonsingular matrixAssume that 119877(119883) sube 119877(119860) 119877(119884) sube 119877(119860119879) and 119866minus1 minus 119884119879119860dagger119883is nonsingular Let Ω = 119860 minus 119883119866119884119879 Then

(119860 minus 119883119866119884119879

)dagger

= 119860dagger

+ 119860dagger

119883(119866minus1

minus 119884119879

119860dagger

119883)minus1

119884119879

119860dagger

(8)

4 Nonnegativity of (119860minus119883119866119884119879)dagger

In this section we consider perturbations of the form 119860 minus

119883119866119884119879 and derive characterizations for (119860 minus 119883119866119884119879)dagger to be

nonnegative when 119860dagger ge 0 119883 ge 0 119884 ge 0 and 119866 ge 0 In orderto motivate the first main result of this paper let us recall thefollowing well known characterization of119872-matrices [5]

Theorem 8 Let 119860 be a 119885-matrix with the representation 119860 =119904119868 minus 119861 where 119861 ge 0 and 119904 ge 0 Then the following statementsare equivalent

(a) 119860minus1 exists and 119860minus1 ge 0

(b) There exists 119909 gt 0 such that 119860119909 gt 0

(c) 120588(119861) lt 119904

Let us prove the first result of this article This extendsTheorem 8 to singular matrices We will be interested inextensions of conditions (a) and (c) only

Theorem 9 Let 119860 = 119880 minus 119881 be a proper splitting of 119860 isin R119898times119899

with 119880dagger ge 0 119880dagger119881 ge 0 and 120588(119880dagger119881) lt 1 Let 119866 isin R119896times119896 benonsingular and nonnegative 119883 isin R119898times119896 and 119884 isin R119899times119896 be

4 Journal of Applied Mathematics

nonnegative such that 119877(119883) sube 119877(119860) 119877(119884) sube 119877(119860119879) and119866minus1 minus119884119879

119860dagger

119883 is nonsingular LetΩ = 119860minus119883119866119884119879Then the followingare equivalent

(a) Ωdagger ge 0(b) (119866minus1 minus 119884119879119860dagger119883)minus1 ge 0(c) 120588(119880dagger119882) lt 1 where119882 = 119881 + 119883119866119884

119879

Proof First we observe that since 119877(119883) sube 119877(119860) 119877(119884) sube119877(119860119879

) and 119866minus1 minus 119884119879119860dagger119883 is nonsingular by Theorem 7 wehave

Ωdagger

= 119860dagger

+ 119860dagger

119883(119866minus1

minus 119884119879

119860dagger

119883)minus1

119884119879

119860dagger

(9)

We thus haveΩΩdagger = 119860119860dagger andΩdaggerΩ = 119860dagger119860 Therefore

119877 (Ω) = 119877 (119860) 119873 (Ω) = 119873 (119860) (10)

Note that the first statement also implies that 119860dagger ge 0 byTheorem 4

(a) rArr (b) By taking 119864 = 119860 and 119865 = 119883119866119884119879 we get

Ω = 119864 minus 119865 as a proper splitting for Ω such that 119864dagger = 119860dagger ge 0and 119864dagger119865 = 119860dagger119883119866119884119879 ge 0 (since 119883119866119884119879 ge 0) Since Ωdagger ge 0by Theorem 4 we have 120588(119860dagger119883119866119884119879) = 120588(119864

dagger

119865) lt 1 Thisimplies that 120588(119884119879119860dagger119883119866) lt 1 We also have 119884119879119860dagger119883119866 ge 0Thus by Lemma 1 (119868minus119884119879119860dagger119883119866)minus1 exists and is nonnegativeBut we have 119868 minus 119884119879119860dagger119883119866 = (119866

minus1

minus 119884119879

119860dagger

119883)119866 Now 0 le(119868 minus 119884

119879

119860dagger

119883119866)minus1

= 119866minus1

(119866minus1

minus 119884119879

119860dagger

119883)minus1 This implies that

(119866minus1

minus 119884119879

119860dagger

119883)minus1

ge 0 since 119866 ge 0 This proves (b)(b) rArr (c) We have119880minus119882 = 119880minus119881minus119883119866119884

119879

= 119860minus119883119884119879

=

Ω Also 119877(Ω) = 119877(119860) = 119877(119880) and 119873(Ω) = 119873(119860) = 119873(119880)So Ω = 119880 minus 119882 is a proper splitting Also 119880dagger ge 0 and119880dagger

119882 = 119880dagger

(119881+119883119866119884119879

) ge 119880dagger

119881 ge 0 Since (119866minus1minus119884119879119860dagger119883)minus1 ge0 it follows from (9) that Ωdagger ge 0 (c) now follows fromTheorem 4

(c) rArr (a) Since119860 = 119880minus119881 we haveΩ = 119880minus119881minus119883119866119884119879 =119880minus119882 Also we have119880dagger ge 0119880dagger119881 ge 0 Thus119880dagger119882 ge 0 since119883119866119884119879

ge 0 Now by Theorem 4 we are done if the splittingΩ = 119880 minus119882 is a proper splitting Since 119860 = 119880 minus 119881 is a propersplitting we have 119877(119880) = 119877(119860) and 119873(119880) = 119873(119860) Nowfrom the conditions in (10) we get that 119877(119880) = 119877(Ω) and119873(119880) = 119873(Ω) Hence Ω = 119880 minus 119882 is a proper splitting andthis completes the proof

The following result is a special case of Theorem 9

Theorem 10 Let119860 = 119880minus119881 be a proper splitting of119860 isin R119898times119899

with 119880dagger ge 0 119880dagger119881 ge 0 and 120588(119880dagger119881) lt 1 Let 119883 isin R119898times119896 and119884 isin R119899times119896 be nonnegative such that 119877(119883) sube 119877(119860) 119877(119884) sube119877(119860119879

) and 119868minus119884119879119860dagger119883 is nonsingular LetΩ = 119860minus119883119884119879 Thenthe following are equivalent

(a) Ωdagger ge 0(b) (119868 minus 119884119879119860dagger119883)minus1 ge 0(c) 120588(119880dagger119882) lt 1 where119882 = 119881 + 119883119884

119879

The following consequence of Theorem 10 appears to benew This gives two characterizations for a perturbed 119872-matrix to be an119872-matrix

Corollary 11 Let119860 = 119904119868minus119861where119861 ge 0 and 120588(119861) lt 119904 (ie119860is an119872-matrix) Let 119883 isin R119898times119896 and 119884 isin R119899times119896 be nonnegativesuch that 119868 minus 119884119879119860dagger119883 is nonsingular Let Ω = 119860 minus 119883119884119879 Thenthe following are equivalent

(a) Ωminus1 ge 0(b) (119868 minus 119884119879119860dagger119883)minus1 ge 0(c) 120588(119861(119868 + 119883119884119879)) lt 119904

Proof From the proof ofTheorem 10 since 119860 is nonsingularit follows that ΩΩdagger = 119868 and ΩdaggerΩ = 119868 This shows that Ωis invertible The rest of the proof is omitted as it is an easyconsequence of the previous result

In the rest of this section we discuss two applicationsof Theorem 10 First we characterize the least element ina polyhedral set defined by a perturbed matrix Next weconsider the following Suppose that the ldquoendpointsrdquo of aninterval matrix satisfy a certain positivity property Then allmatrices of a particular subset of the interval also satisfythat positivity condition The problem now is if that we aregiven a specific structured perturbation of these endpointswhat conditions guarantee that the positivity property for thecorresponding subset remains valid

The first result is motivated by Theorem 12 below Let usrecall that with respect to the usual order an element xlowast isinX sube R119899 is called a least element ofX if it satisfies xlowast le x forall x isin X Note that a nonempty set may not have the leastelement and if it exists then it is unique In this connectionthe following result is known

Theorem 12 (see [11 Theorem 32]) For 119860 isin R119898times119899 and b isinR119898 let

Xb = x isin R119899

119860x + y ge b 119875119873(119860)

x = 0

119875119877(119860)

y = 0 119891119900119903 119904119900119898119890 y isin R119898 (11)

Then a vector xlowast is the least element ofXb if and only if xlowast =119860daggerb isin Xb with 119860dagger ge 0

Now we obtain the nonnegative least element of apolyhedral set defined by a perturbed matrix This is animmediate application of Theorem 10

Theorem 13 Let 119860 isin R119898times119899 be such that 119860dagger ge 0 Let 119883 isin

R119898times119896 and let 119884 isin R119899times119896 be nonnegative such that 119877(119883) sube119877(119860) 119877(119884) sube 119877(119860119879) and 119868 minus 119884119879119860dagger119883 is nonsingular Supposethat (119868 minus 119884119879119860dagger119883)minus1 ge 0 For b isin R119898 119887 ge 0 let

Sb = x isin R119899

Ωx + y ge b 119875119873(Ω)

x = 0

119875119877(Ω)

y = 0 119891119900119903 119904119900119898119890 y isin R119898 (12)

where Ω = 119860 minus 119883119884119879 Then xlowast = (119860 minus 119883119884119879)daggerb is the least

element of S119887

Proof From the assumptions using Theorem 10 it fol-lows that Ωdagger ge 0 The conclusion now follows fromTheorem 12

Journal of Applied Mathematics 5

To state and prove the result for interval matrices let usfirst recall the notion of interval matrices For119860 119861 isin R119898 times 119899an interval (matrix) 119869 = [119860 119861] is defined as 119869 = [119860 119861] = 119862 119860 le 119862 le 119861 The interval 119869 = [119860 119861] is said to be range-kernelregular if 119877(119860) = 119877(119861) and 119873(119860) = 119873(119861) The followingresult [12] provides necessary and sufficient conditions for119862dagger

ge 0 for 119862 isin 119870 where 119870 = 119862 isin 119869 119877(119862) = 119877(119860) =

119877(119861)119873(119862) = 119873(119860) = 119873(119861)

Theorem 14 Let 119869 = [119860 119861] be range-kernel regular Then thefollowing are equivalent

(a) 119862dagger ge 0 whenever 119862 isin 119870(b) 119860dagger ge 0 and 119861dagger ge 0

In such a case we have 119862dagger = suminfin119895=0(119861dagger

(119861 minus 119862))119895

119861dagger

Now we present a result for the perturbation

Theorem 15 Let 119869 = [119860 119861] be range-kernel regular with119860dagger ge0 and 119861dagger ge 0 Let 119883

1 1198832 1198841 and 119884

2be nonnegative matrices

such that 119877(1198831) sube 119877(119860) 119877(119884

1) sube 119877(119860

119879

) 119877(1198832) sube 119877(119861) and

119877(1198842) sube 119877(119861

119879

) Suppose that 119868minus1198841198791119860dagger

1198831and 119868minus119884119879

2119860dagger

1198832are

nonsingular with nonnegative inverses Suppose further that1198832119884119879

2le 1198831119884119879

1 Let 119869 = [119864 119865] where 119864 = 119860 minus 119883

1119884119879

1and

119865 = 119861 minus 1198832119884119879

2 Finally let = 119885 isin 119869 119877(119885) = 119877(119864) =

119877(119865) 119886119899119889 119873(119885) = 119873(119864) = 119873(119865) Then

(a) 119864dagger ge 0 and 119865dagger ge 0(b) 119885dagger ge 0 whenever 119885 isin

In that case 119885dagger = suminfin119895=0(119861dagger

119861 minus 119865dagger

119885)119895

119865dagger

Proof It follows fromTheorem 7 that

119864dagger

= 119860dagger

+ 119860dagger

1198831(119868 minus 119884

119879

1119860dagger

1198831)minus1

119884119879

1119860dagger

119865dagger

= 119861dagger

+ 119861dagger

1198832(119868 minus 119884

119879

2119861dagger

1198832)minus1

119884119879

2119861dagger

(13)

Also we have 119864119864dagger = 119860119860dagger 119864dagger119864 = 119860

dagger

119860 119865dagger119865 = 119861dagger

119861and 119865119865dagger = 119861119861

dagger Hence 119877(119864) = 119877(119860) 119873(119864) = 119873(119860)119877(119865) = 119877(119861) and 119873(119865) = 119873(119861) This implies that theinterval 119869 is range-kernel regular Now since 119864 and 119865 satisfythe conditions of Theorem 10 we have 119864dagger ge 0 and 119865dagger ge 0proving (a) Hence by Theorem 14 119885dagger ge 0 whenever 119885 isin Again by Theorem 14 we have 119885dagger = suminfin

119895=0(119865dagger

(119865 minus 119885))119895

119865dagger

=

suminfin

119895=0(119861dagger

119861 minus 119865dagger

119885)119895

119865dagger

5 Iterations That Preserve Nonnegativity ofthe Moore-Penrose Inverse

In this section we present results that typically provideconditions for iteratively defined matrices to have nonneg-ative Moore-Penrose inverses given that the matrices thatwe start with have this property We start with the followingresult about the rank-one perturbation case which is a directconsequence of Theorem 10

Theorem 16 Let 119860 isin R119899times119899 be such that 119860dagger ge 0 Let x isin R119898

and y isin R119899 be nonnegative vectors such that x isin 119877(119860) y isin119877(119860119879

) and 1 minus y119879119860daggerx = 0 Then (119860 minus xy119879)dagger ge 0 if and only ify119879119860daggerx lt 1

Example 17 Let us consider 119860 = (13) (1 minus1 1

minus1 4 minus1

1 minus1 1

) It can be

verified that 119860 can be written in the form 119860 = (1198682

119890119879

1

) (1198682+

e1119890119879

1)minus1

119862minus1

(1198682+ e1119890119879

1)minus1

(1198682e1) where e

1= (1 0)

119879 and 119862 isthe 2 times 2 circulant matrix generated by the row (1 12) Wehave 119860dagger = (12) ( 2 1 21 2 1

2 1 2

) ge 0 Also the decomposition 119860 =

119880minus119881 where119880 = (13) ( 1 minus1 2minus1 4 minus2

1 minus1 2

) and119881 = (13) ( 0 0 10 2 minus10 0 1

) isa proper splitting of 119860 with 119880dagger ge 0 119880dagger119881 ge 0 and 120588(119880dagger119881) lt1

Let x = y = (13 16 13)119879 Then x and let y are

nonnegative x isin 119877(119860) and y isin 119877(119860119879) We have 1 minus y119879119860daggerx =512 gt 0 and 120588(119880dagger119882) lt 1 for119882 = 119881 + xy119879 Also it can beseen that (119860 minus xy119879)dagger = (120) ( 47 28 4728 32 28

47 28 47

) ge 0 This illustratesTheorem 10

Let x1 x2 x

119896isin R119898 and y

1 y2 y

119896isin R119899 be nonneg-

ative Denote 119883119894= (x1 x2 x

119894) and 119884

119894= (y1 y2 y

119894) for

119894 = 1 2 119896 Then119860minus119883119896119884119879

119896= 119860minussum

119896

119894=1x119894y119879119894 The following

theorem is obtained by a recurring application of the rank-one result of Theorem 16

Theorem 18 Let119860 isin R119898times119899 and let119883119894 119884119894be as above Further

suppose that x119894isin 119877(119860 minus 119883

119894minus1119884119879

119894minus1) y119894isin 119877(119860 minus 119883

119894minus1119884119879

119894minus1)119879 and

1 minus y119879119894(119860 minus 119883

119894minus1119884119879

119894minus1)daggerx119894be nonzero Let (119860 minus 119883

119894minus1119884119879

119894minus1)dagger

ge 0for all 119894 = 1 2 119896 Then (119860 minus 119883

119896119884119879

119896)dagger

ge 0 if and only ify119879119894(119860 minus 119883

119894minus1119884119879

119894minus1)daggerx119894lt 1 where 119860 minus 119883

0119884119879

0is taken as 119860

Proof Set 119861119894= 119860 minus 119883

119894119884119879

119894for 119894 = 0 1 119896 where 119861

0is

identified as 119860 The conditions in the theorem can be writtenas

x119894isin 119877 (119861

119894minus1) y

119894isin 119877 (119861

119879

119894minus1)

1 minus y119879119894119861dagger

119894minus1x119894= 0

forall119894 = 1 2 119896

(14)

Also we have 119861dagger119894ge 0 for all 119894 = 0 1 119896 minus 1

Now assume that 119861dagger119896ge 0 Then by Theorem 16 and the

conditions in (14) for 119894 = 119896 we have y119896119861dagger

119896minus1x119896lt 1 Also

since by assumption 119861dagger119894ge 0 for all 119894 = 0 1 119896 minus 1 we have

y119879119894119861dagger

119894minus1x119894lt 1 for all 119894 = 1 2 119896

The converse part can be proved iteratively Then condi-tion y119879

11198610

daggerx1lt 1 and the conditions in (14) for 119894 = 1 imply

that 1198611

dagger

ge 0 Repeating the argument for 119894 = 2 to 119896 proves theresult

The following result is an extension of Lemma 23 in [1]which is in turn obtained as a corollaryThis corollary will beused in proving another characterization for (119860minus119883

119896119884119879

119896)dagger

ge 0

6 Journal of Applied Mathematics

Theorem 19 Let 119860 isin R119899times119899 bc isin R119899 be such that minusb ge 0minusc ge 0 and 120572(isin R) gt 0 Further suppose that b isin 119877(119860) andc isin 119877(119860119879) Let 119860 = ( 119860 b

c119879 120572 ) isin R(119899+1)times(119899+1) Then 119860dagger ge 0 if andonly if 119860dagger ge 0 and c119879119860daggerb lt 120572

Proof Let us first observe that 119860119860daggerb = b and c119879119860dagger119860 = c119879Set

119883 =(

119860dagger

+119860daggerbc119879119860dagger

120572 minus c119879119860daggerbminus119860daggerb

120572 minus c119879119860daggerbminus

c119879119860dagger

120572 minus c119879119860daggerb1

120572 minus c119879119860daggerb) (15)

It then follows that 119860119883 = ( 119860119860dagger 00119879 1 ) and 119883119860 = (119860dagger119860 0

0119879 1 ) Usingthese two equations it can easily be shown that119883 = 119860dagger

Suppose that 119860dagger ge 0 and 120573 = 120572 minus c119879119860daggerb gt 0 Then119860dagger

ge 0 Conversely suppose that 119860dagger ge 0 Then we musthave 120573 gt 0 Let e

1 e2 e

119899+1 denote the standard basis

of R119899+1 Then for 119894 = 1 2 119899 we have 0 le 119860dagger

(e119894) =

(119860daggere119894+(119860daggerbc119879119860daggere

119894120573) minus(c119879119860daggere

119894120573))119879 Since c le 0 it follows

that 119860daggere119894ge 0 for 119894 = 1 2 119899 Thus 119860dagger ge 0

Corollary 20 (see [1 Lemma 23]) Let 119860 isin R119899times119899 benonsingular Let b c isin R119899 be such that minusb ge 0 minusc ge 0

and 120572(isin R) gt 0 Let 119860 = (119860 bc119879 120572 ) Then 119860minus1 isin R(119899+1)times(119899+1)

is nonsingular with 119860minus1 ge 0 if and only if 119860minus1 ge 0 andc119879119860minus1b lt 120572

Now we obtain another necessary and sufficient condi-tion for (119860 minus 119883

119896119884119879

119896)dagger to be nonnegative

Theorem 21 Let 119860 isin R119898times119899 be such that 119860dagger ge 0 Let x119894 y119894

be nonnegative vectors inR119899 andR119898 respectively for every 119894 =1 119896 such that

x119894isin 119877 (119860 minus 119883

119894minus1119884119879

119894minus1) y

119894isin 119877(119860 minus 119883

119894minus1119884119879

119894minus1)119879

(16)

where 119883119894= (x1 x

119894) 119884119894= (y1 y

119894) and 119860 minus 119883

0119884119879

0is

taken as119860 If119867119894= 119868119894minus119884119879

119894119860dagger

119883119894is nonsingular and 1minus y119879

119894119860daggerx119894

is positive for all 119894 = 1 2 119896 then (119860 minus 119883119896119884119879

119896)dagger

ge 0 if andonly if

y119879119894119860daggerx119894lt 1 minus y119879

119894119860dagger

119883119894minus1119867minus1

119894minus1119884119879

119894minus1119860daggerx119894 forall119894 = 1 119896 (17)

Proof The range conditions in (16) imply that 119877(119883119896) sube 119877(119860)

and 119877(119884119896) sube 119877(119860

119879

) Also from the assumptions it followsthat 119867

119896is nonsingular By Theorem 10 (119860 minus 119883

119896119884119879

119896)dagger

ge 0 ifand only if119867minus1

119896ge 0 Now

119867119896= (

119867119896minus1

minus119884119879

119896minus1119860daggerx119896

minusy119879119896119860dagger

119883119896minus1

1 minus y119879119896119860daggerx119896

) (18)

Since 1minusy119879119896119860daggerx119896gt 0 usingCorollary 20 it follows that119867minus1

119896ge

0 if and only if y119879119896119860dagger

119883119896minus1119867minus1

119896minus1119884119879

119896minus1119860daggerx119896lt 1 minus y119879

119896119860daggerx119896and

119867minus1

119896minus1ge 0

Now applying the above argument to the matrix 119867119896minus1

we have that 119867minus1

119896minus1ge 0 holds if and only if 119867minus1

119896minus2ge 0

holds and y119879119896minus1119860dagger

119883119896minus2(119868119896minus2

minus 119884119879

119896minus2119860dagger

119883119896minus2)minus1

119884119879

119896minus2119860daggerx119896minus1 lt

1minus y119879119896minus1119860daggerx119896minus1 Continuing the above argument we get (119860minus

119883119896y119879119896)dagger

ge 0 if and only if y119879119894119860dagger

119883119894minus1119867minus1

119894minus1119884119879

119894minus1119860daggerx119894lt 1minusy119879

119894119860daggerx119894

forall119894 = 1 119896 This is condition (17)

We conclude the paper by considering an extension ofExample 17

Example 22 For a fixed 119899 let 119862 be the circulant matrixgenerated by the row vector (1 (12) (13) (1(119899 minus 1)))Consider

119860 = (119868

119890119879

1

) (119868 + e1119890119879

1)minus1

119862minus1

(119868 + e1119890119879

1)minus1

(119868 e1) (19)

where 119868 is the identity matrix of order 119899 minus 1 e1is (119899 minus 1) times 1

vector with 1 as the first entry and 0 elsewhere Then 119860 isin

R119899times119899 119860dagger is the 119899 times 119899 nonnegative Toeplitz matrix

119860dagger

=

((((((

(

11

119899 minus 1

1

119899 minus 2

1

21

1

21

1

119899 minus 1

1

3

1

2

1

119899 minus 1

1

119899 minus 2

1

119899 minus 3 1

1

119899 minus 1

11

119899 minus 1

1

119899 minus 2

1

21

))))))

)

(20)

Let 119909119894be the first row of 119861dagger

119894minus1written as a column vector

multiplied by 1119899119894 and let 119910119894be the first column of 119861dagger

119894minus1

multiplied by 1119899119894 where 119861119894= 119860 minus 119883

119894119884119879

119894 with 119861

0being

identified with 119860 119894 = 0 2 119896 We then have x119894ge 0

y119894ge 0 119909

1isin 119877(119860) and 119910

1isin 119877(119860

119879

) Now if 119861dagger119894minus1

ge 0 foreach iteration then the vectors 119909

119894and 119910

119894will be nonnegative

Hence it is enough to check that 1 minus 119910119879119894119861dagger

119894minus1119909119894gt 0 in order

to get 119861dagger119894ge 0 Experimentally it has been observed that for

119899 gt 3 the condition 1 minus 119910119879119894119861dagger

119894minus1119909119894gt 0 holds true for any

iteration However for 119899 = 3 it is observed that 119861dagger1ge 0 119861dagger

2ge

0 1 minus 1199101198791119860dagger

1199091gt 0 and 1 minus 119910dagger

2119861dagger

11199092gt 0 But 1 minus 119910119879

3119861dagger

21199093lt 0

and in that case we observe that 119861dagger3is not nonnegative

References

[1] J Ding W Pye and L Zhao ldquoSome results on structured119872-matrices with an application to wireless communicationsrdquoLinear Algebra and its Applications vol 416 no 2-3 pp 608ndash614 2006

[2] S K Mitra and P Bhimasankaram ldquoGeneralized inverses ofpartitionedmatrices and recalculation of least squares estimatesfor data ormodel changesrdquo Sankhya A vol 33 pp 395ndash410 1971

[3] C Y Deng ldquoA generalization of the Sherman-Morrison-Woodbury formulardquo Applied Mathematics Letters vol 24 no9 pp 1561ndash1564 2011

[4] A Berman M Neumann and R J SternNonnegative Matricesin Dynamical Systems JohnWiley amp Sons New York NY USA1989

Journal of Applied Mathematics 7

[5] A Berman and R J Plemmons Nonnegative Matrices inthe Mathematical Sciences Society for Industrial and AppliedMathematics (SIAM) 1994

[6] L Collatz Functional Analysis and Numerical MathematicsAcademic Press New York NY USA 1966

[7] C A Desoer and B H Whalen ldquoA note on pseudoinversesrdquoSociety for Industrial and Applied Mathematics vol 11 pp 442ndash447 1963

[8] A Ben-Israel and T N E Greville Generalized InversesTheoryand Applications Springer New York NY USA 2003

[9] A Berman and R J Plemmons ldquoCones and iterative methodsfor best least squares solutions of linear systemsrdquo SIAM Journalon Numerical Analysis vol 11 pp 145ndash154 1974

[10] D Mishra and K C Sivakumar ldquoOn splittings of matrices andnonnegative generalized inversesrdquo Operators and Matrices vol6 no 1 pp 85ndash95 2012

[11] D Mishra and K C Sivakumar ldquoNonnegative generalizedinverses and least elements of polyhedral setsrdquo Linear Algebraand its Applications vol 434 no 12 pp 2448ndash2455 2011

[12] M R Kannan and K C Sivakumar ldquoMoore-Penrose inversepositivity of interval matricesrdquo Linear Algebra and its Applica-tions vol 436 no 3 pp 571ndash578 2012

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 320392 7 pageshttpdxdoiorg1011552013320392

Research ArticleA New Construction of Multisender Authentication Codes fromPolynomials over Finite Fields

Xiuli Wang

College of Science Civil Aviation University of China Tianjin 300300 China

Correspondence should be addressed to Xiuli Wang xlwangcauceducn

Received 3 February 2013 Accepted 7 April 2013

Academic Editor Yang Zhang

Copyright copy 2013 Xiuli Wang This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Multisender authentication codes allow a group of senders to construct an authenticated message for a receiver such that thereceiver can verify the authenticity of the received message In this paper we construct one multisender authentication code frompolynomials over finite fields Some parameters and the probabilities of deceptions of this code are also computed

1 Introduction

Multisender authentication code was firstly constructed byGilbert et al [1] in 1974 Multisender authentication systemrefers towho a groupof senders cooperatively send amessageto a receiver then the receiver should be able to ascertainthat the message is authentic About this case many scholarsand researchers had made great contributions to multisenderauthentication codes such as [2ndash6]

In the actual computer network communications mul-tisender authentication codes include sequential model andsimultaneous model Sequential model is that each senderuses his own encoding rules to encode a source state orderlythe last sender sends the encoded message to the receiverand the receiver receives themessage and verifies whether themessage is legal or not Simultaneousmodel is that all sendersuse their own encoding rules to encode a source state andeach sender sends the encoded message to the synthesizerrespectively then the synthesizer forms an authenticatedmessage and verifies whether the message is legal or not Inthis paper we will adopt the second model

In a simultaneous model there are four participants agroup of senders 119880 = 119880

1 1198802 119880

119899 the key distribution

center he is responsible for the key distribution to sendersand receiver including solving the disputes between them areceiver 119877 and a synthesizer where he only runs the trustedsynthesis algorithm The code works as follows each senderand receiver has their own Cartesian authentication code

respectively Let (119878 119864119894 119879119894 119891119894) (119894 = 1 2 119899) be the sendersrsquo

Cartesian authentication code (119878 119864119877 119879 119892) be the receiverrsquos

Cartesian authentication code ℎ 1198791times 1198792times sdot sdot sdot times 119879

119899rarr

119879 be the synthesis algorithm and 120587119894

119864 rarr 119864119894be a

subkey generation algorithm where 119864 is the key set of thekey distribution center When authenticating a message thesenders and the receiver should comply with the protocolThe key distribution center randomly selects an encodingrule 119890 isin 119864 and sends 119890

119894= 120587119894(119890) to the 119894th sender 119880

119894(119894 =

1 2 119899) secretly then he calculates 119890119877by 119890 according to

an effective algorithm and secretly sends 119890119877to the receiver

119877 If the senders would like to send a source state 119904 to thereceiver 119877 119880

119894computes 119905

119894= 119891119894(119904 119890119894) (119894 = 1 2 119899) and

sends 119898119894= (119904 119905

119894) (119894 = 1 2 119899) to the synthesizer through

an open channel The synthesizer receives the message 119898119894=

(119904 119905119894) (119894 = 1 2 119899) and calculates 119905 = ℎ(119905

1 1199052 119905

119899) by the

synthesis algorithm ℎ and then sends message 119898 = (119904 119905) tothe receiver he checks the authenticity by verifying whether119905 = 119892(119904 119890

119877) or not If the equality holds the message is

authentic and is accepted Otherwise the message is rejectedWe assume that the key distribution center is credible and

though he know the sendersrsquo and receiverrsquos encoding rules hewill not participate in any communication activities Whentransmitters and receiver are disputing the key distributioncenter settles it At the same time we assume that the systemfollows the Kerckhoff principle in which except the actualused keys the other information of the whole system ispublic

2 Journal of Applied Mathematics

In a multisender authentication system we assume thatthe whole senders are cooperative to form a valid messagethat is all senders as a whole and receiver are reliable Butthere are some malicious senders who together cheat thereceiver the part of senders and receiver are not credible andthey can take impersonation attack and substitution attackIn the whole system we assume that 119880

1 1198802 119880

119899 are

senders 119877 is a receiver 119864119894is the encoding rules set of the

sender 119880119894 and 119864

119877is the decoding rules set of the receiver

119877 If the source state space 119878 and the key space 119864119877of receiver

119877 are according to a uniform distribution then the messagespace119872 and the tag space119879 are determined by the probabilitydistribution of 119878 and 119864

119877 119871 = 119894

1 1198942 119894

119897 sub 1 2 119899

119897 lt 119899 119880119871

= 1198801198941 1198801198942 119880

119894119897 119864119871

= 1198641198801198941

1198641198801198942

119864119880119894119897

Now consider that let us consider the attacks from maliciousgroups of senders Here there are two kinds of attack

The opponentrsquos impersonation attack to receiver119880119871 after

receiving their secret keys encode a message and send it tothe receiver 119880

119871are successful if the receiver accepts it as

legitimate message Denote by 119875119868the largest probability of

some opponentrsquos successful impersonation attack to receiverit can be expressed as

119875119868= max119898isin119872

1003816100381610038161003816119890119877 isin 119864119877| 119890119877sub 119898

10038161003816100381610038161003816100381610038161003816119864119877

1003816100381610038161003816

(1)

The opponentrsquos substitution attack to the receiver 119880119871

replace 119898 with another message 1198981015840 after they observe a

legitimatemessage119898119880119871are successful if the receiver accepts

it as legitimate message it can be expressed as

119875119878= max119898isin119872

max1198981015840=119898isin119872

10038161003816100381610038161003816119890119877isin 119864119877| 119890119877sub 119898119898

1015840

10038161003816100381610038161003816

1003816100381610038161003816119890119877 isin 119864119877| 119890119877sub 119898

1003816100381610038161003816

(2)

There might be 119897 malicious senders who together cheatthe receiver that is the part of senders and the receiverare not credible and they can take impersonation attackLet 119871 = 119894

1 1198942 119894

119897 sub 1 2 119899 119897 lt 119899 and 119864

119871=

1198641198801198941

1198641198801198942

119864119880119894119897

Assume that 119880119871

= 1198801198941 1198801198942 119880

119894119897

after receiving their secret keys send a message 119898 to thereceiver 119877119880

119871are successful if the receiver accepts it as legiti-

mate message Denote by 119875119880(119871) the maximum probability of

success of the impersonation attack to the receiver It can beexpressed as

119875119880(119871)

=max119890119871isin119864119871

max119890119871isin119890119880

max119898isin119872

1003816100381610038161003816119890119877isin119864119877 | 119890119877sub119898 119901 (119890119877 119890119875) =0

10038161003816100381610038161003816100381610038161003816119890119877 isin 119864

119877| 119901 (119890119877 119890119875) =0

1003816100381610038161003816

(3)

Notes119901(119890119877 119890119875) = 0 implies that any information 119904 encoded by

119890119879can be authenticated by 119890

119877

In [2] Desmedt et al gave two constructions for MRA-codes based on polynomials and finite geometries respec-tively To construct multisender or multireceiver authenti-cation by polynomials over finite fields many researchershave done much work for example [7ndash9] There are other

constructions of multisender authentication codes that aregiven in [3ndash6] The construction of authentication codesis combinational design in its nature We know that thepolynomial over finite fields can provide a better algebrastructure and is easy to count In this paper we constructone multisender authentication code from the polynomialover finite fields Some parameters and the probabilities ofdeceptions of this code are also computed We realize thegeneralization and the application of the similar idea andmethod of the paper [7ndash9]

2 Some Results about Finite Field

Let 119865119902be the finite field with 119902 elements where 119902 is a power

of a prime 119901 and 119865 is a field containing 119865119902 denote by 119865

lowast

119902be

the nonzero elements set of 119865119902 In this paper we will use the

following conclusions over finite fields

Conclusion 1 A generator 120572 of 119865lowast

119902is called a primitive

element of 119865119902

Conclusion 2 Let 120572 isin 119865119902 if some polynomials contain 120572 as

their root and their leading coefficient are 1 over 119865119902 then the

polynomial having least degree among all such polynomialsis called a minimal polynomial over 119865

119902

Conclusion 3 Let |119865| = 119902119899 then 119865 is an 119899-dimensional

vector space over 119865119902 Let 120572 be a primitive element of 119865

119902

and 119892(119909) the minimal polynomial about 120572 over 119865119902 then

dim119892(119909) = 119899 and 1 120572 1205722

120572119899minus1 is a basis of 119865 Further-

more 1 120572 1205722 120572119899minus1 is linear independent and it is equalto 120572 120572

2

120572119899minus1

120572119899 (120572 is a primitive element 120572 = 0) is also

linear independent moreover 120572119901 1205721199012

120572119901119899minus1

120572119901119899

is alsolinear independent

Conclusion 4 Consider (1199091+ 1199092+ sdot sdot sdot + 119909

119899)119898

= (1199091)119898

+

(1199092)119898

+ sdot sdot sdot + (119909119899)119898 where 119909

119894isin 119865119902 (1 le 119894 le 119899) and 119898 is a

nonnegative power of character 119901 of 119865119902

Conclusion 5 Let119898 le 119899 Then the number of119898times119899matricesof rank119898 over 119865

119902is 119902119898(119898minus1)2prod119899

119894=119899minus119898+1(119902119894

minus 1)

More results about finite fields can be found in [10ndash12]

3 Construction

Let the polynomial 119901119895(119909) = 119886

1198951119909119901119899

+ 1198861198952119909119901(119899minus1)

+ sdot sdot sdot +

119886119895119899119909119901

(1 le 119895 le 119896) where the coefficient 119886119894119897

isin 119865119902

(1 le 119897 le 119899) and these vectors by the composition of theircoefficient are linearly independent The set of source states119878 = 119865

119902 the set of 119894th transmitterrsquos encoding rules 119864

119880119894=

1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) 119909119894

isin 119865lowast

119902 (1 le 119894 le 119899) the set

of receiverrsquos encoding rules 119864119877

= 1199011(120572) 1199012(120572) 119901

119896(120572)

where 120572 is a primitive element of 119865119902 the set of 119894th transmit-

terrsquos tags 119879119894= 119905119894| 119905119894isin 119865119902 (1 le 119894 le 119899) the set of receiverrsquos

tags 119879 = 119905 | 119905 isin 119865119902

Journal of Applied Mathematics 3

Define the encoding map 119891119894 119878 times 119864

119880119894rarr 119879119894 119891119894(119904 119890119880119894) =

1199041199011(119909119894) + 1199042

1199012(119909119894) + sdot sdot sdot + 119904

119896

119901119896(119909119894) 1 le 119894 le 119899

The decoding map 119891 119878 times 119864119877

rarr 119879 119891(119904 119890119877) = 119904119901

1(120572) +

1199042

1199012(120572) + sdot sdot sdot + 119904

119896

119901119896(120572)

The synthesizing map ℎ 1198791times 1198792times sdot sdot sdot times 119879

119899rarr 119879

ℎ(1199051 1199052 119905

119899) = 1199051+ 1199052+ sdot sdot sdot + 119905

119899

The code works as followsAssume that 119902 is larger than or equal to the number of

the possible message and 119899 le 119902

31 Key Distribution The key distribution center randomlygenerates 119896 (119896 le 119899) polynomials 119901

1(119909) 1199012(119909) 119901

119896(119909)

where 119901119895(119909) = 119886

1198951119909119901119899

+ 1198861198952119909119901(119899minus1)

+ sdot sdot sdot + 119886119895119899119909119901

(1 le 119895 le 119896)and make these vectors by composed of their coefficient islinearly independent it is equivalent to the column vectors

of the matrix (

11988611 11988621 sdotsdotsdot 1198861198961

11988612 11988622 sdotsdotsdot 1198861198962

1198861119899 1198862119899 sdotsdotsdot 119886119896119899

) is linearly independent He

selects 119899 distinct nonzero elements 1199091 1199092 119909

119899isin 119865119902again

and makes 119909119894(1 le 119894 le 119899) secret then he sends privately

1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) to the sender 119880

119894(1 le 119894 le 119899) The

key distribution center also randomly chooses a primitiveelement 120572 of 119865

119902satisfying 119909

1+ 1199092+ sdot sdot sdot + 119909

119899= 120572 and sends

1199011(120572) 1199012(120572) 119901

119896(120572) to the receiver 119877

32 Broadcast If the senderswant to send a source state 119904 isin 119878

to the receiver 119877 the sender 119880119894calculates 119905

119894= 119891119894(119904 119890119880119894) =

119860119904(119909119894) = 119904119901

1(119909119894)+1199042

1199012(119909119894)+ sdot sdot sdot+119904

119896

119901119896(119909119894) 1 le 119894 le 119899 and then

sends 119860119904(119909119894) = 119905119894to the synthesizer

33 Synthesis After the synthesizer receives 1199051 1199052 119905

119899 he

calculates ℎ(1199051 1199052 119905

119899) = 1199051+ 1199052+ sdot sdot sdot + 119905

119899and then sends

119898 = (119904 119905) to the receiver 119877

34 Verification When the receiver 119877 receives 119898 = (119904 119905) hecalculates 1199051015840 = 119892(119904 119890

119877) = 119860

119904(120572) = 119904119901

1(120572) + 119904

2

1199012(120572) + sdot sdot sdot +

119904119896

119901119896(120572) If 119905 = 119905

1015840 he accepts 119905 otherwise he rejects itNext we will show that the above construction is a well

defined multisender authentication code with arbitration

Lemma 1 Let 119862119894= (119878 119864

119875119894 119879119894 119891119894) then the code is an A-code

1 le 119894 le 119899

Proof (1) For any 119890119880119894

isin 119864119880119894 119904 isin 119878 because 119864

119880119894= 1199011(119909119894)

1199012(119909119894) 119901

119896(119909119894) 119909119894isin 119865lowast

119902 so 119905

119894= 1199041199011(119909119894) + 1199042

1199012(119909119894) + sdot sdot sdot +

119904119896

119901119896(119909119894) isin 119879119894= 119865119902 Conversely for any 119905

119894isin 119879119894 choose 119890

119880119894=

1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) 119909119894isin 119865lowast

119902 where 119901

119895(119909) = 119886

1198951119909119901119899

+

1198861198952119909119901(119899minus1)

+ sdot sdot sdot + 119886119895119899119909119901

(1 le 119895 le 119896) and let 119905119894= 119891119894(119904 119890119880119894) =

1199041199011(119909119894) + 1199042

1199012(119909119894) + sdot sdot sdot + 119904

119896

119901119896(119909119894) it is equivalent to

(119909119901119899

119894 119909119901119899minus1

119894 119909

119901

119894)(

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)(

119904

1199042

119904119896

) = 119905119894

(4)

It follows that

(

(

119909119901119899

1119909119901119899minus1

1sdot sdot sdot 119909119901

1

119909119901n

2119909119901119899minus1

2sdot sdot sdot 119909119901

2

119909119901119899

119899119909119901119899minus1

119899sdot sdot sdot 119909119901

119899

)

)

times (

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)(

119904

1199042

119904119896

) = (

1199051

1199052

119905119899

)

(5)

Denote

119860 = (

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)

119883 = (

119909119901119899

1119909119901119899minus1

1sdot sdot sdot 119909119901

1

119909119901119899

2119909119901119899minus1

2sdot sdot sdot 119909119901

2

119909119901119899

119899119909119901119899minus1

119899sdot sdot sdot 119909119901

119899

)

119878 = (

119904

1199042

119904119896

) 119905 = (

1199051

1199052

119905119899

)

(6)

The above linear equation is equivalent to 119883119860119878 = 119905because the column vectors of 119860 are linearly independent119883 is equivalent to a Vandermonde matrix and 119883 is inversetherefore the above linear equation has a unique solution so119904 is only defined that is 119891

119894(1 le 119894 le 119899) is a surjection

(2) If 1199041015840 isin 119878 is another source state satisfying 1199041199011(119909119894) +

1199042

1199012(119909119894)+sdot sdot sdot+119904

119896

119901119896(119909119894) = 1199041015840

1199011(119909119894)+11990410158402

1199012(119909119894)+sdot sdot sdot+119904

1015840119896

119901119896(119909119894) =

119905119894 and it is equivalent to (119904 minus 119904

1015840

)1199011(119909119894) + (1199042

minus 11990410158402

)1199012(119909119894) + sdot sdot sdot +

(119904119896

minus 1199041015840119896

)119901119896(119909119894) = 0 then

(119909119901119899

119894 119909119901119899minus1

119894 119909

119901

119894)

times (

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)(

119904 minus 1199041015840

1199042

minus 11990410158402

119904119896

minus 1199041015840119896

) = 0

(7)

4 Journal of Applied Mathematics

Thus

(

(

119909119901119899

1119909119901119899minus1

1sdot sdot sdot 119909119901

1

119909119901n

2119909119901119899minus1

2sdot sdot sdot 119909119901

2

119909119901119899

119899119909119901119899minus1

119899sdot sdot sdot 119909119901

119899

)

)

times (

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)(

119904 minus 1199041015840

1199042

minus 11990410158402

119904119896

minus 1199041015840119896

) = (

0

0

0

)

(8)

Similar to (1) we know that the homogeneous linear equation119883119860119878 = 0 has a unique solution that is there is only zerosolution so 119904 = 119904

1015840 So 119904 is the unique source state determinedby 119890119880119894and 119905119894 thus 119862

119894(1 le 119894 le 119899) is an A-code

Lemma 2 Let 119862 = (119878 119864119877 119879 119892) then the code is an A-code

Proof (1) For any 119904 isin 119878 119890119877isin 119864119877 from the definition of 119890

119877

we assume that 119864119877

= 1199011(120572) 1199012(120572) 119901

119896(120572) where 120572 is a

primitive element of 119865119902 119892(119904 119890

119877) = 119904119901

1(120572) + 119904

2

1199012(120572) + sdot sdot sdot +

119904119896

119901119896(120572) isin 119879 = 119865

119902 on the other hand for any 119905 isin 119879 choose

119890119877= 1199011(120572) 1199012(120572) 119901

119896(120572) where 120572 is a primitive element

of 119865119902 119892(119904 119890

119877) = 119904119901

1(120572) + 119904

2

1199012(120572) + sdot sdot sdot + 119904

119896

119901119896(120572) = 119905 it is

equivalent to

(120572119901119899

120572119901119899minus1

sdot sdot sdot 120572119901

)

times (

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)(

119904

1199042

119904119896

) = 119905

119860 = (

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)

(9)

that is (120572119901119899

120572119901119899minus1

120572119901

)119860(

119904

1199042

119904119896

) = 119905 From Conclusion

3 we know that (120572119901119899

120572119901119899minus1

120572119901

) is linearly independentand the column vectors of 119860 are also linearly independenttherefore the above linear equation has unique solution so 119904

is only defined that is 119892 is a surjection

(2) If 1199041015840 is another source state satisfying 119905 = 119892(1199041015840

119890119877) then

(120572119901119899

120572119901119899minus1

120572119901

)119860(

1199041015840

11990410158402

1199041015840119896

)

= (120572119901119899

120572119901119899minus1

120572119901

)119860(

119904

1199042

119904119896

)

(10)

that is (120572119901119899

120572119901119899minus1

120572119901

)119860([[

[

1199041015840

11990410158402

1199041015840119896

]]

]

minus [

[

119904

1199042

119904119896

]

]

) = 0 Sim-

ilar to (1) we get that the homogeneous linear equation(120572119901119899

120572119901119899minus1

120572119901

)119860(1198781015840

minus 119878) = 0 has a unique solution thatis there is only zero solution so 119878 = 119878

1015840 that is 119904 = 1199041015840 So

119904 is the unique source state determined by 119890119877and 119905 thus

119862 = (119878 119864119877 119879 119892) is an A-code

At the same time for any valid119898 = (119904 119905) we have knownthat 120572 = 119909

1+ 1199092+ sdot sdot sdot + 119909

119899 and it follows that 1199051015840 = 119904119901

1(120572) +

1199042

1199012(120572)+sdot sdot sdot+119904

119896

119901119896(120572) = 119904119901

1(1199091+1199092+sdot sdot sdot+119909

119899)+1199042

1199012(1199091+1199092+

sdot sdot sdot + 119909119899) + sdot sdot sdot + 119904

119896

119901119896(1199091+ 1199092+ sdot sdot sdot + 119909

119899) We also have known

that 119901119895(119909) = 119886

1198951119909119901119899

+1198861198952119909119901(119899minus1)

+ sdot sdot sdot + 119886119895119899119909119901

(1 le 119895 le 119896) fromConclusion 4 (119909

1+ 1199092+ sdot sdot sdot + 119909

119899)119901119898

= (1199091)119901119898

+ (1199092)119901119898

+ sdot sdot sdot +

(119909119899)119901119898

where119898 is a nonnegative power of character 119901 of 119865119902

andwe get119901119895(1199091+1199092+sdot sdot sdot+119909

119899) = 119901119895(1199091)+119901119895(1199092)+sdot sdot sdot+119901

119895(119909119899)

therefore 1199051015840 = 1199041199011(120572) + 119904

2

1199012(120572) + sdot sdot sdot + 119904

119896

119901119896(120572) = (119904119901

1(1199091) +

1199042

1199012(1199091)+sdot sdot sdot+119904

119896

119901119896(1199091))+(119904119901

1(1199092)+1199042

1199012(1199092)+sdot sdot sdot+119904

119896

119901119896(1199092))+

sdot sdot sdot+(1199041199011(119909119899)+1199042

1199012(119909119899)+ sdot sdot sdot+119904

119896

119901119896(119909119899)) = 1199051+1199052+sdot sdot sdot+119905

119899= 119905

and the receiver 119877 accepts119898

From Lemmas 1 and 2 we know that such construction ofmultisender authentication codes is reasonable and there are119899 senders in this system Next we compute the parametersof this code and the maximum probability of success inimpersonation attack and substitution attack by the group ofsenders

Theorem 3 Some parameters of this construction are|119878| = 119902 |119864

119880119894| = [119902

119896(119896minus1)2

prod119899

119894=119899minus119896+1(119902119894

minus 1)] (119902minus1

1) =

[119902119896(119896minus1)2

prod119899

119894=119899minus119896+1(119902119894

minus 1)](119902 minus 1) (1 le 119894 le 119899) |119879119894| = 119902 (1 le

119894 le 119899) |119864119877| = [119902

119896(119896minus1)2

prod119899

119894=119899minus119896+1(119902119894

minus 1)]120593(119902 minus 1) |119879| = 119902Where 120593(119902 minus 1) is the 119864119906119897119890119903 119891119906119899119888119905119894119900119899 of 119902 minus 1 it representsthe number of primitive element of 119865

119902here

Proof For |119878| = 119902 |119879119894| = 119902 and |119879| = 119902 the results

are straightforward For119864119880119894 because119864

119880119894=1199011(119909119894) 1199012(119909119894)

119901119896(119909119894) 119909119894isin 119865lowast

119902 where 119901

119895(119909) = 119886

1198951119909119901119899

+ 1198861198952119909119901(119899minus1)

+ sdot sdot sdot +

119886119895119899119909119901

(1 le 119895 le 119896) and these vectors by the composition oftheir coefficient are linearly independent it is equivalent to

the columns of 119860 = (

11988611 11988621 sdotsdotsdot 1198861198961

11988612 11988622 sdotsdotsdot 1198861198962

1198861119899 1198862119899 sdotsdotsdot 119886119896119899

) is linear independent

Journal of Applied Mathematics 5

From Conclusion 5 we can conclude that the number of 119860satisfying the condition is 119902

119896(119896minus1)2

prod119899

119894=119899minus119896+1(119902119894

minus 1) On theother hand the number of distinct nonzero elements 119909

119894(1 le

119894 le 119899) in 119865119902is ( 119902minus11

) = 119902 minus 1 so |119864119880119894| = [119902

119896(119896minus1)2

prod119899

119894=119899minus119896+1(119902119894

minus

1)](119902 minus 1) For 119864119877 119864119877

= 1199011(120572) 1199012(120572) 119901

119896(120572) where 120572

is a primitive element of 119865119902 For 120572 from Conclusion 1 a

generator of119865lowast119902is called a primitive element of119865

119902 |119865lowast119902| = 119902minus1

by the theory of the group we know that the number ofgenerator of 119865lowast

119902is 120593(119902minus1) that is the number of 120572 is 120593(119902minus1)

For 1199011(119909) 1199012(119909) 119901

119896(119909) From above we have confirmed

that the number of these polynomials is 119902119896(119896minus1)2prod119899119894=119899minus119896+1

(119902119894

minus

1) therefore |119864119877| = [119902

119896(119896minus1)2

prod119899

119894=119899minus119896+1(119902119894

minus 1)]120593(119902 minus 1)

Lemma 4 For any 119898 isin 119872 the number of 119890119877contained 119898 is

120593(119902 minus 1)

Proof Let 119898 = (119904 119905) isin 119872 119890119877

= 1199011(120572) 1199012(120572) 119901

119896(120572)

where 120572 is a primitive element of 119865119902 isin 119864

119877 If 119890119877

sub 119898then 119904119901

1(120572) + 119904

2

1199012(120572) + sdot sdot sdot + 119904

119896

119901119896(120572) = 119905 hArr (120572

119901119899

120572119901119899minus1

120572119901

)119860(

119904

1199042

119904119896

) = 119905 For any 120572 suppose that there is

another 1198601015840 such that (120572

119901119899

120572119901119899minus1

120572119901

)1198601015840

(

119904

1199042

119904119896

) = 119905

then (120572119901119899

120572119901119899minus1

120572119901

)(119860 minus 1198601015840

)(

119904

1199042

119904119896

) = 0 because

120572119901119899

120572119901119899minus1

120572119901 is linearly independent so (119860minus119860

1015840

)(

119904

1199042

119904119896

) =

0 but (

119904

1199042

119904119896

) is arbitrarily therefore 119860 minus 1198601015840

= 0 that is

119860 = 1198601015840 and it follows that 119860 is only determined by 120572

Therefore as 120572 isin 119864119877 for any given 119904 and 119905 the number of

119890119877contained in119898 is 120593(119902 minus 1)

Lemma 5 For any119898 = (119904 119905) isin 119872 and1198981015840

= (1199041015840

1199051015840

) isin 119872 with119904 = 1199041015840 the number of 119890

119877contained119898 and119898

1015840 is 1

Proof Assume that 119890119877

= 1199011(120572) 1199012(120572) 119901

119896(120572) where

120572 is a primitive element of 119865119902 isin 119864

119877 If 119890119877

sub 119898 and119890119877

sub 1198981015840 then 119904119901

1(120572) + 119904

2

1199012(120572) + sdot sdot sdot + 119904

119896

119901119896(120572) = 119905 hArr

(120572119901119899

120572119901119899minus1

120572119901

)119860(

119904

1199042

119904119896

) = 119905 11990410158401199011(120572) + 119904

10158402

1199012(120572) + sdot sdot sdot +

1199041015840119896

119901119896(120572) = 119905 hArr (120572

119901119899

120572119901119899minus1

120572119901

)119860(

1199041015840

11990410158402

1199041015840119896

) = 119905 It is

equivalent to (120572119901119899

120572119901119899minus1

120572119901

)119860(

119904minus1199041015840

1199042minus11990410158402

119904119896minus1199041015840119896

) = 119905 minus 1199051015840 because

119904 = 1199041015840 so 119905 = 119905

1015840 otherwise we assume that 119905 = 1199051015840 and

since 120572119901119899

120572119901119899minus1

120572119901 and the column vectors of 119860 both

are linearly independent it forces that 119904 = 1199041015840 this is a

contradiction Therefore we get

(119905 minus 1199051015840

)minus1

[[[[

[

(120572119901119899

120572119901119899minus1

120572119901

)119860 (

119904 minus 1199041015840

1199042

minus 11990410158402

119904119896

minus 1199041015840119896

)

]]]]

]

= 1

(lowast)

since 119905 1199051015840 is given (119905 minus 119905

1015840

)minus1 is unique by equation (lowast) for

any given 119904 1199041015840 and 119905 119905

1015840 we obtain that (120572119901119899

120572119901119899minus1

120572119901

)119860 isonly determined thus the number of 119890

119877contained119898 and119898

1015840

is 1

Lemma6 For any fixed 119890119880= 1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) 119909119894isin

119865lowast

119902 (1 le 119894 le 119899) containing a given 119890

119871 then the number of 119890

119877

which is incidence with 119890119880is 120593(119902 minus 1)

Proof For any fixed 119890119880

= 1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) 119909119894isin

119865lowast

119902 (1 le 119894 le 119899) containing a given 119890

119871 we assume that

119901119895(119909119894) = 1198861198951119909119901119899

119894+1198861198952119909119901(119899minus1)

119894+sdot sdot sdot+119886

119895119899119909119901

119894(1 le 119895 le 119896 1 le 119894 le 119899)

119890119877= 1199011(120572) 1199012(120572) 119901

119896(120572) where 120572 is a primitive element

of 119865119902 From the definitions of 119890

119877and 119890119880and Conclusion 4

we can conclude that 119890119877is incidence with 119890

119880if and only if

1199091+ 1199092+ sdot sdot sdot + 119909

119899= 120572 For any 120572 since Rank(1 1 1) =

Rank(1 1 1 120572) = 1 lt 119899 so the equation1199091+1199092+sdot sdot sdot+119909

119899=

120572 always has a solution From the proof of Theorem 3 weknow the number of 119890

119877which is incident with 119890

119880(ie the

number of all 119864119877) is [119902119896(119896minus1)2prod119899

119894=119899minus119896+1(119902119894

minus 1)] 120593(119902 minus 1)

Lemma 7 For any fixed 119890119880= 1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) 119909119894isin

119865lowast

119902 (1 le 119894 le 119899) containing a given 119890

119871and119898 = (119904 119905) the num-

ber of 119890119877which is incidence with 119890

119880and contained in119898 is 1

Proof For any 119904 isin 119878 119890119877

isin 119864119877 we assume that 119890

119877=

1199011(120572) 1199012(120572) 119901

119896(120572) where 120572 is a primitive element

of 119865119902 Similar to Lemma 6 for any fixed 119890

119880=

1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) 119909119894isin 119865lowast

119902 (1 le 119894 le 119899) containing

a given 119890119871 we have known that 119890

119877is incident with 119890

119880if and

only if

1199091+ 1199092+ sdot sdot sdot + 119909

119899= 120572 (11)

Again with 119890119877sub 119898 we can get

1199041199011(120572) + 119904

2

1199012(120572) + sdot sdot sdot + 119904

119896

119901119896(120572) = 119905 (12)

By (11) and (12) and the property of 119901119895(119909) (1 le 119895 le 119896) we

have the following conclusion

1199041199011(

119899

sum

119894=1

119909119894) + 1199042

1199012(

119899

sum

119894=1

119909119894) + sdot sdot sdot + 119904

119896

119901119896(

119899

sum

119894=1

119909119894)

= 119905 lArrrArr (1199011(

119899

sum

119894=1

119909119894) 1199012(

119899

sum

119894=1

119909119894) 119901

119896(

119899

sum

119894=1

119909119894))

6 Journal of Applied Mathematics

times (

119904

1199042

119904119896

)

= 119905 lArrrArr (

119899

sum

119894=1

1199011(119909119894)

119899

sum

119894=1

1199012(119909119894)

119899

sum

119894=1

119901119896(119909119894))(

119904

s2119904119896

)

=119905lArrrArr( (

119899

sum

119894=1

119909119894)

119899

(

119899

sum

119894=1

119909119894)

119899minus1

(

119899

sum

119894=1

119909119894))119860(

119904

1199042

119904119896

)

= 119905 lArrrArr [(

119899

sum

119894=1

1199011(119909119894)

119899

sum

119894=1

1199012(119909119894)

119899

sum

119894=1

119901119896(119909119894))

minus((

119899

sum

119894=1

119909119894)

119899

(

119899

sum

119894=1

119909119894)

119899minus1

(

119899

sum

119894=1

119909119894))119860]

times (

119904

1199042

119904119896

) = 0

(13)

because 119904 is any given Similar to the proof of Lemma 4we can get (sum

119899

119894=11199011(119909119894) sum119899

119894=11199012(119909119894) sum

119899

119894=1119901119896(119909119894)) minus

((sum119899

119894=1119909119894)119899(sum119899119894=1

119909119894)119899minus1

(sum119899

119894=1119909119894))119860=0 that is ((sum119899

119894=1119909119894)119899

(sum119899

119894=1119909119894)119899minus1

(sum119899

119894=1119909119894))119860 = (sum

119899

119894=11199011(119909119894) sum119899

119894=11199012(119909119894)

sum119899

119894=1119901119896(119909119894)) but 119901

1(119909119894) 1199012(119909119894) 119901

119896(119909119894) and 119909

119894(1 le 119894 le 119899)

also are fixed thus 120572 and 119860are only determined so thenumber of 119890

119877which is incident with 119890

119880and contained in 119898

is 1

Theorem 8 In the constructed multisender authenticationcodes if the sendersrsquo encoding rules and the receiverrsquos decodingrules are chosen according to a uniform probability distribu-tion then the largest probabilities of success for different typesof deceptions respectively are

119875119868=

1

119902119896(119896minus1)2prod119899

119894=119899minus119896+1(119902119894 minus 1)

119875119878=

1

120593 (119902 minus 1)

119875119880(119871) =

1

[119902119896(119896minus1)2prod119899

119894=119899minus119896+1(119902119894 minus 1)] 120593 (119902 minus 1)

(14)

Proof By Theorem 3 and Lemma 4 we get

119875119868= max119898isin119872

1003816100381610038161003816119890119877 isin 119864119877| 119890119877sub 119898

10038161003816100381610038161003816100381610038161003816119864119877

1003816100381610038161003816

=120593 (119902 minus 1)

[119902119896(119896minus1)2prod119899

119894=119899minus119896+1(119902119894 minus 1)] 120593 (119902 minus 1)

=1

119902119896(119896minus1)2prod119899

119894=119899minus119896+1(119902119894 minus 1)

(15)

By Lemmas 4 and 5 we get

119875119878= max119898isin119872

max1198981015840=119898isin119872

10038161003816100381610038161003816119890119877isin 119864119877| 119890119877sub 119898119898

1015840

10038161003816100381610038161003816

1003816100381610038161003816119890119877 isin 119864119877| 119890119877sub 119898

1003816100381610038161003816

=1

120593 (119902 minus 1)

(16)

By Lemmas 6 and 7 we get

119875119880(119871)

=max119890119871isin119864119871

max119890119871isin119890119880

max119898isin119872

1003816100381610038161003816119890119877 isin 119864119877| 119890119877sub119898 119901 (119890

119877 119890119875) =0

10038161003816100381610038161003816100381610038161003816119890119877 isin 119864

119877| 119901 (119890119877 119890119875) =0

1003816100381610038161003816

=1

[119902119896(119896minus1)2prod119899

119894=119899minus119896+1(119902119894 minus 1)] 120593 (119902 minus 1)

(17)

Acknowledgments

This paper is supported by the NSF of China (61179026)and the Fundamental Research of the Central Universi-ties of China Civil Aviation University of Science special(ZXH2012k003)

References

[1] E N Gilbert F J MacWilliams and N J A Sloane ldquoCodeswhich detect deceptionrdquoThe Bell System Technical Journal vol53 pp 405ndash424 1974

[2] Y Desmedt Y Frankel and M Yung ldquoMulti-receivermulti-sender network security efficient authenticated multicastfeedbackrdquo in Proceedings of the the 11th Annual Conference of theIEEEComputer andCommunications Societies (Infocom rsquo92) pp2045ndash2054 May 1992

[3] K Martin and R Safavi-Naini ldquoMultisender authenticationschemes with unconditional securityrdquo in Information and Com-munications Security vol 1334 of Lecture Notes in ComputerScience pp 130ndash143 Springer Berlin Germany 1997

[4] W Ma and X Wang ldquoSeveral new constructions of multitrasmitters authentication codesrdquo Acta Electronica Sinica vol28 no 4 pp 117ndash119 2000

[5] G J Simmons ldquoMessage authentication with arbitration oftransmitterreceiver disputesrdquo in Advances in CryptologymdashEUROCRYPT rsquo87 Workshop on theTheory and Application of ofCryptographic Techniques vol 304 of Lecture Notes in ComputerScience pp 151ndash165 Springer 1988

[6] S Cheng and L Chang ldquoTwo constructions of multi-senderauthentication codes with arbitration based linear codes to bepublished in rdquoWSEAS Transactions on Mathematics vol 11 no12 2012

Journal of Applied Mathematics 7

[7] R Safavi-Naini and H Wang ldquoNew results on multi-receiver authentication codesrdquo in Advances in CryptologymdashEUROCRYPT rsquo98 (Espoo) vol 1403 of Lecture Notes in ComputSci pp 527ndash541 Springer Berlin Germany 1998

[8] R Aparna and B B Amberker ldquoMulti-sender multi-receiverauthentication for dynamic secure group communicationrdquoInternational Journal of Computer Science andNetwork Securityvol 7 no 10 pp 47ndash63 2007

[9] R Safavi-Naini and H Wang ldquoBounds and constructions formultireceiver authentication codesrdquo inAdvances in cryptologymdashASIACRYPTrsquo98 (Beijing) vol 1514 of Lecture Notes in ComputSci pp 242ndash256 Springer Berlin Germany 1998

[10] S Shen and L Chen Information and Coding Theory Sciencepress in China 2002

[11] J J Rotman Advanced Modern Algebra High Education Pressin China 2004

[12] Z Wan Geometry of Classical Group over Finite Field SciencePress in Beijing New York NY USA 2002

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 435730 8 pageshttpdxdoiorg1011552013435730

Research ArticleGeometrical and Spectral Properties of the OrthogonalProjections of the Identity

Luis Gonzaacutelez Antonio Suaacuterez and Dolores Garciacutea

Department of Mathematics Research Institute SIANI University of Las Palmas de Gran Canaria Campus de Tafira35017 Las Palmas de Gran Canaria Spain

Correspondence should be addressed to Luis Gonzalez luisglezdmaulpgces

Received 28 December 2012 Accepted 10 April 2013

Academic Editor K Sivakumar

Copyright copy 2013 Luis Gonzalez et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We analyze the best approximation119860119873 (in the Frobenius sense) to the identitymatrix in an arbitrarymatrix subspace119860119878 (119860 isin R119899times119899

nonsingular 119878 being any fixed subspace of R119899times119899) Some new geometrical and spectral properties of the orthogonal projection 119860119873are derived In particular new inequalities for the trace and for the eigenvalues of matrix119860119873 are presented for the special case that119860119873 is symmetric and positive definite

1 Introduction

The set of all 119899 times 119899 real matrices is denoted by R119899times119899 and119868 denotes the identity matrix of order 119899 In the following119860119879 and tr(119860) denote as usual the transpose and the trace

of matrix 119860 isin R119899times119899 The notations ⟨sdot sdot⟩119865and sdot

119865stand

for the Frobenius inner product and matrix norm definedon the matrix space R119899times119899 Throughout this paper the termsorthogonality angle and cosine will be used in the sense ofthe Frobenius inner product

Our starting point is the linear system

119860119909 = 119887 119860 isin R119899times119899

119909 119887 isin R119899

(1)

where 119860 is a large nonsingular and sparse matrix Theresolution of this system is usually performed by iterativemethods based on Krylov subspaces (see eg [1 2]) Thecoefficient matrix 119860 of the system (1) is often extremelyill-conditioned and highly indefinite so that in this caseKrylov subspace methods are not competitive without agood preconditioner (see eg [2 3]) Then to improve theconvergence of these Krylov methods the system (1) canbe preconditioned with an adequate nonsingular precondi-tioning matrix 119873 transforming it into any of the equivalentsystems

119873119860119909 = 119873119887

119860119873119910 = 119887 119909 = 119873119910

(2)

the so-called left and right preconditioned systems respec-tively In this paper we address only the case of the right-handside preconditioned matrices 119860119873 but analogous results canbe obtained for the left-hand side preconditioned matrices119873119860

The preconditioning of the system (1) is often performedin order to get a preconditioned matrix 119860119873 as close aspossible to the identity in some sense and the preconditioner119873 is called an approximate inverse of119860 The closeness of119860119873to 119868may bemeasured by using a suitablematrix norm like forinstance the Frobenius norm [4] In this way the problemof obtaining the best preconditioner 119873 (with respect to theFrobenius norm) of the system (1) in an arbitrary subspace119878 of R119899times119899 is equivalent to the minimization problem see forexample [5]

min119872isin119878

119860119872 minus 119868119865= 119860119873 minus 119868

119865 (3)

The solution 119873 to the problem (3) will be referred to asthe ldquooptimalrdquo or the ldquobestrdquo approximate inverse of matrix 119860in the subspace 119878 Since matrix119860119873 is the best approximationto the identity in subspace 119860119878 it will be also referred toas the orthogonal projection of the identity matrix onto thesubspace 119860119878 Although many of the results presented in thispaper are also valid for the case thatmatrix119873 is singular fromnow on we assume that the optimal approximate inverse 119873(and thus also the orthogonal projection119860119873) is a nonsingular

2 Journal of Applied Mathematics

matrix The solution 119873 to the problem (3) has been studiedas a natural generalization of the classical Moore-Penroseinverse in [6] where it has been referred to as the 119878-Moore-Penrose inverse of matrix 119860

The main goal of this paper is to derive new geometricaland spectral properties of the best approximations119860119873 (in thesense of formula (3)) to the identity matrix Such propertiescould be used to analyze the quality and theoretical effective-ness of the optimal approximate inverse119873 as preconditionerof the system (1) However it is important to highlightthat the purpose of this paper is purely theoretical and weare not looking for immediate numerical or computationalapproaches (although our theoretical results could be poten-tially applied to the preconditioning problem) In particularthe term ldquooptimal (or best) approximate inverserdquo is used inthe sense of formula (3) and not in any other sense of thisexpression

Among the many different works dealing with practicalalgorithms that can be used to compute approximate inverseswe refer the reader to for example [4 7ndash9] and to thereferences therein In [4] the author presents an exhaustivesurvey of preconditioning techniques and in particulardescribes several algorithms for computing sparse approxi-mate inverses based on Frobenius norm minimization likefor instance the well-known SPAI and FSAI algorithms Adifferent approach (which is also focused on approximateinverses based on minimizing 119860119872 minus 119868

119865) can be found

in [7] where an iterative descent-type method is used toapproximate each column of the inverse and the iterationis done with ldquosparse matrix by sparse vectorrdquo operationsWhen the system matrix is expressed in block-partitionedform some preconditioning options are explored in [8] In[9] the idea of ldquotargetrdquo matrix is introduced in the contextof sparse approximate inverse preconditioners and the gen-eralized Frobenius norms 1198612

119865119867= tr(119861119867119861119879) (119867 symmetric

positive definite) are used for minimization purposes as analternative to the classical Frobenius norm

The last results of our work are devoted to the special casethat matrix 119860119873 is symmetric and positive definite In thissense let us recall that the cone of symmetric and positivedefinite matrices has a rich geometrical structure and in thiscontext the angle that any symmetric and positive definitematrix forms with the identity plays a very important role[10] In this paper the authors extend this geometrical pointof view and analyze the geometrical structure of the subspaceof symmetric matrices of order 119899 including the location of allorthogonal matrices not only the identity matrix

This paper has been organized as follows In Section 2we present some preliminary results required to make thepaper self-contained Sections 3 and 4 are devoted to obtainnew geometrical and spectral relations respectively for theorthogonal projections 119860119873 of the identity matrix FinallySection 5 closes the paper with its main conclusions

2 Some Preliminaries

Now we present some preliminary results concerning theorthogonal projection 119860119873 of the identity onto the matrix

subspace119860119878 sub R119899times119899 For more details about these results andfor their proofs we refer the reader to [5 6 11]

Taking advantage of the prehilbertian character of thematrix Frobenius norm the solution 119873 to the problem (3)can be obtained using the orthogonal projection theoremMore precisely the matrix product 119860119873 is the orthogonalprojection of the identity onto the subspace119860119878 and it satisfiesthe conditions stated by the following lemmas see [5 11]

Lemma 1 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

0 le 1198601198732

119865= tr (119860119873) le 119899 (4)

0 le 119860119873 minus 1198682

119865= 119899 minus tr (119860119873) le 119899 (5)

An explicit formula for matrix119873 can be obtained by ex-pressing the orthogonal projection119860119873 of the identity matrixonto the subspace 119860119878 by its expansion with respect to anorthonormal basis of 119860119878 [5] This is the idea of the followinglemma

Lemma 2 Let 119860 isin R119899times119899 be nonsingular Let 119878 be a linearsubspace of R119899times119899 of dimension 119889 and 119872

1 119872

119889 a basis of 119878

such that 1198601198721 119860119872

119889 is an orthogonal basis of 119860119878 Then

the solution119873 to the problem (3) is

119873 =

119889

sum

119894=1

tr (119860119872119894)

10038171003817100381710038171198601198721198941003817100381710038171003817

2

119865

119872119894 (6)

and the minimum (residual) Frobenius norm is

119860119873 minus 1198682

119865= 119899 minus

119889

sum

119894=1

[tr (119860119872119894)]2

10038171003817100381710038171198601198721198941003817100381710038171003817

2

119865

(7)

Let us mention two possible options both taken from [5]for choosing in practice the subspace 119878 and its correspondingbasis 119872

119894119889

119894=1 The first example consists of considering the

subspace 119878 of 119899times119899matrices with a prescribed sparsity patternthat is

119878 = 119872 isin R119899times119899

119898119894119895= 0 forall (119894 119895) notin 119870

119870 sub 1 2 119899 times 1 2 119899

(8)

Then denoting by119872119894119895 the 119899 times 119899matrix whose only nonzero

entry is 119898119894119895

= 1 a basis of subspace 119878 is clearly 119872119894119895

(119894 119895) isin 119870 and then 119860119872119894119895

(119894 119895) isin 119870 will be a basisof subspace 119860119878 (since we have assumed that matrix 119860 isnonsingular) In general this basis of 119860119878 is not orthogonalso that we only need to use the Gram-Schmidt procedureto obtain an orthogonal basis of 119860119878 in order to apply theorthogonal expansion (6)

For the second example consider a linearly independentset of 119899 times 119899 real symmetric matrices 119875

1 119875

119889 and the

corresponding subspace

1198781015840

= span 1198751119860119879

119875119889119860119879

(9)

Journal of Applied Mathematics 3

which clearly satisfies

1198781015840

sube 119872 = 119875119860119879

119875 isin R119899times119899 119875119879 = 119875

= 119872 isin R119899times119899 (119860119872)119879

= 119860119872

(10)

Hence we can explicitly obtain the solution 119873 to theproblem (3) for subspace 1198781015840 from its basis 119875

1119860119879

119875119889119860119879

as follows If 119860119875

1119860119879

119860119875119889119860119879

is an orthogonal basis ofsubspace 1198601198781015840 then we just use the orthogonal expansion (6)for obtaining119873 Otherwise we use again the Gram-Schmidtprocedure to obtain an orthogonal basis of subspace1198601198781015840 andthenwe apply formula (6)The interest of this second examplestands in the possibility of using the conjugate gradientmethod for solving the preconditioned linear system whenthe symmetric matrix 119860119873 is positive definite For a moredetailed exposition of the computational aspects related tothese two examples we refer the reader to [5]

Now we present some spectral properties of the orthog-onal projection 119860119873 From now on we denote by 120582

119894119899

119894=1and

120590119894119899

119894=1the sets of eigenvalues and singular values respectively

of matrix119860119873 arranged as usual in nonincreasing order thatis

100381610038161003816100381612058211003816100381610038161003816 ge

100381610038161003816100381612058221003816100381610038161003816 ge sdot sdot sdot ge

10038161003816100381610038161205821198991003816100381610038161003816 gt 0

1205901ge 1205902ge sdot sdot sdot ge 120590

119899gt 0

(11)

The following lemma [11] provides some inequalitiesinvolving the eigenvalues and singular values of the precon-ditioned matrix 119860119873

Lemma 3 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Then

119899

sum

119894=1

1205822

119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816

2

le

119899

sum

119894=1

1205902

119894= 119860119873

2

119865

= tr (119860119873) =

119899

sum

119894=1

120582119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816 le

119899

sum

119894=1

120590119894

(12)

The following fact [11] is a direct consequence ofLemma 3

Lemma 4 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Then the smallest singular value and thesmallest eigenvaluersquos modulus of the orthogonal projection 119860119873of the identity onto the subspace 119860119878 are never greater than 1That is

0 lt 120590119899le1003816100381610038161003816120582119899

1003816100381610038161003816 le 1 (13)

The following theorem [11] establishes a tight connectionbetween the closeness of matrix 119860119873 to the identity matrixand the closeness of 120590

119899(|120582119899|) to the unity

Theorem 5 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

(1 minus1003816100381610038161003816120582119899

1003816100381610038161003816)2

le (1 minus 120590119899)2

le 119860119873 minus 1198682

119865

le 119899 (1 minus1003816100381610038161003816120582119899

1003816100381610038161003816

2

) le 119899 (1 minus 1205902

119899)

(14)

Remark 6 Theorem 5 states that the closer the smallestsingular value 120590

119899of matrix 119860119873 is to the unity the closer

matrix 119860119873 will be to the identity that is the smaller119860119873 minus 119868

119865will be and conversely The same happens with

the smallest eigenvaluersquos modulus |120582119899| of matrix119860119873 In other

words we get a good approximate inverse 119873 of 119860 when 120590119899

(|120582119899|) is sufficiently close to 1

To finish this section let us mention that recently lowerand upper bounds on the normalized Frobenius conditionnumber of the orthogonal projection 119860119873 of the identityonto the subspace 119860119878 have been derived in [12] In additionthis work proposes a natural generalization (related to anarbitrary matrix subspace 119878 of R119899times119899) of the normalizedFrobenius condition number of the nonsingular matrix 119860

3 Geometrical Properties

In this section we present some new geometrical propertiesfor matrix 119860119873 119873 being the optimal approximate inverse ofmatrix 119860 defined by (3) Our first lemma states some basicproperties involving the cosine of the angle between matrix119860119873 and the identity that is

cos (119860119873 119868) =⟨119860119873 119868⟩

119865

119860119873119865119868119865

=tr (119860119873)

119860119873119865radic119899

(15)

Lemma 7 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

cos (119860119873 119868) =tr (119860119873)

119860119873119865radic119899

=119860119873

119865

radic119899=radictr (119860119873)

radic119899 (16)

0 le cos (119860119873 119868) le 1 (17)

119860119873 minus 1198682

119865= 119899 (1 minus cos2 (119860119873 119868)) (18)

Proof First using (15) and (4) we immediately obtain (16)As a direct consequence of (16) we derive that cos(119860119873 119868) isalways nonnegative Finally using (5) and (16) we get

119860119873 minus 1198682

119865= 119899 minus tr (119860119873) = 119899 (1 minus cos2 (119860119873 119868)) (19)

and the proof is concluded

Remark 8 In [13] the authors consider an arbitrary approxi-mate inverse119876 of matrix119860 and derive the following equality

119860119876 minus 1198682

119865= (119860119876

119865minus 119868119865)2

+ 2 (1 minus cos (119860119876 119868)) 119860119876119865119868119865

(20)

4 Journal of Applied Mathematics

that is the typical decomposition (valid in any inner productspace) of the strong convergence into the convergence ofthe norms (119860119876

119865minus 119868119865)2 and the weak convergence (1 minus

cos(119860119876 119868))119860119876119865119868119865 Note that for the special case that 119876

is the optimal approximate inverse119873 defined by (3) formula(18) has stated that the strong convergence is reduced justto the weak convergence and indeed just to the cosinecos(119860119873 119868)

Remark 9 More precisely formula (18) states that the closercos(119860119873 119868) is to the unity (ie the smaller the angle ang(119860119873 119868)

is) the smaller 119860119873 minus 119868119865will be and conversely This gives

us a new measure of the quality (in the Frobenius sense) ofthe approximate inverse 119873 of matrix 119860 by comparing theminimum residual norm 119860119873 minus 119868

119865with the cosine of the

angle between 119860119873 and the identity instead of with tr(119860119873)119860119873

119865(Lemma 1) or 120590

119899 |120582119899| (Theorem 5) So for a fixed

nonsingular matrix 119860 isin R119899times119899 and for different subspaces119878 sub R119899times119899 we have

tr (119860119873) 119899 lArrrArr 119860119873119865 radic119899 lArrrArr 120590

119899 1 lArrrArr

10038161003816100381610038161205821198991003816100381610038161003816 1

lArrrArr cos (119860119873 119868) 1 lArrrArr 119860119873 minus 119868119865 0

(21)

Obviously the optimal theoretical situation corresponds tothe case

tr (119860119873) = 119899 lArrrArr 119860119873119865= radic119899 lArrrArr 120590

119899= 1 lArrrArr 120582

119899= 1

lArrrArr cos (119860119873 119868) = 1 lArrrArr 119860119873 minus 119868119865= 0

lArrrArr 119873 = 119860minus1

lArrrArr 119860minus1

isin 119878

(22)

Remark 10 Note that the ratio between cos(119860119873 119868) andcos(119860 119868) is independent of the order 119899 of matrix 119860 Indeedassuming that tr(119860) = 0 andusing (16) we immediately obtain

cos (119860119873 119868)

cos (119860 119868)=

119860119873119865 radic119899

tr (119860) 119860119865radic119899

= 119860119873119865

119860119865

tr (119860) (23)

The following lemma compares the trace and the Frobe-nius norm of the orthogonal projection 119860119873

Lemma 11 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

tr (119860119873) le 119860119873119865lArrrArr 119860119873

119865le 1 lArrrArr tr (119860119873) le 1

lArrrArr cos (119860119873 119868) le1

radic119899

(24)

tr (119860119873) ge 119860119873119865lArrrArr 119860119873

119865ge 1 lArrrArr tr (119860119873) ge 1

lArrrArr cos (119860119873 119868) ge1

radic119899

(25)

Proof Using (4) we immediately obtain the four leftmostequivalences Using (16) we immediately obtain the tworightmost equivalences

The next lemma provides us with a relationship betweenthe Frobenius norms of the inverses of matrices119860 and its bestapproximate inverse119873 in subspace 119878

Lemma 12 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

10038171003817100381710038171003817119860minus110038171003817100381710038171003817119865

10038171003817100381710038171003817119873minus110038171003817100381710038171003817119865

ge 1 (26)

Proof Using (4) we get

119860119873119865le radic119899 = 119868

119865=10038171003817100381710038171003817(119860119873) (119860119873)

minus110038171003817100381710038171003817119865

le 119860119873119865

10038171003817100381710038171003817(119860119873)minus110038171003817100381710038171003817119865

997904rArr10038171003817100381710038171003817(119860119873)minus110038171003817100381710038171003817119865

ge 1

(27)

and hence10038171003817100381710038171003817119860minus110038171003817100381710038171003817119865

10038171003817100381710038171003817119873minus110038171003817100381710038171003817119865

ge10038171003817100381710038171003817119873minus1

119860minus110038171003817100381710038171003817119865

=10038171003817100381710038171003817(119860119873)minus110038171003817100381710038171003817119865

ge 1 (28)

and the proof is concluded

The following lemma compares the minimum residualnorm 119860119873 minus 119868

119865with the distance (with respect to the

Frobenius norm) 119860minus1 minus 119873119865between the inverse of 119860 and

the optimal approximate inverse 119873 of 119860 in any subspace119878 sub R119899times119899 First note that for any two matrices 119860 119861 isin R119899times119899

(119860 nonsingular) from the submultiplicative property of theFrobenius norm we immediately get

119860119861 minus 1198682

119865=10038171003817100381710038171003817119860 (119861 minus 119860

minus1

)10038171003817100381710038171003817

2

119865

le 1198602

119865

10038171003817100381710038171003817119861 minus 119860

minus110038171003817100381710038171003817

2

119865

(29)

However for the special case that 119861 = 119873 (the solution tothe problem (3)) we also get the following inequality

Lemma 13 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

119860119873 minus 1198682

119865le 119860

119865

10038171003817100381710038171003817119873 minus 119860

minus110038171003817100381710038171003817119865 (30)

Proof Using the Cauchy-Schwarz inequality and (5) we get10038161003816100381610038161003816⟨119860minus1

minus 119873119860119879

⟩119865

10038161003816100381610038161003816le10038171003817100381710038171003817119860minus1

minus 11987310038171003817100381710038171003817119865

1003817100381710038171003817100381711986011987910038171003817100381710038171003817119865

997904rArr10038161003816100381610038161003816tr ((119860minus1 minus 119873)119860)

10038161003816100381610038161003816le10038171003817100381710038171003817119860minus1

minus 11987310038171003817100381710038171003817119865

1003817100381710038171003817100381711986011987910038171003817100381710038171003817119865

997904rArr10038161003816100381610038161003816tr (119860 (119860

minus1

minus 119873))10038161003816100381610038161003816le10038171003817100381710038171003817119873 minus Aminus110038171003817100381710038171003817119865119860119865

997904rArr |tr (119868 minus 119860119873)| le10038171003817100381710038171003817119873 minus 119860

minus110038171003817100381710038171003817119865119860119865

997904rArr 119899 minus tr (119860119873) le10038171003817100381710038171003817119873 minus 119860

minus110038171003817100381710038171003817119865119860119865

997904rArr 119860119873 minus 1198682

119865le 119860

119865

10038171003817100381710038171003817119873 minus 119860

minus110038171003817100381710038171003817119865

(31)

Journal of Applied Mathematics 5

The following extension of the Cauchy-Schwarz inequal-ity in a real or complex inner product space (119867 ⟨sdot sdot⟩) wasobtained by Buzano [14] For all 119886 119909 119887 isin 119867 we have

|⟨119886 119909⟩ sdot ⟨119909 119887⟩| le1

2(119886 119887 + |⟨119886 119887⟩|) 119909

2

(32)

Thenext lemmaprovides uswith lower and upper boundson the inner product ⟨119860119873 119861⟩

119865 for any 119899 times 119899 real matrix 119861

Lemma 14 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then for every 119861 isin R119899times119899 we have

1003816100381610038161003816⟨119860119873 119861⟩119865

1003816100381610038161003816 le1

2(radic119899119861

119865+ |tr (119861)|) (33)

Proof Using (32) for 119886 = 119868 119909 = 119860119873 119887 = 119861 and (4) we get

1003816100381610038161003816⟨119868 119860119873⟩119865sdot ⟨119860119873 119861⟩

119865

1003816100381610038161003816 le1

2(119868119865119861119865+1003816100381610038161003816⟨119868 119861⟩119865

1003816100381610038161003816) 1198601198732

119865

997904rArr1003816100381610038161003816tr (119860119873) sdot ⟨119860119873 119861⟩

119865

1003816100381610038161003816

le1

2(radic119899119861

119865+ |tr (119861)|) 119860119873

2

119865

997904rArr1003816100381610038161003816⟨119860119873 119861⟩

119865

1003816100381610038161003816 le1

2(radic119899119861

119865+ |tr (119861)|)

(34)

The next lemma provides an upper bound on the arith-metic mean of the squares of the 1198992 terms in the orthogonalprojection 119860119873 By the way it also provides us with an upperbound on the arithmetic mean of the 119899 diagonal terms inthe orthogonal projection 119860119873 These upper bounds (validfor any matrix subspace 119878) are independent of the optimalapproximate inverse119873 and thus they are independent of thesubspace 119878 and only depend on matrix 119860

Lemma 15 Let 119860 isin R119899times119899 be nonsingular with tr(119860) = 0 andlet 119878 be a linear subspace of R119899times119899 Let119873 be the solution to theproblem (3) Then

1198601198732

119865

1198992le

1198602

119865

[tr (119860)]2

tr (119860119873)

119899le

1198991198602

119865

[tr (119860)]2

(35)

Proof Using (32) for 119886 = 119860119879 119909 = 119868 and 119887 = 119860119873 and the

Cauchy-Schwarz inequality for ⟨119860119879 119860119873⟩119865and (4) we get

10038161003816100381610038161003816⟨119860119879

119868⟩119865

sdot ⟨119868 119860119873⟩119865

10038161003816100381610038161003816

le1

2(1003817100381710038171003817100381711986011987910038171003817100381710038171003817119865119860119873

119865+10038161003816100381610038161003816⟨119860119879

119860119873⟩119865

10038161003816100381610038161003816) 1198682

119865

997904rArr |tr (119860) sdot tr (119860119873)|

le119899

2(119860119865119860119873

119865+10038161003816100381610038161003816⟨119860119879

119860119873⟩119865

10038161003816100381610038161003816)

le119899

2(119860119865119860119873

119865+1003817100381710038171003817100381711986011987910038171003817100381710038171003817119865119860119873

119865)

= 119899119860119865119860119873

119865

997904rArr |tr (119860)| 1198601198732

119865le 119899119860

119865119860119873

119865

997904rArr119860119873

119865

119899le

119860119865

|tr (119860)|

997904rArr119860119873

2

119865

1198992le

1198602

119865

[tr (119860)]2997904rArr

tr (119860119873)

119899le

1198991198602

119865

[tr (119860)]2

(36)

Remark 16 Lemma 15 has the following interpretation interms of the quality of the optimal approximate inverse119873 ofmatrix 119860 in subspace 119878 The closer the ratio 119899119860

119865| tr(119860)|

is to zero the smaller tr(119860119873) will be and thus due to (5)the larger 119860119873 minus 119868

119865will be and this happens for any matrix

subspace 119878

Remark 17 By the way from Lemma 15 we obtain thefollowing inequality for any nonsingular matrix 119860 isin R119899times119899Consider any matrix subspace 119878 st 119860minus1 isin 119878 Then119873 = 119860

minus1and using Lemma 15 we get

1198601198732

119865

1198992=1198682

119865

1198992=1

nle

1198602

119865

[tr (119860)]2

997904rArr |tr (119860)| le radic119899119860119865

(37)

4 Spectral Properties

In this section we present some new spectral propertiesfor matrix 119860119873 119873 being the optimal approximate inverseof matrix 119860 defined by (3) Mainly we focus on the casethat matrix 119860119873 is symmetric and positive definite This hasbeenmotivated by the following reasonWhen solving a largenonsymmetric linear system (1) by using Krylov methodsa possible strategy consists of searching for an adequateoptimal preconditioner 119873 such that the preconditionedmatrix119860119873 is symmetric positive definite [5]This enables oneto use the conjugate gradientmethod (CG-method) which isin general a computationally efficient method for solving thenew preconditioned system [2 15]

Our starting point is Lemma 3 which has established thatthe sets of eigenvalues and singular values of any orthogonalprojection 119860119873 satisfy

119899

sum

119894=1

1205822

119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816

2

le

119899

sum

119894=1

1205902

119894

=

119899

sum

119894=1

120582119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816 le

119899

sum

119894=1

120590119894

(38)

Let us particularize (38) for some special cases

6 Journal of Applied Mathematics

First note that if 119860119873 is normal (ie for all 1 le 119894 le 119899120590119894= |120582119894| [16]) then (38) becomes

119899

sum

119894=1

1205822

119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816

2

=

119899

sum

119894=1

1205902

119894

=

119899

sum

119894=1

120582119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816 =

119899

sum

119894=1

120590119894

(39)

In particular if 119860119873 is symmetric (120590119894= |120582119894| = plusmn120582

119894isin R) then

(38) becomes119899

sum

119894=1

1205822

119894=

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816

2

=

119899

sum

119894=1

1205902

119894

=

119899

sum

119894=1

120582119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816 =

119899

sum

119894=1

120590119894

(40)

In particular if 119860119873 is symmetric and positive definite (120590119894=

|120582119894| = 120582119894isin R+) then the equality holds in all (38) that is

119899

sum

119894=1

1205822

119894=

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816

2

=

119899

sum

119894=1

1205902

119894

=

119899

sum

119894=1

120582119894=

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816 =

119899

sum

119894=1

120590119894

(41)

The next lemma compares the traces of matrices 119860119873 and(119860119873)2

Lemma 18 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

(i) for any orthogonal projection 119860119873

tr ((119860119873)2

) le 1198601198732

119865= tr (119860119873) (42)

(ii) for any symmetric orthogonal projection 11986011987310038171003817100381710038171003817(119860119873)210038171003817100381710038171003817119865

le tr ((119860119873)2

) = 1198601198732

119865= tr (119860119873) (43)

(iii) for any symmetric positive definite orthogonal projec-tion 119860119873

10038171003817100381710038171003817(119860119873)210038171003817100381710038171003817119865

le tr ((119860119873)2

) = 1198601198732

119865= tr (119860119873) le [tr (119860119873)]

2

(44)

Proof (i) Using (38) we get119899

sum

119894=1

1205822

119894le

119899

sum

119894=1

1205902

119894=

119899

sum

119894=1

120582119894 (45)

(ii) It suffices to use the obvious fact that (119860119873)2

119865

le

1198601198732

119865and the following equalities taken from (40)

119899

sum

119894=1

1205822

119894=

119899

sum

119894=1

1205902

119894=

119899

sum

119894=1

120582119894 (46)

(iii) It suffices to use (43) and the fact that (see eg [1718]) if 119875 and 119876 are symmetric positive definite matrices thentr(119875119876) le tr(119875) tr(119876) for 119875 = 119876 = 119860119873

The rest of the paper is devoted to obtain new propertiesabout the eigenvalues of the orthogonal projection119860119873 for thespecial case that this matrix is symmetric positive definite

First let us recall that the smallest singular value and thesmallest eigenvaluersquos modulus of the orthogonal projection119860119873 are never greater than 1 (see Lemma 4) The followingtheorem establishes the dual result for the largest eigenvalueof matrix 119860119873 (symmetric positive definite)

Theorem 19 Let 119860 isin R119899times119899 be nonsingular and let 119878 be alinear subspace of R119899times119899 Let 119873 be the solution to the problem(3) Suppose thatmatrix119860119873 is symmetric and positive definiteThen the largest eigenvalue of the orthogonal projection119860119873 ofthe identity onto the subspace 119860119878 is never less than 1 That is

1205901= 1205821ge 1 (47)

Proof Using (41) we get

119899

sum

119894=1

1205822

119894=

119899

sum

119894=1

120582119894997904rArr 1205822

119899minus 120582119899=

119899minus1

sum

119894=1

(120582119894minus 1205822

119894) (48)

Now since 120582119899le 1 (Lemma 4) then 1205822

119899minus 120582119899le 0 This implies

that at least one summand in the rightmost sum in (48) mustbe less than or equal to zero Suppose that such summand isthe 119896th one (1 le 119896 le 119899 minus 1) Since 119860119873 is positive definitethen 120582

119896gt 0 and thus

120582119896minus 1205822

119896le 0 997904rArr 120582

119896le 1205822

119896997904rArr 120582

119896ge 1 997904rArr 120582

1ge 1 (49)

and the proof is concluded

InTheorem 19 the assumption that matrix119860119873 is positivedefinite is essential for assuring that |120582

1| ge 1 as the following

simple counterexample shows Moreover from Lemma 4 andTheorem 19 respectively we have that the smallest and largesteigenvalues of 119860119873 (symmetric positive definite) satisfy 120582

119899le

1 and 1205821ge 1 respectively Nothing can be asserted about

the remaining eigenvalues of the symmetric positive definitematrix 119860119873 which can be greater than equal to or less thanthe unity as the same counterexample also shows

Example 20 For 119899 = 3 let

119860119896= (

3 0 0

0 119896 0

0 0 1

) 119896 isin R (50)

let 1198683be identitymatrix of order 3 and let 119878 be the subspace of

all 3times3 scalarmatrices that is 119878 = span1198683Then the solution

119873119896to the problem (3) for subspace 119878 can be immediately

obtained by using formula (6) as follows

119873119896=tr (119860119896)

10038171003817100381710038171198601198961003817100381710038171003817

2

119865

1198683=

119896 + 4

1198962 + 101198683 (51)

and then we get

119860119896119873119896=

119896 + 4

1198962 + 10119860119896=

119896 + 4

1198962 + 10(

3 0 0

0 119896 0

0 0 1

) (52)

Journal of Applied Mathematics 7

Let us arrange the eigenvalues and singular values ofmatrix 119860

119896119873119896 as usual in nonincreasing order (as shown in

(11))On one hand for 119896 = minus2 we have

119860minus2119873minus2

=1

7(

3 0 0

0 minus2 0

0 0 1

) (53)

and then

1205901=10038161003816100381610038161205821

1003816100381610038161003816 =3

7

ge 1205902=10038161003816100381610038161205822

1003816100381610038161003816 =2

7

ge 1205903=10038161003816100381610038161205823

1003816100381610038161003816 =1

7

(54)

Hence 119860minus2119873minus2

is indefinite and 1205901= |1205821| = 37 lt 1

On the other hand for 1 lt 119896 lt 3 we have (see matrix(52))

1205901= 1205821= 3

119896 + 4

1198962 + 10

ge 1205902= 1205822= 119896

119896 + 4

1198962 + 10

ge 1205903= 1205823=

119896 + 4

1198962 + 10

(55)

and then

119896 = 2 1205901= 1205821=9

7gt 1

1205902= 1205822=6

7lt 1

1205903= 1205823=3

7lt 1

119896 =5

2 1205901= 1205821=6

5gt 1

1205902= 1205822= 1

1205903= 1205823=2

5lt 1

119896 =8

3 1205901= 1205821=90

77gt 1

1205902= 1205822=80

77gt 1

1205903= 1205823=30

77lt 1

(56)

Hence for 119860119896119873119896positive definite we have (depending on 119896)

1205822lt 1 120582

2= 1 or 120582

2gt 1

The following corollary improves the lower bound zeroon both tr(119860119873) given in (4) and cos(119860119873 119868) given in (17)

Corollary 21 Let 119860 isin R119899times119899 be nonsingular and let 119878

be a linear subspace of R119899times119899 Let 119873 be the solution to the

problem (3) Suppose thatmatrix119860119873 is symmetric and positivedefinite Then

1 le 119860119873119865le tr (119860119873) = 119860119873

2

119865le 119899 (57)

cos (119860119873 119868) ge1

radic119899 (58)

Proof Denote by sdot 2the spectral norm Using the well-

known inequality sdot 2le sdot

119865[19] Theorem 19 and (4) we

get

119860119873119865ge 119860119873

2= 1205901= 1205821ge 1

997904rArr 1 le 119860119873119865le tr (119860119873) = 119860119873

2

119865le 119899

(59)

Finally (58) follows immediately from (57) and (25)

Let usmention that an upper bound on all the eigenvaluesmoduli and on all singular values of any orthogonal projec-tion 119860119873 can be immediately obtained from (38) and (4) asfollows

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816

2

le

119899

sum

119894=1

1205902

119894= 119860119873

2

119865le 119899

997904rArr1003816100381610038161003816120582119894

1003816100381610038161003816 120590119894le radic119899 forall119894 = 1 2 119899

(60)

Our last theorem improves the upper bound given in(60) for the special case that the orthogonal projection 119860119873

is symmetric positive definite

Theorem 22 Let 119860 isin R119899times119899 be nonsingular and let 119878 be alinear subspace of R119899times119899 Let 119873 be the solution to the problem(3) Suppose thatmatrix119860119873 is symmetric and positive definiteThen all the eigenvalues of matrix 119860119873 satisfy

120590119894= 120582119894le1 + radic119899

2forall119894 = 1 2 119899 (61)

Proof First note that the assertion is obvious for the smallestsingular value since |120582

119899| le 1 for any orthogonal projection

119860119873 (Lemma 4) For any eigenvalue of 119860119873 we use the factthat 119909 minus 119909

2

le 14 for all 119909 gt 0 Then from (41) we get119899

sum

119894=1

1205822

119894=

119899

sum

119894=1

120582119894

997904rArr 1205822

1minus 1205821=

119899

sum

119894=2

(120582119894minus 1205822

119894) le

119899 minus 1

4

997904rArr 1205821le1 + radic119899

2997904rArr 120582

119894le1 + radic119899

2

forall119894 = 1 2 119899

(62)

5 Conclusion

In this paper we have considered the orthogonal projection119860119873 (in the Frobenius sense) of the identity matrix onto an

8 Journal of Applied Mathematics

arbitrary matrix subspace 119860119878 (119860 isin R119899times119899 nonsingular 119878 sub

R119899times119899) Among other geometrical properties of matrix 119860119873we have established a strong relation between the qualityof the approximation 119860119873 asymp 119868 and the cosine of the angleang(119860119873 119868) Also the distance between119860119873 and the identity hasbeen related to the ratio 119899119860

119865| tr(119860)| (which is independent

of the subspace 119878) The spectral analysis has provided lowerand upper bounds on the largest eigenvalue of the symmetricpositive definite orthogonal projections of the identity

Acknowledgments

Theauthors are grateful to the anonymous referee for valuablecomments and suggestions which have improved the earlierversion of this paperThisworkwas partially supported by theldquoMinisterio de Economıa y Competitividadrdquo (Spanish Gov-ernment) and FEDER through Grant Contract CGL2011-29396-C03-01

References

[1] O Axelsson Iterative Solution Methods Cambridge UniversityPress Cambridge Mass USA 1994

[2] Y Saad Iterative Methods for Sparse Linear Systems PWS Pub-lishing Boston Mass USA 1996

[3] S-L Wu F Chen and X-Q Niu ldquoA note on the eigenvalueanalysis of the SIMPLE preconditioning for incompressibleflowrdquo Journal of Applied Mathematics vol 2012 Article ID564132 7 pages 2012

[4] M Benzi ldquoPreconditioning techniques for large linear systemsa surveyrdquo Journal of Computational Physics vol 182 no 2 pp418ndash477 2002

[5] G Montero L Gonzalez E Florez M D Garcıa and ASuarez ldquoApproximate inverse computation using Frobenius in-ner productrdquoNumerical Linear Algebra with Applications vol 9no 3 pp 239ndash247 2002

[6] A Suarez and L Gonzalez ldquoA generalization of the Moore-Penrose inverse related to matrix subspaces of 119862119899times119898rdquo AppliedMathematics andComputation vol 216 no 2 pp 514ndash522 2010

[7] E Chow and Y Saad ldquoApproximate inverse preconditioners viasparse-sparse iterationsrdquo SIAM Journal on Scientific Computingvol 19 no 3 pp 995ndash1023 1998

[8] E Chow and Y Saad ldquoApproximate inverse techniques forblock-partitioned matricesrdquo SIAM Journal on Scientific Com-puting vol 18 no 6 pp 1657ndash1675 1997

[9] R M Holland A J Wathen and G J Shaw ldquoSparse approxi-mate inverses and target matricesrdquo SIAM Journal on ScientificComputing vol 26 no 3 pp 1000ndash1011 2005

[10] R Andreani M Raydan and P Tarazaga ldquoOn the geometricalstructure of symmetric matricesrdquo Linear Algebra and Its Appli-cations vol 438 no 3 pp 1201ndash1214 2013

[11] L Gonzalez ldquoOrthogonal projections of the identity spectralanalysis and applications to approximate inverse precondition-ingrdquo SIAM Review vol 48 no 1 pp 66ndash75 2006

[12] A Suarez and L Gonzalez ldquoNormalized Frobenius conditionnumber of the orthogonal projections of the identityrdquo Journalof Mathematical Analysis and Applications vol 400 no 2 pp510ndash516 2013

[13] J-P Chehab and M Raydan ldquoGeometrical properties of theFrobenius condition number for positive definite matricesrdquo

Linear Algebra and Its Applications vol 429 no 8-9 pp 2089ndash2097 2008

[14] M L Buzano ldquoGeneralizzazione della diseguaglianza di Cau-chy-Schwarzrdquo Rendiconti del Seminario Matematico Universitae Politecnico di Torino vol 31 pp 405ndash409 1974 (Italian)

[15] M R Hestenes and E Stiefel ldquoMethods of conjugate gradientsfor solving linear systemsrdquo Journal of Research of the NationalBureau of Standards vol 49 no 6 pp 409ndash436 1952

[16] R A Horn and C R Johnson Topics in Matrix Analysis Cam-bridge University Press Cambridge UK 1991

[17] E H Lieb andWThirring ldquoInequalities for themoments of theeigenvalues of the Schrodinger Hamiltonian and their relationto Sobolev inequalitiesrdquo in Studies in Mathematical PhysicsEssays in Honor of Valentine Bargmann E Lieb B Simon andA Wightman Eds pp 269ndash303 Princeton University PressPrinceton NJ USA 1976

[18] Z Ulukok and R Turkmen ldquoOn some matrix trace inequali-tiesrdquo Journal of Inequalities and Applications vol 2010 ArticleID 201486 8 pages 2010

[19] R A Horn andC R JohnsonMatrix Analysis CambridgeUni-versity Press Cambridge UK 1985

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 489295 6 pageshttpdxdoiorg1011552013489295

Research ArticleA Parameterized Splitting Preconditioner for GeneralizedSaddle Point Problems

Wei-Hua Luo12 and Ting-Zhu Huang1

1 School of Mathematical Sciences University of Electronic Science and Technology of China Chengdu Sichuan 611731 China2 Key Laboratory of Numerical Simulation of Sichuan Province University Neijiang Normal University Neijiang Sichuan 641112 China

Correspondence should be addressed to Ting-Zhu Huang tingzhuhuang126com

Received 4 January 2013 Revised 31 March 2013 Accepted 1 April 2013

Academic Editor P N Shivakumar

Copyright copy 2013 W-H Luo and T-Z Huang This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

By using Sherman-Morrison-Woodbury formula we introduce a preconditioner based on parameterized splitting idea forgeneralized saddle point problems which may be singular and nonsymmetric By analyzing the eigenvalues of the preconditionedmatrix we find that when 120572 is big enough it has an eigenvalue at 1 with multiplicity at least n and the remaining eigenvalues are alllocated in a unit circle centered at 1 Particularly when the preconditioner is used in general saddle point problems it guaranteeseigenvalue at 1 with the same multiplicity and the remaining eigenvalues will tend to 1 as the parameter 120572 rarr 0 Consequently thiscan lead to a good convergence when some GMRES iterative methods are used in Krylov subspace Numerical results of Stokesproblems and Oseen problems are presented to illustrate the behavior of the preconditioner

1 Introduction

In some scientific and engineering applications such as finiteelement methods for solving partial differential equations [12] and computational fluid dynamics [3 4] we often considersolutions of the generalized saddle point problems of the form

(119860 119861119879

minus119861 119862)(

119909

119910) = (

119891

119892) (1)

where 119860 isin R119899times119899 119861 isin R119898times119899 (119898 le 119899) and 119862 isin R119898times119898

are positive semidefinite 119909 119891 isin R119899 and 119910 119892 isin R119898 When119862 = 0 (1) is a general saddle point problem which is also aresearching object for many authors

It is well known that when the matrices 119860 119861 and 119862

are large and sparse the iterative methods are more efficientand attractive than direct methods assuming that (1) has agood preconditioner In recent years a lot of preconditioningtechniques have arisen for solving linear system for exampleSaad [5] and Chen [6] have comprehensively surveyed someclassical preconditioning techniques including ILU pre-conditioner triangular preconditioner SPAI preconditionermultilevel recursive Schur complements preconditioner and

sparse wavelet preconditioner Particularly many precondi-tioning methods for saddle problems have been presentedrecently such as dimensional splitting (DS) [7] relaxeddimensional factorization (RDF) [8] splitting preconditioner[9] and Hermitian and skew-Hermitian splitting precondi-tioner [10]

Among these results Cao et al [9] have used splitting ideato give a preconditioner for saddle point problems where thematrix 119860 is symmetric and positive definite and 119861 is of fullrow rank According to his preconditioner the eigenvaluesof the preconditioned matrix would tend to 1 when theparameter 119905 rarr infin Consequently just as we have seen fromthose examples of [9] preconditioner has guaranteed a goodconvergence when some iterative methods were used

In this paper being motivated by [9] we use the splittingidea to present a preconditioner for the system (1) where 119860may be nonsymmetric and singular (when rank(119861) lt 119898) Wefind that when the parameter is big enough the precondi-tioned matrix has the eigenvalue at 1 with multiplicity at least119899 and the remaining eigenvalues are all located in a unit circlecentered at 1 Particularly when the precondidtioner is usedin some general saddle point problems (namely 119862 = 0) we

2 Journal of Applied Mathematics

see that the multiplicity of the eigenvalue at 1 is also at least 119899but the remaining eigenvalues will tend to 1 as the parameter120572 rarr 0

The remainder of the paper is organized as followsIn Section 2 we present our preconditioner based on thesplitting idea and analyze the bound of eigenvalues of thepreconditioned matrix In Section 3 we use some numericalexamples to show the behavior of the new preconditionerFinally we draw some conclusions and outline our futurework in Section 4

2 A Parameterized Splitting Preconditioner

Now we consider using splitting idea with a variable param-eter to present a preconditioner for the system (1)

Firstly it is evident that when 120572 = 0 the system (1) isequivalent to

(

119860 119861119879

minus119861

120572

119862

120572

)(119909

119910) = (

119891

119892

120572

) (2)

Let

119872 = (

119860 119861119879

minus119861

120572119868

) 119873 = (

0 0

0 119868 minus119862

120572

)

119883 = (119909

119910) 119865 = (

119891

119892

120572

)

(3)

Then the coefficient matrix of (2) can be expressed by119872minus119873Multiplying both sides of system (2) from the left with matrix119872minus1 we have

(119868 minus119872minus1

119873)119883 = 119872minus1

119865 (4)

Hence we obtain a preconditioned linear system from (1)using the idea of splitting and the corresponding precondi-tioner is

119867 = (

119860 119861119879

minus119861

120572119868

)

minus1

(

119868 0

0119868

120572

) (5)

Nowwe analyze the eigenvalues of the preconditioned system(4)

Theorem 1 The preconditioned matrix 119868 minus 119872minus1

119873 has aneigenvalue at 1 with multiplicity at least 119899 The remainingeigenvalues 120582 satisfy

120582 =1199041+ 1199042

1199041+ 120572

(6)

where 1199041= 120596119879

119861119860minus1

119861119879

120596 1199042= 120596119879

119862120596 and 120596 isin C119898 satisfies

(119861119860minus1

119861119879

120572+119862

120572)120596 = 120582(119868 +

119861119860minus1

119861119879

120572)120596 120596 = 1 (7)

Proof Because

119872minus1

=(

(119860 +119861119879

119861

120572)

minus1

minus119860minus1

119861119879

(119868 +119861119860minus1

119861119879

120572)

minus1

119861

120572(119860 +

119861119879

119861

120572)

minus1

(119868 +119861119860minus1

119861119879

120572)

minus1)

(8)

we can easily get

119868 minus119872minus1

119873 =(

119868 119860minus1

119861119879

(119868 +119861119860minus1

119861119879

120572)

minus1

(119868 minus119862

120572)

0 (119868 +119861119860minus1

119861119879

120572)

minus1

(119861119860minus1

119861119879

120572+119862

120572)

)

(9)

which implies the preconditioned matrix 119868 minus 119872minus1119873 has aneigenvalue at 1 with multiplicity at least 119899

For the remaining eigenvalues let

(119868 +119861119860minus1

119861119879

120572)

minus1

(119861119860minus1

119861119879

120572+119862

120572)120596 = 120582120596 (10)

with 120596 = 1 then we have

(119861119860minus1

119861119879

120572+119862

120572)120596 = 120582(119868 +

119861119860minus1

119861119879

120572)120596 (11)

By multiplying both sides of this equality from the left with120596119879 we can get

120582 =1199041+ 1199042

1199041+ 120572

(12)

This completes the proof of Theorem 1

Remark 2 FromTheorem 1 we can get that when parameter120572 is big enough the modulus of nonnil eigenvalues 120582 will belocated in interval (0 1)

Remark 3 InTheorem 1 if the matrix 119862 = 0 then for nonnileigenvalues we have

lim120572rarr0

120582 = 1 (13)

Figures 1 2 and 3 are the eigenvalues plots of the pre-conditioned matrices obtained with our preconditioner Aswe can see in the following numerical experiments thisgood phenomenon is useful for accelerating convergence ofiterative methods in Krylov subspace

Additionally for the purpose of practically executing ourpreconditioner

119867 =(

(119860 +119861119879

119861

120572)

minus1

minus119860minus1

119861119879

120572(119868 +

119861119860minus1

119861119879

120572)

minus1

119861

120572(119860 +

119861119879

119861

120572)

minus1

1

120572(119868 +

119861119860minus1

119861119879

120572)

minus1)

(14)

Journal of Applied Mathematics 3

0 02 04 06 08 12 141

0

1

2

3

4

minus1

minus2

minus3

times10minus4

= 01 120572 = 00001

(a)

0

002

004

006

008

0 02 04 06 08 12 141minus008

minus006

minus004

minus002

= 01 120572 = norm(119862 fro)20

(b)

Figure 1 Spectrum of the preconditioned steady Oseen matrix with viscosity coefficient V = 01 32 times 32 grid (a) Q2-Q1 FEM 119862 = 0 (b)Q1-P0 FEM 119862 = 0

0

0 02 04 06 08 121

02

04

06

08

1

minus02

minus04

minus06

minus08

minus1

minus02

times10minus3

= 001 120572 = 00001

(a)

0 02 04 06 08 12 141minus02

minus015

minus01

minus005

0

005

01

015

02 = 001 120572 = norm(119862 fro)20

(b)

Figure 2 Spectrum of the preconditioned steady Oseen matrix with viscosity coefficient V = 001 32 times 32 grid (a) Q2-Q1 FEM 119862 = 0 (b)Q1-P0 FEM 119862 = 0

0

2

4

6

8

0 02 04 06 08 12 141

times10minus4

minus8

minus6

minus4

minus2

= 0001 120572 = 00001

(a)

0

002

004

006

008

minus008

minus006

minus004

minus002

0 02 04 06 08 121minus02

= 0001 120572 = norm(119862 fro)20

(b)

Figure 3 Spectrum of the preconditioned steady Oseen matrix with viscosity coefficient V = 0001 32 times 32 grid (a) Q2-Q1 FEM 119862 = 0 (b)Q1-P0 FEM 119862 = 0

4 Journal of Applied Mathematics

Table 1 Preconditioned GMRES(20) on Stokes problems withdifferent grid sizes (uniform grids)

Grid 120572 its LU time its time Total time16 times 16 00001 4 00457 00267 0072432 times 32 00001 6 03912 00813 0472564 times 64 00001 9 54472 05519 59991128 times 128 00001 14 914698 57732 972430

Table 2 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 01

Grid 120572 its LU time its time Total time16 times 16 00001 3 00479 00244 0072332 times 32 00001 3 06312 00696 0700964 times 64 00001 4 132959 03811 136770128 times 128 00001 6 1305463 37727 1343190

Table 3 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 001

Grid 120572 its LU time its time Total time16 times 16 00001 2 00442 00236 0067832 times 32 00001 3 04160 00557 0471764 times 64 00001 3 73645 02623 76268128 times 128 00001 4 1694009 31709 1725718

Table 4 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 0001

Grid 120572 its LU time its time Total time16 times 16 00001 2 00481 00206 0068732 times 32 00001 3 04661 00585 0524664 times 64 00001 3 66240 02728 68969128 times 128 00001 4 1773130 29814 1802944

Table 5 Preconditioned GMRES(20) on Stokes problems withdifferent grid sizes (uniform grids)

Grid 120572 its LU time its time Total time16 times 16 10000 5 00471 00272 0074332 times 32 10000 7 03914 00906 0482064 times 64 10000 9 56107 04145 60252128 times 128 10000 14 927154 48908 976062

Table 6 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 01

Grid 120572 its LU time its time Total time16 times 16 10000 4 00458 00223 0068132 times 32 10000 4 05748 00670 0641864 times 64 10000 5 122642 04179 126821128 times 128 10000 7 1282758 16275 1299033

Table 7 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 001

Grid 120572 its LU time its time Total time16 times 16 10000 3 00458 00224 0068232 times 32 10000 3 04309 00451 0476064 times 64 10000 4 76537 01712 78249128 times 128 10000 5 1751587 34554 1786141

Table 8 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 0001

Grid 120572 its LU time its time Total time16 times 16 10000 3 00507 00212 0071932 times 32 10000 3 04735 00449 0518464 times 64 10000 4 66482 01645 68127128 times 128 10000 4 1720516 21216 1741732

Table 9 Preconditioned GMRES(20) on Stokes problems withdifferent grid sizes (uniform grids)

Grid 120572 its LU time its time Total time16 times 16 norm(119862 fro) 21 00240 00473 0071332 times 32 norm(119862 fro) 20 00890 01497 0238764 times 64 norm(119862 fro) 20 08387 06191 14578128 times 128 norm(119862 fro) 20 68917 30866 99783

Table 10 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 01

Grid 120572 its LU time its time Total time16 times 16 norm(119862 fro)20 10 00250 00302 0055232 times 32 norm(119862 fro)20 10 00816 00832 0164864 times 64 norm(119862 fro)20 12 08466 03648 12114128 times 128 norm(119862 fro)20 14 69019 20398 89417

Table 11 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 001

Grid 120572 its LU time its time Total time16 times 16 norm(119862 fro)20 7 00238 00286 0052432 times 32 norm(119862 fro)20 7 00850 00552 0140264 times 64 norm(119862 fro)20 11 08400 03177 11577128 times 128 norm(119862 fro)20 16 69537 22306 91844

Table 12 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 0001

Grid 120572 its LU time its time Total time16 times 16 norm(119862 fro)20 5 00245 00250 0049532 times 32 norm(119862 fro)20 5 00905 00587 0149264 times 64 norm(119862 fro)20 8 12916 03200 16116128 times 128 norm(119862 fro)20 15 104399 27468 131867

Journal of Applied Mathematics 5

Table 13 Size and number of non-nil elements of the coefficient matrix generalized by using Q2-Q1 FEM in steady Stokes problems

Grid dim(119860) nnz(119860) dim(119861) nnz(119861) nnz(119860 + (1120572)119861119879119861)16 times 16 578 times 578 6178 81 times 578 2318 3690432 times 32 2178 times 2178 28418 289 times 2178 10460 19322064 times 64 8450 times 8450 122206 1089 times 8450 44314 875966128 times 128 33282 times 33282 506376 4225 times 33282 184316 3741110

we should efficiently deal with the computation of (119868 +(119861119860minus1

119861119879

120572))minus1 This can been tackled by the well-known

Sherman-Morrison-Woodbury formula

(1198601+ 11988311198602119883119879

2)minus1

= 119860minus1

1minus 119860minus1

11198831(119860minus1

2+ 119883119879

2119860minus1

11198831)minus1

119883119879

2119860minus1

1

(15)

where 1198601isin R119899times119899 and 119860

2isin R1199031times1199031 are invertible matrices

1198831isin R119899times1199031 and 119883

2isin R119899times1199031 are any matrices and 119899 119903

1are

any positive integersFrom (15) we immediately get

(119868 +119861119860minus1

119861119879

120572)

minus1

= 119868 minus 119861(120572119860 + 119861119879

119861)minus1

119861119879

(16)

In the following numerical examples we will always use (16)to compute (119868 + (119861119860minus1119861119879120572))minus1 in (14)

3 Numerical Examples

In this sectionwe give numerical experiments to illustrate thebehavior of our preconditioner The numerical experimentsare done by using MATLAB 71 The linear systems areobtained by using finite element methods in the Stokes prob-lems and steady Oseen problems and they are respectivelythe cases of

(1) 119862 = 0 which is caused by using Q2-Q1 FEM(2) 119862 = 0 which is caused by using Q1-P0 FEM

Furthermore we compare our preconditionerwith that of[9] in the case of general saddle point problems (namely 119862 =0) For the general saddle point problem [9] has presentedthe preconditioner

= (

119860 + 119905119861119879

119861 0

minus2119861119868

119905

)

minus1

(17)

with 119905 as a parameter and has proved that when 119860 issymmetric positive definite the preconditioned matrix hasan eigenvalue 1 with multiplicity at 119899 and the remainingeigenvalues satisfy

120582 =1199051205902

119894

1 + 1199051205902

119894

(18)

lim119905rarrinfin

120582 = 1 (19)

where 120590119894 119894 = 1 2 119898 are119898 positive singular values of the

matrix 119861119860minus12All these systems can be generalized by

using IFISS software package [11] (this is a freepackage that can be downloaded from the sitehttpwwwmathsmanchesteracuksimdjsifiss) We userestarted GMRES(20) as the Krylov subspace method andwe always take a zero initial guess The stopping criterion is

100381710038171003817100381711990311989610038171003817100381710038172

1003817100381710038171003817119903010038171003817100381710038172

le 10minus6

(20)

where 119903119896is the residual vector at the 119896th iteration

In the whole course of computation we always replace(119868 + (119861119860

minus1

119861119879

120572))minus1 in (14) with (16) and use the 119871119880 factor-

ization of 119860 + (119861119879

119861120572) to tackle (119860 + (119861119879

119861120572))minus1V where

V is a corresponding vector in the iteration Concretely let119860 + (119861

119879

119861120572) = 119871119880 then we complete the matrix-vectorproduct (119860 + (119861

119879

119861120572))minus1V by 119880 119871 V in MATLAB term

In the following tables the denotation norm (119862 fro) meansthe Frobenius form of the matrix 119862 The total time is the sumof LU time and iterative time and the LU time is the time tocompute LU factorization of 119860 + (119861119879119861120572)

Case 1 (for our preconditioner) 119862 = 0 (using Q2-Q1 FEM inStokes problems and steady Oseen problems with differentviscosity coefficients The results are in Tables 1 2 3 4 )

Case 1rsquo (for preconditioner of [9]) 119862 = 0 (using Q2-Q1 FEMin Stokes problems and steady Oseen problems with differentviscosity coefficients The results are in Tables 5 6 7 8)

Case 2 119862 = 0 (using Q1-P0 FEM in Stokes problems andsteady Oseen problems with different viscosity coefficientsThe results are in Tables 9 10 11 12)

From Tables 1 2 3 4 5 6 7 and 8 we can see that theseresults are in agreement with the theoretical analyses (13) and(19) respectively Additionally comparing with the results inTables 9 10 11 and 12 we find that although the iterationsused in Case 1 (either for the preconditioner of [9] or ourpreconditioner) are less than those in Case 2 the time spentby Case 1 ismuchmore than that of Case 2This is because thedensity of the coefficient matrix generalized by Q2-Q1 FEMis much larger than that generalized by Q1-P0 FEMThis canbe partly illustrated by Tables 13 and 14 and the others can beillustrated similarly

6 Journal of Applied Mathematics

Table 14 Size and number of non-nil elements of the coefficient matrix generalized by using Q1-P0 FEM in steady Stokes problems

Grid dim(119860) nnz(119860) dim(119861) nnz(119861) nnz(119860 + (1120572)119861119879119861)16 times 16 578 times 578 3826 256 times 578 1800 707632 times 32 2178 times 2178 16818 1024 times 2178 7688 3187464 times 64 8450 times 8450 70450 4096 times 8450 31752 136582128 times 128 33282 times 33282 288306 16384 times 33282 129032 567192

4 Conclusions

In this paper we have introduced a splitting preconditionerfor solving generalized saddle point systems Theoreticalanalysis showed the modulus of eigenvalues of the precon-ditioned matrix would be located in interval (0 1) when theparameter is big enough Particularly when the submatrix119862 = 0 the eigenvalues will tend to 1 as the parameter 120572 rarr 0These performances are tested by some examples and theresults are in agreement with the theoretical analysis

There are still some future works to be done how to prop-erly choose a parameter 120572 so that the preconditioned matrixhas better propertiesHow to further precondition submatrix(119868 + (119861119860

minus1

119861119879

120572))minus1

((119861119860minus1

119861119879

120572) + (119862120572)) to improve ourpreconditioner

Acknowledgments

The authors would express their great thankfulness to thereferees and the editor Professor P N Shivakumar for theirhelpful suggestions for revising this paperThe authors wouldlike to thank H C Elman A Ramage and D J Silvester fortheir free IFISS software package This research is supportedby Chinese Universities Specialized Research Fund for theDoctoral Program (20110185110020) Sichuan Province Sci ampTech Research Project (2012GZX0080) and the Fundamen-tal Research Funds for the Central Universities

References

[1] F Brezzi andM FortinMixed and Hybrid Finite ElementMeth-ods vol 15 of Springer Series in Computational MathematicsSpringer New York NY USA 1991

[2] Z Chen Q Du and J Zou ldquoFinite element methods withmatching and nonmatching meshes for Maxwell equationswith discontinuous coefficientsrdquo SIAM Journal on NumericalAnalysis vol 37 no 5 pp 1542ndash1570 2000

[3] H C Elman D J Silvester and A J Wathen Finite Elementsand Fast Iterative Solvers With Applications in IncompressibleFluid Dynamics Numerical Mathematics and Scientific Com-putation Oxford University Press New York NY USA 2005

[4] C Cuvelier A Segal and A A van Steenhoven Finite ElementMethods and Navier-Stokes Equations vol 22 of Mathematicsand its Applications D Reidel Dordrecht The Netherlands1986

[5] Y Saad Iterative Methods for Sparse Linear Systems Society forIndustrial and Applied Mathematics Philadelphia Pa USA2nd edition 2003

[6] K Chen Matrix Preconditioning Techniques and Applicationsvol 19 ofCambridgeMonographs onApplied andComputational

Mathematics Cambridge University Press Cambridge UK2005

[7] M Benzi and X-P Guo ldquoA dimensional split preconditionerfor Stokes and linearized Navier-Stokes equationsrdquo AppliedNumerical Mathematics vol 61 no 1 pp 66ndash76 2011

[8] M Benzi M Ng Q Niu and Z Wang ldquoA relaxed dimensionalfactorization preconditioner for the incompressible Navier-Stokes equationsrdquo Journal of Computational Physics vol 230no 16 pp 6185ndash6202 2011

[9] YCaoM-Q Jiang andY-L Zheng ldquoA splitting preconditionerfor saddle point problemsrdquo Numerical Linear Algebra withApplications vol 18 no 5 pp 875ndash895 2011

[10] V Simoncini and M Benzi ldquoSpectral properties of the Her-mitian and skew-Hermitian splitting preconditioner for saddlepoint problemsrdquo SIAM Journal on Matrix Analysis and Applica-tions vol 26 no 2 pp 377ndash389 2004

[11] H C Elman A Ramage and D J Silvester ldquoAlgorithm 886IFISS a Matlab toolbox for modelling incompressible flowrdquoACM Transactions on Mathematical Software vol 33 no 2article 14 2007

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 593549 11 pageshttpdxdoiorg1011552013593549

Research ArticleSolving Optimization Problems on Hermitian Matrix Functionswith Applications

Xiang Zhang and Shu-Wen Xiang

Department of Computer Science and Information Guizhou University Guiyang 550025 China

Correspondence should be addressed to Xiang Zhang zxjnsc163com

Received 17 October 2012 Accepted 20 March 2013

Academic Editor K Sivakumar

Copyright copy 2013 X Zhang and S-W Xiang This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

We consider the extremal inertias and ranks of the matrix expressions 119891(119883 119884) = 1198603minus 119861

3119883 minus (119861

3119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast where

1198603= 119860

lowast

3 119861

3 119862

3 and 119863

3are known matrices and 119884 and 119883 are the solutions to the matrix equations 119860

1119884 = 119862

1 119884119861

1= 119863

1 and

1198602119883 = 119862

2 respectively As applications we present necessary and sufficient condition for the previous matrix function 119891(119883 119884)

to be positive (negative) non-negative (positive) definite or nonsingular We also characterize the relations between the Hermitianpart of the solutions of the above-mentioned matrix equations Furthermore we establish necessary and sufficient conditions forthe solvability of the system of matrix equations 119860

1119884 = 119862

1 119884119861

1= 119863

1 119860

2119883 = 119862

2 and 119861

3119883 + (119861

3119883)

lowast

+ 1198623119884119863

3+ (119862

3119884119863

3)lowast

= 1198603

and give an expression of the general solution to the above-mentioned system when it is solvable

1 Introduction

Throughout we denote the field of complex numbers by Cthe set of all 119898 times 119899 matrices over C by C119898times119899 and the set ofall119898 times119898Hermitian matrices by C119898times119898

ℎ The symbols 119860lowast and

R(119860) stand for the conjugate transpose the column space ofa complex matrix119860 respectively 119868

119899denotes the 119899 times 119899 identity

matrix TheMoore-Penrose inverse [1] 119860dagger of 119860 is the uniquesolution119883 to the four matrix equations

(i) 119860119883119860 = 119860

(ii) 119883119860119883 = 119883

(iii) (119860119883)lowast = 119860119883

(iv) (119883119860)lowast = 119883119860

(1)

Moreover 119871119860and 119877

119860stand for the projectors 119871

119860= 119868 minus

119860dagger

119860 119877119860= 119868 minus 119860119860

dagger induced by 119860 It is well known that theeigenvalues of a Hermitian matrix 119860 isin C119899times119899 are real and theinertia of 119860 is defined to be the triplet

I119899(119860) = 119894

+(119860) 119894

minus(119860) 119894

0(119860) (2)

where 119894+(119860) 119894

minus(119860) and 119894

0(119860) stand for the numbers of

positive negative and zero eigenvalues of119860 respectivelyThe

symbols 119894+(119860) and 119894

minus(119860) are called the positive index and

the negative index of inertia respectively For two Hermitianmatrices 119860 and 119861 of the same sizes we say 119860 ge 119861 (119860 le 119861)in the Lowner partial ordering if 119860 minus 119861 is positive (negative)semidefinite The Hermitian part of 119883 is defined as 119867(119883) =119883 + 119883

lowast We will say that 119883 is Re-nnd (Re-nonnegativesemidefinite) if 119867(119883) ge 0 119883 is Re-pd (Re-positive definite)if119867(119883) gt 0 and119883 is Re-ns if119867(119883) is nonsingular

It is well known that investigation on the solvabilityconditions and the general solution to linearmatrix equationsis very active (eg [2ndash9]) In 1999 Braden [10] gave thegeneral solution to

119861119883 + (119861119883)lowast

= 119860 (3)

In 2007 Djordjevic [11] considered the explicit solution to (3)for linear bounded operators on Hilbert spaces MoreoverCao [12] investigated the general explicit solution to

119861119883119862 + (119861119883119862)lowast

= 119860 (4)

Xu et al [13] obtained the general expression of the solutionof operator equation (4) In 2012 Wang and He [14] studied

2 Journal of Applied Mathematics

some necessary and sufficient conditions for the consistenceof the matrix equation

1198601119883 + (119860

1119883)

lowast

+ 1198611119884119862

1+ (119861

1119884119862

1)lowast

= 1198641

(5)

and presented an expression of the general solution to (5)Note that (5) is a special case of the following system

1198601119884 = 119862

1 119884119861

1= 119863

1 119860

2119883 = 119862

2

1198613119883 + (119861

3119883)

lowast

+ 1198623119884119863

3+ (119862

3119884119863

3)lowast

= 1198603

(6)

To our knowledge there has been little information about (6)One goal of this paper is to give some necessary and sufficientconditions for the solvability of the system of matrix (6) andpresent an expression of the general solution to system (6)when it is solvable

In order to find necessary and sufficient conditions for thesolvability of the system of matrix equations (6) we need toconsider the extremal ranks and inertias of (10) subject to (13)and (11)

There have been many papers to discuss the extremalranks and inertias of the following Hermitian expressions

119901 (119883) = 1198603minus 119861

3119883 minus (119861

3119883)

lowast

(7)

119892 (119884) = 119860 minus 119861119884119862 minus (119861119884119862)lowast

(8)

ℎ (119883 119884) = 1198601minus 119861

1119883119861

lowast

1minus 119862

1119884119862

lowast

1 (9)

119891 (119883 119884) = 1198603minus 119861

3119883 minus (119861

3119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

(10)

Tian has contributed much in this field One of his works[15] considered the extremal ranks and inertias of (7) Heand Wang [16] derived the extremal ranks and inertias of (7)subject to119860

1119883 = 119862

1119860

2119883119861

2= 119862

2 Liu and Tian [17] studied

the extremal ranks and inertias of (8) Chu et al [18] and Liuand Tian [19] derived the extremal ranks and inertias of (9)Zhang et al [20] presented the extremal ranks and inertias of(9) where119883 and 119884 are Hermitian solutions of

1198602119883 = 119862

2 (11)

1198841198612= 119863

2 (12)

respectively He and Wang [16] derived the extremal ranksand inertias of (10)We consider the extremal ranks and iner-tias of (10) subject to (11) and

1198601119884 = 119862

1 119884119861

1= 119863

1 (13)

which is not only the generalization of the abovematrix func-tions but also can be used to investigate the solvability con-ditions for the existence of the general solution to the system(6) Moreover it can be applied to characterize the relationsbetween Hermitian part of the solutions of (11) and (13)

The remainder of this paper is organized as follows InSection 2 we consider the extremal ranks and inertias of(10) subject to (11) and (13) In Section 3 we characterize therelations between the Hermitian part of the solution to (11)and (13) In Section 4 we establish the solvability conditionsfor the existence of a solution to (6) and obtain an expressionof the general solution to (6)

2 Extremal Ranks and Inertias of HermitianMatrix Function (10) with Some Restrictions

In this section we consider formulas for the extremal ranksand inertias of (10) subject to (11) and (13) We begin with thefollowing Lemmas

Lemma 1 (see [21]) (a) Let1198601119862

1 119861

1 and119863

1be givenThen

the following statements are equivalent

(1) system (13) is consistent(2)

1198771198601119862

1= 0 119863

1119871

1198611= 0 119860

1119863

1= 119862

11198611 (14)

(3)

119903 [1198601119862

1] = 119903 (119860

1)

[119863

1

1198611

] = 119903 (1198611)

1198601119863

1= 119862

11198611

(15)

In this case the general solution can be written as

119884 = 119860dagger

1119862

1+ 119871

1198601119863

1119861dagger

1+ 119871

1198601119881119877

1198611 (16)

where 119881 is arbitrary(b) Let 119860

2and 119862

2be given Then the following statements

are equivalent

(1) equation (11) is consistent(2)

1198771198602119862

2= 0 (17)

(3)

119903 [1198602119862

2] = 119903 (119860

2) (18)

In this case the general solution can be written as

119883 = 119860dagger

119862 + 119871119860119882 (19)

where119882 is arbitrary

Lemma 2 ([22 Lemma 15 Theorem 23]) Let 119860 isin C119898times119898

ℎ

119861 isin C119898times119899 and119863 isin C119899times119899

ℎ and denote that

119872 = [119860 119861

119861lowast

0]

119873 = [0 119876

119876lowast

0]

119871 = [119860 119861

119861lowast

119863]

119866 = [119875 119872119871

119873

119871119873119872

lowast

0]

(20)

Journal of Applied Mathematics 3

Then one has the following

(a) the following equalities hold

119894plusmn(119872) = 119903 (119861) + 119894

plusmn(119877

119861119860119877

119861) (21)

119894plusmn(119873) = 119903 (119876) (22)

(b) if R(119861) sube R(119860) then 119894plusmn(119871) = 119894

plusmn(119860) + 119894

plusmn(119863 minus

119861lowast

119860dagger

119861)) Thus 119894plusmn(119871) = 119894

plusmn(119860) if and only if R(119861) sube

R(119860) and 119894plusmn(119863 minus 119861

lowast

119860dagger

119861) = 0(c)

119894plusmn(119866) = [

[

119875 119872 0

119872lowast

0 119873lowast

0 119873 0

]

]

minus 119903 (119873) (23)

Lemma 3 (see [23]) Let 119860 isin C119898times119899 119861 isin C119898times119896 and 119862 isin C119897times119899Then they satisfy the following rank equalities

(a) 119903[119860 119861] = 119903(119860) + 119903(119864119860119861) = 119903(119861) + 119903(119864

119861119860)

(b) 119903 [ 119860119862] = 119903(119860) + 119903(119862119865

119860) = 119903(119862) + 119903(119860119865

119862)

(c) 119903 [ 119860 119861

119862 0] = 119903(119861) + 119903(119862) + 119903(119864

119861119860119865

119862)

(d) 119903[119861 119860119865119862] = 119903 [

119861 119860

0 119862] minus 119903(119862)

(e) 119903 [ 119862

119864119861119860] = 119903 [

119862 0

119860 119861] minus 119903(119861)

(f) 119903 [ 119860 119861119865119863

119864119864119862 0] = 119903 [

119860 119861 0

119862 0 119864

0 119863 0

] minus 119903(119863) minus 119903(119864)

Lemma 4 (see [15]) Let119860 isin C119898times119898

ℎ 119861 isin C119898times119899119862 isin C119899times119899

ℎ119876 isin

C119898times119899 and 119875 isin C119901times119899 be given and 119879 isin C119898times119898 be nonsingularThen one has the following

(1) 119894plusmn(119879119860119879

lowast

) = 119894plusmn(119860)

(2) 119894plusmn[119860 0

0 119862] = 119894

plusmn(119860) + 119894

plusmn(119862)

(3) 119894plusmn[

0 119876

119876lowast

0] = 119903(119876)

(4) 119894plusmn[

119860 119861119871119875

119871119875119861lowast

0] + 119903(119875) = 119894

plusmn[

119860 119861 0

119861lowast

0 119875lowast

0 119875 0

]

Lemma 5 (see [22 Lemma 14]) Let 119878 be a set consisting of(square) matrices over C119898times119898 and let 119867 be a set consisting of(square) matrices overC119898times119898

ℎ ThenThen one has the following

(a) 119878 has a nonsingularmatrix if and only ifmax119883isin119878119903(119883) =

119898(b) any 119883 isin 119878 is nonsingular if and only if min

119883isin119878119903(119883) =

119898(c) 0 isin 119878 if and only ifmin

119883isin119878119903(119883) = 0

(d) 119878 = 0 if and only ifmax119883isin119878119903(119883) = 0

(e) 119867 has a matrix 119883 gt 0 (119883 lt 0) if and only ifmax

119883isin119867119894+(119883) = 119898(max

119883isin119867119894minus(119883) = 119898)

(f) any 119883 isin 119867 satisfies 119883 gt 0 (119883 lt 0) if and only ifmin

119883isin119867119894+(119883) = 119898 (min

119883isin119867119894minus(119883) = 119898)

(g) 119867 has a matrix 119883 ge 0 (119883 le 0) if and only ifmin

119883isin119867119894minus(119883) = 0 (min

119883isin119867119894+(119883) = 0)

(h) any 119883 isin 119867 satisfies 119883 ge 0 (119883 le 0) if and only ifmax

119883isin119867119894minus(119883) = 0 (max

119883isin119867119894+(119883) = 0)

Lemma 6 (see [16]) Let 119901(119883 119884) = 119860minus119861119883minus (119861119883)lowast minus119862119884119863minus(119862119884119863)

lowast where119860119861119862 and119863 are givenwith appropriate sizesand denote that

1198721= [

[

119860 119861 119862

119861lowast

0 0

119862lowast

0 0

]

]

1198722= [

[

119860 119861 119863lowast

119861lowast

0 0

119863 0 0

]

]

1198723= [

119860 119861 119862 119863lowast

119861lowast

0 0 0]

1198724= [

[

119860 119861 119862 119863lowast

119861lowast

0 0 0

119862lowast

0 0 0

]

]

1198725= [

[

119860 119861 119862 119863lowast

119861lowast

0 0 0

119863 0 0 0

]

]

(24)

Then one has the following

(1) the maximal rank of 119901(119883 119884) is

max119883isinC119899times119898 119884

119903 [119901 (119883 119884)] = min 119898 119903 (1198721) 119903 (119872

2) 119903 (119872

3)

(25)

(2) the minimal rank of 119901(119883 119884) is

min119883isinC119899times119898 119884

119903 [119901 (119883 119884)]

= 2119903 (1198723) minus 2119903 (119861)

+max 119906++ 119906

minus V

++ V

minus 119906

++ V

minus 119906

minus+ V

+

(26)

(3) the maximal inertia of 119901(119883 119884) is

max119883isinC119899times119898119884

119894plusmn[119901 (119883 119884)] = min 119894

plusmn(119872

1) 119894

plusmn(119872

2) (27)

(4) the minimal inertias of 119901(119883 119884) is

min119883isinC119899times119898119884

119894plusmn[119901 (119883 119884)] = 119903 (119872

3) minus 119903 (119861)

+max 119894plusmn(119872

1) minus 119903 (119872

4)

119894plusmn(119872

2) minus 119903 (119872

5)

(28)

where

119906plusmn= 119894

plusmn(119872

1) minus 119903 (119872

4) V

plusmn= 119894

plusmn(119872

2) minus 119903 (119872

5) (29)

Now we present the main theorem of this section

4 Journal of Applied Mathematics

Theorem7 Let1198601isin C119898times119899119862

1isin C119898times119896119861

1isin C119896times119897119863

1isin C119899times119897

1198602isin C119905times119902 119862

2isin C119905times119901119860

3isin C

119901times119901

ℎ 119861

3isin C119901times119902 119862

3isin C119901times119899

and1198633isin C119901times119899 be given and suppose that the system of matrix

equations (13) and (11) is consistent respectively Denote the setof all solutions to (13) by 119878 and (11) by 119866 Put

1198641=

[[[[[

[

1198603

1198623119863

lowast

3119862

lowast

11198613119862

lowast

2

119862lowast

30 119860

lowast

10 0

1198621119863

3119860

10 0 0

119861lowast

30 0 0 119860

lowast

2

1198622

0 0 11986020

]]]]]

]

1198642= 119903

[[[[[

[

1198603

119863lowast

3119862

3119863

11198613119862

lowast

2

1198633

0 1198611

0 0

119863lowast

1119862

lowast

3119861lowast

10 0 0

119861lowast

30 0 0 119860

lowast

2

1198622

0 0 11986020

]]]]]

]

1198643=

[[[[[

[

1198603

1198613119862

lowast

3119863

lowast

3119862

lowast

2

119861lowast

30 0 0 119860

lowast

2

119863lowast

1119862

lowast

30 0 119861

lowast

10

1198621119863

30 119860

10 0

1198622

11986020 0 0

]]]]]

]

1198644=

[[[[[[[

[

1198603

1198613119862

3119863

lowast

3119862

lowast

2119863

lowast

3119862

lowast

1

119861lowast

30 0 0 119860

lowast

20

119862lowast

30 0 0 0 119860

lowast

1

0 0 0 119861lowast

10 0

1198621119863

30 119860

10 0 0

1198622

11986020 0 0 0

]]]]]]]

]

1198645=

[[[[[[[

[

1198603

1198613119862

3119863

lowast

3119862

lowast

2119862

3119863

1

119861lowast

30 0 0 119860

lowast

20

1198633

0 0 0 0 1198611

119863lowast

1119862

lowast

30 0 119860

lowast

10 0

0 0 11986010 0 0

1198622

11986020 0 0 0

]]]]]]]

]

(30)

Then one has the following(a) the maximal rank of (10) subject to (13) and (11) is

max119883isin119866119884isin119878

119903 [119891 (119883 119884)]

= min 119901 119903 (1198641) minus 2119903 (119860

1) minus 2119903 (119860

2)

119903 (1198642) minus 2119903 (119861

1) minus 2119903 (119860

2)

119903 (1198643) minus 2119903 (119860

2) minus 119903 (119860

1) minus 119903 (119861

1)

(31)

(b) the minimal rank of (10) subject to (13) and (11) is

min119883isin119866119884isin119878

119903 [119891 (119883 119884)]

= 2119903 (1198643) minus 2119903 [

1198613

1198602

]

+max 119903 (1198641) minus 2119903 (119864

4) 119903 (119864

2) minus 2119903 (119864

5)

119894+(119864

1) + 119894

minus(119864

2) minus 119903 (119864

4) minus 119903 (119864

5)

119894minus(119864

1) + 119894

+(119864

2) minus 119903 (119864

4) minus 119903 (119864

5)

(32)

(c) the maximal inertia of (10) subject to (13) and (11) is

max119883isin119866119884isin119878119909

119894plusmn[119891 (119883 119884)] = min 119894

plusmn(119864

1) minus 119903 (119860

1) minus 119903 (119860

2)

119894plusmn(119864

2) minus 119903 (119861

1) minus 119903 (119860

2)

(33)

(d) the minimal inertia of (10) subject to (13) and (11) is

min119883isin119866119884isin119878

119894plusmn[119891 (119883 119884)] = 119903 (119864

3) minus 119903 [

1198613

1198602

]

+max 119894plusmn(119864

1) minus 119903 (119864

4)

119894plusmn(119864

2) minus 119903 (119864

5)

(34)

Proof By Lemma 1 the general solutions to (13) and (11) canbe written as

119883 = 119860dagger

2119862

2+ 119871

1198602119882

119884 = 119860dagger

1119862

1+ 119871

1198601119863

1119861dagger

1+ 119871

1198601119885119877

1198611

(35)

where119882 and 119885 are arbitrary matrices with appropriate sizesPut

119876 = 1198613119871

1198602 119879 = 119862

3119871

1198601 119869 = 119877

1198611119863

3

119875 = 1198603minus 119861

3119860

dagger

2119862

2minus (119861

3119860

dagger

2119862

2)lowast

minus 1198623(119860

dagger

1119862

1+ 119871

1198601119863

1119861dagger

1)119863

3

minus (1198623(119860

dagger

1119862

1+ 119871

1198601119863

1119861dagger

1)119863

3)lowast

(36)

Substituting (36) into (10) yields

119891 (119883 119884) = 119875 minus 119876119882 minus (119876119882)lowast

minus 119879119885119869 minus (119879119885119869)lowast

(37)

Clearly 119875 is Hermitian It follows from Lemma 6 that

max119883isin119866119884isin119878

119903 [119891 (119883 119884)]

= max119882119885

119903 (119875 minus 119876119882 minus (119876119882)lowast

minus 119879119885119869 minus (119879119885119869)lowast

)

= min 119898 119903 (1198731) 119903 (119873

2) 119903 (119873

3)

(38)

min119883isin119866119884isin119878

119903 [119891 (119883 119884)]

= max119882119885

119903 (119875 minus 119876119882 minus (119876119882)lowast

minus 119879119885119869 minus (119879119885119869)lowast

)

= 2119903 (1198733) minus 2119903 (119876)

+max 119904++ 119904

minus 119905

++ 119905

minus 119904

++ 119905

minus 119904

minus+ 119905

+

(39)

Journal of Applied Mathematics 5

max119883isin119866119884isin119878

119894plusmn[119891 (119883 119884)]

= max119882119885

119903 (119875 minus 119876119882 minus (119876119882)lowast

minus 119879119885119869 minus (119879119885119869)lowast

)

= min 119894plusmn(119873

1) 119894

plusmn(119873

2)

(40)

min119883isin119866119884isin119878

119894plusmn[119891 (119883 119884)]

= max119882119885

119903 (119875 minus 119876119882 minus (119876119882)lowast

minus 119879119885119869 minus (119879119885119869)lowast

)

= 119903 (1198733) minus 119903 (119876) +max 119904

plusmn 119905

plusmn

(41)

where

1198731= [

[

119875 119876 119879

119876lowast

0 0

119879lowast

0 0

]

]

1198732= [

[

119875 119876 119869lowast

119876lowast

0 0

119869 0 0

]

]

1198733= [

119875 119876 119879 119869lowast

119876lowast

0 0 0]

1198734= [

[

119875 119876 119879 119869lowast

119876lowast

0 0 0

119879lowast

0 0 0

]

]

1198735= [

[

119875 119876 119879 119869lowast

119876lowast

0 0 0

119869 0 0 0

]

]

119904plusmn= 119894

plusmn(119873

1) minus 119903 (119873

4) 119905

plusmn= 119894

plusmn(119873

2) minus 119903 (119873

5)

(42)

Now we simplify the ranks and inertias of block matrices in(38)ndash(41)

By Lemma 4 blockGaussian elimination and noting that

119871lowast

119878= (119868 minus 119878

dagger

119878)lowast

= 119868 minus 119878lowast

(119878lowast

)dagger

= 119877119878lowast (43)

we have the following

119903 (1198731) = 119903[

[

119875 119876 119879

119876lowast

0 0

119879lowast

0 0

]

]

= 119903

[[[[[

[

1198603

1198623119863

lowast

3119862

lowast

11198613119862

lowast

2

119862lowast

30 119860

lowast

10 0

1198621119863

3119860

10 0 0

119861lowast

30 0 0 119860

lowast

2

1198622

0 0 11986020

]]]]]

]

minus 2119903 (1198601) minus 2119903 (119860

2)

(44)

By 11986211198611= 119860

1119863

1 we obtain

119903 (1198732) = 119903

[[

[

119875 119876 119869lowast

119876lowast

0 0

119869 0 0

]]

]

= 119903

[[[[[[[[

[

1198603

119863lowast

3119862

3119863

11198613119862

lowast

2

1198633

0 1198611

0 0

119863lowast

1119862

lowast

3119861lowast

10 0 0

119861lowast

30 0 0 119860

lowast

2

1198622

0 0 11986020

]]]]]]]]

]

minus 2119903 (1198611) minus 2119903 (119860

2)

119903 (1198733) = 119903 [

119875 119876 119879 119869lowast

119876lowast

0 0 0]

= 119903

[[[[[[[[

[

1198603

1198613119862

lowast

3119863

lowast

3119862

lowast

2

119861lowast

30 0 0 119860

lowast

2

119863lowast

1119862

lowast

30 0 119861

lowast

10

1198621119863

30 119860

10 0

1198622

11986020 0 0

]]]]]]]]

]

minus 119903 (1198611) minus 2119903 (119860

2) minus 119903 (119860

1)

119903 (1198734) = 119903

[[

[

119875 119876 119879 119869lowast

119876lowast

0 0 0

119879lowast

0 0 0

]]

]

= 119903

[[[[[[[

[

1198603

1198613119862

3119863

lowast

3119862

lowast

2119863

lowast

3119862

lowast

1

119861lowast

30 0 0 119860

lowast

20

119862lowast

30 0 0 0 119860

lowast

1

0 0 0 119861lowast

10 0

1198621119863

30 119860

10 0 0

1198622

11986020 0 0 0

]]]]]]]

]

minus 119903 (1198611) minus 2119903 (119860

2) minus 2119903 (119860

1)

119903 (1198735) = 119903[

[

119875 119876 119879 119869lowast

119876lowast

0 0 0

119879lowast

0 0 0

]

]

= 119903

[[[[[[[

[

1198603

1198613119862

3119863

lowast

3119862

lowast

2119862

3119863

1

119861lowast

30 0 0 119860

lowast

20

1198633

0 0 0 0 1198611

119863lowast

1119862

lowast

30 0 119860

lowast

10 0

0 0 11986010 0 0

1198622

11986020 0 0 0

]]]]]]]

]

minus 2119903 (1198611) minus 2119903 (119860

2) minus 119903 (119860

1)

(45)

By Lemma 2 we can get the following

119894plusmn(119873

1) = 119894

plusmn

[

[

119875 119876 119879

119876lowast

0 0

119879lowast

0 0

]

]

6 Journal of Applied Mathematics

= 119894plusmn

[[[[[

[

1198603

1198623119863

lowast

3119862

lowast

11198613119862

lowast

2

119862lowast

30 119860

lowast

10 0

1198621119863

3119860

10 0 0

119861lowast

30 0 0 119860

lowast

2

1198622

0 0 11986020

]]]]]

]

minus 119903 (1198601) minus 119903 (119860

2)

(46)

119894plusmn(119873

2) = 119894

plusmn

[

[

119875 119876 119869lowast

119876lowast

0 0

119869 0 0

]

]

= 119894plusmn

[[[[[

[

1198603

119863lowast

3119862

3119863

11198613119862

lowast

2

1198633

0 1198611

0 0

119863lowast

1119862

lowast

3119861lowast

10 0 0

119861lowast

30 0 0 119860

lowast

2

1198622

0 0 11986020

]]]]]

]

minus 119903 (1198611) minus 119903 (119860

2)

(47)

Substituting (44)-(47) into (38) and (41) yields (31)ndash(34)respectively

Corollary 8 Let 1198601 119862

1 119861

1 119863

1 119860

2 119862

2 119860

3 119861

3 119862

3 119863

3

and 119864119894 (119894 = 1 2 5) be as in Theorem 7 and suppose

that the system of matrix equations (13) and (11) is consistentrespectively Denote the set of all solutions to (13) by 119878 and (11)by 119866 Then one has the following

(a) there exist 119883 isin 119866 and 119884 isin 119878 such that 1198603minus 119861

3119883 minus

(1198613119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

gt 0 if and only if

119894+(119864

1) minus 119903 (119860

1) minus 119903 (119860

2) ge 119901

119894+(119864

2) minus 119903 (119861

1) minus 119903 (119860

2) ge 119901

(48)

(b) there exist 119883 isin 119866 and 119884 isin 119878 such that 1198603minus 119861

3119883 minus

(1198613119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

lt 0 if and only if

119894minus(119864

1) minus 119903 (119860

1) minus 119903 (119860

2) ge 119901

119894minus(119864

2) minus 119903 (119861

1) minus 119903 (119860

2) ge 119901

(49)

(c) there exist 119883 isin 119866 and 119884 isin 119878 such that 1198603minus 119861

3119883 minus

(1198613119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

ge 0 if and only if

119903 (1198643) minus 119903 [

1198613

1198602

] + 119894minus(119864

1) minus 119903 (119864

4) le 0

119903 (1198643) minus 119903 [

1198613

1198602

] + 119894minus(119864

2) minus 119903 (119864

5) le 0

(50)

(d) there exist 119883 isin 119866 and 119884 isin 119878 such that 1198603minus 119861

3119883 minus

(1198613119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

le 0 if and only if

119903 (1198643) minus 119903 [

1198613

1198602

] + 119894+(119864

1) minus 119903 (119864

4) le 0

119903 (1198643) minus 119903 [

1198613

1198602

] + 119894+(119864

2) minus 119903 (119864

5) le 0

(51)

(e) 1198603minus 119861

3119883 minus (119861

3119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

gt 0 for all119883 isin 119866 and 119884 isin 119878 if and only if

119903 (1198643) minus 119903 [

1198613

1198602

] + 119894+(119864

1) minus 119903 (119864

4) = 119901

119900119903 119903 (1198643) minus 119903 [

1198613

1198602

] + 119894+(119864

2) minus 119903 (119864

5) = 119901

(52)

(f) 1198603minus 119861

3119883 minus (119861

3119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

lt 0 for all119883 isin 119866 and 119884 isin 119878 if and only if

119903 (1198643) minus 119903 [

1198613

1198602

] + 119894minus(119864

1) minus 119903 (119864

4) = 119901

119900119903 119903 (1198643) minus 119903 [

1198613

1198602

] + 119894minus(119864

2) minus 119903 (119864

5) = 119901

(53)

(g) 1198603minus 119861

3119883 minus (119861

3119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

ge 0 for all119883 isin 119866 and 119884 isin 119878 if and only if

119894minus(119864

1) minus 119903 (119860

1) minus 119903 (119860

2) le 0

119900119903 119894minus(119864

2) minus 119903 (119861

1) minus 119903 (119860

2) le 0

(54)

(h) 1198603minus 119861

3119883 minus (119861

3119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

le 0 for all119883 isin 119866 and 119884 isin 119878 if and only if

119894+(119864

1) minus 119903 (119860

1) minus 119903 (119860

2) le 0

119900119903 119894+(119864

2) minus 119903 (119861

1) minus 119903 (119860

2) le 0

(55)

(i) there exist 119883 isin 119866 and 119884 isin 119878 such that 1198603minus 119861

3119883 minus

(1198613119883)

lowast

minus1198623119884119863

3minus(119862

3119884119863

3)lowast is nonsingular if and only

if

119903 (1198641) minus 2119903 (119860

1) minus 2119903 (119860

2) ge 119901

119903 (1198642) minus 2119903 (119861

1) minus 2119903 (119860

2) ge 119901

119903 (1198643) minus 2119903 (119860

2) minus 119903 (119860

1) minus 119903 (119861

1) ge 119901

(56)

3 Relations between the Hermitian Part ofthe Solutions to (13) and (11)

Now we consider the extremal ranks and inertias of thedifference between the Hermitian part of the solutions to (13)and (11)

Theorem 9 Let 1198601isin C119898times119901 119862

1isin C119898times119901 119861

1isin C119901times119897 119863

1isin

C119901times119897 1198602isin C119905times119901 and 119862

2isin C119905times119901 be given Suppose that

the system of matrix equations (13) and (11) is consistent

Journal of Applied Mathematics 7

respectively Denote the set of all solutions to (13) by 119878 and (11)by 119866 Put

1198671=

[[[[[

[

0 119868 119862lowast

1minus119868 119862

lowast

2

119868 0 119860lowast

10 0

1198621119860

10 0 0

minus119868 0 0 0 119860lowast

2

11986220 0 119860

20

]]]]]

]

1198672= 119903

[[[[[

[

0 119868 1198631minus119868 119862

lowast

2

119868 0 1198611

0 0

119863lowast

1119861lowast

10 0 0

minus119868 0 0 0 119860lowast

2

11986220 0 119860

20

]]]]]

]

1198673=

[[[[[

[

0 minus119868 119868 119868 119862lowast

2

minus119868 0 0 0 119860lowast

2

119863lowast

10 0 119861

lowast

10

1198621

0 11986010 0

1198622119860

20 0 0

]]]]]

]

1198674=

[[[[[[[

[

0 minus119868 119868 119868 119862lowast

2119862

lowast

1

minus119868 0 0 0 119860lowast

20

119868 0 0 0 0 119860lowast

1

0 0 0 119861lowast

10 0

11986210 119860

10 0 0

1198622119860

20 0 0 0

]]]]]]]

]

1198675=

[[[[[[[

[

0 minus119868 119868 119868 119862lowast

2119863

1

minus119868 0 0 0 119860lowast

20

119868 0 0 0 0 1198611

119863lowast

10 0 119860

lowast

10 0

0 0 11986010 0 0

1198622119860

20 0 0 0

]]]]]]]

]

(57)

Then one has the following

max119883isin119866119884isin119878

119903 [(119883 + 119883lowast

) minus (119884 + 119884lowast

)]

= min 119901 119903 (1198671) minus 2119903 (119860

1) minus 2119903 (119860

2) 119903 (119867

2) minus 2119903 (119861

1)

minus2119903 (1198602) 119903 (119867

3) minus 2119903 (119860

2) minus 119903 (119860

1) minus 119903 (119861

1)

min119883isin119866119884isin119878

119903 [(119883 + 119883lowast

) minus (119884 + 119884lowast

)]

= 2119903 (1198673) minus 2119901

+max 119903 (1198671) minus 2119903 (119867

4) 119903 (119867

2) minus 2119903 (119867

5)

119894+(119867

1) + 119894

minus(119867

2) minus 119903 (119867

4) minus 119903 (119867

5)

119894minus(119867

1) + 119894

+(119867

2) minus 119903 (119867

4) minus 119903 (119867

5)

max119883isin119866119884isin119878

119894plusmn[(119883 + 119883

lowast

) minus (119884 + 119884lowast

)]

= min 119894plusmn(119867

1) minus 119903 (119860

1) minus 119903 (119860

2)

119894plusmn(119867

2) minus 119903 (119861

1) minus 119903 (119860

2)

min119883isin119866119884isin119878

119894plusmn[(119883 + 119883

lowast

) minus (119884 + 119884lowast

)]

= 119903 (1198643) minus 119901 +max 119894

plusmn(119867

1) minus 119903 (119867

4) 119894

plusmn(119867

2) minus 119903 (119867

5)

(58)

Proof By letting 1198603= 0 119861

3= minus119868 119862

3= 119868 and 119863

3= 119868 in

Theorem 7 we can get the results

Corollary 10 Let 1198601 119862

1 119861

1 119863

1 119860

2 119862

2 and 119867

119894 (119894 =

1 2 5) be as in Theorem 9 and suppose that the system ofmatrix equations (13) and (11) is consistent respectively Denotethe set of all solutions to (13) by 119878 and (11) by 119866 Then one hasthe following

(a) there exist 119883 isin 119866 and 119884 isin 119878 such that (119883 + 119883lowast

) gt

(119884 + 119884lowast

) if and only if

119894+(119867

1) minus 119903 (119860

1) minus 119903 (119860

2) ge 119901

119894+(119867

2) minus 119903 (119861

1) minus 119903 (119860

2) ge 119901

(59)

(b) there exist 119883 isin 119866 and 119884 isin 119878 such that (119883 + 119883lowast

) lt

(119884 + 119884lowast

) if and only if

119894minus(119867

1) minus 119903 (119860

1) minus 119903 (119860

2) ge 119901

119894minus(119867

2) minus 119903 (119861

1) minus 119903 (119860

2) ge 119901

(60)

(c) there exist 119883 isin 119866 and 119884 isin 119878 such that (119883 + 119883lowast

) ge

(119884 + 119884lowast

) if and only if

119903 (1198673) minus 119901 + 119894

minus(119867

1) minus 119903 (119867

4) le 0

119903 (1198673) minus 119901 + 119894

minus(119867

1) minus 119903 (119867

4) le 0

(61)

(d) there exist 119883 isin 119866 and 119884 isin 119878 such that (119883 + 119883lowast

) le

(119884 + 119884lowast

) if and only if

119903 (1198673) minus 119901 + 119894

+(119867

1) minus 119903 (119867

4) le 0

119903 (1198673) minus 119901 + 119894

+(119867

2) minus 119903 (119867

5) le 0

(62)

(e) (119883 + 119883lowast

) gt (119884 + 119884lowast

) for all 119883 isin 119866 and 119884 isin 119878 if andonly if

119903 (1198673) minus 119901 + 119894

+(119867

1) minus 119903 (119867

4) = 119901

119900119903 119903 (1198673) minus 119901 + 119894

+(119867

2) minus 119903 (119867

5) = 119901

(63)

(f) (119883 + 119883lowast

) lt (119884 + 119884lowast

) for all 119883 isin 119866 and 119884 isin 119878 if andonly if

119903 (1198673) minus 119901 + 119894

minus(119867

1) minus 119903 (119867

4) = 119901

119900119903 119903 (1198673) minus 119901 + 119894

minus(119867

2) minus 119903 (119867

5) = 119901

(64)

(g) (119883 + 119883lowast

) ge (119884 + 119884lowast

) for all 119883 isin 119866 and 119884 isin 119878 if andonly if

119894minus(119867

1) minus 119903 (119860

1) minus 119903 (119860

2) le 0

119900119903 119894minus(119867

2) minus 119903 (119861

1) minus 119903 (119860

2) le 0

(65)

8 Journal of Applied Mathematics

(h) (119883 + 119883lowast

) le (119884 + 119884lowast

) for all 119883 isin 119866 and 119884 isin 119878 if andonly if

119894+(119867

1) minus 119903 (119860

1) minus 119903 (119860

2) le 0

119900119903 119894+(119867

2) minus 119903 (119861

1) minus 119903 (119860

2) le 0

(66)

(i) there exist119883 isin 119866 and 119884 isin 119878 such that (119883 +119883lowast

) minus (119884 +

119884lowast

) is nonsingular if and only if119903 (119867

1) minus 2119903 (119860

1) minus 2119903 (119860

2) ge 119901

119903 (1198672) minus 2119903 (119861

1) minus 2119903 (119860

2) ge 119901

119903 (1198673) minus 2119903 (119860

2) minus 119903 (119860

1) minus 119903 (119861

1) ge 119901

(67)

4 The Solvability Conditions and the GeneralSolution to System (6)

We now turn our attention to (6) We in this section useTheorem 9 to give some necessary and sufficient conditionsfor the existence of a solution to (6) and present an expressionof the general solution to (6) We begin with a lemma whichis used in the latter part of this section

Lemma 11 (see [14]) Let 1198601isin C119898times1198991 119861

1isin C119898times1198992 119862

1isin

C119902times119898 and 1198641isin C119898times119898

ℎbe given Let 119860 = 119877

11986011198611 119861 = 119862

1119877

1198601

119864 = 1198771198601119864

1119877

1198601119872 = 119877

119860119861lowast119873 = 119860

lowast

119871119861 and 119878 = 119861lowast

119871119872 Then

the following statements are equivalent(1) equation (5) is consistent(2)119877

119872119877

119860119864 = 0 119877

119860119864119877

119860= 0 119871

119861119864119871

119861= 0 (68)

(3)

119903 [119864

11198611119862

lowast

1119860

1

119860lowast

10 0 0

] = 119903 [1198611119862

lowast

1119860

1] + 119903 (119860

1)

119903 [

[

11986411198611119860

1

119860lowast

10 0

119861lowast

10 0

]

]

= 2119903 [1198611119860

1]

119903 [

[

1198641119862

lowast

1119860

1

119860lowast

10 0

11986210 0

]

]

= 2119903 [119862lowast

1119860

1]

(69)

In this case the general solution of (5) can be expressed as

119884 =1

2[119860

dagger

119864119861dagger

minus 119860dagger

119861lowast

119872dagger

119864119861dagger

minus 119860dagger

119878(119861dagger

)lowast

119864119873dagger

119860lowast

119861dagger

+119860dagger

119864(119872dagger

)lowast

+ (119873dagger

)lowast

119864119861dagger

119878dagger

119878] + 1198711198601198811+ 119881

2119877

119861

+ 1198801119871

119878119871

119872+ 119877

119873119880

lowast

2119871

119872minus 119860

dagger

1198781198802119877

119873119860

lowast

119861dagger

119883 = 119860dagger

1[119864

1minus 119861

1119884119862

1minus (119861

1119884119862

1)lowast

]

minus1

2119860

dagger

1[119864

1minus 119861

1119884119862

1minus (119861

1119884119862

1)lowast

] 1198601119860

dagger

1

minus 119860dagger

1119882

1119860

lowast

1+119882

lowast

1119860

1119860

dagger

1+ 119871

1198601119882

2

(70)

where 1198801 119880

2 119881

1 119881

2119882

1 and119882

2are arbitrary matrices over

C with appropriate sizes

Now we give the main theorem of this section

Theorem 12 Let 119860119894 119862

119894 (119894 = 1 2 3) 119861

119895 and 119863

119895 (119895 = 1 3) be

given Set

119860 = 1198613119871

1198602 119861 = 119862

3119871

1198601

119862 = 1198771198611119863

3 119865 = 119877

119860119861

119866 = 119862119877119860 119872 = 119877

119865119866

lowast

119873 = 119865lowast

119871119866 119878 = 119866

lowast

119871119872

(71)

119863 = 1198603minus 119861

3119860

dagger

2119862

2minus (119861

3119860

dagger

2119862

2)lowast

minus 1198623(119860

dagger

1119862

1+ 119871

1198601119863

1119861dagger

1)119863

3

minus 119863lowast

3(119860

dagger

1119862

1+ 119871

1198601119863

1119861dagger

1)lowast

119862lowast

3

(72)

119864 = 119877119860119863119877

119860 (73)

Then the following statements are equivalent

(1) system (6) is consistent

(2) the equalities in (14) and (17) hold and

119877119872119877

119865119864 = 0 119877

119865119864119877

119865= 0 119871

119866119864119871

119866= 0 (74)

(3) the equalities in (15) and (18) hold and

119903

[[[[[

[

1198603

1198623119863

lowast

31198613119862

lowast

2

119861lowast

30 0 0 119860

lowast

2

1198621119863

3119860

10 0 0

119863lowast

1119862

lowast

30 119861

lowast

10 0

1198622

0 0 11986020

]]]]]

]

= 119903

[[[

[

1198623119863

lowast

31198613

11986010 0

0 119861lowast

10

0 0 1198602

]]]

]

+ 119903 [119860

2

1198613

]

119903

[[[[[

[

1198603

11986231198613119863

lowast

3119862

lowast

1119862

lowast

2

119862lowast

30 0 119860

lowast

10

119861lowast

30 0 0 119860

lowast

2

1198621119863

3119860

10 0 0

1198622

0 1198602

0 0

]]]]]

]

= 2119903[

[

11986231198613

11986010

0 1198602

]

]

119903

[[[[[

[

1198603

119863lowast

31198613119862

3119863

1119862

lowast

2

1198633

0 0 1198611

0

119861lowast

30 0 0 119860

lowast

2

119863lowast

1119862

lowast

3119861lowast

10 0 0

1198622

0 1198602

0 0

]]]]]

]

= 2119903[

[

119863lowast

31198613

119861lowast

10

0 1198602

]

]

(75)

In this case the general solution of system (6) can be expressedas

119883 = 119860dagger

2119862

2+ 119871

1198602119880

119884 = 119860dagger

1119862

1+ 119871

1198601119863

1119861dagger

1+ 119871

1198601119881119877

1198611

(76)

Journal of Applied Mathematics 9

where

119881 =1

2[119865

dagger

119864119866dagger

minus 119865dagger

119866lowast

119872dagger

119864119866dagger

minus 119865dagger

119878(119866dagger

)lowast

119864119873dagger

119865lowast

119866dagger

+119865dagger

119864(119872dagger

)lowast

+ (119873dagger

)lowast

119864119866dagger

119878dagger

119878] + 1198711198651198811

+ 1198812119877

119866+ 119880

1119871

119878119871

119872+ 119877

119873119880

lowast

2119871

119872minus 119865

dagger

1198781198802119877

119873119865

lowast

119866dagger

119880 = 119860dagger

[119863 minus 119861119881119862 minus (119861119881119862)lowast

]

minus1

2119860

dagger

[119863 minus 119861119881119862 minus (119861119881119862)lowast

] 119860119860dagger

minus 119860dagger

1198821119860

lowast

+119882lowast

1119860119860

dagger

+ 119871119860119882

2

(77)

where 1198801 119880

2 119881

1 119881

2119882

1 and119882

2are arbitrary matrices over

C with appropriate sizes

Proof (2)hArr (3) Applying Lemma 3 and Lemma 11 gives

119877119872119877

119865119864 = 0 lArrrArr 119903 (119877

119872119877

119865119864)

= 0 lArrrArr 119903[119863 119861 119862

lowast

119860

119860lowast

0 0 0]

= 119903 [119861 119862lowast

119860] + 119903 (119860)

lArrrArr 119903

[[[[[

[

119863 1198623119863

lowast

31198613

0

119861lowast

30 0 0 119860

lowast

2

0 11986010 0 0

0 0 119861lowast

10 0

0 0 0 11986020

]]]]]

]

= 119903

[[[

[

1198623119863

lowast

31198613

11986010 0

0 119861lowast

10

0 0 1198602

]]]

]

+ 119903 [119860

2

1198613

]

lArrrArr 119903

[[[[[

[

1198603

1198623119863

lowast

31198613119862

lowast

2

119861lowast

30 0 0 119860

lowast

2

1198621119863

3119860

10 0 0

119863lowast

1119862

lowast

30 119861

lowast

10 0

1198622

0 0 11986020

]]]]]

]

= 119903

[[[

[

1198623119863

lowast

31198613

11986010 0

0 119861lowast

10

0 0 1198602

]]]

]

+ 119903 [119860

2

1198613

]

(78)

By a similar approach we can obtain that

119877119865119864119877

119865= 0 lArrrArr 119903

[[[[[

[

1198603

11986231198613119863

lowast

3119862

lowast

1119862

lowast

2

119862lowast

30 0 119860

lowast

10

119861lowast

30 0 0 119860

lowast

2

1198621119863

3119860

10 0 0

1198622

0 1198602

0 0

]]]]]

]

= 2119903[

[

11986231198613

11986010

0 1198602

]

]

119871119866119864119871

119866= 0 lArrrArr 119903

[[[[[[[[

[

1198603

119863lowast

31198613119862

3119863

1119862

lowast

2

1198633

0 0 1198611

0

119861lowast

30 0 0 119860

lowast

2

119863lowast

1119862

lowast

3119861lowast

10 0 0

1198622

0 1198602

0 0

]]]]]]]]

]

= 2119903[[

[

119863lowast

31198613

119861lowast

10

0 1198602

]]

]

(79)

(1)hArr (2) We separate the four equations in system (6) intothree groups

1198601119884 = 119862

1 119884119861

1= 119863

1 (80)

1198602119883 = 119862

2 (81)

1198613119883 + (119861

3119883)

lowast

+ 1198623119884119863

3+ (119862

3119884119863

3)lowast

= 1198603 (82)

By Lemma 1 we obtain that system (80) is solvable if andonly if (14) (81) is consistent if and only if (17) The generalsolutions to system (80) and (81) can be expressed as (16) and(19) respectively Substituting (16) and (19) into (82) yields

119860119880 + (119860119880)lowast

+ 119861119881119862 + (119861119881119862)lowast

= 119863 (83)

Hence the system (5) is consistent if and only if (80) (81) and(83) are consistent respectively It follows fromLemma 11 that(83) is solvable if and only if

119877119872119877

119865119864 = 0 119877

119865119864119877

119865= 0 119871

119866119864119871

119866= 0 (84)

We know by Lemma 11 that the general solution of (83) canbe expressed as (77)

InTheorem 12 let 1198601and119863

1vanishThen we can obtain

the general solution to the following system

1198602119883 = 119862

2 119884119861

1= 119863

1

1198613119883 + (119861

3119883)

lowast

+ 1198623119884119863

3+ (119862

3119884119863

3)lowast

= 1198603

(85)

Corollary 13 Let 1198602 119862

2 119861

1 119863

1 119861

3 119862

3 119863

3 and 119860

3= 119860

lowast

3

be given Set

119860 = 1198613119871

1198602 119862 = 119877

1198611119863

3

119865 = 119877119860119862

3 119866 = 119862119877

119860

119872 = 119877119865119866

lowast

119873 = 119865lowast

119871119866

119878 = 119866lowast

119871119872

119863 = 1198603minus 119861

3119860

dagger

2119862

2minus (119861

3119860

dagger

2119862

2)lowast

minus 1198623119863

1119861dagger

1119863

3minus (119862

3119863

1119861dagger

1119863

3)lowast

119864 = 119877119860119863119877

119860

(86)

10 Journal of Applied Mathematics

Then the following statements are equivalent

(1) system (85) is consistent(2)

1198771198602119862

2= 0 119863

1119871

1198611= 0 119877

119872119877

119865119864 = 0 (87)

119877119865119864119877

119865= 0 119871

119866119864119871

119866= 0 (88)

(3)

119903 [1198602119862

2] = 119903 (119860

2) [

1198631

1198611

] = 119903 (1198611)

119903

[[[

[

1198603

1198623119863

lowast

31198613119862

lowast

2

119861lowast

30 0 0 119860

lowast

2

119863lowast

1119862

lowast

30 119861

lowast

10 0

1198622

0 0 11986020

]]]

]

= 119903[

[

1198623119863

lowast

31198613

0 119861lowast

10

0 0 1198602

]

]

+ 119903 [119860

2

1198613

]

119903

[[[

[

1198603119862

31198613119862

lowast

2

119862lowast

30 0 0

119861lowast

30 0 119860

lowast

2

11986220 119860

20

]]]

]

= 2119903 [119862

31198613

0 1198602

]

119903

[[[[[

[

1198603

119863lowast

31198613119862

3119863

1119862

lowast

2

1198633

0 0 1198611

0

119861lowast

30 0 0 119860

lowast

2

119863lowast

1119862

lowast

3119861lowast

10 0 0

1198622

0 1198602

0 0

]]]]]

]

= 2119903[

[

119863lowast

31198613

119861lowast

10

0 1198602

]

]

(89)

In this case the general solution of system (6) can be expressedas

119883 = 119860dagger

2119862

2+ 119871

1198602119880

119884 = 1198631119861dagger

1+ 119881119877

1198611

(90)

where

119881 =1

2[119865

dagger

119864119866dagger

minus 119865dagger

119866lowast

119872dagger

119864119866dagger

minus 119865dagger

119878(119866dagger

)lowast

119864119873dagger

119865lowast

119866dagger

+119865dagger

119864(119872dagger

)lowast

+ (119873dagger

)lowast

119864119866dagger

119878dagger

119878] + 1198711198651198811

+ 1198812119877

119866+ 119880

1119871

119878119871

119872+ 119877

119873119880

lowast

2119871

119872minus 119865

dagger

1198781198802119877

119873119865

lowast

119866dagger

119880 = 119860dagger

[119863 minus 1198623119881119862 minus (119862

3119881119862)

lowast

]

minus1

2119860

dagger

[119863 minus 1198623119881119862 minus (119862

3119881119862)

lowast

] 119860119860dagger

minus 119860dagger

1198821119860

lowast

+119882lowast

1119860119860

dagger

+ 119871119860119882

2

(91)

where 1198801 119880

2 119881

1 119881

2119882

1 and119882

2are arbitrary matrices over

C with appropriate sizes

Acknowledgments

The authors would like to thank Dr Mohamed Dr Sivaku-mar and a referee very much for their valuable suggestions

and comments which resulted in a great improvement of theoriginal paper This research was supported by the Grantsfrom the National Natural Science Foundation of China(NSFC (11161008)) and Doctoral Program Fund of Ministryof Education of PRChina (20115201110002)

References

[1] H Zhang ldquoRank equalities for Moore-Penrose inverse andDrazin inverse over quaternionrdquo Annals of Functional Analysisvol 3 no 2 pp 115ndash127 2012

[2] L Bao Y Lin and YWei ldquoA new projectionmethod for solvinglarge Sylvester equationsrdquo Applied Numerical Mathematics vol57 no 5ndash7 pp 521ndash532 2007

[3] MDehghan andMHajarian ldquoOn the generalized reflexive andanti-reflexive solutions to a system of matrix equationsrdquo LinearAlgebra and Its Applications vol 437 no 11 pp 2793ndash2812 2012

[4] F O Farid M S Moslehian Q-W Wang and Z-C Wu ldquoOnthe Hermitian solutions to a system of adjointable operatorequationsrdquo Linear Algebra and Its Applications vol 437 no 7pp 1854ndash1891 2012

[5] Z H He and Q W Wang ldquoA real quaternion matrix equationwith with applicationsrdquo Linear Multilinear Algebra vol 61 pp725ndash740 2013

[6] Q-W Wang and Z-C Wu ldquoCommon Hermitian solutionsto some operator equations on Hilbert 119862119909-modulesrdquo LinearAlgebra and Its Applications vol 432 no 12 pp 3159ndash3171 2010

[7] Q-W Wang and C-Z Dong ldquoPositive solutions to a systemof adjointable operator equations over Hilbert 119862119909-modulesrdquoLinear Algebra and Its Applications vol 433 no 7 pp 1481ndash14892010

[8] Q-W Wang J W van der Woude and H-X Chang ldquoA systemof real quaternion matrix equations with applicationsrdquo LinearAlgebra and Its Applications vol 431 no 12 pp 2291ndash2303 2009

[9] Q W Wang H-S Zhang and G-J Song ldquoA new solvable con-dition for a pair of generalized Sylvester equationsrdquo ElectronicJournal of Linear Algebra vol 18 pp 289ndash301 2009

[10] H W Braden ldquoThe equations 119860119879

119883 plusmn 119883119879

119860 = 119861rdquo SIAM Journalon Matrix Analysis and Applications vol 20 no 2 pp 295ndash3021999

[11] D S Djordjevic ldquoExplicit solution of the operator equation119860

lowast

119883 + 119883plusmn

119860 = 119861rdquo Journal of Computational and Applied Math-ematics vol 200 no 2 pp 701ndash704 2007

[12] W Cao ldquoSolvability of a quaternion matrix equationrdquo AppliedMathematics vol 17 no 4 pp 490ndash498 2002

[13] Q Xu L Sheng and Y Gu ldquoThe solutions to some operatorequationsrdquo Linear Algebra and Its Applications vol 429 no 8-9pp 1997ndash2024 2008

[14] Q-W Wang and Z-H He ldquoSome matrix equations with appli-cationsrdquo Linear and Multilinear Algebra vol 60 no 11-12 pp1327ndash1353 2012

[15] Y Tian ldquoMaximization and minimization of the rank andinertia of the Hermitian matrix expression 119860 minus 119861119883 minus (119861119883)

119909rdquoLinear Algebra and Its Applications vol 434 no 10 pp 2109ndash2139 2011

[16] Z-H He and Q-WWang ldquoSolutions to optimization problemson ranks and inertias of a matrix function with applicationsrdquoAppliedMathematics andComputation vol 219 no 6 pp 2989ndash3001 2012

[17] Y Liu andY Tian ldquoMax-min problems on the ranks and inertiasof the matrix expressions 119860 minus 119861119883119862 plusmn (119861119883119862)

lowastrdquo Journal of

Journal of Applied Mathematics 11

Optimization Theory and Applications vol 148 no 3 pp 593ndash622 2011

[18] DChu Y SHung andH JWoerdeman ldquoInertia and rank cha-racterizations of some matrix expressionsrdquo SIAM Journal onMatrix Analysis and Applications vol 31 no 3 pp 1187ndash12262009

[19] Y Liu and Y Tian ldquoA simultaneous decomposition of a mat-rix triplet with applicationsrdquoNumerical Linear Algebra with Ap-plications vol 18 no 1 pp 69ndash85 2011

[20] X Zhang Q-W Wang and X Liu ldquoInertias and ranks of someHermitian matrix functions with applicationsrdquo Central Euro-pean Journal of Mathematics vol 10 no 1 pp 329ndash351 2012

[21] Q-W Wang ldquoThe general solution to a system of real quater-nion matrix equationsrdquo Computers amp Mathematics with Appli-cations vol 49 no 5-6 pp 665ndash675 2005

[22] Y Tian ldquoEqualities and inequalities for inertias of Hermitianmatrices with applicationsrdquo Linear Algebra and Its Applicationsvol 433 no 1 pp 263ndash296 2010

[23] G Marsaglia and G P H Styan ldquoEqualities and inequalities forranks of matricesrdquo Linear and Multilinear Algebra vol 2 pp269ndash292 1974

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 230408 6 pageshttpdxdoiorg1011552013230408

Research ArticleLeft and Right Inverse Eigenpairs Problemfor 120581-Hermitian Matrices

Fan-Liang Li1 Xi-Yan Hu2 and Lei Zhang2

1 Institute of Mathematics and Physics School of Sciences Central South University of Forestry and TechnologyChangsha 410004 China

2 College of Mathematics and Econometrics Hunan University Changsha 410082 China

Correspondence should be addressed to Fan-Liang Li lfl302tomcom

Received 6 December 2012 Accepted 20 March 2013

Academic Editor Panayiotis J Psarrakos

Copyright copy 2013 Fan-Liang Li et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Left and right inverse eigenpairs problem for 120581-hermitian matrices and its optimal approximate problem are considered Based onthe special properties of 120581-hermitian matrices the equivalent problem is obtained Combining a new inner product of matricesthe necessary and sufficient conditions for the solvability of the problem and its general solutions are derived Furthermore theoptimal approximate solution and a calculation procedure to obtain the optimal approximate solution are provided

1 Introduction

Throughout this paper we use some notations as follows Let119862119899times119898 be the set of all 119899times119898 complex matrices119880119862

119899times119899119867119862119899times119899

119878119867119862119899times119899 denote the set of all 119899times119899 unitary matrices hermitian

matrices skew-hermitian matrices respectively Let 119860 119860119867and119860

+ be the conjugate conjugate transpose and theMoore-Penrose generalized inverse of 119860 respectively For 119860 119861 isin

119862119899times119898 ⟨119860 119861⟩ = re(tr(119861119867119860)) where re(tr(119861119867119860)) denotes the

real part of tr(119861119867119860) the inner product of matrices 119860 and 119861The induced matrix norm is called Frobenius norm That is119860 = ⟨119860 119860⟩

12

= (tr(119860119867119860))12

Left and right inverse eigenpairs problem is a specialinverse eigenvalue problem That is giving partial left andright eigenpairs (eigenvalue and corresponding eigenvector)(120582119894 119909119894) 119894 = 1 ℎ (120583

119895 119910119895) 119895 = 1 119897 a special matrix set

119878 finding a matrix 119860 isin 119878 such that

119860119909119894= 120582119894119909119894 119894 = 1 ℎ

119910119879

119895119860 = 120583

119895119910119879

119895 119895 = 1 119897

(1)

This problem which usually arises in perturbation analysisof matrix eigenvalues and in recursive matters has profoundapplication background [1ndash6] When the matrix set 119878 isdifferent it is easy to obtain different left and right inverse

eigenpairs problem For example we studied the left andright inverse eigenpairs problem of skew-centrosymmetricmatrices and generalized centrosymmetric matrices respec-tively [5 6] Based on the special properties of left and righteigenpairs of these matrices we derived the solvability condi-tions of the problem and its general solutions In this papercombining the special properties of 120581-hermitianmatrices anda new inner product ofmatrices we first obtain the equivalentproblem then derive the necessary and sufficient conditionsfor the solvability of the problem and its general solutions

Hill and Waters [7] introduced the following matrices

Definition 1 Let 120581 be a fixed product of disjoint transposi-tions and let119870 be the associated permutation matrix that is119870 = 119870

119867

= 119870 1198702 = 119868119899 a matrix 119860 isin 119862

119899times119899 is said to be 120581-hermitian matrices (skew 120581-hermitian matrices) if and onlyif 119886119894119895= 119886119896(119895)119896(119894)

(119886119894119895= minus119886119896(119895)119896(119894)

) 119894 119895 = 1 119899 We denote theset of 120581-hermitian matrices (skew 120581-hermitian matrices) by119870119867119862119899times119899

(119878119870119867119862119899times119899

)

FromDefinition 1 it is easy to see that hermitianmatricesand perhermitian matrices are special cases of 120581-hermitianmatrices with 119896(119894) = 119894 and 119896(119894) = 119899 minus 119894 + 1 respectivelyHermitian matrices and perhermitian matrices which areone of twelve symmetry patterns of matrices [8] are appliedin engineering statistics and so on [9 10]

2 Journal of Applied Mathematics

From Definition 1 it is also easy to prove the followingconclusions

(1) 119860 isin 119870119867119862119899times119899 if and only if 119860 = 119870119860

119867

119870(2) 119860 isin 119878119870119867119862

119899times119899 if and only if 119860 = minus119870119860119867

119870(3) If 119870 is a fixed permutation matrix then 119870119867119862

119899times119899 and119878119870119867119862

119899times119899 are the closed linear subspaces of 119862119899times119899 andsatisfy

119862119899times119899

= 119870119867119862119899times119899

⨁119878119870119867119862119899times119899

(2)

The notation 1198811oplus 1198812stands for the orthogonal direct sum of

linear subspace 1198811and 119881

2

(4) 119860 isin 119870119867119862119899times119899 if and only if there is a matrix isin

119867119862119899times119899 such that = 119870119860

(5) 119860 isin 119878119870119867119862119899times119899 if and only if there is a matrix isin

119878119867119862119899times119899 such that = 119870119860

Proof (1) From Definition 1 if 119860 = (119886119894119895) isin 119870119867119862

119899times119899 then119886119894119895= 119886119896(119895)119896(119894)

this implies119860 = 119870119860119867

119870 for119870119860119867

119870 = (119886119896(119895)119896(119894)

)(2)With the samemethod we can prove (2) So the proof

is omitted

(3) (a) For any 119860 isin 119862119899times119899 there exist 119860

1isin 119870119867119862

119899times1198991198602isin 119878119870119867119862

119899times119899 such that

119860 = 1198601+ 1198602 (3)

where 1198601= (12)(A + 119870119860

119867

119870) 1198602= (12)(119860 minus

119870119860119867

119870)(b) If there exist another 119860

1isin 119870119867119862

119899times119899 1198602

isin

119878119870119867119862119899times119899 such that

119860 = 1198601+ 1198602 (4)

(3)-(4) yields

1198601minus 1198601= minus (119860

2minus 1198602) (5)

Multiplying (5) on the left and on the right by119870 respectively and according to (1) and (2) weobtain

1198601minus 1198601= 1198602minus 1198602 (6)

Combining (5) and (6) gives1198601= 11986011198602= 1198602

(c) For any1198601isin 119870119867119862

119899times1198991198602isin 119878119870119867119862

119899times119899 we have

⟨1198601 1198602⟩ = re (tr (119860119867

21198601)) = re (tr (119870119860

119867

21198701198701198601119870))

= re (tr (minus11986011986721198601)) = minus ⟨119860

1 1198602⟩

(7)

This implies ⟨1198601 1198602⟩ = 0 Combining (a) (b)

and (c) gives (3)

(4) Let = 119870119860 if 119860 isin 119870119867119862119899times119899 then

119867

= isin 119867119862119899times119899

If 119867 = isin 119867119862119899timesn then 119860 = 119870 and 119870119860

119867

119870 = 119870119867

119870119870 =

119870 = 119860 isin 119870119867119862119899times119899

(5)With the samemethod we can prove (5) So the proofis omitted

In this paper we suppose that 119870 is a fixed permutationmatrix and assume (120582

119894 119909119894) 119894 = 1 ℎ be right eigenpairs

of 119860 (120583119895 119910119895) 119895 = 1 119897 be left eigenpairs of 119860 If we let

119883 = (1199091 119909

ℎ) isin 119862

119899timesℎ Λ = diag (1205821 120582

ℎ) isin 119862

ℎtimesℎ119884 = (119910

1 119910

119897) isin 119862119899times119897 Γ = diag(120583

1 120583

119897) isin 119862119897times119897 then the

problems studied in this paper can be described as follows

Problem 2 Giving 119883 isin 119862119899timesℎ Λ = diag(120582

1 120582

ℎ) isin 119862

ℎtimesℎ119884 isin 119862

119899times119897 Γ = diag(1205831 120583

119897) isin 119862119897times119897 find 119860 isin 119870119867119862

119899times119899 suchthat

119860119883 = 119883Λ

119884119879

119860 = Γ119884119879

(8)

Problem 3 Giving 119861 isin 119862119899times119899 find isin 119878

119864such that

10038171003817100381710038171003817119861 minus

10038171003817100381710038171003817= minforall119860isin119878119864

119861 minus 119860 (9)

where 119878119864is the solution set of Problem 2

This paper is organized as follows In Section 2 we firstobtain the equivalent problemwith the properties of119870119867119862

119899times119899

and then derive the solvability conditions of Problem 2 and itsgeneral solutionrsquos expression In Section 3 we first attest theexistence and uniqueness theorem of Problem 3 then presentthe unique approximation solution Finally we provide acalculation procedure to compute the unique approximationsolution and numerical experiment to illustrate the resultsobtained in this paper correction

2 Solvability Conditions of Problem 2

We first discuss the properties of119870119867119862119899times119899

Lemma 4 Denoting 119872 = 119870119864119870119866119864 and 119864 isin 119867119862119899times119899 one has

the following conclusions(1) If 119866 isin 119870119867119862

119899times119899 then119872 isin 119870119867119862119899times119899

(2) If 119866 isin 119878119870119867119862119899times119899 then119872 isin 119878119870119867119862

119899times119899(3) If 119866 = 119866

1+ 1198662 where 119866

1isin 119870119867119862

119899times119899 1198662isin 119878119870119867119862

119899times119899then 119872 isin 119870119867119862

119899times119899 if and only if 1198701198641198701198662119864 = 0 In

addition one has119872 = 1198701198641198701198661119864

Proof (1) 119870119872119867

119870 = 119870119864119866119867

119870119864119870119870 = 119870119864(119870119866119870)119870119864 =

119870119864119870119866119864 = 119872Hence we have119872 isin 119870119867119862

119899times119899(2) 119870119872

119867

119870 = 119870119864119866119867

119870119864119870119870 = 119870119864(minus119870119866119870)119870119864 =

minus119870119864119870119866119864 = minus119872Hence we have119872 isin 119878119870119867119862

119899times119899(3) 119872 = 119870119864119870(119866

1+ 1198662)119864 = 119870119864119870119866

1119864 + 119870119864119870119866

2119864 we

have 1198701198641198701198661119864 isin 119870119867119862

119899times119899 1198701198641198701198662119864 isin 119878119870119867119862

119899times119899 from (1)and (2) If 119872 isin 119870119867119862

119899times119899 then 119872 minus 1198701198641198701198661119864 isin 119870119867119862

119899times119899while119872minus119870119864119870119866

1119864 = 119870119864119870119866

2119864 isin 119878119870119867119862

119899times119899Therefore fromthe conclusion (3) of Definition 1 we have119870119864119870119866

2119864 = 0 that

is119872 = 1198701198641198701198661119864 On the contrary if119870119864119870119866

2119864 = 0 it is clear

that119872 = 1198701198641198701198661119864 isin 119870119867119862

119899times119899 The proof is completed

Lemma 5 Let 119860 isin 119870119867119862119899times119899 if (120582 119909) is a right eigenpair of 119860

then (120582 119870119909) is a left eigenpair of 119860

Journal of Applied Mathematics 3

Proof If (120582 119909) is a right eigenpair of 119860 then we have

119860119909 = 120582119909 (10)

From the conclusion (1) of Definition 1 it follows that

119870119860119867

119870119909 = 120582119909 (11)

This implies

(119870119909)119879

119860 = 120582(119870119909)119879

(12)

So (120582 119870119909) is a left eigenpair of 119860

From Lemma 5 without loss of the generality we mayassume that Problem 2 is as follows

119883 isin 119862119899timesℎ

Λ = diag (1205821 120582

ℎ) isin 119862ℎtimesℎ

119884 = 119870119883 isin 119862119899timesℎ

Γ = Λ isin 119862ℎtimesℎ

(13)

Combining (13) and the conclusion (4) of Definition 1 itis easy to derive the following lemma

Lemma 6 If 119883 Λ 119884 Γ are given by (13) then Problem 2 isequivalent to the following problem If 119883 Λ 119884 Γ are given by(13) find 119870119860 isin 119867119862

119899times119899 such that

119870119860119883 = 119870119883Λ (14)

Lemma 7 (see [11]) If giving119883 isin 119862119899timesℎ119861 isin 119862

119899timesℎ thenmatrixequation 119883 = 119861 has solution isin 119867119862

119899times119899 if and only if

119861 = 119861119883+

119883 119861119867

119883 = 119883119867

119861 (15)

Moreover the general solution can be expressed as

= 119861119883+

+ (119861119883+

)119867

(119868119899minus 119883119883

+

)

+ (119868119899minus 119883119883

+

) (119868119899minus 119883119883

+

) forall isin 119867119862119899times119899

(16)

Theorem 8 If119883Λ119884 Γ are given by (13) then Problem 2 hasa solution in 119870119867119862

119899times119899 if and only if

119883119867

119870119883Λ = Λ119883119867

119870119883 119883Λ = 119883Λ119883+

119883 (17)

Moreover the general solution can be expressed as

119860 = 1198600+ 119870119864119870119866119864 forall119866 isin 119870119867119862

119899times119899

(18)

where

1198600= 119883Λ119883

+

+ 119870(119883Λ119883+

)119867

119870119864 119864 = 119868119899minus 119883119883

+

(19)

Proof Necessity If there is a matrix 119860 isin 119870119867119862119899times119899 such that

(119860119883 = 119883Λ 119884119879119860 = Γ119884119879

) then from Lemma 6 there exists amatrix 119870119860 isin 119867119862

119899times119899 such that 119870119860119883 = 119870119883Λ and accordingto Lemma 7 we have

119870119883Λ = 119870119883Λ119883+

119883 (119870119883Λ)119867

119883 = 119883119867

(119870119883Λ) (20)

It is easy to see that (20) is equivalent to (17)

Sufficiency If (17) holds then (20) holds Hence matrixequation119870119860119883 = 119870119883Λ has solution119870119860 isin 119867119862

119899times119899 Moreoverthe general solution can be expressed as follows

119870119860 = 119870119883Λ119883+

+ (119870119883Λ119883+

)119867

(119868119899minus 119883119883

+

)

+ (119868119899minus 119883119883

+

) (119868119899minus 119883119883

+

) forall isin 119867119862119899times119899

(21)

Let

1198600= 119883Λ119883

+

+ 119870(119883Λ119883+

)119867

119870119864 119864 = 119868119899minus 119883119883

+

(22)

This implies 119860 = 1198600+ 119870119864119864 Combining the definition of

119870 119864 and the first equation of (17) we have

119870119860119867

0119870 = 119870(119883Λ119883

+

)119867

119870 + 119883Λ119883+

minus 119870(119883119883+

)119867

119870(119883Λ119883+

)

= 119883Λ119883+

+ 119870(119883Λ119883+

)119867

119870 minus 119870(119883Λ119883+

)119879

119870119883119883+

= 1198600

(23)

Hence1198600isin 119870119867119862

119899times119899 Combining the definition of119870 119864 (13)and (17) we have

1198600119883 = 119883Λ119883

+

119883 + 119870(119883Λ119883+

)119867

119870(119868119899minus 119883119883

+

)119883 = 119883Λ

119884119879

1198600= 119883119867

119870119883Λ119883+

+ 119883119867

119870119870(119883Λ119883+

)119867

119870119864

= Λ119883119867

119870119883119883+

+ (119870119883Λ119883+

119883)119867

(119868119899minus 119883119883

+

)

= Λ119883119867

119870119883119883+

+ Λ119883119867

119870(119868119899minus 119883119883

+

)

= Λ119883119867

119870 = Γ119884119879

(24)

Therefore 1198600is a special solution of Problem 2 Combining

the conclusion (4) of Definition 1 Lemma 4 and 119864 = 119868119899minus

119883119883+

isin 119867119862119899times119899 it is easy to prove that 119860 = 119860

0+ 119870119864119870119866119864 isin

119870119867119862119899times119899 if and only if 119866 isin 119870119867119862

119899times119899 Hence the solution setof Problem 2 can be expressed as (18)

3 An Expression of the Solution of Problem 3

From (18) it is easy to prove that the solution set 119878119864of

Problem 2 is a nonempty closed convex set if Problem 2 hasa solution in 119870119867119862

119899times119899 We claim that for any given 119861 isin 119877119899times119899

there exists a unique optimal approximation for Problem 3

Theorem9 Giving119861 isin 119862119899times119899 if the conditions of119883119884Λ Γ are

the same as those in Theorem 8 then Problem 3 has a uniquesolution isin 119878

119864 Moreover can be expressed as

= 1198600+ 119870119864119870119861

1119864 (25)

where 1198600 119864 are given by (19) and 119861

1= (12)(119861 + 119870119861

119867

119870)

Proof Denoting1198641= 119868119899minus119864 it is easy to prove thatmatrices119864

and1198641are orthogonal projectionmatrices satisfying119864119864

1= 0

It is clear that matrices 119870119864119870 and 1198701198641119870 are also orthogonal

4 Journal of Applied Mathematics

projection matrices satisfying (119870119864119870)(1198701198641119870) = 0 According

to the conclusion (3) of Definition 1 for any 119861 isin 119862119899times119899 there

exists unique

1198611isin 119870119867119862

119899times119899

1198612isin 119878119870119867119862

119899times119899 (26)

such that

119861 = 1198611+ 1198612 ⟨119861

1 1198612⟩ = 0 (27)

where

1198611=

1

2(119861 + 119870119861

119867

119870) 1198612=

1

2(119861 minus 119870119861

119867

119870) (28)

CombiningTheorem 8 for any 119860 isin 119878119864 we have

119861 minus 1198602

=1003817100381710038171003817119861 minus 119860

0minus 119870119864119870119866119864

1003817100381710038171003817

2

=10038171003817100381710038171198611 + 119861

2minus 1198600minus 119870119864119870119866119864

1003817100381710038171003817

2

=10038171003817100381710038171198611 minus 119860

0minus 119870119864119870119866119864

1003817100381710038171003817

2

+10038171003817100381710038171198612

1003817100381710038171003817

2

=1003817100381710038171003817(1198611 minus 119860

0) (119864 + 119864

1) minus 119870119864119870119866119864

1003817100381710038171003817

2

+10038171003817100381710038171198612

1003817100381710038171003817

2

=1003817100381710038171003817(1198611 minus 119860

0) 119864 minus 119870119864119870119866119864

1003817100381710038171003817

2

+1003817100381710038171003817(1198611 minus 119860

0) 1198641

1003817100381710038171003817

2

+10038171003817100381710038171198612

1003817100381710038171003817

2

=1003817100381710038171003817119870 (119864 + 119864

1)119870 (119861

1minus 1198600) 119864 minus 119870119864119870119866119864

1003817100381710038171003817

2

+1003817100381710038171003817(1198611 minus 119860

0) 1198641

1003817100381710038171003817

2

+10038171003817100381710038171198612

1003817100381710038171003817

2

=1003817100381710038171003817119870119864119870 (119861

1minus 1198600) 119864 minus 119870119864119870119866119864

1003817100381710038171003817

2

+10038171003817100381710038171198701198641119870(1198611minus 1198600) 119864

1003817100381710038171003817

2

+1003817100381710038171003817(1198611 minus 119860

0)1198641

1003817100381710038171003817

2

+10038171003817100381710038171198612

1003817100381710038171003817

2

(29)

It is easy to prove that 1198701198641198701198600119864 = 0 according to the

definitions of 1198600 119864 So we have

119861 minus 1198602

=1003817100381710038171003817119870119864119870119861

1119864 minus 119870119864119870119866119864

1003817100381710038171003817

2

+10038171003817100381710038171198701198641119870(1198611minus 1198600) 119864

1003817100381710038171003817

2

+1003817100381710038171003817(1198611 minus 119860

0)1198641

1003817100381710038171003817

2

+10038171003817100381710038171198612

1003817100381710038171003817

2

(30)

Obviously min119860isin119878119864

119861 minus 119860 is equivalent to

min119866isin119870119867119862

119899times119899

10038171003817100381710038171198701198641198701198611119864 minus 119870119864119870119866119864

1003817100381710038171003817 (31)

Since 1198641198641

= 0 (119870119864119870)(1198701198641119870) = 0 it is clear that 119866 =

1198611+ 11987011986411198701198641 for any isin 119870119867119862

119899times119899 is a solution of (31)Substituting this result to (18) we can obtain (25)

Algorithm 10 (1) Input 119883 Λ 119884 Γ according to (13) (2)Compute 119883

119867

119870119883Λ Λ119883119867

119870119883 119883Λ119883+

119883 119883Λ if (17) holdsthen continue otherwise stop (3) Compute 119860

0according to

(19) and compute 1198611according to (28) (4) According to (25)

calculate

Example 11 (119899 = 8 ℎ = 119897 = 4)

119883 =

((((

(

05661 minus02014 minus 01422119894 01446 + 02138119894 0524

minus02627 + 01875119894 05336 minus02110 minus 04370119894 minus00897 + 03467119894

minus04132 + 02409119894 00226 minus 00271119894 minus01095 + 02115119894 minus03531 minus 00642119894

minus00306 + 02109119894 minus03887 minus 00425119894 02531 + 02542119894 00094 + 02991119894

00842 minus 01778119894 minus00004 minus 03733119894 03228 minus 01113119894 01669 + 01952119894

00139 minus 03757119894 minus02363 + 03856119894 02583 + 00721119894 01841 minus 02202119894

00460 + 03276119894 minus01114 + 00654119894 minus00521 minus 02556119894 minus02351 + 03002119894

00085 minus 01079119894 00974 + 03610119894 05060 minus02901 minus 00268119894

))))

)

Λ = (

minus03967 minus 04050119894 0 0 0

0 minus03967 + 04050119894 0 0

0 0 00001 0

0 0 0 minus00001119894

)

119870 =

((((

(

0 0 0 0 0 0 0 1

0 0 0 0 0 0 1 0

0 0 0 0 0 1 0 0

0 0 0 1 0 0 0 0

0 0 0 0 1 0 0 0

0 0 1 0 0 0 0 0

0 1 0 0 0 0 0 0

1 0 0 0 0 0 0 0

))))

)

119884 = 119870119883 Γ = Λ

(32)

Journal of Applied Mathematics 5

119861 =

From the first column to the fourth column

((((

(

minus05218 + 00406119894 02267 minus 00560119894 minus01202 + 00820119894 minus00072 minus 03362119894

03909 minus 03288119894 02823 minus 02064119894 minus00438 minus 00403119894 02707 + 00547119894

02162 minus 01144119894 minus04307 + 02474119894 minus00010 minus 00412119894 02164 minus 01314119894

minus01872 minus 00599119894 minus00061 + 04698119894 03605 minus 00247119894 04251 + 01869119894

minus01227 minus 00194119894 02477 minus 00606119894 03918 + 06340119894 01226 + 00636119894

minus00893 + 04335119894 00662 + 00199119894 minus00177 minus 01412119894 04047 + 02288119894

01040 minus 02015119894 01840 + 02276119894 02681 minus 03526119894 minus05252 + 01022119894

01808 + 02669119894 02264 + 03860119894 minus01791 + 01976119894 minus00961 minus 00117119894

(33)

From the fifth column to the eighth column

minus02638 minus 04952119894 minus00863 minus 01664119894 02687 + 01958119894 minus02544 minus 01099119894

minus02741 minus 01656119894 minus00227 + 02684119894 01846 + 02456119894 minus00298 + 05163119894

minus01495 minus 03205119894 01391 + 02434119894 01942 minus 05211119894 minus03052 minus 01468119894

minus02554 + 02690119894 minus04222 minus 01080119894 02232 + 00774119894 00965 minus 00421119894

03856 minus 00619119894 01217 minus 00270119894 01106 minus 03090119894 minus01122 + 02379119894

minus01130 + 00766119894 07102 minus 00901119894 01017 + 01397119894 minus00445 + 00038119894

01216 + 00076119894 02343 minus 01772119894 05242 minus 00089119894 minus00613 + 00258119894

minus00750 minus 03581119894 00125 + 00964119894 00779 minus 01074119894 06735 minus 00266119894

))))

)

(34)

It is easy to see that matrices 119883 Λ 119884 Γ satisfy (17) Hencethere exists the unique solution for Problem 3 Using the

software ldquoMATLABrdquo we obtain the unique solution ofProblem 3

From the first column to the fourth column

((((

(

minus01983 + 00491119894 01648 + 00032119894 00002 + 01065119894 01308 + 02690119894

01071 minus 02992119894 02106 + 02381119894 minus00533 minus 03856119894 minus00946 + 00488119894

01935 minus 01724119894 minus00855 minus 00370119894 00200 minus 00665119894 minus02155 minus 00636119894

00085 minus 02373119894 minus00843 minus 01920119894 00136 + 00382119894 minus00328 minus 00000119894

minus00529 + 01703119894 01948 minus 00719119894 01266 + 01752119894 00232 minus 02351119894

00855 + 01065119894 00325 minus 02068119894 02624 + 00000119894 00136 minus 00382119894

01283 minus 01463119894 minus00467 + 00000119894 00325 + 02067119894 minus00843 + 01920119894

02498 + 00000119894 01283 + 01463119894 00855 minus 01065119894 00086 + 02373119894

(35)

From the fifth column to the eighth column

02399 minus 01019119894 01928 minus 01488119894 minus03480 minus 02574119894 01017 minus 00000119894

minus01955 + 00644119894 02925 + 01872119894 03869 + 00000119894 minus03481 + 02574119894

00074 + 00339119894 minus03132 minus 00000119894 02926 minus 01872119894 01928 + 01488119894

00232 + 02351119894 minus02154 + 00636119894 minus00946 minus 00489119894 01309 minus 02691119894

minus00545 minus 00000119894 00074 minus 00339119894 minus01955 minus 00643119894 02399 + 01019119894

01266 minus 01752119894 00200 + 00665119894 minus00533 + 03857119894 00002 minus 01065119894

01949 + 00719119894 minus00855 + 00370119894 02106 minus 02381119894 01648 minus 00032119894

minus00529 minus 01703119894 01935 + 01724119894 01071 + 02992119894 minus01983 minus 00491119894

))))

)

(36)

6 Journal of Applied Mathematics

Conflict of Interests

There is no conflict of interests between the authors

Acknowledgments

This research was supported by National Natural ScienceFoundation of China (31170532)The authors are very gratefulto the referees for their valuable comments and also thank theeditor for his helpful suggestions

References

[1] J H Wilkinson The Algebraic Eigenvalue Problem OxfordUniversity Press Oxford UK 1965

[2] A Berman and R J Plemmons Nonnegative Matrices in theMathematical Sciences vol 9 ofClassics in AppliedMathematicsSIAM Philadelphia Pa USA 1994

[3] MAravDHershkowitz VMehrmann andH Schneider ldquoTherecursive inverse eigenvalue problemrdquo SIAM Journal on MatrixAnalysis and Applications vol 22 no 2 pp 392ndash412 2000

[4] R Loewy and V Mehrmann ldquoA note on the symmetricrecursive inverse eigenvalue problemrdquo SIAM Journal on MatrixAnalysis and Applications vol 25 no 1 pp 180ndash187 2003

[5] F L Li X YHu and L Zhang ldquoLeft and right inverse eigenpairsproblem of skew-centrosymmetric matricesrdquo Applied Mathe-matics and Computation vol 177 no 1 pp 105ndash110 2006

[6] F L Li X Y Hu and L Zhang ldquoLeft and right inverseeigenpairs problem of generalized centrosymmetric matricesand its optimal approximation problemrdquo Applied Mathematicsand Computation vol 212 no 2 pp 481ndash487 2009

[7] R D Hill and S R Waters ldquoOn 120581-real and 120581-Hermitianmatricesrdquo Linear Algebra and its Applications vol 169 pp 17ndash29 1992

[8] S P Irwin ldquoMatrices with multiple symmetry propertiesapplications of centro-Hermitian and per-Hermitian matricesrdquoLinear Algebra and its Applications vol 284 no 1ndash3 pp 239ndash258 1998

[9] L C Biedenharn and J D Louck Angular Momentum inQuantum Physics vol 8 of Encyclopedia of Mathematics and itsApplications Addison-Wesley Reading Mass USA 1981

[10] WM Gibson and B R Pollard Symmetry Principles in Elemen-tary Particle Physics Cambridge University Press CambridgeUK 1976

[11] W F Trench ldquoHermitian Hermitian R-symmetric and Her-mitian R-skew symmetric Procrustes problemsrdquo Linear Algebraand its Applications vol 387 pp 83ndash98 2004

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 271978 5 pageshttpdxdoiorg1011552013271978

Research ArticleCompleting a 2 times 2 Block Matrix of Real Quaternions witha Partial Specified Inverse

Yong Lin12 and Qing-Wen Wang1

1 Department of Mathematics Shanghai University Shanghai 200444 China2 School of Mathematics and Statistics Suzhou University Suzhou 234000 China

Correspondence should be addressed to Qing-Wen Wang wqw858yahoocomcn

Received 4 December 2012 Revised 23 February 2013 Accepted 20 March 2013

Academic Editor K Sivakumar

Copyright copy 2013 Y Lin and Q-W Wang This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

This paper considers a completion problem of a nonsingular 2 times 2 block matrix over the real quaternion algebra H Let1198981 1198982 1198991 1198992be nonnegative integers119898

1+ 1198982= 1198991+ 1198992= 119899 gt 0 and 119860

12isin H1198981times1198992 119860

21isin H1198982times1198991 119860

22isin H1198982times1198992 119861

11isin H1198991times1198981

be given We determine necessary and sufficient conditions so that there exists a variant block entry matrix11986011isin H1198981times1198991 such that

119860 = (11986011 11986012

11986021 11986022) isin H119899times119899 is nonsingular and 119861

11is the upper left block of a partitioning of 119860minus1 The general expression for 119860

11is also

obtained Finally a numerical example is presented to verify the theoretical findings

1 Introduction

The problem of completing a block-partitioned matrix of aspecified type with some of its blocks given has been studiedby many authors Fiedler and Markham [1] considered thefollowing completion problem over the real number field RSuppose119898

1 1198982 1198991 1198992are nonnegative integers119898

1+1198982=

1198991+1198992= 119899 gt 0 119860

11isin R1198981times1198991 119860

12isin R1198981times1198992 119860

21isin R1198982times1198991

and 11986122isin R1198992times1198982 Determine a matrix 119860

22isin R1198982times1198992 such

that

119860 = (11986011

11986012

11986021

11986022

) (1)

is nonsingular and 11986122is the lower right block of a partition-

ing of 119860minus1 This problem has the form of

(11986011

11986012

11986021

)

minus1

= (

11986122

) (2)

and the solution and the expression for 11986022were obtained in

[1] Dai [2] considered this form of completion problemswithsymmetric and symmetric positive definite matrices over R

Some other particular forms for 2times2 block matrices overR have also been examined (see eg [3]) such as

(11986011

11986012

11986021

)

minus1

= (11986111

)

(11986011

)

minus1

= (

11986122

)

(11986011

11986022

)

minus1

= ( 11986112

11986121

)

(3)

The real quaternion matrices play a role in computerscience quantum physics and so on (eg [4ndash6]) Quaternionmatrices are receiving much attention as witnessed recently(eg [7ndash9]) Motivated by the work of [1 10] and keepingsuch applications of quaternionmatrices in view in this paperwe consider the following completion problem over the realquaternion algebra

H = 1198860+ 1198861119894 + 1198862119895 + 1198863119896 |

1198942

= 1198952

= 1198962

= 119894119895119896 = minus1 and 1198860 1198861 1198862 1198863isin R

(4)

Problem 1 Suppose 1198981 1198982 1198991 1198992are nonnegative inte-

gers 1198981+ 1198982= 1198991+ 1198992= 119899 gt 0 and 119860

12isin H1198981times1198992

2 Journal of Applied Mathematics

11986021isin H1198982times1198991 119860

22isin 1198771198982times1198992 119861

11isin H1198991times1198981 Find a matrix

11986011isin H1198981times1198991 such that

119860 = (11986011

11986012

11986021

11986022

) isin H119899times119899 (5)

is nonsingular and11986111is the upper left block of a partitioning

of 119860minus1 That is

( 119860

12

11986021

11986022

)

minus1

= (11986111

) (6)

where H119898times119899 denotes the set of all 119898 times 119899matrices over H and119860minus1 denotes the inverse matrix of 119860

Throughout over the real quaternion algebra H wedenote the identity matrix with the appropriate size by 119868 thetranspose of 119860 by 119860119879 the rank of 119860 by 119903(119860) the conjugatetranspose of119860 by119860lowast = (119860)119879 a reflexive inverse of a matrix119860over H by 119860+ which satisfies simultaneously 119860119860+119860 = 119860 and119860+

119860119860+

= 119860+Moreover119871

119860= 119868minus119860

+

119860 119877119860= 119868minus119860119860

+ where119860+ is an arbitrary but fixed reflexive inverse of 119860 Clearly 119871

119860

and119877119860are idempotent and each is a reflexive inverse of itself

R(119860) denotes the right column space of the matrix 119860The rest of this paper is organized as follows In Section 2

we establish some necessary and sufficient conditions to solveProblem 1 over H and the general expression for 119860

11is also

obtained In Section 3 we present a numerical example toillustrate the developed theory

2 Main Results

In this section we begin with the following lemmas

Lemma 1 (singular-value decomposition [9]) Let 119860 isin H119898times119899

be of rank 119903 Then there exist unitary quaternion matrices 119880 isin

H119898times119898 and 119881 isin H119899times119899 such that

119880119860119881 = (1198631199030

0 0) (7)

where 119863119903= diag(119889

1 119889

119903) and the 119889

119895rsquos are the positive

singular values of 119860

Let H119899119888denote the collection of column vectors with 119899

components of quaternions and 119860 be an 119898 times 119899 quaternionmatrix Then the solutions of 119860119909 = 0 form a subspace of H119899

119888

of dimension 119899(119860) We have the following lemma

Lemma 2 Let

(11986011

11986012

11986021

11986022

) (8)

be a partitioning of a nonsingular matrix 119860 isin H119899times119899 and let

(11986111

11986112

11986121

11986122

) (9)

be the corresponding (ie transpose) partitioning of 119860minus1 Then119899(11986011) = 119899(119861

22)

Proof It is readily seen that

(11986122

11986121

11986112

11986111

)

(11986022

11986021

11986012

11986011

)

(10)

are inverse to each other so we may suppose that 119899(11986011) lt

119899(11986122)

If 119899(11986122) = 0 necessarily 119899(119860

11) = 0 and we are finished

Let 119899(11986122) = 119888 gt 0 then there exists a matrix 119865 with 119888 right

linearly independent columns such that 11986122119865 = 0 Then

using

1198601111986112+ 1198601211986122= 0 (11)

we have

1198601111986112119865 = 0 (12)

From

1198602111986112+ 1198602211986122= 119868 (13)

we have

1198602111986112119865 = 119865 (14)

It follows that the rank 119903(11986112119865) ge 119888 In view of (12) this

implies

119899 (11986011) ge 119903 (119861

12119865) ge 119888 = 119899 (119861

22) (15)

Thus

119899 (11986011) = 119899 (119861

22) (16)

Lemma 3 (see [10]) Let 119860 isin H119898times119899 119861 isin H119901times119902 119863 isin H119898times119902 beknown and 119883 isin H119899times119901 unknown Then the matrix equation

119860119883119861 = 119863 (17)

is consistent if and only if

119860119860+

119863119861+

119861 = 119863 (18)

In that case the general solution is

119883 = 119860+

119863119861+

+ 1198711198601198841+ 1198842119877119861 (19)

where11988411198842are any matrices with compatible dimensions over

H

By Lemma 1 let the singular value decomposition of thematrix 119860

22and 119861

11in Problem 1 be

11986022= 119876(

Λ 0

0 0)119877lowast

(20)

11986111= 119880(

Σ 0

0 0)119881lowast

(21)

Journal of Applied Mathematics 3

where Λ = diag(1205821 1205822 120582

119904) is a positive diagonal matrix

120582119894= 0 (119894 = 1 119904) are the singular values of 119860

22 119904 =

119903(11986022) Σ = diag(120590

1 1205902 120590

119903) is a positive diagonal matrix

120590119894= 0 (119894 = 1 119903) are the singular values of 119861

11and 119903 =

119903(11986111)119876 = (119876

11198762) isin H1198982times1198982 119877 = (119877

11198772) isin H1198992times1198992

119880 = (11988011198802) isin H1198991times1198991 119881 = (119881

11198812) isin H1198981times1198981 are unitary

quaternion matrices where 1198761isin H1198982times119904 119877

1isin H1198992times119904 119880

1isin

H1198991times119903 and 1198811isin H1198981times119903

Theorem 4 Problem 1 has a solution if and only if thefollowing conditions are satisfied

(a) 119903 ( 1198601211986022

) = 1198992

(b) 1198992minus 119903(119860

22) = 119898

1minus 119903(11986111) that is 119899

2minus 119904 = 119898

1minus 119903

(c) R(1198602111986111) subR(119860

22)

(d) R(119860lowast

12119861lowast

11) subR(119860

lowast

22)

In that case the general solution has the form of

11986011= 119861+

11+ 11986012119877(

Λminus1

119876lowast

1119860211198801Σ 0

119867 minus(119881lowast

2119860121198772)minus1)

times 119881lowast

119861+

11+ 119884 minus 119884119861

11119861+

11

(22)

where 119867 is an arbitrary matrix in H(1198992minus119904)times119903 and 119884 is anarbitrary matrix in H1198981times1198991

Proof If there exists an 1198981times 1198991matrix 119860

11such that 119860 is

nonsingular and 11986111is the corresponding block of 119860minus1 then

(a) is satisfied From 119860119861 = 119861119860 = 119868 we have that

1198602111986111+ 1198602211986121= 0

1198611111986012+ 1198611211986022= 0

(23)

so that (c) and (d) are satisfiedBy (11) we have

119903 (11986022) + 119899 (119860

22) = 1198992 119903 (119861

11) + 119899 (119861

11) = 119898

1 (24)

From Lemma 2 Notice that ( 11986011 1198601211986021 11986022

) is the correspondingpartitioning of 119861minus1 we have

119899 (11986111) = 119899 (119860

22) (25)

implying that (b) is satisfiedConversely from (c) we know that there exists a matrix

119870 isin H1198992times1198981 such that

1198602111986111= 11986022119870 (26)

Let

11986121= minus119870 (27)

From (20) (21) and (26) we have

11986021119880(

Σ 0

0 0)119881lowast

= 119876(Λ 0

0 0)119877lowast

119870 (28)

It follows that

119876lowast

11986021119880(

Σ 0

0 0)119881lowast

119881 = 119876lowast

119876(Λ 0

0 0)119877lowast

119870119881 (29)

This implies that

(

119876lowast

1119860211198801119876lowast

1119860211198802

119876lowast

2119860211198801119876lowast

2119860211198802

)(Σ 0

0 0)

= (Λ 0

0 0)(

119877lowast

11198701198811119877lowast

11198701198812

119877lowast

21198701198811119877lowast

21198701198812

)

(30)

Comparing corresponding blocks in (30) we obtain

119876lowast

2119860211198801= 0 (31)

Let 119877lowast119870119881 = From (29) (30) we have

= (Λminus1

119876lowast

1119860211198801Σ 0

119867 11987022

)

119867 isin H(1198992minus119904)times119903 119870

22isin H(1198992minus119904)times(1198981minus119903)

(32)

In the same way from (d) we can obtain

119881lowast

1119860121198772= 0 (33)

Notice that ( 1198601211986022

) in (a) is a full column rank matrix By (20)(21) and (33) we have

(0 119876lowast

119881lowast

0)(

11986012

11986022

)119877 = (

Λ 0

0 0

119881lowast

1119860121198771119881lowast

1119860121198772

119881lowast

2119860121198771119881lowast

2119860121198772

) (34)

so that

1198992= 119903(

11986012

11986022

) = 119903((0 119876lowast

119881lowast

0)(

11986012

11986022

)119877)

= 119903(

Λ 0

0 0

119881lowast

1119860121198771119881lowast

1119860121198772

119881lowast

2119860121198771119881lowast

2119860121198772

)

= 119903 (Λ) + 119903 (119881lowast

2119860121198772)

= 119904 + 119903 (119881lowast

2119860121198772)

(35)

It follows from (b) and (35) that 1198811198792119860121198772is a full column

rank matrix so it is nonsingularFrom 119860119861 = 119868 we have the following matrix equation

1198601111986111+ 1198601211986121= 119868 (36)

that is

1198601111986111= 119868 minus 119860

1211986121 119868 isin H

1198981times1198981 (37)

4 Journal of Applied Mathematics

where 11986111 11986012

were given 11986121

= minus119870 (from (27)) ByLemma 3 the matrix equation (37) has a solution if and onlyif

(119868 minus 1198601211986121) 119861+

1111986111= 119868 minus 119860

1211986121 (38)

By (21) (27) (32) and (33) we have that (38) is equivalent to

(119868 + 11986012119870)119881(

Σminus1

0

0 0)119880lowast

119880(Σ 0

0 0)119881lowast

= 119868 + 11986012119870 (39)

We simplify the equation aboveThe left hand side reduces to(119868 + 119860

12119870)1198811119881lowast

1and so we have

119860121198701198811119881lowast

1minus 11986012119870 = 119868 minus 119881

1119881lowast

1 (40)

So

11986012119877119881lowast

1198811119881lowast

1minus 11986012119877119881lowast

= (11988111198812) (119881lowast

1

119881lowast

2

) minus 1198811119881lowast

1

(41)

This implies that

11986012119877(

119881lowast

11198811

119881lowast

21198811

)119881lowast

1minus 11986012119877(

119881lowast

1

119881lowast

2

) = 1198812119881lowast

2 (42)

so that

11986012119877(

119868

0)119881lowast

1minus 11986012119877(

119881lowast

1

119881lowast

2

) = 1198812119881lowast

2 (43)

So

minus11986012119877(

0

119881lowast

2

) = 1198812119881lowast

2 (44)

and hence

minus (119860121198771119860121198772) (Λminus1

119876lowast

1119860211198801Σ 0

119867 11987022

)(0

119881lowast

2

) = 1198812119881lowast

2

(45)

Finally we obtain

11986012119877211987022119881lowast

2= minus1198812119881lowast

2 (46)

Multiplying both sides of (46) by 119881lowast from the left consider-ing (33) and the fact that 119881lowast

2119860121198772is nonsingular we have

11987022= minus(119881

lowast

2119860121198772)minus1

(47)

From Lemma 3 (38) (47) Problem 1 has a solution and thegeneral solution is

11986011= 119861+

11+ 11986012119877(

Λminus1

119876lowast

1119860211198801Σ 0

119867 minus(119881lowast

2119860121198772)minus1)

times 119881lowast

119861+

11+ 119884 minus 119884119861

11119861+

11

(48)

where 119867 is an arbitrary matrix in H(1198992minus119904)times119903 and 119884 is anarbitrary matrix in H1198981times1198991

3 An Example

In this section we give a numerical example to illustrate thetheoretical results

Example 5 Consider Problem 1 with the parameter matricesas follows

11986012= (

2 + 1198951

2119896

minus119896 1 +1

2119895

)

11986021= (

3

2+1

2119894 minus

1

2119895 minus

1

2119896

1

2119895 +

1

2119896

3

2+1

2119894

)

11986022= (

2 119894

2119895 119896) 119861

11= (

1 119894

119895 119896)

(49)

It is easy to show that (c) (d) are satisfied and that

1198992= 119903(

11986012

11986022

) = 2

1198992minus 119903 (119860

22) = 119898

1minus 119903 (119861

11) = 0

(50)

so (a) (b) are satisfied too Therefore we have

119861+

11= (

1

2minus1

2119895

minus1

2119894 minus

1

2119896

)

11986022= 119876(

Λ 0

0 0)119877lowast

11986111= 119880(

Σ 0

0 0)119881lowast

(51)

where

119876 =1

radic2

(1 119894

119895 119896) Λ = (

2radic2 0

0 radic2)

119877 = (1 0

0 1) 119880 =

1

radic2

(1 119894

119895 119896)

Σ = (radic2 0

0 radic2) 119881 = (

1 0

0 1)

(52)

We also have

1198761=

1

radic2

(1 119894

119895 119896) 119877

1= (

1 0

0 1)

1198801=

1

radic2

(1 119894

119895 119896) 119881

1= (

1 0

0 1)

(53)

Journal of Applied Mathematics 5

By Theorem 4 for an arbitrary matrices 119884 isin H2times2 wehave

11986011= 119861+

11+ 11986012119877 (Λminus1

119876lowast

1119860211198801Σ)119881lowast

119861+

11+ 119884 minus 119884119861

11119861+

11

= (

3

2+1

4119895 +

1

4119896

3

4+1

4119894 minus

3

2119895

1

2minus 119894 +

1

4119895 minus

1

41198961

4minus3

4119894 minus

1

2119895 minus 119896

)

(54)

it follows that

119860 =((

(

3

2

+

1

4

119895 +

1

4

119896

3

4

+

1

4

119894 minus

3

2

119895 2 + 119895

1

2

119896

1

2

minus 119894 +

1

4

119895 minus

1

4

119896

1

4

minus

3

4

119894 minus

1

2

119895 minus 119896 minus119896 1 +

1

2

119895

3

2

+

1

2

119894 minus

1

2

119895 minus

1

2

119896 2 119894

1

2

119895 +

1

2

119896

3

2

+

1

2

119894 2119895 119896

))

)

119860minus1

=(

1 119894 minus1 minus1

119895 119896 0 minus1

minus1 03

4

1

2minus3

4119895

minus1 minus11

2minus 119894

1

2minus1

2119894 minus

1

2119895 minus 119896

)

(55)

The results verify the theoretical findings of Theorem 4

Acknowledgments

The authors would like to give many thanks to the refereesand Professor K C Sivakumar for their valuable suggestionsand comments which resulted in a great improvement ofthe paper This research was supported by Grants from theKey Project of Scientific Research Innovation Foundation ofShanghai Municipal Education Commission (13ZZ080) theNational Natural Science Foundation of China (11171205) theNatural Science Foundation of Shanghai (11ZR1412500) theDiscipline Project at the corresponding level of Shanghai (A13-0101-12-005) and Shanghai Leading Academic DisciplineProject (J50101)

References

[1] M Fiedler and T L Markham ldquoCompleting a matrix whencertain entries of its inverse are specifiedrdquo Linear Algebra andIts Applications vol 74 pp 225ndash237 1986

[2] H Dai ldquoCompleting a symmetric 2 times 2 block matrix and itsinverserdquo Linear Algebra and Its Applications vol 235 pp 235ndash245 1996

[3] W W Barrett M E Lundquist C R Johnson and H JWoerdeman ldquoCompleting a block diagonal matrix with apartially prescribed inverserdquoLinearAlgebra and Its Applicationsvol 223224 pp 73ndash87 1995

[4] S L Adler Quaternionic Quantum Mechanics and QuantumFields Oxford University Press New York NY USA 1994

[5] A Razon and L P Horwitz ldquoUniqueness of the scalar productin the tensor product of quaternion Hilbert modulesrdquo Journalof Mathematical Physics vol 33 no 9 pp 3098ndash3104 1992

[6] J W Shuai ldquoThe quaternion NN model the identification ofcolour imagesrdquo Acta Comput Sinica vol 18 no 5 pp 372ndash3791995 (Chinese)

[7] F Zhang ldquoOn numerical range of normal matrices of quater-nionsrdquo Journal of Mathematical and Physical Sciences vol 29no 6 pp 235ndash251 1995

[8] F Z Zhang Permanent Inequalities and Quaternion Matrices[PhD thesis] University of California Santa Barbara CalifUSA 1993

[9] F Zhang ldquoQuaternions and matrices of quaternionsrdquo LinearAlgebra and Its Applications vol 251 pp 21ndash57 1997

[10] Q-W Wang ldquoThe general solution to a system of real quater-nion matrix equationsrdquo Computers amp Mathematics with Appli-cations vol 49 no 5-6 pp 665ndash675 2005

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 178209 7 pageshttpdxdoiorg1011552013178209

Research ArticleFuzzy Approximate Solution of Positive Fully Fuzzy LinearMatrix Equations

Xiaobin Guo1 and Dequan Shang2

1 College of Mathematics and Statistics Northwest Normal University Lanzhou 730070 China2Department of Public Courses Gansu College of Traditional Chinese Medicine Lanzhou 730000 China

Correspondence should be addressed to Dequan Shang gxbglz163com

Received 22 November 2012 Accepted 19 January 2013

Academic Editor Panayiotis J Psarrakos

Copyright copy 2013 X Guo and D ShangThis is an open access article distributed under theCreativeCommonsAttribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

The fuzzy matrix equations otimesotimes = in which and are119898times119898 119899times119899 and119898times119899 nonnegative LR fuzzy numbers matricesrespectively are investigated The fuzzy matrix systems is extended into three crisp systems of linear matrix equations accordingto arithmetic operations of LR fuzzy numbers Based on pseudoinverse of matrix the fuzzy approximate solution of original fuzzysystems is obtained by solving the crisp linear matrix systems In addition the existence condition of nonnegative fuzzy solution isdiscussed Two examples are calculated to illustrate the proposed method

1 Introduction

Since many real-world engineering systems are too com-plex to be defined in precise terms imprecision is ofteninvolved in any engineering design process Fuzzy systemshave an essential role in this fuzzy modeling which canformulate uncertainty in actual environment Inmanymatrixequations some or all of the system parameters are vagueor imprecise and fuzzy mathematics is a better tool thancrisp mathematics for modeling these problems and hencesolving a fuzzy matrix equation is becomingmore importantThe concept of fuzzy numbers and arithmetic operationswith these numbers were first introduced and investigatedby Zadeh [1 2] Dubois and Prade [3] and Nahmias [4]A different approach to fuzzy numbers and the structureof fuzzy number spaces was given by Puri and Ralescu [5]Goetschel Jr and Voxman [6] and Wu and Ma [7 8]

In the past decades many researchers have studied thefuzzy linear equations such as fuzzy linear systems (FLS) dualfuzzy linear systems (DFLS) general fuzzy linear systems(GFLS) fully fuzzy linear systems (FFLS) dual fully fuzzylinear systems (DFFLS) and general dual fuzzy linear systems(GDFLS) These works were performed mainly by Friedmanet al [9 10] Allahviranloo et al [11ndash17] Abbasbandy et al

[18ndash21]Wang andZheng [22 23] andDehghan et al [24 25]The general method they applied is the fuzzy linear equationswere converted to a crisp function system of linear equationswith high order according to the embedding principlesand algebraic operations of fuzzy numbers Then the fuzzysolution of the original fuzzy linear systems was derived fromsolving the crisp function linear systemsHowever for a fuzzymatrix equationwhich always has awide use in control theoryand control engineering few works have been done in thepast In 2009 Allahviranloo et al [26] discussed the fuzzy lin-ear matrix equations (FLME) of the form 119860119861 = in whichthematrices119860 and119861 are known119898times119898 and 119899times119899 real matricesrespectively is a given 119898 times 119899 fuzzy matrix By using theparametric form of fuzzy number they derived necessaryand sufficient conditions for the existence condition of fuzzysolutions and designed a numerical procedure for calculatingthe solutions of the fuzzy matrix equations In 2011 Guoet al [27ndash29] investigated a class of fuzzy matrix equations119860 = by means of the block Gaussian elimination methodand studied the least squares solutions of the inconsistentfuzzy matrix equation 119860 = by using generalized inversesof the matrix and discussed fuzzy symmetric solutions offuzzy matrix equations 119860 = What is more there are two

2 Journal of Applied Mathematics

shortcomings in the above fuzzy systems The first is that inthese fuzzy linear systems and fuzzy matrix systems the fuzzyelements were denoted by triangular fuzzy numbers so theextended model equations always contain parameter 119903 0 le

119903 le 1 which makes their computation especially numericalimplementation inconvenient in some sense The other oneis that the weak fuzzy solution of fuzzy linear systems119860 =

does not exist sometimes see [30]To make the multiplication of fuzzy numbers easy and

handle the fully fuzzy systems Dubois and Prade [3] intro-duced the LR fuzzy number in 1978 We know that triangularfuzzy numbers and trapezoidal fuzzy numbers [31] are justspecious cases of LR fuzzy numbers In 2006 Dehghan etal [25] discussed firstly computational methods for fullyfuzzy linear systems = whose coefficient matrix andthe right-hand side vector are LR fuzzy numbers In 2012Allahviranloo et al studied LR fuzzy linear systems [32]by the linear programming with equality constraints Otadiand Mosleh considered the nonnegative fuzzy solution [33]of fully fuzzy matrix equations = by employinglinear programming with equality constraints at the sameyear Recently Guo and Shang [34] investigated the fuzzyapproximate solution of LR fuzzy Sylvester matrix equations119860 + 119861 = In this paper we propose a general model forsolving the fuzzy linearmatrix equation otimesotimes = where and are119898times119898 119899times119899 and119898times119899 nonnegative LR fuzzynumbersmatrices respectivelyThemodel is proposed in thisway that is we extend the fuzzy linear matrix system intoa system of linear matrix equations according to arithmeticoperations of LR fuzzy numbers The LR fuzzy solution ofthe original matrix equation is derived from solving crispsystems of linearmatrix equationsThe structure of this paperis organized as follows

In Section 2 we recall the LR fuzzy numbers andpresent the concept of fully fuzzy linear matrix equationThecomputing model to the positive fully fuzzy linear matrixequation is proposed in detail and the fuzzy approximatesolution of the fuzzy linear matrix equation is obtained byusing pseudo-inverse in Section 3 Some examples are givento illustrate our method in Section 4 and the conclusion isdrawn in Section 5

2 Preliminaries

21 The LR Fuzzy Number

Definition 1 (see [1]) A fuzzy number is a fuzzy set like 119906

119877 rarr 119868 = [0 1] which satisfies the following

(1) 119906 is upper semicontinuous(2) 119906 is fuzzy convex that is 119906(120582119909 + (1 minus 120582)119910) ge

min119906(119909) 119906(119910) for all 119909 119910 isin 119877 120582 isin [0 1](3) 119906 is normal that is there exists 119909

0isin 119877 such that

119906(1199090) = 1

(4) supp 119906 = 119909 isin 119877 | 119906(119909) gt 0 is the support of the 119906and its closure cl(supp 119906) is compact

Let 1198641 be the set of all fuzzy numbers on 119877

Definition 2 (see [3]) A fuzzy number is said to be an LRfuzzy number if

120583(119909) =

119871(119898 minus 119909

120572) 119909 le 119898 120572 gt 0

119877(119909 minus 119898

120573) 119909 ge 119898 120573 gt 0

(1)

where 119898 120572 and 120573 are called the mean value left and rightspreads of respectively The function 119871(sdot) which is calledleft shape function satisfies

(1) 119871(119909) = 119871(minus119909)(2) 119871(0) = 1 and 119871(1) = 0(3) 119871(119909) is nonincreasing on [0infin)

The definition of a right shape function 119877(sdot) is similar tothat of 119871(sdot)

Clearly = (119898 120572 120573)LR is positive (negative) if and onlyif119898 minus 120572 gt 0 (119898 + 120573 lt 0)

Also two LR fuzzy numbers = (119898 120572 120573)LR and =

(119899 120574 120575)LR are said to be equal if and only if 119898 = 119899 120572 = 120574and 120573 = 120575

Definition 3 (see [5]) For arbitrary LR fuzzy numbers =

(119898 120572 120573)LR and = (119899 120574 120575)LR we have the following

(1) Addition

oplus = (119898 120572 120573)LR oplus (119899 120574 120575)LR = (119898 + 119899 120572 + 120574 120573 + 120575)LR

(2)

(2) Multiplication(i) If gt 0 and gt 0 then

otimes = (119898 120572 120573)LR otimes (119899 120574 120575)LR

cong (119898119899119898120574 + 119899120572119898120575 + 119899120573)LR(3)

(ii) if lt 0 and gt 0 then

otimes = (119898 120572 120573)RL otimes (119899 120574 120575)LR

cong (119898119899 119899120572 minus 119898120575 119899120573 minus 119898120574)RL(4)

(iii) if lt 0 and lt 0 then

otimes = (119898 120572 120573)RL otimes (119899 120574 120575)LR

cong (119898119899 minus119898120575 minus 119899120573 minus119898120574 minus 119899120572)RL(5)

(3) Scalar multiplication

120582 times = 120582 times (119898 120572 120573)LR

= (120582119898 120582120572 120582120573)LR 120582 ge 0

(120582119898 minus120582120573 minus120582120572)RL 120582 lt 0

(6)

Journal of Applied Mathematics 3

22 The LR Fuzzy Matrix

Definition 4 Amatrix = (119894119895) is called an LR fuzzy matrix

if each element 119894119895of is an LR fuzzy number

will be positive (negative) and denoted by gt 0 ( lt

0) if each element 119894119895of is positive (negative) where

119894119895=

(119886119894119895 119886119897

119894119895 119886119903

119894119895)LR Up to the rest of this paper we use positive

LR fuzzy numbers and formulas given in Definition 3 Forexample we represent 119898 times 119899 LR fuzzy matrix = (

119894119895) that

119894119895= (119886119894119895 120572119894119895 120573119894119895)LR with new notation = (119860119872119873) where

119860 = (119886119894119895) 119872 = (120572

119894119895) and 119873 = (120573

119894119895) are three 119898 times 119899 crisp

matrices In particular an 119899 dimensions LR fuzzy numbersvector can be denoted by (119909 119909119897 119909119903) where 119909 = (119909

119894) 119909119897 =

(119909119897

119894) and 119909119903 = (119909

119903

119894) are three 119899 dimensions crisp vectors

Definition 5 Let = (119894119895) and = (

119894119895) be two 119898 times 119899 and

119899 times 119901 fuzzy matrices we define otimes = = (119894119895) which is an

119898 times 119901 fuzzy matrix where

119894119895=

oplus

sum

119896=12119899

119894119896otimes 119887119896119895 (7)

23 The Fully Fuzzy Linear Matrix Equation

Definition 6 Thematrix system

(

11

12

sdot sdot sdot 1119898

21

22

sdot sdot sdot 2119898

1198981

1198982

sdot sdot sdot 119898119898

)otimes(

11

12

sdot sdot sdot 1119899

21

22

sdot sdot sdot 2119899

1198981

1198982

sdot sdot sdot 119898119899

)

otimes(

11

11988712

sdot sdot sdot 1198871119899

21

11988722

sdot sdot sdot 1198872119899

1198991

1198992

sdot sdot sdot 119899119899

)

=(

11

12

sdot sdot sdot 1119899

21

22

sdot sdot sdot 2119899

1198981

1198982

sdot sdot sdot 119898119899

)

(8)

where 119894119895 1 le 119894 119895 le 119898

119894119895 1 le 119894 119895 le 119899 and

119894119895 1 le 119894 le 119898

1 le 119895 le 119899 are LR fuzzy numbers is called an LR fully fuzzylinear matrix equation (FFLME)

Using matrix notation we have

otimes otimes = (9)

A fuzzy numbers matrix

= (119909119894119895 119910119894119895 119911119894119895)LR 1 le 119894 le 119898 1 le 119895 le 119899 (10)

is called an LR fuzzy approximate solution of fully fuzzy linearmatrix equation (8) if satisfies (9)

Up to the rest of this paper we will discuss the nonnega-tive solution = (119883 119884 119885) ge 0 of FFLME otimes otimes =

where = (119860119872119873) ge 0 = (119861 119864 119865) ge 0 and =

(119862 119866119867) ge 0

3 Method for Solving FFLME

First we extend the fuzzy linear matrix system (9) into threesystems of linear matrix equations according to the LR fuzzynumber and its arithmetic operations

Theorem 7 The fuzzy linear matrix system (9) can beextended into the following model

119860119883119861 = 119862

119860119883119864 + 119860119884119861 +119872119883119861 = 119866

119860119883119865 + 119860119885119861 + 119873119883119861 = 119867

(11)

Proof We denote = (119860119872119873) = (119861 119864 119865) and =

(119862 119866119867) and assume = (119883 119884 119885) ge 0 then

= (119860119872119873) otimes (119883 119884 119885) otimes (119861 119864 119865)

= (119860119883119860119884 +119872119883119860119885 + 119873119883) otimes (119861 119864 119865)

= (119860119883119861119860119883119864 + 119860119884119861 +119872119883119861119860119883119865 + 119860119885119861 + 119873119883119861)

= (119862 119866119867)

(12)

according tomultiplication of nonnegative LR fuzzy numbersof Definition 2 Thus we obtain a model for solving FFLME(9) as follows

119860119883119861 = 119862

119860119883119864 + 119860119884119861 +119872119883119861 = 119866

119860119883119865 + 119860119885119861 + 119873119883119861 = 119867

(13)

Secondly in order to solve the fuzzy linear matrix equa-tion (9) we need to consider the crisp systems of linearmatrix equation (11) Supposing119860 and119861 are nonsingular crispmatrices we have

119883 = 119860minus1

119862119861minus1

119884 = 119860minus1

119866 minus 119883119864119861minus1

minus 119860minus1

119872119883

119885 = 119860minus1

119867 minus 119883119865119861minus1

minus 119860minus1

119873119883

(14)

that is

119883 = 119860minus1

119862119861minus1

119884 = 119860minus1

119866 minus 119860minus1

119862119861minus1

119864119861minus1

minus 119860minus1

119872119860minus1

119862119861minus1

119885 = 119860minus1

119867 minus 119860minus1

119862119861minus1

119865119861minus1

minus 119860minus1

119873119860minus1

119862119861minus1

(15)

Definition 8 Let = (119883 119884 119885) be an LR fuzzy matrix If(119883 119884 119885) is an exact solution of (11) such that 119883 ge 0 119884 ge 0119885 ge 0 and 119883 minus 119884 ge 0 we call = (119883 119884 119885) a nonnegativeLR fuzzy approximate solution of (9)

4 Journal of Applied Mathematics

Theorem 9 Let = (119860119872119873) = (119861 119864 119865) and =

(119862 119866119867) be three nonnegative fuzzymatrices respectively andlet 119860 and 119861 be the product of a permutation matrix by adiagonal matrix with positive diagonal entries Moreover let119866119861 ge 119862119861

minus1

119864 + 119872119860minus1

119862 119867119861 ge 119862119861minus1

119865 + 119873119860minus1

119862 and119862 + 119862119861

minus1

119864 +119872119860minus1

119862 ge 119866119861 Then the systems otimes otimes =

has nonnegative fuzzy solutions

Proof Our hypotheses on 119860 and 119861 imply that 119860minus1 and119861minus1 exist and they are nonnegative matrices Thus 119883 =

119860minus1

119862119861minus1

ge 0On the other hand because 119866119861 ge 119862119861

minus1

119864 + 119872119860minus1

119862 and119867119861 ge 119862119861

minus1

119865 + 119873119860minus1

119862 so with 119884 = 119860minus1

(119866119861 minus 119862119861minus1

119864 minus

119872119860minus1

119862)119861minus1 and 119885 = 119860

minus1

(119867119861 minus 119862119861minus1

119865 minus 119873119860minus1

119862)119861minus1 we

have 119884 ge 0 and 119885 ge 0 Thus = (119883 119884 119885) is a fuzzy matrixwhich satisfies otimes otimes = Since119883 minus 119884 = 119860

minus1

(119862 minus 119866119861 +

119862119861minus1

119864 + 119872119860minus1

119862)119861minus1 the positivity property of can be

obtained from the condition 119862+119862119861minus1119864+119872119860minus1

119862 ge 119866119861

When 119860 or 119861 is a singular crisp matrix the followingresult is obvious

Theorem 10 (see [35]) For linearmatrix equations119860119883119861 = 119862where 119860 isin 119877

119898times119898 119861 isin 119877119899times119899 and 119862 isin 119877

119898times119899 Then

119883 = 119860dagger

119862119861dagger (16)

is its minimal norm least squares solutionBy the pseudoinverse of matrices we solve model (11) and

obtain its minimal norm least squares solution as follows

119883 = 119860dagger

119862119861dagger

119884 = 119860dagger

119866 minus 119883119864119861dagger

minus 119860dagger

119872119883

119885 = 119860dagger

119867 minus 119883119865119861dagger

minus 119860dagger

119873119883

(17)

that is

119883 = 119860dagger

119862119861dagger

119884 = 119860dagger

119866 minus 119860dagger

119862119861dagger

119864119861dagger

minus 119860dagger

119872119860dagger

119862119861dagger

119885 = 119860dagger

119867 minus 119860dagger

119862119861dagger

119865119861dagger

minus 119860dagger

119873119860dagger

119862119861dagger

(18)

Definition 11 Let = (119883 119884 119885) be an LR fuzzy matrix If(119883 119884 119885) is a minimal norm least squares solution of (11) suchthat119883 ge 0119884 ge 0119885 ge 0 and119883minus119884 ge 0 we call = (119883 119884 119885)

a nonnegative LR fuzzy minimal norm least squares solutionof (9)

At last we give a sufficient condition for nonnegativefuzzy minimal norm least squares solution of FFLME (9) inthe same way

Theorem 12 Let 119860dagger and 119861dagger be nonnegative matrices More-over let 119866119861 ge 119862119861

dagger

119864 + 119872119860dagger

119862 119867119861 ge 119862119861dagger

119865 + 119873119860dagger

119862 and119862+119862119861

dagger

119864+119872119860dagger

119862 ge 119866119861 Then the systems otimes otimes = hasnonnegative fuzzy minimal norm least squares solutions

Proof Since 119860dagger and 119861dagger are nonnegative matrices we have

119883 = 119860dagger

119862119861dagger

ge 0Now that 119866119861 ge 119862119861

dagger

119864+119872119860dagger

119862 and119867119861 ge 119862119861dagger

119865+119873119860dagger

119862therefore with 119884 = 119860

dagger

(119866119861 minus 119862119861dagger

119864 minus 119872119860dagger

119862)119861dagger and 119885 =

119860dagger

(119867119861 minus 119862119861dagger

119865 minus 119873119860dagger

119862)119861dagger we have 119884 ge 0 and 119885 ge 0 Thus

= (119883 119884 119885) is a fuzzy matrix which satisfies otimes otimes =

Since 119883 minus 119884 = 119860dagger

(119862 minus 119866119861 + 119862119861dagger

119864 + 119872119860dagger

119862)119861dagger

the nonnegativity property of can be obtained from thecondition 119862 + 119862119861

dagger

119864 +119872119860dagger

119862 ge 119866119861

The following Theorems give some results for such 119878minus1

and 119878dagger to be nonnegative As usual (sdot)⊤ denotes the transposeof a matrix (sdot)

Theorem 13 (see [36]) The inverse of a nonnegative matrix119860 is nonnegative if and only if 119860 is a generalized permutationmatrix

Theorem 14 (see [37]) Let119860 be an119898 times119898 nonnegativematrixwith rank 119903 Then the following assertions are equivalent

(a) 119860dagger ge 0

(b) There exists a permutation matrix 119875 such that 119875119860 hasthe form

119875119860 =(

1198781

1198782

119878119903

119874

) (19)

where each 119878119894has rank 1 and the rows of 119878

119894are

orthogonal to the rows of 119878119895 whenever 119894 = 119895 the zero

matrix may be absent

(c) 119860dagger = (119870119875⊤119870119876⊤

119870119875⊤119870119875⊤ ) for some positive diagonal matrix 119870

In this case

(119875 + 119876)dagger

= 119870(119875 + 119876)⊤

(119875 minus 119876)dagger

= 119870(119875 minus 119876)⊤

(20)

4 Numerical Examples

Example 15 Consider the fully fuzzy linear matrix system

((2 1 1)LR (1 0 1)LR(1 0 0)LR (2 1 0)LR

) otimes (11

12

13

21

22

23

)

otimes (

(1 0 1)LR (2 1 1)LR (1 1 0)LR(2 1 0)LR (1 0 0)LR (2 1 1)LR(1 0 0)LR (2 1 1)LR (1 0 1)LR

)

= ((17 12 23)LR (22 22 29)LR (17 16 28)LR(19 15 13)LR (23 23 16)LR (19 16 17)LR

)

(21)

Journal of Applied Mathematics 5

ByTheorem 7 the model of the above fuzzy linear matrixsystem is made of the following three crisp systems of linearmatrix equations

(2 1

1 2)(

11990911

11990912

11990913

11990921

11990922

11990923

)(

1 2 1

2 1 2

1 2 1

) = (17 22 17

19 23 19)

(2 1

1 2)(

11990911

11990912

11990913

11990921

11990922

11990923

)(

0 1 1

1 0 1

0 1 0

)

+ (2 1

1 2)(

11991011

11991012

11991013

11991021

11991022

11991023

)(

1 2 1

2 1 2

1 2 1

)

+ (1 0

0 1)(

11990911

11990912

11990913

11990921

11990922

11990923

)(

1 2 1

2 1 2

1 2 1

) = (12 22 16

15 23 20)

(2 1

1 2)(

11990911

11990912

11990913

11990921

11990922

11990923

)(

1 1 0

0 0 1

0 1 1

)

+ (2 1

1 2)(

11991111

11991112

11991113

11991121

11991122

11991123

)(

1 2 1

2 1 2

1 2 1

)

+ (1 1

0 0)(

11990911

11990912

11990913

11990921

11990922

11990923

)(

1 2 1

2 1 2

1 2 1

) = (23 29 28

13 16 17)

(22)

Now that the matrix 119861 is singular according to formula(18) the solutions of the above three systems of linear matrixequations are as follows

119883 = 119860dagger

119862119861dagger

= (2 1

1 2)

dagger

(17 22 17

19 23 19)(

1 2 1

2 1 2

1 2 1

)

dagger

= (15000 10000 15000

15000 20000 15000)

119884 = 119860dagger

119866 minus 119883119864119861dagger

minus 119860dagger

119872119883 = (2 1

1 2)

dagger

(12 22 16

19 23 19)

minus 119883(

0 1 1

1 0 1

0 1 0

)(

1 2 1

2 1 2

1 2 1

)

dagger

minus (2 1

1 2)

dagger

(1 0

0 1)119883

= (10333 06667 08725

11250 16553 12578)

119885 = 119860dagger

119867 minus 119883119865119861dagger

minus 119860dagger

119873119883 = (2 1

1 2)

dagger

(23 29 28

13 16 17)

minus 119883(

1 1 0

0 0 1

0 1 1

)(

1 2 1

2 1 2

1 2 1

)

dagger

minus (2 1

1 2)

dagger

(1 1

0 0)119883

= (83333 116667 93333

14167 13333 24167)

(23)

By Definition 11 we know that the original fuzzy linearmatrix equations have a nonnegative LR fuzzy solution

= (11

12

13

21

22

23

) = ((15000 10333 83333)LR (10000 06667 116667)LR (15000 08725 93333)LR(15000 11250 14167)LR (20000 16553 13333)LR (15000 12578 24167)LR

) (24)

since119883 ge 0 119884 ge 0 119885 ge 0 and119883 minus 119884 ge 0

Example 16 Consider the following fuzzy matrix system

((1 1 0)LR (2 0 1)LR(2 0 1)LR (1 0 0)LR

)(11

12

21

22

)((1 0 1)LR (2 1 0)LR(2 1 1)LR (3 1 2)LR

)

= ((15 14 18)LR (25 24 24)LR(12 10 12)LR (20 17 15)LR

)

(25)

ByTheorem 7 the model of the above fuzzy linear matrixsystem is made of following three crisp systems of linear

matrix equations

(2 1

1 2)(

11990911

11990912

11990921

11990922

)(1 2

2 3) = (

15 25

10 20)

(2 1

1 2)(

11990911

11990912

11990921

11990922

)(0 1

1 1) + (

2 1

1 2)(

11991011

11991012

11991021

11991022

)(1 2

2 3)

+ (1 0

0 0)(

11990911

11990912

11990921

11990922

)(1 2

2 3) = (

14 24

10 17)

(2 1

1 2)(

11990911

11990912

11990921

11990922

)(1 0

1 2) + (

2 1

1 2)(

11991111

11991112

11991121

11991122

)(1 2

2 3)

+ (0 1

1 0)(

11990911

11990912

11990921

11990922

)(1 2

2 3) = (

18 24

12 15)

(26)

6 Journal of Applied Mathematics

By the same way we obtain that the solutions of the abovethree systems of linear matrix equations are as follows

119883 = 119860minus1

119862119861minus1

= (1 1

2 2)

119884 = 119860minus1

119866 minus 119883119864119861minus1

minus 119860minus1

119872119883 = (0 1

0 1)

119885 = 119860minus1

119867 minus 119883119865119861minus1

minus 119860minus1

119873119883 = (0 0

1 0)

(27)

Since 119883 ge 0 119884 ge 0 119885 ge 0 and 119883 minus 119884 ge 0 we know thatthe original fuzzy linear matrix equations have a nonnegativeLR fuzzy solution given by

= (11

12

21

22

)

= ((1000 1000 0000)LR (1000 1000 0000)LR(2000 0000 1000)LR (2000 1000 0000)LR

)

(28)

5 Conclusion

In this work we presented a model for solving fuzzy linearmatrix equations otimes otimes = in which and are119898 times 119898 and 119899 times 119899 fuzzy matrices respectively and isan 119898 times 119899 arbitrary LR fuzzy numbers matrix The modelwas made of three crisp systems of linear equations whichdetermined the mean value and the left and right spreads ofthe solution The LR fuzzy approximate solution of the fuzzylinear matrix equation was derived from solving the crispsystems of linear matrix equations In addition the existencecondition of strong LR fuzzy solution was studied Numericalexamples showed that ourmethod is feasible to solve this typeof fuzzy matrix equations Based on LR fuzzy numbers andtheir operations we can investigate all kinds of fully fuzzymatrix equations in future

Acknowledgments

The work is supported by the Natural Scientific Funds ofChina (no 71061013) and the Youth Research Ability Projectof Northwest Normal University (NWNU-LKQN-1120)

References

[1] L A Zadeh ldquoFuzzy setsrdquo Information and Computation vol 8pp 338ndash353 1965

[2] L A Zadeh ldquoThe concept of a linguistic variable and itsapplication to approximate reasoning Irdquo Information Sciencevol 8 pp 199ndash249 1975

[3] D Dubois and H Prade ldquoOperations on fuzzy numbersrdquoInternational Journal of Systems Science vol 9 no 6 pp 613ndash626 1978

[4] S Nahmias ldquoFuzzy variablesrdquo Fuzzy Sets and Systems AnInternational Journal in Information Science and Engineeringvol 1 no 2 pp 97ndash110 1978

[5] M L Puri and D A Ralescu ldquoDifferentials of fuzzy functionsrdquoJournal of Mathematical Analysis and Applications vol 91 no 2pp 552ndash558 1983

[6] R Goetschel Jr and W Voxman ldquoElementary fuzzy calculusrdquoFuzzy Sets and Systems vol 18 no 1 pp 31ndash43 1986

[7] C X Wu and M Ma ldquoEmbedding problem of fuzzy numberspace Irdquo Fuzzy Sets and Systems vol 44 no 1 pp 33ndash38 1991

[8] C X Wu and M Ma ldquoEmbedding problem of fuzzy numberspace IIIrdquo Fuzzy Sets and Systems vol 46 no 2 pp 281ndash2861992

[9] M Friedman M Ming and A Kandel ldquoFuzzy linear systemsrdquoFuzzy Sets and Systems vol 96 no 2 pp 201ndash209 1998

[10] M Ma M Friedman and A Kandel ldquoDuality in fuzzy linearsystemsrdquo Fuzzy Sets and Systems vol 109 no 1 pp 55ndash58 2000

[11] T Allahviranloo and M Ghanbari ldquoOn the algebraic solutionof fuzzy linear systems based on interval theoryrdquo AppliedMathematical Modelling vol 36 no 11 pp 5360ndash5379 2012

[12] T Allahviranloo and S Salahshour ldquoFuzzy symmetric solutionsof fuzzy linear systemsrdquo Journal of Computational and AppliedMathematics vol 235 no 16 pp 4545ndash4553 2011

[13] T Allahviranloo and M Ghanbari ldquoA new approach to obtainalgebraic solution of interval linear systemsrdquo Soft Computingvol 16 no 1 pp 121ndash133 2012

[14] T Allahviranloo ldquoNumericalmethods for fuzzy systemof linearequationsrdquo Applied Mathematics and Computation vol 155 no2 pp 493ndash502 2004

[15] T Allahviranloo ldquoSuccessive over relaxation iterative methodfor fuzzy system of linear equationsrdquo Applied Mathematics andComputation vol 162 no 1 pp 189ndash196 2005

[16] T Allahviranloo ldquoThe Adomian decomposition method forfuzzy system of linear equationsrdquo Applied Mathematics andComputation vol 163 no 2 pp 553ndash563 2005

[17] RNuraei T Allahviranloo andMGhanbari ldquoFinding an innerestimation of the so- lution set of a fuzzy linear systemrdquoAppliedMathematical Modelling vol 37 no 7 pp 5148ndash5161 2013

[18] S Abbasbandy R Ezzati and A Jafarian ldquoLU decompositionmethod for solving fuzzy system of linear equationsrdquo AppliedMathematics and Computation vol 172 no 1 pp 633ndash6432006

[19] S Abbasbandy A Jafarian and R Ezzati ldquoConjugate gradientmethod for fuzzy symmetric positive definite system of linearequationsrdquo Applied Mathematics and Computation vol 171 no2 pp 1184ndash1191 2005

[20] S Abbasbandy M Otadi andM Mosleh ldquoMinimal solution ofgeneral dual fuzzy linear systemsrdquo Chaos Solitons amp Fractalsvol 37 no 4 pp 1113ndash1124 2008

[21] B Asady S Abbasbandy and M Alavi ldquoFuzzy general linearsystemsrdquo Applied Mathematics and Computation vol 169 no 1pp 34ndash40 2005

[22] K Wang and B Zheng ldquoInconsistent fuzzy linear systemsrdquoApplied Mathematics and Computation vol 181 no 2 pp 973ndash981 2006

[23] B Zheng and K Wang ldquoGeneral fuzzy linear systemsrdquo AppliedMathematics and Computation vol 181 no 2 pp 1276ndash12862006

[24] M Dehghan and B Hashemi ldquoIterative solution of fuzzy linearsystemsrdquo Applied Mathematics and Computation vol 175 no 1pp 645ndash674 2006

[25] M Dehghan B Hashemi and M Ghatee ldquoSolution of the fullyfuzzy linear systems using iterative techniquesrdquo Chaos Solitonsamp Fractals vol 34 no 2 pp 316ndash336 2007

Journal of Applied Mathematics 7

[26] T Allahviranloo N Mikaeilvand and M Barkhordary ldquoFuzzylinear matrix equationrdquo Fuzzy Optimization and Decision Mak-ing vol 8 no 2 pp 165ndash177 2009

[27] X Guo and Z Gong ldquoBlock Gaussian elimination methodsfor fuzzy matrix equationsrdquo International Journal of Pure andApplied Mathematics vol 58 no 2 pp 157ndash168 2010

[28] Z Gong and X Guo ldquoInconsistent fuzzy matrix equationsand its fuzzy least squares solutionsrdquo Applied MathematicalModelling vol 35 no 3 pp 1456ndash1469 2011

[29] X Guo and D Shang ldquoFuzzy symmetric solutions of fuzzymatrix equationsrdquo Advances in Fuzzy Systems vol 2012 ArticleID 318069 9 pages 2012

[30] T Allahviranloo M Ghanbari A A Hosseinzadeh E Haghiand R Nuraei ldquoA note on lsquoFuzzy linear systemsrsquordquo Fuzzy Sets andSystems vol 177 pp 87ndash92 2011

[31] C-T Yeh ldquoWeighted semi-trapezoidal approximations of fuzzynumbersrdquo Fuzzy Sets and Systems vol 165 pp 61ndash80 2011

[32] T Allahviranloo F H Lotfi M K Kiasari and M KhezerlooldquoOn the fuzzy solution of LR fuzzy linear systemsrdquo AppliedMathematical Modelling vol 37 no 3 pp 1170ndash1176 2013

[33] MOtadi andMMosleh ldquoSolving fully fuzzymatrix equationsrdquoApplied Mathematical Modelling vol 36 no 12 pp 6114ndash61212012

[34] X B Guo and D Q Shang ldquoApproximate solutions of LR fuzzySylvester matrix equationsrdquo Journal of Applied Mathematics Inpress

[35] A Ben-Israel and T N E Greville Generalized Inverses Theoryand Applications Springer-Verlag New York NY USA 2ndedition 2003

[36] A Berman and R J Plemmons Nonnegative Matrices in theMathematical Sciences Academic Press New York NY USA1979

[37] R J Plemmons ldquoRegular nonnegative matricesrdquo Proceedings ofthe American Mathematical Society vol 39 pp 26ndash32 1973

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 514984 9 pageshttpdxdoiorg1011552013514984

Research ArticleRanks of a Constrained Hermitian MatrixExpression with Applications

Shao-Wen Yu

Department of Mathematics East China University of Science and Technology Shanghai 200237 China

Correspondence should be addressed to Shao-Wen Yu yushaowenecusteducn

Received 12 November 2012 Accepted 4 January 2013

Academic Editor Yang Zhang

Copyright copy 2013 Shao-Wen YuThis is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We establish the formulas of the maximal and minimal ranks of the quaternion Hermitian matrix expression 1198624minus 1198604119883119860lowast

4where

119883 is a Hermitian solution to quaternion matrix equations 1198601119883 = 119862

1 1198831198611= 1198622 and 119860

3119883119860lowast

3= 1198623 As applications we give a new

necessary and sufficient condition for the existence of Hermitian solution to the system of matrix equations 1198601119883 = 119862

11198831198611= 1198622

1198603119883119860lowast

3= 1198623 and 119860

4119883119860lowast

4= 1198624 which was investigated by Wang and Wu 2010 by rank equalities In addition extremal ranks of

the generalized Hermitian Schur complement 1198624minus 1198604119860sim

3119860lowast

4with respect to a Hermitian g-inverse 119860sim

3of 1198603 which is a common

solution to quaternion matrix equations 1198601119883 = 119862

1and119883119861

1= 1198622 are also considered

1 Introduction

Throughout this paper we denote the real number field byRthe complex number field by C the set of all 119898 times 119899 matricesover the quaternion algebra

H = 1198860+ 1198861119894 + 1198862119895 + 1198863119896 | 1198942

= 1198952

= 1198962

= 119894119895119896 = minus1 1198860 1198861 1198862 1198863isin R

(1)

by H119898times119899 the identity matrix with the appropriate size by 119868the column right space the row left space of a matrix 119860 overH by R(119860) N(119860) respectively the dimension of R(119860) bydimR(119860) a Hermitian g-inverse of a matrix 119860 by 119883 = 119860

∽

which satisfies 119860119860∽119860 = 119860 and 119883 = 119883lowast and the Moore-

Penrose inverse of matrix119860 overH by119860dagger which satisfies fourPenrose equations 119860119860dagger119860 = 119860 119860

dagger

119860119860dagger

= 119860dagger

(119860119860dagger

)lowast

=

119860119860dagger

and (119860dagger

119860)lowast

= 119860dagger

119860 In this case 119860dagger is unique and(119860dagger

)lowast

= (119860lowast

)dagger Moreover 119877

119860and 119871

119860stand for the two

projectors 119871119860

= 119868 minus 119860dagger

119860 119877119860

= 119868 minus 119860119860dagger induced by 119860

Clearly119877119860and 119871

119860are idempotent Hermitian and119877

119860= 119871119860lowast

By [1] for a quaternion matrix 119860 dimR(119860) = dimN(119860)dimR(119860) is called the rank of a quaternion matrix 119860 anddenoted by 119903(119860)

Mitra [2] investigated the system of matrix equations

1198601119883 = 119862

1 119883119861

1= 1198622 (2)

Khatri and Mitra [3] gave necessary and sufficient con-ditions for the existence of the common Hermitian solutionto (2) and presented an explicit expression for the generalHermitian solution to (2) by generalized inverses Using thesingular value decomposition (SVD) Yuan [4] investigatedthe general symmetric solution of (2) over the real numberfield R By the SVD Dai and Lancaster [5] considered thesymmetric solution of equation

119860119883119860lowast

= 119862 (3)

over R which was motivated and illustrated with an inverseproblem of vibration theory Groszlig [6] Tian and Liu [7]gave the solvability conditions for Hermitian solution and itsexpressions of (3) over C in terms of generalized inversesrespectively Liu Tian and Takane [8] investigated ranksof Hermitian and skew-Hermitian solutions to the matrixequation (3) By using the generalized SVD Chang andWang[9] examined the symmetric solution to the matrix equations

1198603119883119860lowast

3= 1198623 119860

4119883119860lowast

4= 1198624

(4)

2 Journal of Applied Mathematics

over R Note that all the matrix equations mentioned aboveare special cases of

1198601119883 = 119862

1 119883119861

1= 1198622

1198603119883119860lowast

3= 1198623 119860

4119883119860lowast

4= 1198624

(5)

Wang and Wu [10] gave some necessary and sufficientconditions for the existence of the common Hermitiansolution to (5) for operators between Hilbert Clowast-modulesby generalized inverses and range inclusion of matrices Inview of the complicated computations of the generalizedinverses of matrices we naturally hope to establish a morepractical necessary and sufficient condition for system (5)over quaternion algebra to have Hermitian solution by rankequalities

As is known to us solutions tomatrix equations and ranksof solutions to matrix equations have been considered previ-ously by many authors [10ndash34] and extremal ranks of matrixexpressions can be used to characterize their rank invariancenonsingularity range inclusion and solvability conditionsof matrix equations Tian and Cheng [35] investigated themaximal and minimal ranks of 119860 minus 119861119883119862 with respect to 119883with applications Tian [36] gave the maximal and minimalranks of 119860

1minus 11986111198831198621subject to a consistent matrix equation

11986121198831198622= 1198602 Tian and Liu [7] established the solvability

conditions for (4) to have a Hermitian solution over C bythe ranks of coefficientmatricesWang and Jiang [20] derivedextreme ranks of (skew)Hermitian solutions to a quaternionmatrix equation 119860119883119860

lowast

+ 119861119884119861lowast

= 119862 Wang Yu and Lin[31] derived the extremal ranks of 119862

4minus 11986041198831198614subject to a

consistent system of matrix equations

1198601119883 = 119862

1 119883119861

1= 1198622 119860

31198831198613= 1198623

(6)

over H and gave a new solvability condition to system

1198601119883 = 119862

1 119883119861

1= 1198622

11986031198831198613= 1198623 119860

41198831198614= 1198624

(7)

In matrix theory and its applications there are manymatrix expressions that have symmetric patterns or involveHermitian (skew-Hermitian) matrices For example

119860 minus 119861119883119861lowast

119860 minus 119861119883 plusmn 119883lowast

119861lowast

119860 minus 119861119883119861lowast

minus 119862119884119862lowast

119860 minus 119861119883119862 plusmn (119861119883119862)lowast

(8)

where 119860 = plusmn119860lowast

119861 and 119862 are given and 119883 and 119884 arevariable matrices In recent papers [7 8 37 38] Liu and Tianconsidered some maximization and minimization problemson the ranks of Hermitian matrix expressions (8)

Define a Hermitian matrix expression

119891 (119883) = 1198624minus 1198604119883119860lowast

4 (9)

where 1198624= 119862lowast

4 we have an observation that by investigating

extremal ranks of (9) where 119883 is a Hermitian solution to asystem of matrix equations

1198601119883 = 119862

1 119883119861

1= 1198622 119860

3119883119860lowast

3= 1198623 (10)

A new necessary and sufficient condition for system (5) tohave Hermitian solution can be given by rank equalitieswhich is more practical than one given by generalizedinverses and range inclusion of matrices

It is well known that Schur complement is one of themostimportant matrix expressions in matrix theory there havebeen many results in the literature on Schur complementsand their applications [39ndash41] Tian [36 42] has investigatedthe maximal and minimal ranks of Schur complements withapplications

Motivated by the workmentioned above we in this paperinvestigate the extremal ranks of the quaternion Hermitianmatrix expression (9) subject to the consistent system ofquaternion matrix equations (10) and its applications InSection 2 we derive the formulas of extremal ranks of (9)with respect to Hermitian solution of (10) As applications inSection 3 we give a new necessary and sufficient conditionfor the existence of Hermitian solution to system (5) byrank equalities In Section 4 we derive extremal ranks ofgeneralized Hermitian Schur complement subject to (2) Wealso consider the rank invariance problem in Section 5

2 Extremal Ranks of (9) Subject to System (10)Corollary 8 in [10] over Hilbert Clowast-modules can be changedinto the following lemma over H

Lemma 1 Let 1198601 1198621

isin H119898times119899 1198611 1198622

isin H119899times119904 1198603

isin

H119903times119899 1198623isin H119903times119903 be given and 119865 = 119861

lowast

11198711198601 119872 = 119878119871

119865 119878 =

11986031198711198601 119863 = 119862

lowast

2minus119861lowast

1119860dagger

11198621 119869 = 119860

dagger

11198621+119865dagger

119863 119866 = 1198623minus1198603(119869+

1198711198601119871lowast

119865119869lowast

)119860lowast

3 then the following statements are equivalent

(1) the system (10) have a Hermitian solution(2) 1198623= 119862lowast

3

11986011198622= 11986211198611 119860

1119862lowast

1= 1198621119860lowast

1 119861

lowast

11198622= 119862lowast

21198611 (11)

11987711986011198621= 0 119877

119865119863 = 0 119877

119872119866 = 0 (12)

(3) 1198623= 119862lowast

3 the equalities in (11) hold and

119903 [11986011198621] = 119903 (119860

1) 119903 [

11986011198621

119861lowast

1119862lowast

2

] = 119903 [1198601

119861lowast

1

]

119903[[

[

11986011198621119860lowast

3

119861lowast

1119862lowast

2119860lowast

3

1198603

1198623

]]

]

= 119903[

[

1198601

119861lowast

1

1198603

]

]

(13)

In that case the general Hermitian solution of (10) can beexpressed as

119883 = 119869 + 1198711198601119871119865119869lowast

+ 1198711198601119871119865119872dagger

119866(119872dagger

)lowast

1198711198651198711198601

+ 11987111986011198711198651198711198721198811198711198651198711198601+ 1198711198601119871119865119881lowast

1198711198721198711198651198711198601

(14)

where 119881 is Hermitian matrix over H with compatible size

Journal of Applied Mathematics 3

Lemma 2 (see Lemma 24 in [24]) Let 119860 isin H119898times119899 119861 isin

H119898times119896 119862 isin H119897times119899 119863 isin H119895times119896 and 119864 isin H119897times119894 Then the followingrank equalities hold

(a) 119903(119862119871119860) = 119903 [

119860

119862] minus 119903(119860)

(b) 119903 [ 119861 119860119871119862 ] = 119903 [119861 119860

0 119862] minus 119903(119862)

(c) 119903 [ 119862119877119861119860

] = 119903 [119862 0

119860 119861] minus 119903(119861)

(d) 119903 [ 119860 119861119871119863119877119864119862 0

] = 119903 [119860 119861 0

119862 0 119864

0 119863 0

] minus 119903(119863) minus 119903(119864)

Lemma 2 plays an important role in simplifying ranks ofvarious block matrices

Liu and Tian [38] has given the following lemma over afield The result can be generalized to H

Lemma 3 Let 119860 = plusmn119860lowast

isin H119898times119898 119861 isin H119898times119899 and 119862 isin H119901times119898

be given then

max119883isinH119899times119901

119903 [119860 minus 119861119883119862 ∓ (119861119883119862)lowast

]

= min119903 [119860 119861 119862lowast

] 119903 [119860 119861

119861lowast

0] 119903 [

119860 119862lowast

119862 0]

min119883isinH119899times119901

119903 [119860 minus 119861119883119862 ∓ (119861119883119862)lowast

]

= 2119903 [119860 119861 119862lowast

] +max 1199041 1199042

(15)

where

1199041= 119903 [

119860 119861

119861lowast

0] minus 2119903 [

119860 119861 119862lowast

119861lowast

0 0]

1199042= 119903 [

119860 119862lowast

119862 0] minus 2119903 [

119860 119861 119862lowast

119862 0 0]

(16)

IfR(119861) sube R(119862lowast

)

max119883

119903 [119860 minus 119861119883119862 minus (119861119883119862)lowast

] = min119903 [119860 119862lowast

] 119903 [119860 119861

119861lowast

0]

max119883

119903 [119860 minus 119861119883119862 minus (119861119883119862)lowast

] = min119903 [119860 119862lowast

] 119903 [119860 119861

119861lowast

0]

(17)

Now we consider the extremal ranks of the matrixexpression (9) subject to the consistent system (10)

Theorem 4 Let 1198601 1198621 1198611 1198622 1198603 and 119862

3be defined as

Lemma 11198624isin H119905times119905 and 119860

4isin H119905times119899Then the extremal ranks

of the quaternionmatrix expression119891(119883) defined as (9) subjectto system (10) are the following

max 119903 [119891 (119883)] = min 119886 119887 (18)

where

119886 = 119903

[[[[

[

1198624

1198604

119862lowast

2119860lowast

4119861lowast

1

1198621119860lowast

41198601

]]]]

]

minus 119903 [119861lowast

1

1198601

]

119887 = 119903

[[[[[[[[

[

0 119860lowast

4119860lowast

31198611

119860lowast

1

11986041198624

0 0 0

1198603

0 minus1198623

minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]]

]

minus 2119903[

[

1198603

119861lowast

1

1198601

]

]

(19)

min 119903 [119891 (119883)] = 2119903

[[[[

[

1198624

1198604

119862lowast

2119860lowast

4119861lowast

1

1198621119860lowast

41198601

]]]]

]

+ 119903

[[[[[[[[[

[

0 119860lowast

4119860lowast

31198611

119860lowast

1

11986041198624

0 0 0

1198603

0 minus1198623

minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]]]

]

minus 2119903

[[[[[[[[[

[

0 119860lowast

41198611

119860lowast

1

11986041198624

0 0

1198603

0 minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus11986211198611minus1198621119860lowast

1

]]]]]]]]]

]

(20)

Proof By Lemma 1 the general Hermitian solution of thesystem (10) can be expressed as

119883 = 119869 + 1198711198601119871119865119869lowast

+ 1198711198601119871119865119872dagger

119866(119872dagger

)lowast

1198711198651198711198601

+ 11987111986011198711198651198711198721198811198711198651198711198601+ 1198711198601119871119865119881lowast

1198711198721198711198651198711198601

(21)

where 119881 is Hermitian matrix over H with appropriate sizeSubstituting (21) into (9) yields

119891 (119883) = 1198624minus 1198604(119869 + 119871

1198601119871119865119869lowast

+ 1198711198601119871119865119872dagger

119866(119872dagger

)lowast

1198711198651198711198601)119860

lowast

4

minus 119860411987111986011198711198651198711198721198811198711198651198711198601119860lowast

4

minus 11986041198711198601119871119865119881lowast

1198711198721198711198651198711198601119860lowast

4

(22)

4 Journal of Applied Mathematics

Put

1198624minus 1198604(119869 + 119871

1198601119871119865119869lowast

+ 1198711198601119871119865119872dagger

119866(119872dagger

)lowast

1198711198651198711198601)119860lowast

4= 119860

119869 + 1198711198601119871119865119869lowast

+ 1198711198601119871119865119872dagger

119866(119872dagger

)lowast

1198711198651198711198601

= 1198691015840

11986041198711198601119871119865119871119872= 119873

1198711198651198711198601119860lowast

4= 119875

(23)

then

119891 (119883) = 119860 minus 119873119881119875 minus (119873119881119875)lowast

(24)

Note that119860 = 119860lowast andR(119873) sube R(119875

lowast

) Thus applying (17) to(24) we get the following

max 119903 [119891 (119883)] = max119881

119903 (119860 minus 119873119881119875 minus (119873119881119875)lowast

)

= min119903 [119860 119875lowast

] 119903 [119860 119873

119873lowast

0]

min 119903 [119891 (119883)] = min119881

119903 (119860 minus 119873119881119875 minus (119873119881119875)lowast

)

= 2119903 [119860 119875lowast

] + 119903 [119860 119873

119873lowast

0] minus 2119903 [

119860 119873

119875 0]

(25)

Now we simplify the ranks of block matrices in (25)In view of Lemma 2 block Gaussian elimination (11)

(12) and (23) we have the following

119903 (119865) = 119903 (119861lowast

11198711198601) = 119903 [

119861lowast

1

1198601

] minus 119903 (1198601)

119903 (119872) = 119903 (119878119871119865) = 119903 [

119878

119865] minus 119903 (119865)

= 119903 [

11986031198711198601

119861lowast

11198711198601

] minus 119903 (119865)

= 119903[

[

1198603

119861lowast

1

1198601

]

]

minus 119903 (1198601) minus 119903 (119865)

119903 [119860 119875lowast

] = 119903 [1198624minus 1198604119869119860lowast

4119875lowast

]

= 119903 [1198624minus 1198604119869119860lowast

411986041198711198601

0 119865] minus 119903 (119865)

= 119903[

[

1198624minus 1198604119869119860lowast

41198604

0 119861lowast

1

0 1198601

]

]

minus 119903 (119865) minus 119903 (1198601)

= 119903

[[[

[

1198624

1198604

119862lowast

2119860lowast

4119861lowast

1

1198621119860lowast

41198601

]]]

]

minus 119903 [119861lowast

1

1198601

]

119903 [119860 119873

119873lowast

0] = 119903 [

1198624minus 11986041198691015840

119860lowast

411986041198711198601119871119865119871119872

119877119872lowast119877119865lowast119877119860lowast

1

119860lowast

40

]

= 119903

[[[[[[

[

1198624minus 11986041198691015840

119860lowast

41198604

0 0 0

119860lowast

40 119860lowast

31198611119860lowast

1

0 1198603

0 0 0

0 119861lowast

10 0 0

0 1198601

0 0 0

]]]]]]

]

minus 2119903 (119872) minus 2119903 (119865) minus 2119903 (1198601)

= 119903

[[[[[[[[[[[[

[

1198624

1198604

0 0 0

119860lowast

40 119860lowast

31198611119860lowast

1

11986031198691015840

119860lowast

41198603

0 0 0

119861lowast

11198691015840

119860lowast

4119861lowast

10 0 0

11986011198691015840

119860lowast

41198601

0 0 0

]]]]]]]]]]]]

]

minus 2119903[

[

1198603

119861lowast

1

1198601

]

]

= 119903

[[[[[[[[[

[

11986241198604

0 0 0

119860lowast

40 119860

lowast

31198611

119860lowast

1

0 1198603

minus1198623

minus11986031198622minus1198603119862lowast

1

0 119861lowast

1minus119862lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

0 1198601minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]]]

]

minus 2119903[

[

1198603

119861lowast

1

1198601

]

]

= 119903

[[[[[[[[[

[

0 119860lowast

4119860lowast

31198611

119860lowast

1

11986041198624

0 0 0

1198603

0 minus1198623

minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]]]

]

minus 2119903[

[

1198603

119861lowast

1

1198601

]

]

119903 [119860 119873

119875 0] = 119903 [

1198624minus 11986041198691015840

119860lowast

411986041198711198601119871119865119871119872

119877119865lowast119877119860lowast

1

119860lowast

40

]

= 119903

[[[[[[[[[[

[

11986241198604

0 0

119860lowast

40 119861

1119860lowast

1

0 1198603minus11986031198622minus1198603119862lowast

1

0 119861lowast

1minus119862lowast

21198611minus119862lowast

2119860lowast

1

0 1198601minus11986211198611minus1198621119860lowast

1

]]]]]]]]]]

]

minus 119903[

[

1198603

119861lowast

1

1198601

]

]

minus 119903 [119861lowast

1

1198601

]

Journal of Applied Mathematics 5

= 119903

[[[[[[[[[[

[

0 119860lowast

41198611

119860lowast

1

11986041198624

0 0

1198603

0 minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus11986211198611minus1198621119860lowast

1

]]]]]]]]]]

]

minus 119903[[

[

1198603

119861lowast

1

1198601

]]

]

minus 119903 [

119861lowast

1

1198601

]

(26)

Substituting (26) into (25) yields (18) and (20)

In Theorem 4 letting 1198624vanish and 119860

4be 119868 with

appropriate size respectively we have the following

Corollary 5 Assume that 1198601 1198621

isin H119898times119899 11986111198622

isin

H119899times119904 1198603isin H119903times119899 and 119862

3isin H119903times119903 are given then the maximal

and minimal ranks of the Hermitian solution 119883 to the system(10) can be expressed as

max 119903 (119883) = min 119886 119887 (27)

where

119886 = 119899 + 119903 [119862lowast

2

1198621

] minus 119903 [119861lowast

1

1198601

]

119887 = 2119899 + 119903

[[[[

[

1198623

119860311986221198603119862lowast

1

119862lowast

2119860lowast

3119862lowast

21198611119862lowast

2119860lowast

1

1198621119860lowast

3119862111986111198621119860lowast

1

]]]]

]

minus 2119903[

[

1198603

119861lowast

1

1198601

]

]

min 119903 (119883) = 2119903 [

119862lowast

2

1198621

]

+ 119903

[[[

[

1198623

119860311986221198603119862lowast

1

119862lowast

2119860lowast

3119862lowast

21198611119862lowast

2119860lowast

1

1198621119860lowast

3119862111986111198621119860lowast

1

]]]

]

minus 2119903

[[[[

[

119860311986221198603119862lowast

1

119862lowast

21198611119862lowast

2119860lowast

1

119862111986111198621119860lowast

1

]]]]

]

(28)

InTheorem 4 assuming that 1198601 1198611 1198621 and 119862

2vanish

we have the following

Corollary 6 Suppose that the matrix equation 1198603119883119860lowast

3= 1198623

is consistent then the extremal ranks of the quaternion matrixexpression 119891(119883) defined as (9) subject to 119860

3119883119860lowast

3= 1198623are the

following

max 119903 [119891 (119883)]

= min

119903 [11986241198604] 119903 [

[

0 119860lowast

4119860lowast

3

11986041198624

0

1198603

0 minus1198623

]

]

minus 2119903 (1198603)

min 119903 [119891 (119883)] = 2119903 [11986241198604]

+ 119903[

[

0 119860lowast

4119860lowast

3

11986041198624

0

1198603

0 minus1198623

]

]

minus 2119903[

[

0 119860lowast

4

11986041198624

1198603

0

]

]

(29)

3 A Practical Solvability Condition forHermitian Solution to System (5)

In this section we use Theorem 4 to give a necessary andsufficient condition for the existence of Hermitian solutionto system (5) by rank equalities

Theorem 7 Let 1198601 1198621

isin H119898times119899 11986111198622

isin H119899times119904 1198603

isin

H119903times119899 1198623isin H119903times119903 119860

4isin H119905times119899 and 119862

4isin H119905times119905be given then

the system (5) have Hermitian solution if and only if 1198623= 119862lowast

3

(11) (13) hold and the following equalities are all satisfied

119903 [11986041198624] = 119903 (119860

4) (30)

119903

[[[

[

1198624

1198604

119862lowast

2119860lowast

4119861lowast

1

1198621119860lowast

41198601

]]]

]

= 119903[

[

1198604

119861lowast

1

1198601

]

]

(31)

119903

[[[[[[[[

[

0 119860lowast

4119860lowast

31198611

119860lowast

1

11986041198624

0 0 0

1198603

0 minus1198623

minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]]

]

= 2119903

[[[[

[

1198604

1198603

119861lowast

1

1198601

]]]]

]

(32)

Proof It is obvious that the system (5) have Hermitiansolution if and only if the system (10) haveHermitian solutionand

min 119903 [119891 (119883)] = 0 (33)

where 119891(119883) is defined as (9) subject to system (10) Let1198830be

aHermitian solution to the system (5) then1198830is aHermitian

solution to system (10) and1198830satisfies119860

41198830119860lowast

4= 1198624 Hence

Lemma 1 yields 1198623

= 119862lowast

3 (11) (13) and (30) It follows

6 Journal of Applied Mathematics

from

[[[[[[

[

119868 0 0 0 0

0 119868 0 0 0

119860311988300 119868 0 0

119861lowast

111988300 0 119868 0

119860111988300 0 0 119868

]]]]]]

]

times

[[[[[[[

[

0 119860lowast

4119860lowast

31198611

119860lowast

1

11986041198624

0 0 0

1198603

0 minus1198623

minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]

]

times

[[[[[

[

119868 minus1198830119860lowast

40 0 0

0 119868 0 0 0

0 0 119868 0 0

0 0 0 119868 0

0 0 0 0 119868

]]]]]

]

=

[[[[[

[

0 119860lowast

4119860lowast

31198611119860lowast

1

1198604

0 0 0 0

1198603

0 0 0 0

119861lowast

10 0 0 0

1198601

0 0 0 0

]]]]]

]

(34)

that (32) holds Similarly we can obtain (31)Conversely assume that 119862

3= 119862lowast

3 (11) (13) hold then by

Lemma 1 system (10) have Hermitian solution By (20) (31)-(32) and

119903

[[[[[[[

[

0 119860lowast

41198611

119860lowast

1

11986041198624

0 0

1198603

0 minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus11986211198611minus1198621119860lowast

1

]]]]]]]

]

ge 119903

[[[

[

1198604

1198603

119861lowast

1

1198601

]]]

]

+ 119903[

[

1198604

119861lowast

1

1198601

]

]

(35)

we can get

min 119903 [119891 (119883)] le 0 (36)

However

min 119903 [119891 (119883)] ge 0 (37)

Hence (33) holds implying that the system (5) have Hermi-tian solution

ByTheorem 7 we can also get the following

Corollary 8 Suppose that 1198603 1198623 1198604 and 119862

4are those in

Theorem 7 then the quaternion matrix equations 1198603119883119860lowast

3=

1198623and 119860

4119883119860lowast

4= 1198624have common Hermitian solution if and

only if (30) hold and the following equalities are satisfied

119903 [11986031198623] = 119903 (119860

3)

119903 [

[

0 119860lowast

4119860lowast

3

11986041198624

0

1198603

0 minus1198623

]

]

= 2119903 [1198603

1198604

]

(38)

Corollary 9 Suppose that11986011198621isin H119898times119899119861

11198622isin H119899times119904 and

119860 119861 isin H119899times119899 are Hermitian Then 119860 and 119861 have a common

Hermitian g-inverse which is a solution to the system (2) if andonly if (11) holds and the following equalities are all satisfied

119903[[

[

11986011198621119860

119861lowast

1119862lowast

2119860

119860 119860

]]

]

= 119903[

[

1198601

119861lowast

1

119860

]

]

119903[[

[

11986011198621119861

119861lowast

1119862lowast

2119861

119861 119861

]]

]

= 119903[

[

1198601

119861lowast

1

119861

]

]

(39)

119903

[[[[[[[

[

0 119861 119860 1198611

119860lowast

1

119861 119861 0 0 0

119860 0 minus119860 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

2119860 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

1119860 minus119862

11198611minus1198621119860lowast

1

]]]]]]]

]

= 2119903

[[[[

[

119861

119860

119861lowast

1

1198601

]]]]

]

(40)

4 Extremal Ranks of Schur ComplementSubject to (2)

As is well known for a given block matrix

119872 = [119860 119861

119861lowast

119863] (41)

where 119860 and 119863 are Hermitian quaternion matrices withappropriate sizes then the Hermitian Schur complement of119860 in119872 is defined as

119878119860= 119863 minus 119861

lowast

119860sim

119861 (42)

where 119860sim is a Hermitian g-inverse of 119860 that is 119860sim isin 119883 |

119860119883119860 = 119860119883 = 119883lowast

Now we use Theorem 4 to establish the extremal ranks

of 119878119860given by (42) with respect to 119860sim which is a solution to

system (2)

Theorem 10 Suppose 1198601 1198621isin H119898times119899 119861

1 1198622isin H119899times119904 119863 isin

H119905times119905 119861 isin H119899times119905 and 119860 isin H119899times119899 are given and system (2)is consistent then the extreme ranks of 119878

119860given by (42) with

respect to 119860sim which is a solution of (2) are the following

max1198601119860sim=1198621

119860sim1198611=1198622

119903 (119878119860) = min 119886 119887

(43)

Journal of Applied Mathematics 7

where

119886 = 119903[

[

119863 119861lowast

119862lowast

2119861 119861lowast

1

1198621119861 1198601

]

]

minus 119903 [119861lowast

1

1198601

]

119887 = 119903

[[[[[[

[

0 119861 119860 1198611

119860lowast

1

119861lowast

119863 0 0 0

119860 0 minus119860 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

2119860 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

1119860 minus119862

11198611minus1198621119860lowast

1

]]]]]]

]

minus 2119903[

[

119860

119861lowast

1

1198601

]

]

min1198601119860sim=1198621

119860sim1198611=1198622

119903 (119878119860) = 2119903[

[

119863 119861lowast

119862lowast

2119861 119861lowast

1

1198621119861 1198601

]

]

+ 119903

[[[[[[[[

[

0 119861 119860 1198611

119860lowast

1

119861lowast

119863 0 0 0

119860 0 minus119860 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

2119860 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

1119860 minus119862

11198611minus1198621119860lowast

1

]]]]]]]]

]

minus 2119903

[[[[[[[[

[

0 119861 1198611

119860lowast

1

119861lowast

119863 0 0

119860 0 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

11198611minus1198621119860lowast

1

]]]]]]]]

]

(44)

Proof It is obvious that

max1198601119860sim=1198621 119860

sim1198611=1198622

119903 (119863 minus 119861lowast

119860sim

119861)

= max1198601119883=1198621 1198831198611=1198622 119860119883119860=119860

119903 (119863 minus 119861lowast

119883119861)

min1198601119860sim=1198621 119860

sim1198611=1198622

119903 (119863 minus 119861lowast

119860sim

119861)

= min1198601119883=1198621 1198831198611=1198622 119860119883119860=119860

119903 (119863 minus 119861lowast

119883119861)

(45)

Thus in Theorem 4 and its proof letting 1198603= 119860lowast

3= 1198623= 119860

1198604= 119861lowast

and 1198624= 119863 we can easily get the proof

In Theorem 10 let 1198601 1198621 1198611 and 119862

2vanish Then we

can easily get the following

Corollary 11 The extreme ranks of 119878119860given by (42) with

respect to 119860sim are the following

max119860sim119903 (119878119860) = min

119903 [119863 119861lowast

] 119903 [

[

0 119861 119860

119861lowast

119863 0

119860 0 minus119860

]

]

minus 2119903 (119860)

min119860sim119903 (119878119860) = 2119903 [119863 119861

lowast

] + 119903[

[

0 119861 119860

119861lowast

119863 0

119860 0 minus119860

]

]

minus 2119903[

[

0 119861

119861lowast

119863

119860 0

]

]

(46)

5 The Rank Invariance of (9)As another application of Theorem 4 we in this sectionconsider the rank invariance of thematrix expression (9) withrespect to the Hermitian solution of system (10)

Theorem 12 Suppose that (10) have Hermitian solution thenthe rank of 119891(119883) defined by (9) with respect to the Hermitiansolution of (10) is invariant if and only if

119903

[[[[[[[[

[

0 119860lowast

41198611

119860lowast

1

11986041198624

0 0

1198603

0 minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus11986211198611minus1198621119860lowast

1

]]]]]]]]

]

= 119903[[

[

1198624

1198604

119862lowast

2119860lowast

4119861lowast

1

1198621119860lowast

41198601

]]

]

+ 119903[

[

1198603

119861lowast

1

1198601

]

]

119903

[[[[[[[[

[

0 119860lowast

4119860lowast

31198611

119860lowast

1

11986041198624

0 0 0

1198603

0 minus1198623

minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]]

]

+ 119903 [119861lowast

1

1198601

]

= 119903

[[[[[[

[

0 119860lowast

41198611

119860lowast

1

11986041198624

0 0

1198603

0 minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus11986211198611minus1198621119860lowast

1

]]]]]]

]

+ 119903[

[

1198603

119861lowast

1

1198601

]

]

(47)

or

119903

[[[[[[[

[

0 119860lowast

41198611

119860lowast

1

11986041198624

0 0

1198603

0 minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus11986211198611minus1198621119860lowast

1

]]]]]]]

]

= 119903[[

[

1198624

1198604

119862lowast

2119860lowast

4119861lowast

1

1198621119860lowast

41198601

]]

]

+ 119903[[

[

1198603

119861lowast

1

1198601

]]

]

(48)

Proof It is obvious that the rank of 119891(119883) with respect toHermitian solution of system (10) is invariant if and only if

max 119903 [119891 (119883)] minusmin 119903 [119891 (119883)] = 0 (49)

By (49) Theorem 4 and simplifications we can get (47)and (48)

8 Journal of Applied Mathematics

Corollary 13 The rank of 119878119860defined by (42) with respect to

119860sim which is a solution to system (2) is invariant if and only if

119903

[[[[[[

[

0 119861 1198611

119860lowast

1

119861lowast

119863 0 0

119860 0 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

11198611minus1198621119860lowast

1

]]]]]]

]

= 119903[

[

119863 119861lowast

119862lowast

2119861 119861lowast

1

1198621119861 1198601

]

]

+ 119903[

[

119860

119861lowast

1

1198601

]

]

119903

[[[[[[[[

[

0 119861 119860 1198611

119860lowast

1

119861lowast

119863 0 0 0

119860 0 minus119860 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

2119860 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

1119860 minus119862

11198611minus1198621119860lowast

1

]]]]]]]]

]

+ 119903 [119861lowast

1

1198601

]

= 119903

[[[[[[

[

0 119861 1198611

119860lowast

1

119861lowast

119863 0 0

119860 0 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

11198611minus1198621119860lowast

1

]]]]]]

]

+ 119903[

[

119860

119861lowast

1

1198601

]

]

(50)

or

119903

[[[[[[

[

0 119861 1198611

119860lowast

1

119861lowast

119863 0 0

119860 0 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

11198611minus1198621119860lowast

1

]]]]]]

]

= 119903[

[

119863 119861lowast

119862lowast

2119861 119861lowast

1

1198621119861 1198601

]

]

+ 119903[

[

119860

119861lowast

1

1198601

]

]

(51)

Acknowledgments

This research was supported by the National Natural ScienceFoundation of China Tian Yuan Foundation (11226067) theFundamental Research Funds for the Central Universities(WM1214063) and China Postdoctoral Science Foundation(2012M511014)

References

[1] T W Hungerford Algebra Springer New York NY USA 1980[2] S K Mitra ldquoThe matrix equations 119860119883 = 119862 119883119861 = 119863rdquo Linear

Algebra and its Applications vol 59 pp 171ndash181 1984[3] C G Khatri and S K Mitra ldquoHermitian and nonnegative

definite solutions of linear matrix equationsrdquo SIAM Journal onApplied Mathematics vol 31 no 4 pp 579ndash585 1976

[4] Y X Yuan ldquoOn the symmetric solutions of matrix equation(119860119883119883119862) = (119861119863)rdquo Journal of East China Shipbuilding Institutevol 15 no 4 pp 82ndash85 2001 (Chinese)

[5] H Dai and P Lancaster ldquoLinear matrix equations from aninverse problem of vibration theoryrdquo Linear Algebra and itsApplications vol 246 pp 31ndash47 1996

[6] J Groszlig ldquoA note on the general Hermitian solution to 119860119883119860lowast =119861rdquo Malaysian Mathematical Society vol 21 no 2 pp 57ndash621998

[7] Y G Tian and Y H Liu ldquoExtremal ranks of some symmetricmatrix expressions with applicationsrdquo SIAM Journal on MatrixAnalysis and Applications vol 28 no 3 pp 890ndash905 2006

[8] Y H Liu Y G Tian and Y Takane ldquoRanks of Hermitian andskew-Hermitian solutions to the matrix equation 119860119883119860lowast = 119861rdquoLinear Algebra and its Applications vol 431 no 12 pp 2359ndash2372 2009

[9] X W Chang and J S Wang ldquoThe symmetric solution of thematrix equations 119860119883 + 119884119860 = 119862 119860119883119860119879 + 119861119884119861

119879

= 119862 and(119860119879

119883119860 119861119879

119883119861) = (119862119863)rdquo Linear Algebra and its Applicationsvol 179 pp 171ndash189 1993

[10] Q-W Wang and Z-C Wu ldquoCommon Hermitian solutionsto some operator equations on Hilbert 119862lowast-modulesrdquo LinearAlgebra and its Applications vol 432 no 12 pp 3159ndash3171 2010

[11] F O Farid M S Moslehian Q-W Wang and Z-C Wu ldquoOnthe Hermitian solutions to a system of adjointable operatorequationsrdquo Linear Algebra and its Applications vol 437 no 7pp 1854ndash1891 2012

[12] Z-H He and Q-WWang ldquoSolutions to optimization problemson ranks and inertias of a matrix function with applicationsrdquoAppliedMathematics andComputation vol 219 no 6 pp 2989ndash3001 2012

[13] Q-W Wang ldquoThe general solution to a system of real quater-nion matrix equationsrdquo Computers amp Mathematics with Appli-cations vol 49 no 5-6 pp 665ndash675 2005

[14] Q-W Wang ldquoBisymmetric and centrosymmetric solutions tosystems of real quaternion matrix equationsrdquo Computers ampMathematics with Applications vol 49 no 5-6 pp 641ndash6502005

[15] Q W Wang ldquoA system of four matrix equations over vonNeumann regular rings and its applicationsrdquo Acta MathematicaSinica vol 21 no 2 pp 323ndash334 2005

[16] Q-W Wang ldquoA system of matrix equations and a linear matrixequation over arbitrary regular rings with identityrdquo LinearAlgebra and its Applications vol 384 pp 43ndash54 2004

[17] Q-WWang H-X Chang andQNing ldquoThe common solutionto six quaternion matrix equations with applicationsrdquo AppliedMathematics and Computation vol 198 no 1 pp 209ndash2262008

[18] Q-W Wang H-X Chang and C-Y Lin ldquoP-(skew)symmetriccommon solutions to a pair of quaternion matrix equationsrdquoApplied Mathematics and Computation vol 195 no 2 pp 721ndash732 2008

[19] Q W Wang and Z H He ldquoSome matrix equations withapplicationsrdquo Linear and Multilinear Algebra vol 60 no 11-12pp 1327ndash1353 2012

[20] Q W Wang and J Jiang ldquoExtreme ranks of (skew-)Hermitiansolutions to a quaternion matrix equationrdquo Electronic Journal ofLinear Algebra vol 20 pp 552ndash573 2010

[21] Q-W Wang and C-K Li ldquoRanks and the least-norm of thegeneral solution to a system of quaternion matrix equationsrdquoLinear Algebra and its Applications vol 430 no 5-6 pp 1626ndash1640 2009

[22] Q-W Wang X Liu and S-W Yu ldquoThe common bisymmetricnonnegative definite solutions with extreme ranks and inertiasto a pair of matrix equationsrdquo Applied Mathematics and Com-putation vol 218 no 6 pp 2761ndash2771 2011

Journal of Applied Mathematics 9

[23] Q-W Wang F Qin and C-Y Lin ldquoThe common solution tomatrix equations over a regular ring with applicationsrdquo IndianJournal of Pure andAppliedMathematics vol 36 no 12 pp 655ndash672 2005

[24] Q-W Wang G-J Song and C-Y Lin ldquoExtreme ranks of thesolution to a consistent system of linear quaternion matrixequations with an applicationrdquo Applied Mathematics and Com-putation vol 189 no 2 pp 1517ndash1532 2007

[25] Q-W Wang G-J Song and C-Y Lin ldquoRank equalities relatedto the generalized inverse 119860

(2)

119879119878with applicationsrdquo Applied

Mathematics and Computation vol 205 no 1 pp 370ndash3822008

[26] Q W Wang G J Song and X Liu ldquoMaximal and minimalranks of the common solution of some linear matrix equationsover an arbitrary division ring with applicationsrdquo AlgebraColloquium vol 16 no 2 pp 293ndash308 2009

[27] Q-W Wang Z-C Wu and C-Y Lin ldquoExtremal ranks ofa quaternion matrix expression subject to consistent systemsof quaternion matrix equations with applicationsrdquo AppliedMathematics and Computation vol 182 no 2 pp 1755ndash17642006

[28] Q W Wang and S W Yu ldquoRanks of the common solution tosome quaternion matrix equations with applicationsrdquo Bulletinof Iranian Mathematical Society vol 38 no 1 pp 131ndash157 2012

[29] Q W Wang S W Yu and W Xie ldquoExtreme ranks of realmatrices in solution of the quaternion matrix equation 119860119883119861 =

119862with applicationsrdquoAlgebra Colloquium vol 17 no 2 pp 345ndash360 2010

[30] Q-W Wang S-W Yu and Q Zhang ldquoThe real solutions toa system of quaternion matrix equations with applicationsrdquoCommunications in Algebra vol 37 no 6 pp 2060ndash2079 2009

[31] Q-W Wang S-W Yu and C-Y Lin ldquoExtreme ranks of alinear quaternionmatrix expression subject to triple quaternionmatrix equations with applicationsrdquo Applied Mathematics andComputation vol 195 no 2 pp 733ndash744 2008

[32] Q-W Wang and F Zhang ldquoThe reflexive re-nonnegativedefinite solution to a quaternion matrix equationrdquo ElectronicJournal of Linear Algebra vol 17 pp 88ndash101 2008

[33] Q W Wang X Zhang and Z H He ldquoOn the Hermitianstructures of the solution to a pair of matrix equationsrdquo Linearand Multilinear Algebra vol 61 no 1 pp 73ndash90 2013

[34] X Zhang Q-W Wang and X Liu ldquoInertias and ranks ofsome Hermitian matrix functions with applicationsrdquo CentralEuropean Journal of Mathematics vol 10 no 1 pp 329ndash3512012

[35] Y G Tian and S Z Cheng ldquoThemaximal andminimal ranks of119860 minus 119861119883119862 with applicationsrdquo New York Journal of Mathematicsvol 9 pp 345ndash362 2003

[36] Y G Tian ldquoUpper and lower bounds for ranks of matrixexpressions using generalized inversesrdquo Linear Algebra and itsApplications vol 355 pp 187ndash214 2002

[37] Y H Liu and Y G Tian ldquoMore on extremal ranks of thematrix expressions119860minus119861119883plusmn119883

lowast

119861lowast with statistical applicationsrdquo

Numerical Linear Algebra with Applications vol 15 no 4 pp307ndash325 2008

[38] Y H Liu and Y G Tian ldquoMax-min problems on the ranksand inertias of the matrix expressions 119860 minus 119861119883119862 plusmn (119861119883119862)

lowast withapplicationsrdquo Journal of Optimization Theory and Applicationsvol 148 no 3 pp 593ndash622 2011

[39] T Ando ldquoGeneralized Schur complementsrdquo Linear Algebra andits Applications vol 27 pp 173ndash186 1979

[40] D Carlson E Haynsworth and T Markham ldquoA generalizationof the Schur complement by means of the Moore-Penroseinverserdquo SIAM Journal onAppliedMathematics vol 26 pp 169ndash175 1974

[41] M Fiedler ldquoRemarks on the Schur complementrdquo Linear Algebraand its Applications vol 39 pp 189ndash195 1981

[42] Y G Tian ldquoMore on maximal and minimal ranks of Schurcomplements with applicationsrdquo Applied Mathematics andComputation vol 152 no 3 pp 675ndash692 2004

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 217540 14 pageshttpdxdoiorg1011552013217540

Research ArticleAn Efficient Algorithm for the Reflexive Solution of theQuaternion Matrix Equation 119860119883119861 + 119862119883119867119863 = 119865

Ning Li12 Qing-Wen Wang1 and Jing Jiang3

1 Department of Mathematics Shanghai University Shanghai 200444 China2 School of Mathematics and Quantitative Economics Shandong University of Finance and Economics Jinan 250002 China3Department of Mathematics Qilu Normal University Jinan 250013 China

Correspondence should be addressed to Qing-Wen Wang wqw858yahoocomcn

Received 3 October 2012 Accepted 4 December 2012

Academic Editor P N Shivakumar

Copyright copy 2013 Ning Li et al This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We propose an iterative algorithm for solving the reflexive solution of the quaternion matrix equation 119860119883119861 + 119862119883119867

119863 = 119865 Whenthe matrix equation is consistent over reflexive matrix X a reflexive solution can be obtained within finite iteration steps in theabsence of roundoff errors By the proposed iterative algorithm the least Frobenius norm reflexive solution of the matrix equationcan be derived when an appropriate initial iterative matrix is chosen Furthermore the optimal approximate reflexive solution to agiven reflexive matrix119883

0can be derived by finding the least Frobenius norm reflexive solution of a new corresponding quaternion

matrix equation Finally two numerical examples are given to illustrate the efficiency of the proposed methods

1 Introduction

Throughout the paper the notations R119898times119899 and H119898times119899 repre-sent the set of all 119898 times 119899 real matrices and the set of all 119898 times 119899

matrices over the quaternion algebraH = 1198861+ 119886

2119894+ 119886

3119895+ 119886

4119896 |

1198942

= 1198952

= 1198962

= 119894119895119896 = minus1 1198861 119886

2 119886

3 119886

4isin R We denote the

identity matrix with the appropriate size by 119868 We denote theconjugate transpose the transpose the conjugate the tracethe column space the real part the 119898119899 times 1 vector formedby the vertical concatenation of the respective columnsof a matrix 119860 by 119860

119867

119860119879

119860 tr(119860) 119877(119860)Re(119860) and vec(119860)respectivelyThe Frobenius norm of119860 is denoted by 119860 thatis 119860 = radictr(119860119867119860) Moreover119860otimes119861 and119860⊙119861 stand for theKronecker matrix product and Hadmard matrix product ofthe matrices 119860 and 119861

Let 119876 isin H119899times119899 be a generalized reflection matrix that is119876

2

= 119868 and 119876119867

= 119876 A matrix 119860 is called reflexive withrespect to the generalized reflection matrix 119876 if 119860 = 119876119860119876It is obvious that any matrix is reflexive with respect to 119868Let RH119899times119899

(119876) denote the set of order 119899 reflexive matriceswith respect to 119876 The reflexive matrices with respect to ageneralized reflection matrix 119876 have been widely used inengineering and scientific computations [1 2]

In the field of matrix algebra quaternion matrix equa-tions have received much attention Wang et al [3] gavenecessary and sufficient conditions for the existence andthe representations of P-symmetric and P-skew-symmetricsolutions of quaternion matrix equations 119860

119886119883 = 119862

119886and

119860119887119883119861

119887= 119862

119887 Yuan and Wang [4] derived the expressions of

the least squares 120578-Hermitian solution with the least normand the expressions of the least squares anti-120578-Hermitiansolution with the least norm for the quaternion matrixequation 119860119883119861 + 119862119883119863 = 119864 Jiang and Wei [5] derivedthe explicit solution of the quaternion matrix equation 119883 minus

119860119883119861 = 119862 Li and Wu [6] gave the expressions of symmetricand skew-antisymmetric solutions of the quaternion matrixequations 119860

1119883 = 119862

1and 119883119861

3= 119862

3 Feng and Cheng [7]

gave a clear description of the solution set of the quaternionmatrix equation 119860119883 minus 119883119861 = 0

The iterative method is a very important method tosolve matrix equations Peng [8] constructed a finite iterationmethod to solve the least squares symmetric solutions oflinear matrix equation 119860119883119861 = 119862 Also Peng [9ndash11] presentedseveral efficient iteration methods to solve the constrainedleast squares solutions of linear matrix equations 119860119883119861 =

119862 and 119860119883119861 + 119862119884119863 = 119864 by using Paigersquos algorithm [12]

2 Journal of Applied Mathematics

as the frame method Duan et al [13ndash17] proposed iterativealgorithms for the (Hermitian) positive definite solutionsof some nonlinear matrix equations Ding et al proposedthe hierarchical gradient-based iterative algorithms [18] andhierarchical least squares iterative algorithms [19] for solvinggeneral (coupled) matrix equations based on the hierarchi-cal identification principle [20] Wang et al [21] proposedan iterative method for the least squares minimum-normsymmetric solution of 119860X119861 = 119864 Dehghan and Hajarianconstructed finite iterative algorithms to solve several linearmatrix equations over (anti)reflexive [22ndash24] generalizedcentrosymmetric [25 26] and generalized bisymmetric [2728] matrices Recently Wu et al [29ndash31] proposed iterativealgorithms for solving various complex matrix equations

However to the best of our knowledge there has beenlittle information on iterative methods for finding a solutionof a quaternionmatrix equation Due to the noncommutativemultiplication of quaternions the study of quaternionmatrixequations is more complex than that of real and complexequations Motivated by the work mentioned above andkeeping the interests and wide applications of quaternionmatrices in view (eg [32ndash45]) we in this paper consideran iterative algorithm for the following two problems

Problem 1 For given matrices 119860119862 isin H119898times119899

119861 119863 isin

H119899times119901

119865 isin H119898times119901 and the generalized reflectionmatrix119876 find119883 isin RH119899times119899

(119876) such that

119860119883119861 + 119862119883119867

119863 = 119865 (1)

Problem 2 When Problem 1 is consistent let its solutionset be denoted by 119878

119867 For a given reflexive matrix 119883

0isin

RH119899times119899

(119876) find119883 isin RH119899times119899

(119876) such that10038171003817100381710038171003817119883 minus 119883

0

10038171003817100381710038171003817= min

119883isin119878119867

1003817100381710038171003817119883 minus 1198830

1003817100381710038171003817 (2)

The remainder of this paper is organized as followsIn Section 2 we give some preliminaries In Section 3 weintroduce an iterative algorithm for solving Problem 1 Thenwe prove that the given algorithm can be used to obtain areflexive solution for any initial matrix within finite stepsin the absence of roundoff errors Also we prove that theleast Frobenius norm reflexive solution can be obtained bychoosing a special kind of initial matrix In addition theoptimal reflexive solution of Problem 2 by finding the leastFrobenius norm reflexive solution of a new matrix equationis given In Section 4 we give two numerical examples toillustrate our results In Section 5 we give some conclusionsto end this paper

2 Preliminary

In this section we provide some results which will playimportant roles in this paper First we give a real innerproduct for the space H119898times119899 over the real field R

Theorem 3 In the space H119898times119899 over the field R a real innerproduct can be defined as

⟨119860 119861⟩ = Re [tr (119861119867

119860)] (3)

for 119860 119861 isin H119898times119899 This real inner product space is denoted as(H119898times119899

R ⟨sdot sdot⟩)

Proof (1) For 119860 isin H119898times119899 let 119860 = 1198601+ 119860

2119894 + 119860

3119895 + 119860

4119896 then

⟨119860 119860⟩ = Re [tr (119860119867

119860)]

= tr (119860119879

1119860

1+ 119860

119879

2119860

2+ 119860

119879

3119860

3+ 119860

119879

4119860

4)

(4)

It is obvious that ⟨119860 119860⟩ gt 0 and ⟨119860 119860⟩ = 0 hArr 119860 = 0(2) For 119860 119861 isin H119898times119899 let 119860 = 119860

1+ 119860

2119894 + 119860

3119895 + 119860

4119896 and

119861 = 1198611+ 119861

2119894 + 119861

3119895 + 119861

4119896 then we have

⟨119860 119861⟩ = Re [tr (119861119867

119860)]

= tr (119861119879

1119860

1+ 119861

119879

2119860

2+ 119861

119879

3119860

3+ 119861

119879

4119860

4)

= tr (119860119879

1119861

1+ 119860

119879

2119861

2+ 119860

119879

3119861

3+ 119860

119879

4119861

4)

= Re [tr (119860119867

119861)] = ⟨119861 119860⟩

(5)

(3) For 119860 119861 119862 isin H119898times119899

⟨119860 + 119861 119862⟩ = Re tr [119862119867

(119860 + 119861)]

= Re [tr (119862119867

119860 + 119862119867

119861)]

= Re [tr (119862119867

119860)] + Re [tr (119862119867

119861)]

= ⟨119860 119862⟩ + ⟨119861 119862⟩

(6)

(4) For 119860 119861 isin H119898times119899 and 119886 isin R

⟨119886119860 119861⟩ = Re tr [119861119867

(119886119860)] = Re [tr (119886119861119867

119860)]

= 119886Re [tr (119861119867

119860)] = 119886 ⟨119860 119861⟩

(7)

All the above arguments reveal that the space H119898times119899 overfield R with the inner product defined in (3) is an innerproduct space

Let sdot Δrepresent the matrix norm induced by the inner

product ⟨sdot sdot⟩ For an arbitrary quaternion matrix 119860 isin H119898times119899it is obvious that the following equalities hold

119860Δ= radic⟨119860119860⟩ = radicRe [tr (119860119867119860)] = radictr (119860119867119860) = 119860

(8)

which reveals that the induced matrix norm is exactly theFrobenius norm For convenience we still use sdot to denotethe induced matrix norm

Let 119864119894119895

denote the 119898 times 119899 quaternion matrix whose(119894 119895) entry is 1 and the other elements are zeros In innerproduct space (H119898times119899

R ⟨sdot sdot⟩) it is easy to verify that 119864119894119895 119864

119894119895119894

119864119894119895119895 119864

119894119895119896 119894 = 1 2 119898 119895 = 1 2 119899 is an orthonormal

basis which reveals that the dimension of the inner productspace (H119898times119899

R ⟨sdot sdot⟩) is 4119898119899Next we introduce a real representation of a quaternion

matrix

Journal of Applied Mathematics 3

For an arbitrary quaternion matrix 119872 = 1198721+ 119872

2119894 +

1198723119895+ 119872

4119896 a map 120601(sdot) fromH119898times119899 toR4119898times4119899 can be defined

as

120601 (119872) =

[[[

[

1198721

minus1198722

minus1198723

minus1198724

1198722

1198721

minus1198724

1198723

1198723

1198724

1198721

minus1198722

1198724

minus1198723

1198722

1198721

]]]

]

(9)

Lemma4 (see [41]) Let119872 and119873 be two arbitrary quaternionmatrices with appropriate size The map 120601(sdot) defined by (9)satisfies the following properties

(a) 119872 = 119873 hArr 120601(119872) = 120601(119873)

(b) 120601(119872 + 119873) = 120601(119872) + 120601(119873) 120601(119872119873) = 120601(119872)120601(119873)

120601(119896119872) = 119896120601(119872) 119896 isin R

(c) 120601(119872119867

) = 120601119879

(119872)

(d) 120601(119872) = 119879minus1

119898120601(119872)119879

119899= 119877

minus1

119898120601(119872)119877

119899= 119878

minus1

119898120601(119872)119878

119899

where

119879119905=

[[[

[

0 minus119868119905

0 0

119868119905

0 0 0

0 0 0 119868119905

0 0 minus119868119905

0

]]]

]

119877119905= [

0 minus1198682119905

1198682119905

0]

119878119905=

[[[

[

0 0 0 minus119868119905

0 0 119868119905

0

0 minus119868119905

0 0

119868119905

0 0 0

]]]

]

119905 = 119898 119899

(10)

By (9) it is easy to verify that

1003817100381710038171003817120601 (119872)1003817100381710038171003817 = 2 119872 (11)

Finally we introduce the commutation matrixA commutation matrix 119875(119898 119899) is a 119898119899 times 119898119899 matrix

which has the following explicit form

119875 (119898 119899) =

119898

sum

119894=1

119899

sum

119895=1

119864119894119895otimes 119864

119879

119894119895= [119864

119879

119894119895] 119894=1119898

119895=1119899

(12)

Moreover 119875(119898 119899) is a permutation matrix and 119875(119898 119899) =

119875119879

(119899119898) = 119875minus1

(119899119898) We have the following lemmas on thecommutation matrix

Lemma 5 (see [46]) Let 119860 be a 119898 times 119899 matrix There is acommutation matrix 119875(119898 119899) such that

vec (119860119879

) = 119875 (119898 119899) vec (119860) (13)

Lemma 6 (see [46]) Let 119860 be a 119898 times 119899 matrix and 119861 a 119901 times

119902 matrix There exist two commutation matrices 119875(119898 119901) and119875(119899 119902) such that

119861 otimes 119860 = 119875119879

(119898 119901) (119860 otimes 119861) 119875 (119899 119902) (14)

3 Main Results

31 The Solution of Problem 1 In this subsection we willconstruct an algorithm for solving Problem 1 Then somelemmaswill be given to analyse the properties of the proposedalgorithm Using these lemmas we prove that the proposedalgorithm is convergent

Algorithm 7 (Iterative algorithm for Problem 1)

(1) Choose an initial matrix119883(1) isin RH119899times119899

(119876)(2) Calculate

119877 (1) = 119865 minus 119860119883 (1) 119861 minus 119862119883119867

(1)119863

119875 (1) =1

2(119860

119867

119877 (1) 119861119867

+ 119863119877119867

(1) 119862

+ 119876119860119867

119877 (1) 119861119867

119876 + 119876119863119877119867

(1) 119862119876)

119896 = 1

(15)

(3) If 119877(119896) = 0 then stop and 119883(119896) is the solution ofProblem 1 else if 119877(119896) =0 and 119875(119896) = 0 then stopand Problem 1 is not consistent else 119896 = 119896 + 1

(4) Calculate

119883 (119896) = 119883 (119896 minus 1) +119877 (119896 minus 1)

2

119875 (119896 minus 1)2119875 (119896 minus 1)

119877 (119896) = 119877 (119896 minus 1) minus119877 (119896 minus 1)

2

119875 (119896 minus 1)2

times (119860119875 (119896 minus 1) 119861 + 119862119875119867

(119896 minus 1)119863)

119875 (119896) =1

2(119860

119867

119877 (119896) 119861119867

+ 119863119877119867

(119896) 119862

+ 119876119860119867

119877 (119896) 119861119867

119876 +119876119863119877119867

(119896) 119862119876)

+119877 (119896)

2

119877 (119896 minus 1)2119875 (119896 minus 1)

(16)

(5) Go to Step (3)

Lemma 8 Assume that the sequences 119877(119894) and 119875(119894)

are generated by Algorithm 7 then ⟨119877(119894) 119877(119895)⟩ =

0 and ⟨119875(119894) 119875(119895)⟩ = 0 for 119894 119895 = 1 2

119894 =119895

Proof Since ⟨119877(119894) 119877(119895)⟩ = ⟨119877(119895) 119877(119894)⟩ and⟨119875(119894) 119875(119895)⟩ = ⟨119875(119895) 119875(119894)⟩ for 119894 119895 = 1

2 we only need to prove that ⟨119877(119894)

119877(119895)⟩ = 0 and ⟨119875(119894) 119875(119895)⟩ = 0 for 1 le 119894 lt 119895Now we prove this conclusion by induction

Step 1 We show that

⟨119877 (119894) 119877 (119894 + 1)⟩ = 0

⟨119875 (119894) 119875 (119894 + 1)⟩ = 0 for 119894 = 1 2

(17)

4 Journal of Applied Mathematics

We also prove (17) by induction When 119894 = 1 we have

⟨119877 (1) 119877 (2)⟩

= Re tr [119877119867

(2) 119877 (1)]

= Retr[(119877 (1) minus119877 (1)

2

119875 (1)2(119860119875 (1) 119861 + 119862119875

119867

(1)119863))

119867

times 119877 (1) ]

= 119877 (1)2

minus119877 (1)

2

119875 (1)2

times Re tr [119875119867

(1) (119860119867

119877 (1) 119861119867

+ 119863119877119867

(1) 119862)]

= 119877(1)2

minus119877(1)

2

119875(1)2

timesRetr [119875119867

(1) ((119860119867

119877 (1) 119861119867

+119863119877119867

(1) 119862+119876119860119867

119877 (1)

times 119861119867

119876 + 119876119863119877119867

(1) 119862119876) times (2)minus1

)]

= 119877 (1)2

minus119877 (1)

2

119875 (1)2119875 (1)

2

= 0

(18)

Also we can write

⟨119875 (1) 119875 (2)⟩

= Re tr [119875119867

(2) 119875 (1)]

= Retr[(1

2(119860

119867

119877 (2) 119861119867

+ 119863119877119867

(2) 119862

+ 119876119860119867

119877 (2) 119861119867

119876 + 119876119863119877119867

(2) 119862119876)

+119877 (2)

2

119877 (1)2119875 (1))

119867

119875 (1)]

=119877 (2)

2

119877 (1)2119875 (1)

2

+ Re tr [119875119867

(1) times ((119860119867

119877 (2) 119861119867

+ 119863119877119867

(2) 119862

+ 119876119860119867

119877 (2) 119861119867

119876

+119876119863119877119867

(2) 119862119876) times (2)minus1

)]

=119877 (2)

2

119877 (1)2119875 (1)

2

+ Re tr [119875119867

(1) (119860119867

119877 (2) 119861119867

+ 119863119877119867

(2) 119862)]

=119877 (2)

2

119877 (1)2119875 (1)

2

+ Re tr [119877119867

(2) (119860119875 (1) 119861 + 119862119875119867

(1)119863)]

=119877 (2)

2

119877 (1)2119875 (1)

2

+119875 (1)

2

119877 (1)2Re tr [119877119867

(2) (119877 (1) minus 119877 (2))]

=119877 (2)

2

119877 (1)2119875 (1)

2

minus119875 (1)

2

119877 (1)2119877 (2)

2

= 0

(19)

Now assume that conclusion (17) holds for 1 le 119894 le 119904 minus 1 then

⟨119877 (119904) 119877 (119904 + 1)⟩

= Re tr [119877119867

(119904 + 1) 119877 (119904)]

= Retr[(119877 (119904) minus119877 (119904)

2

119875 (119904)2(119860119875 (119904) 119861 + 119862119875

119867

(119904)119863))

119867

times 119877 (119904) ]

= 119877 (s)2 minus119877 (119904)

2

119875 (119904)2

times Re tr [119875119867

(119904) (119860119867

119877 (119904) 119861119867

+ 119863119877119867

(119904) 119862)]

= 119877 (119904)2

minus119877 (119904)

2

119875 (119904)2

times Re tr [119875119867

(119904) times ((119860119867

119877 (119904) 119861119867

+ 119863119877119867

(119904) 119862

+ 119876119860119867

119877 (119904) 119861119867

119876

+ 119876119863119877119867

(119904) 119862119876) times (2)minus1

)]

= 119877 (119904)2

minus119877 (119904)

2

119875 (119904)2

times Retr[119875119867

(119904) (119875 (119904) minus119877 (119904)

2

119877 (119904 minus 1)2119875 (119904 minus 1))]

= 0

(20)

Journal of Applied Mathematics 5

And it can also be obtained that

⟨119875 (119904) 119875 (119904 + 1)⟩

= Re tr [119875119867

(119904 + 1) 119875 (119904)]

= Retr[(((119860119867

119877 (119904 + 1) 119861119867

+ 119863119877119867

(119904 + 1) 119862

+ 119876119860119867

119877 (119904 + 1) 119861119867

119876

+ 119876119863119877119867

(119904 + 1) 119862119876) times (2)minus1

)

+119877 (119904 + 1)

2

119877 (119904)2

119875(119904))

119867

119875 (119904)]

=119877 (119904 + 1)

2

119877 (119904)2

119875 (119904)2

+ Re tr [119875119867

(119904) (119860119867

119877 (119904 + 1) 119861119867

+ 119863119877119867

(119904 + 1) 119862)]

=119877 (119904 + 1)

2

119877 (119904)2

119875 (119904)2

+ Re tr [119877119867

(119904 + 1) (119860119875 (119904) 119861 + 119862119875119867

(119904)119863)]

=119877 (119904 + 1)

2

119877 (119904)2

119875 (119904)2

+119875 (119904)

2

119877 (119904)2Re tr [119877119867

(119904 + 1) (119877 (119904) minus 119877 (119904 + 1))]

= 0

(21)

Therefore the conclusion (17) holds for 119894 = 1 2 Step 2 Assume that ⟨119877(119894) 119877(119894+119903)⟩ = 0 and ⟨119875(119894) 119875(119894+119903)⟩ = 0

for 119894 ge 1 and 119903 ge 1 We will show that

⟨119877 (119894) 119877 (119894 + 119903 + 1)⟩ = 0 ⟨119875 (119894) 119875 (119894 + 119903 + 1)⟩ = 0

(22)

We prove the conclusion (22) in two substepsSubstep 21 In this substep we show that

⟨119877 (1) 119877 (119903 + 2)⟩ = 0 ⟨119875 (1) 119875 (119903 + 2)⟩ = 0 (23)

It follows from Algorithm 7 that

⟨119877 (1) 119877 (119903 + 2)⟩

= Re tr [119877119867

(119903 + 2) 119877 (1)]

= Retr[(119877 (119903 + 1) minus119877 (119903 + 1)

2

119875 (119903 + 1)2

times (119860119875 (119903 + 1) 119861+119862119875119867

(119903 + 1)119863))

119867

119877 (1)]

= Re tr [119877119867

(119903 + 1) 119877 (1)] minus119877 (119903 + 1)

2

119875 (119903 + 1)2

times Re tr [119875119867

(119903 + 1) (119860119867

119877 (1) 119861119867

+ 119863119877119867

(1) 119862)]

= Re tr [119877119867

(119903 + 1) 119877 (1)] minus119877 (119903 + 1)

2

119875 (119903 + 1)2

times Re tr [119875119867

(119903 + 1) ((119860119867

119877 (1) 119861119867

+ 119863119877119867

(1) 119862

+ 119876119860119867

119877 (1) 119861119867

119876

+ 119876119863119877119867

(1) 119862119876) times (2)minus1

)]

= Re tr [119877119867

(119903 + 1) 119877 (1)]

minus119877 (119903 + 1)

2

119875 (119903 + 1)2Re tr [119875119867

(119903 + 1) 119875 (1)]

= 0

(24)

Also we can write

⟨119875 (1) 119875 (119903 + 2)⟩

= Re tr [119875119867

(119903 + 2) 119875 (1)]

= Retr[( ((119860119867

119877 (119903 + 2) 119861119867

+ 119863119877119867

(119903 + 2) 119862

+119876119860119867

119877 (119903 + 2) 119861119867

119876 + 119876119863119877119867

(119903 + 2) 119862119876)

times (2)minus1

) +119877(119903 + 2)

2

119877(119903 + 1)2119875(119903 + 1))

119867

119875 (1)]

= Re tr [(119860119867

119877 (119903 + 2) 119861119867

+ 119863119877119867

(119903 + 2) 119862)119867

119875 (1)]

+119877 (119903 + 2)

2

119877 (119903 + 1)2Re tr [119875119867

(119903 + 1) 119875 (1)]

= Re tr [119877119867

(119903 + 2) (119860119875 (1) 119861 + 119862119875119867

(1)119863)]

+119877 (119903 + 2)

2

119877 (119903 + 1)2Re tr [119875119867

(119903 + 1) 119875 (1)]

6 Journal of Applied Mathematics

=119875 (1)

2

119877 (1)2Re tr [119877119867

(119903 + 2) (119877 (1) minus 119877 (2))]

+119877 (119903 + 2)

2

119877 (119903 + 1)2Re tr [119875119867

(119903 + 1) 119875 (1)]

= 0

(25)

Substep 22 In this substep we prove the conclusion (22) inStep 2 It follows from Algorithm 7 that

⟨119877 (119894) 119877 (119894 + 119903 + 1)⟩

= Re tr [119877119867

(119894 + 119903 + 1) 119877 (119894)]

= Retr[(119877 (119894 + 119903) minus119877 (119894 + 119903)

2

119875 (119894 + 119903)2

times (119860119875 (119894 + 119903) 119861 + 119862119875119867

(119894 + 119903)119863))

119867

119877 (119894) ]

= Re tr [119877119867

(119894 + 119903) 119877 (119894)] minus119877 (119894 + 119903)

2

119875 (119894 + 119903)2

times Re tr [119875119867

(119894 + 119903) (119860119867

119877 (119894) 119861119867

+ 119863119877119867

(119894) 119862)]

= minus119877 (119894 + 119903)

2

119875 (119894 + 119903)2

times Re tr [119875119867

(119894 + 119903) ((119860119867

119877 (119894) 119861119867

+ 119863119877119867

(119894) 119862

+ 119876119860119867

119877 (119894) 119861119867

119876

+ 119876119863119877119867

(119894) 119862119876) times (2)minus1

)]

= minus119877 (119894 + 119903)

2

119875 (119894 + 119903)2

times Retr[119875119867

(119894 + 119903) (119875 (119894) minus119877 (119894)

2

119877 (119894 minus 1)2119875 (119894 minus 1))]

=119877 (119894 + 119903)

2

119877 (119894)2

119875 (119894 + 119903)2

119877 (119894 minus 1)2Re tr [119875119867

(119894 + 119903) 119875 (119894 minus 1)]

⟨119875 (119894) 119875 (119894 + 119903 + 1)⟩

= Re tr [119875119867

(119894 + 119903 + 1) 119875 (119894)]

= Retr[(1

2(119860

119867

119877 (119894 + 119903 + 1) 119861119867

+ 119863119877119867

(119894 + 119903 + 1) 119862

+ 119876119860119867

119877 (119894 + 119903 + 1) 119861119867

119876

+ 119876119863119877119867

(119894 + 119903 + 1) 119862119876)

+119877(119894 + 119903 + 1)

2

119877(119894 + 119903)2

119875(119894 + 119903))

119867

119875 (119894)]

= Re tr [(119860119867

119877 (119894 + 119903 + 1) 119861119867

+119863119877119867

(119894+119903+1) 119862)119867

119875 (119894)]

= Re tr [119877119867

(119894 + 119903 + 1) (119860119875 (119894)B + 119862119875119867

(119894) 119863)]

=119875 (119894)

2

119877 (119894)2Re tr [119877119867

(119894 + 119903 + 1) (119877 (119894) minus 119877 (119894 + 1))]

=119875 (119894)

2

119877 (119894)2Re tr [119877119867

(119894 + 119903 + 1) 119877 (119894)]

minus119875 (119894)

2

119877 (119894)2Re tr [119877119867

(119894 + 119903 + 1) 119877 (119894 + 1)]

=119875 (119894)

2

119877 (119894 + 119903)2

119877 (119894)2

119877 (119894)2

119875 (119894 + 119903)2

119877 (119894 minus 1)2

times Re tr [119875119867

(119894 + 119903) 119875 (119894 minus 1)]

(26)

Repeating the above process (26) we can obtain

⟨119877 (119894) 119877 (119894 + 119903 + 1)⟩ = sdot sdot sdot = 120572Re tr [119875119867

(119903 + 2) 119875 (1)]

⟨119875 (119894) 119875 (119894 + 119903 + 1)⟩ = sdot sdot sdot = 120573Re tr [119875119867

(119903 + 2) 119875 (1)]

(27)

Combining these two relations with (24) and (25) it impliesthat (22) holds So by the principle of induction we knowthat Lemma 8 holds

Lemma 9 Assume that Problem 1 is consistent and let 119883 isin

RH119899times119899

(119876) be its solution Then for any initial matrix 119883(1) isin

RH119899times119899

(119876) the sequences 119877(119894) 119875(119894) and 119883(119894) generatedby Algorithm 7 satisfy

⟨119875 (119894) 119883 minus 119883 (119894)⟩ = 119877 (119894)2

119894 = 1 2 (28)

Proof We also prove this conclusion by inductionWhen 119894 = 1 it follows from Algorithm 7 that

⟨119875 (1) 119883 minus 119883 (1)⟩

= Re tr [(119883 minus 119883 (1))119867

119875 (1)]

= Re tr [(119883 minus 119883 (1))119867

times ((119860119867

119877 (1) 119861119867

+ 119863119877119867

(1) 119862+119876119860119867

119877 (1) 119861119867

119876

+ 119876119863119877119867

(1) 119862119876) times (2)minus1

) ]

Journal of Applied Mathematics 7

= Re tr [(119883 minus 119883 (1))119867

(119860119867

119877 (1) 119861119867

+ 119863119877119867

(1) 119862)]

= Retr[119877119867

(1) (119860 (119883 minus 119883 (1)) 119861+119862 (119883 minus 119883 (1))119867

119863)]

= Re tr [119877119867

(1) 119877 (1)]

= 119877 (1)2

(29)

This implies that (28) holds for 119894 = 1Now it is assumed that (28) holds for 119894 = 119904 that is

⟨119875 (119904) 119883 minus 119883 (119904)⟩ = 119877 (119904)2

(30)

Then when 119894 = 119904 + 1

⟨119875 (119904 + 1) 119883 minus 119883 (119904 + 1)⟩

= Re tr [(119883 minus 119883 (119904 + 1))119867

119875 (119904 + 1)]

= Retr[(119883 minus 119883 (119904 + 1))119867

times (1

2(119860

119867

119877 (119904 + 1) 119861119867

+ 119863119877119867

(119904 + 1) 119862

+ 119876119860119867

119877 (119904 + 1) 119861119867

119876

+ 119876119863119877119867

(119904 + 1) 119862119876)

+119877(119904 + 1)

2

119877(119904)2

119875 (119904))]

= Re tr [(119883 minus 119883 (119904 + 1))119867

times (119860119867

119877 (119904 + 1) 119861119867

+ 119863119877119867

(119904 + 1) 119862) ]

+119877 (119904 + 1)

2

119877 (119904)2

Re tr [(119883 minus 119883 (119904 + 1))119867

119875 (119904)]

= Re tr [119877119867

(119904 + 1)

times (119860 (119883 minus 119883 (119904 + 1)) 119861

+ 119862 (119883 minus 119883 (119904 + 1))119867

119863)]

+119877 (119904 + 1)

2

119877 (119904)2

Re tr [(119883 minus 119883 (119904))119867

119875 (119904)]

minus Re tr [(119883 (119904 + 1) minus 119883 (119904))119867

119875 (119904)]

= 119877 (119904 + 1)2

+119877 (119904 + 1)

2

119877 (119904)2

times 119877 (119904)2

minus119877 (119904)

2

119875 (119904)2Re tr [119875119867

(119904) 119875 (119904)]

= 119877 (119904 + 1)2

(31)

Therefore Lemma 9 holds by the principle of induction

From the above two lemmas we have the followingconclusions

Remark 10 If there exists a positive number 119894 such that119877(119894) =0 and 119875(119894) = 0 then we can get from Lemma 9 thatProblem 1 is not consistent Hence the solvability of Problem1 can be determined by Algorithm 7 automatically in theabsence of roundoff errors

Theorem 11 Suppose that Problem 1 is consistentThen for anyinitial matrix119883(1) isin RH119899times119899

(119876) a solution of Problem 1 can beobtained within finite iteration steps in the absence of roundofferrors

Proof In Section 2 it is known that the inner productspace (H119898times119901

119877 ⟨sdot sdot⟩) is 4119898119901-dimensional According toLemma 9 if 119877(119894) =0 119894 = 1 2 4119898119901 then we have119875(119894) =0 119894 = 1 2 4119898119901 Hence 119877(4119898119901+1) and 119875(4119898119901+1)

can be computed From Lemma 8 it is not difficult to get

⟨119877 (119894) 119877 (119895)⟩ = 0 for 119894 119895 = 1 2 4119898119901 119894 =119895 (32)

Then 119877(1) 119877(2) 119877(4119898119901) is an orthogonal basis of theinner product space (H119898times119901

119877 ⟨sdot sdot⟩) In addition we can getfrom Lemma 8 that

⟨119877 (119894) 119877 (4119898119901 + 1)⟩ = 0 for 119894 = 1 2 4119898119901 (33)

It follows that 119877(4119898119901+1) = 0 which implies that119883(4119898119901+1)

is a solution of Problem 1

32 The Solution of Problem 2 In this subsection firstly weintroduce some lemmas Then we will prove that the leastFrobenius norm reflexive solution of (1) can be derived bychoosing a suitable initial iterative matrix Finally we solveProblem 2 by finding the least Frobenius norm reflexivesolution of a new-constructed quaternion matrix equation

Lemma 12 (see [47]) Assume that the consistent system oflinear equations119872119910 = 119887 has a solution 119910

0isin 119877(119872

119879

) then 1199100is

the unique least Frobenius norm solution of the system of linearequations

Lemma 13 Problem 1 is consistent if and only if the system ofquaternion matrix equations

119860119883119861 + 119862119883119867

119863 = 119865

119860119876119883119876119861 + 119862119876119883119867

119876119863 = 119865

(34)

8 Journal of Applied Mathematics

is consistent Furthermore if the solution sets of Problem 1 and(34) are denoted by 119878

119867and 119878

1

119867 respectively then we have 119878

119867sube

1198781

119867

Proof First we assume that Problem 1 has a solution 119883 By119860119883119861 + 119862119883

119867

119863 = 119865 and 119876119883119876 = 119883 we can obtain 119860119883119861 +

119862119883119867

119863 = 119865 and 119860119876119883119876119861 + 119862119876119883119867

119876119863 = 119865 which impliesthat 119883 is a solution of quaternion matrix equations (34) and119878119867

sube 1198781

119867

Conversely suppose (34) is consistent Let119883 be a solutionof (34) Set 119883

119886= (119883 + 119876119883119876)2 It is obvious that 119883

119886isin

RH119899times119899

(119876) Now we can write

119860119883119886119861 + 119862119883

119867

119886119863

=1

2[119860 (119883 + 119876119883119876)119861 + 119862(119883 + 119876119883119876)

119867

119863]

=1

2[119860119883119861 + 119862119883

119867

119863 + 119860119876119883119876119861 + 119862119876119883119867

119876119863]

=1

2[119865 + 119865]

= 119865

(35)

Hence 119883119886is a solution of Problem 1 The proof is completed

Lemma 14 The system of quaternion matrix equations (34) isconsistent if and only if the system of real matrix equations

120601 (119860) [119883119894119895]4 times 4

120601 (119861) + 120601 (119862) [119883119894119895]119879

4 times 4

120601 (119863) = 120601 (119865)

120601 (119860) 120601 (119876) [119883119894119895]4 times 4

120601 (119876) 120601 (119861)

+ 120601 (119862) 120601 (119876) [119883119894119895]119879

4 times 4

120601 (119876) 120601 (119863) = 120601 (119865)

(36)

is consistent where 119883119894119895

isin H119899times119899

119894 119895 = 1 2 3 4 are submatri-ces of the unknown matrix Furthermore if the solution sets of(34) and (36) are denoted by 119878

1

119867and 119878

2

119877 respectively then we

have 120601(1198781

119867) sube 119878

2

119877

Proof Suppose that (34) has a solution

119883 = 1198831+ 119883

2119894 + 119883

3119895 + 119883

4119896 (37)

Applying (119886) (119887) and (119888) in Lemma 4 to (34) yields

120601 (119860) 120601 (119883) 120601 (119861) + 120601 (119862) 120601119879

(119883) 120601 (119863) = 120601 (119865)

120601 (119860) 120601 (119876) 120601 (119883) 120601 (119876) 120601 (119861)

+ 120601 (119862) 120601 (119876) 120601119879

(119883) 120601 (119876) 120601 (119863) = 120601 (119865)

(38)

which implies that 120601(119883) is a solution of (36) and 120601(1198781

119867) sube 119878

2

119877

Conversely suppose that (36) has a solution

119883 = [119883119894119895]4 times 4

(39)

By (119889) in Lemma 4 we have that

119879minus1

119898120601 (119860) 119879

119899119883119879

minus1

119899120601 (119861) 119879

119901

+ 119879minus1

119898120601 (119862) 119879

119899119883

119879

119879minus1

119899120601 (119863)119879

119901= 119879

minus1

119898120601 (119865) 119879

119901

119877minus1

119898120601 (119860) 119877

119899119883119877

minus1

119899120601 (119861) 119877

119901

+ 119877minus1

119898120601 (119862) 119877

119899119883

119879

119877minus1

119899120601 (119863) 119877

119901= 119877

minus1

119898120601 (119865) 119877

119901

119878minus1

119898120601 (119860) 119878

119899119883119878

minus1

119899120601 (119861) 119878

119901

+ Sminus1

119898120601 (119862) 119878

119899119883

119879

119878minus1

119899120601 (119863) 119878

119901= 119878

minus1

119898120601 (119865) 119878

119901

119879minus1

119898120601 (119860) 120601 (119876) 119879

119899119883119879

minus1

119899120601 (119876) 120601 (119861) 119879

119901

+ 119879minus1

119898120601 (119862) 120601 (119876) 119879

119899119883

119879

119879minus1

119899120601 (119876) 120601 (119863) 119879

119901= 119879

minus1

119898120601 (119865) 119879

119901

119877minus1

119898120601 (119860) 120601 (119876) 119877

119899119883119877

minus1

119899120601 (119876) 120601 (119861) 119877

119901

+ 119877minus1

119898120601 (119862) 120601 (119876) 119877

119899119883

119879

119877minus1

119899120601 (119876) 120601 (119863) 119877

119901= 119877

minus1

119898120601 (119865) 119877

119901

119878minus1

119898120601 (119860) 120601 (119876) 119878

119899119883119878

minus1

119899120601 (119876) 120601 (119861) 119878

119901

+ 119878minus1

119898120601 (119862) 120601 (119876) 119878

119899119883

119879

119878minus1

119899120601 (119876) 120601 (119863) 119878

119901= 119878

minus1

119898120601 (119865) 119878

119901

(40)

Hence

120601 (119860) 119879119899119883119879

minus1

119899120601 (119861) + 120601 (119862) (119879

119899119883119879

minus1

119899)119879

120601 (119863) = 120601 (119865)

120601 (119860) 119877119899119883119877

minus1

119899120601 (119861) + 120601 (119862) (R

119899119883119877

minus1

119899)119879

120601 (119863) = 120601 (119865)

120601 (119860) 119878119899119883119878

minus1

119899120601 (119861) + 120601 (119862) (119878

119899119883119878

minus1

119899)119879

120601 (119863) = 120601 (119865)

120601 (119860) 120601 (119876) 119879119899119883119879

minus1

119899120601 (119876) 120601 (119861)

+ 120601 (119862) 120601 (119876) (119879119899119883119879

minus1

119899)119879

120601 (119876) 120601 (119863) = 120601 (119865)

120601 (119860) 120601 (119876) 119877119899119883119877

minus1

119899120601 (119876) 120601 (119861)

+ 120601 (119862) 120601 (119876) (119877119899119883119877

minus1

119899)119879

120601 (119876) 120601 (119863) = 120601 (119865)

120601 (119860) 120601 (119876) 119878119899119883119878

minus1

119899120601 (119876) 120601 (119861)

+ 120601 (119862) 120601 (119876) (119878119899119883119878

minus1

119899)119879

120601 (119876) 120601 (119863) = 120601 (119865)

(41)

which implies that 119879119899119883119879

minus1

119899 119877

119899119883119877

minus1

119899 and 119878

119899119883119878

minus1

119899are also

solutions of (36) Thus

1

4(119883 + 119879

119899119883119879

minus1

119899+ 119877

119899119883119877

minus1

119899+ 119878

119899119883119878

minus1

119899) (42)

Journal of Applied Mathematics 9

is also a solution of (36) where

119883 + 119879119899119883119879

minus1

119899+ 119877

119899119883119877

minus1

119899+ 119878

119899119883119878

minus1

119899

= [119883119894119895]4 times 4

119894 119895 = 1 2 3 4

11988311

= 11988311

+ 11988322

+ 11988333

+ 11988344

11988312

= 11988312

minus 11988321

+ 11988334

minus 11988343

11988313

= 11988313

minus 11988324

minus 11988331

+ 11988342

11988314

= 11988314

+ 11988323

minus 11988332

minus 11988341

11988321

= minus11988312

+ 11988321

minus 11988334

+ 11988343

11988322

= 11988311

+ 11988322

+ 11988333

+ 11988344

11988323

= 11988314

+ 11988323

minus 11988332

minus 11988341

11988324

= minus11988313

+ 11988324

+ 11988331

minus 11988342

11988331

= minus11988313

+ 11988324

+ 11988331

minus 11988342

11988332

= minus11988314

minus 11988323

+ 11988332

+ 11988341

11988333

= 11988311

+ 11988322

+ 11988333

+ 11988344

11988334

= 11988312

minus 11988321

+ 11988334

minus 11988343

11988341

= minus11988314

minus 11988323

+ 11988332

+ 11988341

11988342

= 11988313

minus 11988324

minus 11988331

+ 11988342

11988343

= minus11988312

+ 11988321

minus 11988334

+ 11988343

11988344

= 11988311

+ 11988322

+ 11988333

+ 11988344

(43)

Let

119883 =1

4(119883

11+ 119883

22+ 119883

33+ 119883

44)

+1

4(minus119883

12+ 119883

21minus 119883

34+ 119883

43) 119894

+1

4(minus119883

13+ 119883

24+ 119883

31minus 119883

42) 119895

+1

4(minus119883

14minus 119883

23+ 119883

32+ 119883

41) 119896

(44)

Then it is not difficult to verify that

120601 (119883) =1

4(119883 + 119879

119899119883119879

minus1

119899+ 119877

119899119883119877

minus1

119899+ 119878

119899119883119878

minus1

119899) (45)

We have that 119883 is a solution of (34) by (119886) in Lemma 4 Theproof is completed

0 5 10 15 20

0

2

4

Iteration number k

minus12

minus10

minus8

minus6

minus4

minus2

log10r (k)

Figure 1 The convergence curve for the Frobenius norm of theresiduals from Example 18

0 5 10 15 20

0

2

4

Iteration number k

minus12

minus10

minus8

minus6

minus4

minus2

log10r (k)

Figure 2 The convergence curve for the Frobenius norm of theresiduals from Example 19

Lemma 15 There exists a permutation matrix P(4n4n) suchthat (36) is equivalent to

[120601

119879

(119861) otimes 120601 (119860) + (120601119879

(119863) otimes 120601 (119862)) 119875 (4119899 4119899)

120601119879

(119861) 120601 (119876) otimes 120601 (119860) 120601 (119876) + (120601119879

(119863) 120601 (119876) otimes 120601 (119862) 120601 (119876)) 119875 (4119899 4119899)

]

times vec ([119883119894119895]4 times 4

) = [vec (120601 (119865))

vec (120601 (119865))]

(46)

Lemma 15 is easily proven through Lemma 5 So we omitit here

10 Journal of Applied Mathematics

Theorem 16 When Problem 1 is consistent let its solution setbe denoted by 119878

119867 If

∘

119883 isin 119878119867 and

∘

119883 can be expressed as

∘

119883 = 119860119867

119866119861119867

+ 119863119866119867

119862 + 119876119860119867

119866119861119867

119876

+ 119876119863119866119867

119862119876 119866 isin H119898times119901

(47)

then∘

119883 is the least Frobenius norm solution of Problem 1

Proof By (119886) (119887) and (119888) in Lemmas 4 5 and 6 we have that

vec(120601(

∘

119883))

= vec (120601119879

(119860) 120601 (119866) 120601119879

(119861) + 120601 (119863) 120601119879

(119866) 120601 (119862)

+ 120601 (119876) 120601119879

(119860) 120601 (119866) 120601119879

(119861) 120601 (119876)

+ 120601 (119876) 120601 (119863) 120601119879

(119866) 120601 (119862) 120601 (119876))

= [120601 (119861) otimes 120601119879

(119860) + (120601119879

(119862) otimes 120601 (119863)) 119875 (4119898 4119901)

120601 (119876) 120601 (119861) otimes 120601 (119876) 120601119879

(119860)

+ (120601 (119876) 120601119879

(119862) otimes 120601 (119876) 120601 (119863)) 119875 (4119898 4119901)]

times [vec (120601 (119866))

vec (120601 (119866))]

= [120601

119879

(119861) otimes 120601(119860) + (120601119879

(119863) otimes 120601(119862))119875(4119899 4119899)

120601119879

(119861)120601(119876) otimes 120601(119860)120601(119876) + (120601119879

(119863)120601(119876) otimes 120601(119862)120601(119876))119875(4119899 4119899)

]

119879

times [vec (120601 (119866))

vec (120601 (119866))]

isin 119877 [120601

119879

(119861) otimes 120601 (119860)

120601119879

(119861) 120601 (119876) otimes 120601 (119860) 120601 (119876)

+ (120601119879

(119863) otimes 120601 (119862)) 119875 (4119899 4119899)

+ (120601119879

(119863) 120601 (119876) otimes 120601 (119862) 120601 (119876)) 119875 (4119899 4119899)]

119879

(48)

By Lemma 12 120601(∘

119883) is the least Frobenius norm solution ofmatrix equations (46)

Noting (11) we derive from Lemmas 13 14 and 15 that∘

119883

is the least Frobenius norm solution of Problem 1

From Algorithm 7 it is obvious that if we consider

119883(1) = 119860119867

119866119861119867

+ 119863119866119867

119862 + 119876119860119867

119866119861119867

119876

+ 119876119863119866119867

119862119876 119866 isin H119898times119901

(49)

then all119883(119896) generated by Algorithm 7 can be expressed as

119883 (119896) = 119860119867

119866119896119861

119867

+ 119863119866119867

119896119862 + 119876119860

119867

119866119896119861

119867

119876

+ 119876119863119866119867

119896119862119876 119866

119896isin H

119898times119901

(50)

Using the above conclusion and consideringTheorem 16we propose the following theorem

Theorem 17 Suppose that Problem 1 is consistent Let theinitial iteration matrix be

119883 (1) = 119860119867

119866119861H+ 119863119866

119867

119862 + 119876119860119867

119866119861119867

119876 + 119876119863119866119867

119862119876 (51)

where 119866 is an arbitrary quaternion matrix or especially119883(1) = 0 then the solution 119883

lowast generated by Algorithm 7 isthe least Frobenius norm solution of Problem 1

Now we study Problem 2 When Problem 1 is consistentthe solution set of Problem 1 denoted by 119878

119867is not empty

Then For a given reflexive matrix1198830isin RH119899times119899

(119876)

119860119883119861 + 119862119883119867

119863 = 119865 lArrrArr 119860(119883 minus 1198830) 119861 + 119862(119883 minus 119883

0)119867

119863

= 119865 minus 1198601198830119861 minus 119862119883

119867

0119863

(52)

Let 119883 = 119883 minus 1198830and 119865 = 119865 minus 119860119883

0119861 minus 119862119883

119867

0119863 then Problem

2 is equivalent to finding the least Frobenius norm reflexivesolution of the quaternion matrix equation

119860119883119861 + 119862119883119867

119863 = 119865 (53)

By using Algorithm 7 let the initial iteration matrix 119883(1) =

119860119867

119866119861119867

+ 119863119866119867

119862 + 119876119860119867

119866119861119867

119876 + 119876119863119866119867

119862Q where 119866 is anarbitrary quaternion matrix in H119898times119901 or especially 119883(1) = 0we can obtain the least Frobenius norm reflexive solution119883

lowast

of (53) Then we can obtain the solution of Problem 2 whichis

119883 = 119883lowast

+ 1198830 (54)

Journal of Applied Mathematics 11

4 Examples

In this section we give two examples to illustrate the effi-ciency of the theoretical results

Example 18 Consider the quaternion matrix equation

119860119883119861 + 119862119883119867

119863 = 119865 (55)with

119860 = [1 + 119894 + 119895 + 119896 2 + 2119895 + 2119896 119894 + 3119895 + 3119896 2 + 119894 + 4119895 + 4119896

3 + 2 119894 + 2 119895 + 119896 3 + 3119894 + 119895 1 + 7119894 + 3119895 + 119896 6 minus 7119894 + 2119895 minus 4119896]

119861 =

[[[

[

minus2 + 119894 + 3119895 + 119896 3 + 3119894 + 4119895 + 119896

5 minus 4119894 minus 6119895 minus 5119896 minus1 minus 5119894 + 4119895 + 3119896

minus1 + 2119894 + 119895 + 119896 2 + 2119895 + 6119896

3 minus 2119894 + 119895 + 119896 4 + 119894 minus 2119895 minus 4119896

]]]

]

119862 = [1 + 119894 + 119895 + 119896 2 + 2119894 + 2119895 119896 2 + 2119894 + 2119895 + 119896

minus1 + 119894 + 3119895 + 3119896 minus2 + 3119894 + 4119895 + 2119896 6 minus 2119894 + 6119895 + 4119896 3 + 119894 + 119895 + 5119896]

119863 =

[[[

[

minus1 + 4119895 + 2119896 1 + 2119894 + 3119895 + 3119896

7 + 6119894 + 5119895 + 6119896 3 + 7119894 minus 119895 + 9119896

4 + 119894 + 6119895 + 119896 7 + 2119894 + 9119895 + 119896

1 + 119894 + 3119895 minus 3119896 1 + 3119894 + 2119895 + 2119896

]]]

]

119865 = [1 + 3119894 + 2119895 + 119896 2 minus 2119894 + 3119895 + 3119896

minus1 + 4119894 + 2119895 + 119896 minus2 + 2119894 minus 1119895]

(56)

We apply Algorithm 7 to find the reflexive solution with res-pect to the generalized reflection matrix

119876 =

[[[

[

028 0 096119896 0

0 minus1 00 0

minus096119896 0 minus028 0

0 0 0 minus1

]]]

]

(57)

For the initial matrix119883(1) = 119876 we obtain a solution that is

119883lowast

= 119883 (20)

=[[

[

027257 minus 023500119894 + 0034789119895 + 0054677119896 0085841 minus 0064349119894 + 010387119895 minus 026871119896

0028151 minus 0013246119894 minus 0091097119895 + 0073137119896 minus046775 + 0048363119894 minus 0015746119895 + 021642119896

0043349 + 0079178119894 + 0085167119895 minus 084221119896 035828 minus 013849119894 minus 0085799119895 + 011446119896

013454 minus 0028417119894 minus 0063010119895 minus 010574119896 010491 minus 0039417119894 + 018066119895 minus 0066963119896

minus0043349 + 0079178119894 + 0085167119895 + 084221119896 minus0014337 minus 014868119894 minus 0011146119895 minus 000036397119896

0097516 + 012146119894 minus 0017662119895 minus 0037534119896 minus0064483 + 0070885119894 minus 0089331119895 + 0074969119896

minus021872 + 018532119894 + 0011399119895 + 0029390119896 000048529 + 0014861119894 minus 019824119895 minus 0019117119896

minus014099 + 0084013119894 minus 0037890119895 minus 017938119896 minus063928 minus 0067488119894 + 0042030119895 + 010106119896

]]

]

(58)

with corresponding residual 119877(20) = 22775 times 10minus11 The

convergence curve for the Frobenius norm of the residuals119877(119896) is given in Figure 1 where 119903(119896) = 119877(119896)

Example 19 In this example we choose the matrices119860 119861 119862 119863 119865 and 119876 as same as in Example 18 Let

1198830= [

[

1 minus036119894 minus 075119896 0 minus05625119894 minus 048119896

036119894 128 048119895 minus096119895

0 1 minus 048119895 1 064 minus 075119895

048119896 096119895 064 072

]

]

isin RH119899times119899

(119876)

(59)

In order to find the optimal approximation reflexive solutionto the given matrix 119883

0 let 119883 = 119883 minus 119883

0and 119865 = 119865 minus

1198601198830119861 minus 119862119883

119867

0119863 Now we can obtain the least Frobenius

norm reflexive solution119883lowast of the quaternionmatrix equation

119860119883119861 + 119862 119883119867

119863 = 119865 by choosing the initial matrix 119883(1) = 0that is

12 Journal of Applied Mathematics

119883lowast

= 119883 (20)

=

[[[

[

minus047933 + 0021111119894 + 011703119895 minus 0066934119896 minus052020 minus 019377119894 + 017420119895 minus 059403119896

minus031132 minus 011705119894 minus 033980119895 + 011991119896 minus086390 minus 021715119894 minus 0037089119895 + 040609119896

minus024134 minus 0025941119894 minus 031930119895 minus 026961119896 minus044552 + 013065119894 + 014533119895 + 039015119896

minus011693 minus 027043119894 + 022718119895 minus 00000012004119896 minus044790 minus 031260119894 minus 10271119895 + 027275119896

024134 minus 0025941119894 minus 031930119895 + 026961119896 0089010 minus 067960119894 + 043044119895 minus 0045772119896

minus0089933 minus 025485119894 + 0087788119895 minus 023349119896 minus017004 minus 037655119894 + 080481119895 + 037636119896

minus063660 + 016515119894 minus 013216119895 + 0073845119896 minus0034329 + 032283119894 + 050970119895 minus 0066757119896

000000090029 + 017038119894 + 020282119895 minus 0087697119896 minus044735 + 021846119894 minus 021469119895 minus 015347119896

]]]

]

(60)

with corresponding residual 119877(20) = 43455 times 10minus11 The

convergence curve for the Frobenius norm of the residuals119877(119896) is given in Figure 2 where 119903(119896) = 119877(119896)

Therefore the optimal approximation reflexive solutionto the given matrix119883

0is

119883 = 119883lowast

+ 1198830

=

[[[

[

052067 + 0021111119894 + 011703119895 minus 0066934119896 minus052020 minus 055377119894 + 017420119895 minus 13440119896

minus031132 + 024295119894 minus 033980119895 + 011991119896 041610 minus 021715119894 minus 0037089119895 + 040609119896

minus024134 minus 0025941119894 minus 031930119895 minus 026961119896 055448 + 013065119894 minus 033467119895 + 039015119896

minus011693 minus 027043119894 + 022718119895 + 048000119896 minus044790 minus 031260119894 minus 0067117119895 + 027275119896

024134 minus 0025941119894 minus 031930119895 + 026961119896 0089010 minus 12421119894 + 043044119895 minus 052577119896

minus0089933 minus 025485119894 + 056779119895 minus 023349119896 minus017004 minus 037655119894 minus 015519119895 + 037636119896

036340 + 016515119894 minus 013216119895 + 0073845119896 060567 + 032283119894 minus 024030119895 minus 0066757119896

064000 + 017038119894 + 020282119895 minus 0087697119896 027265 + 021846119894 minus 021469119895 minus 015347119896

]]]

]

(61)

The results show that Algorithm 7 is quite efficient

5 Conclusions

In this paper an algorithm has been presented for solving thereflexive solution of the quaternion matrix equation 119860119883119861 +

119862119883119867

119863 = 119865 By this algorithm the solvability of the problemcan be determined automatically Also when the quaternionmatrix equation 119860119883119861 + 119862119883

119867

119863 = 119865 is consistent overreflexive matrix 119883 for any reflexive initial iterative matrixa reflexive solution can be obtained within finite iterationsteps in the absence of roundoff errors It has been proventhat by choosing a suitable initial iterative matrix we canderive the least Frobenius norm reflexive solution of thequaternion matrix equation 119860119883119861 + 119862119883

119867

119863 = 119865 throughAlgorithm 7 Furthermore by using Algorithm 7 we solvedProblem 2 Finally two numerical examples were given toshow the efficiency of the presented algorithm

Acknowledgments

This research was supported by Grants from InnovationProgram of Shanghai Municipal Education Commission(13ZZ080) the National Natural Science Foundation ofChina (11171205) theNatural Science Foundation of Shanghai

(11ZR1412500) the Discipline Project at the correspondinglevel of Shanghai (A 13-0101-12-005) and Shanghai LeadingAcademic Discipline Project (J50101)

References

[1] H C Chen and A H Sameh ldquoAmatrix decompositionmethodfor orthotropic elasticity problemsrdquo SIAM Journal on MatrixAnalysis and Applications vol 10 no 1 pp 39ndash64 1989

[2] H C Chen ldquoGeneralized reflexive matrices special propertiesand applicationsrdquo SIAM Journal on Matrix Analysis and Appli-cations vol 19 no 1 pp 140ndash153 1998

[3] Q W Wang H X Chang and C Y Lin ldquo119875-(skew)symmetriccommon solutions to a pair of quaternion matrix equationsrdquoApplied Mathematics and Computation vol 195 no 2 pp 721ndash732 2008

[4] S F Yuan and Q W Wang ldquoTwo special kinds of least squaressolutions for the quaternionmatrix equation119860119883119861+119862119883119863 = 119864rdquoElectronic Journal of Linear Algebra vol 23 pp 257ndash274 2012

[5] T S Jiang and M S Wei ldquoOn a solution of the quaternionmatrix equation 119883 minus 119860119883119861 = 119862 and its applicationrdquo ActaMathematica Sinica (English Series) vol 21 no 3 pp 483ndash4902005

[6] Y T Li and W J Wu ldquoSymmetric and skew-antisymmetricsolutions to systems of real quaternion matrix equationsrdquoComputers amp Mathematics with Applications vol 55 no 6 pp1142ndash1147 2008

Journal of Applied Mathematics 13

[7] L G Feng and W Cheng ldquoThe solution set to the quaternionmatrix equation119860119883minus119883119861 = 0rdquo Algebra Colloquium vol 19 no1 pp 175ndash180 2012

[8] Z Y Peng ldquoAn iterative method for the least squares symmetricsolution of the linear matrix equation 119860119883119861 = 119862rdquo AppliedMathematics and Computation vol 170 no 1 pp 711ndash723 2005

[9] Z Y Peng ldquoA matrix LSQR iterative method to solve matrixequation 119860119883119861 = 119862rdquo International Journal of ComputerMathematics vol 87 no 8 pp 1820ndash1830 2010

[10] Z Y Peng ldquoNew matrix iterative methods for constraintsolutions of the matrix equation 119860119883119861 = 119862rdquo Journal ofComputational and Applied Mathematics vol 235 no 3 pp726ndash735 2010

[11] Z Y Peng ldquoSolutions of symmetry-constrained least-squaresproblemsrdquo Numerical Linear Algebra with Applications vol 15no 4 pp 373ndash389 2008

[12] C C Paige ldquoBidiagonalization of matrices and solutions of thelinear equationsrdquo SIAM Journal on Numerical Analysis vol 11pp 197ndash209 1974

[13] X F Duan A P Liao and B Tang ldquoOn the nonlinear matrixequation 119883 minus sum

119898

119894=1119860

119909

2119886119894119883

120575119894119860119894= 119876rdquo Linear Algebra and Its

Applications vol 429 no 1 pp 110ndash121 2008[14] X F Duan and A P Liao ldquoOn the existence of Hermitian

positive definite solutions of thematrix equation119883119904

+119860lowast

119883minus119905

119860 =

119876rdquo Linear Algebra and Its Applications vol 429 no 4 pp 673ndash687 2008

[15] X FDuan andA P Liao ldquoOn the nonlinearmatrix equation119883+

119860lowast

119883minus119902

119860 = 119876(119902 ge 1)rdquo Mathematical and Computer Modellingvol 49 no 5-6 pp 936ndash945 2009

[16] X F Duan and A P Liao ldquoOn Hermitian positive definitesolution of the matrix equation119883minussum

119898

119894=1119860

lowast

119894119883

119903

119860119894= 119876rdquo Journal

of Computational and Applied Mathematics vol 229 no 1 pp27ndash36 2009

[17] X F Duan CM Li and A P Liao ldquoSolutions and perturbationanalysis for the nonlinear matrix equation119883+sum

119898

119894=1119860

lowast

119894119883

minus1

119860119894=

119868rdquo Applied Mathematics and Computation vol 218 no 8 pp4458ndash4466 2011

[18] F Ding P X Liu and J Ding ldquoIterative solutions of thegeneralized Sylvester matrix equations by using the hierarchicalidentification principlerdquo Applied Mathematics and Computa-tion vol 197 no 1 pp 41ndash50 2008

[19] F Ding and T Chen ldquoIterative least-squares solutions ofcoupled Sylvester matrix equationsrdquo Systems amp Control Lettersvol 54 no 2 pp 95ndash107 2005

[20] FDing andTChen ldquoHierarchical gradient-based identificationof multivariable discrete-time systemsrdquo Automatica vol 41 no2 pp 315ndash325 2005

[21] M H Wang M S Wei and S H Hu ldquoAn iterative method forthe least-squares minimum-norm symmetric solutionrdquo CMESComputer Modeling in Engineering amp Sciences vol 77 no 3-4pp 173ndash182 2011

[22] M Dehghan and M Hajarian ldquoAn iterative algorithm for thereflexive solutions of the generalized coupled Sylvester matrixequations and its optimal approximationrdquoAppliedMathematicsand Computation vol 202 no 2 pp 571ndash588 2008

[23] M Dehghan and M Hajarian ldquoFinite iterative algorithmsfor the reflexive and anti-reflexive solutions of the matrixequation119860

1119883

1119861

1+119860

2119883

2119861

2= 119862rdquoMathematical and Computer

Modelling vol 49 no 9-10 pp 1937ndash1959 2009[24] M Hajarian and M Dehghan ldquoSolving the generalized

Sylvester matrix equation119901

sum

119894=1

119860119894119883119861

119894+

119902

sum

119895=1

119862119895119884119863

119895= 119864 over

reflexive and anti-reflexive matricesrdquo International Journal ofControl and Automation vol 9 no 1 pp 118ndash124 2011

[25] M Dehghan and M Hajarian ldquoAn iterative algorithm forsolving a pair of matrix equations 119860119884119861 = 119864 119862119884119863 = 119865

over generalized centro-symmetric matricesrdquo Computers ampMathematics with Applications vol 56 no 12 pp 3246ndash32602008

[26] M Dehghan and M Hajarian ldquoAnalysis of an iterative algo-rithm to solve the generalized coupled Sylvester matrix equa-tionsrdquo Applied Mathematical Modelling vol 35 no 7 pp 3285ndash3300 2011

[27] M Dehghan and M Hajarian ldquoThe general coupled matrixequations over generalized bisymmetric matricesrdquo Linear Alge-bra and Its Applications vol 432 no 6 pp 1531ndash1552 2010

[28] M Dehghan and M Hajarian ldquoAn iterative method for solvingthe generalized coupled Sylvester matrix equations over gener-alized bisymmetric matricesrdquo Applied Mathematical Modellingvol 34 no 3 pp 639ndash654 2010

[29] A G Wu B Li Y Zhang and G R Duan ldquoFinite iterativesolutions to coupled Sylvester-conjugate matrix equationsrdquoApplied Mathematical Modelling vol 35 no 3 pp 1065ndash10802011

[30] A G Wu L L Lv and G R Duan ldquoIterative algorithmsfor solving a class of complex conjugate and transpose matrixequationsrdquo Applied Mathematics and Computation vol 217 no21 pp 8343ndash8353 2011

[31] AGWu L L Lv andMZHou ldquoFinite iterative algorithms forextended Sylvester-conjugate matrix equationsrdquo Mathematicaland Computer Modelling vol 54 no 9-10 pp 2363ndash2384 2011

[32] N L Bihan and J Mars ldquoSingular value decomposition ofmatrices of Quaternions matrices a new tool for vector-sensorsignal processingrdquo Signal Process vol 84 no 7 pp 1177ndash11992004

[33] N L Bihan and S J Sangwine ldquoJacobi method for quaternionmatrix singular value decompositionrdquoAppliedMathematics andComputation vol 187 no 2 pp 1265ndash1271 2007

[34] F O Farid Q W Wang and F Zhang ldquoOn the eigenvalues ofquaternion matricesrdquo Linear and Multilinear Algebra vol 59no 4 pp 451ndash473 2011

[35] S de Leo and G Scolarici ldquoRight eigenvalue equation inquaternionic quantummechanicsrdquo Journal of Physics A vol 33no 15 pp 2971ndash2995 2000

[36] S J Sangwine and N L Bihan ldquoQuaternion singular valuedecomposition based on bidiagonalization to a real or com-plex matrix using quaternion Householder transformationsrdquoApplied Mathematics and Computation vol 182 no 1 pp 727ndash738 2006

[37] C C Took D P Mandic and F Zhang ldquoOn the unitarydiagonalisation of a special class of quaternion matricesrdquoApplied Mathematics Letters vol 24 no 11 pp 1806ndash1809 2011

[38] QWWang H X Chang and Q Ning ldquoThe common solutionto six quaternion matrix equations with applicationsrdquo AppliedMathematics and Computation vol 198 no 1 pp 209ndash2262008

[39] Q W Wang J W van der Woude and H X Chang ldquoA systemof real quaternion matrix equations with applicationsrdquo LinearAlgebra and Its Applications vol 431 no 12 pp 2291ndash2303 2009

[40] Q W Wang S W Yu and W Xie ldquoExtreme ranks of realmatrices in solution of the quaternion matrix equation 119860119883119861 =

119862with applicationsrdquoAlgebra Colloquium vol 17 no 2 pp 345ndash360 2010

14 Journal of Applied Mathematics

[41] Q W Wang and J Jiang ldquoExtreme ranks of (skew-)Hermitiansolutions to a quaternion matrix equationrdquo Electronic Journal ofLinear Algebra vol 20 pp 552ndash573 2010

[42] Q W Wang X Liu and S W Yu ldquoThe common bisymmetricnonnegative definite solutions with extreme ranks and inertiasto a pair of matrix equationsrdquo Applied Mathematics and Com-putation vol 218 no 6 pp 2761ndash2771 2011

[43] Q W Wang Y Zhou and Q Zhang ldquoRanks of the commonsolution to six quaternion matrix equationsrdquo Acta Mathemati-cae Applicatae Sinica (English Series) vol 27 no 3 pp 443ndash4622011

[44] F Zhang ldquoGersgorin type theorems for quaternionic matricesrdquoLinear Algebra and Its Applications vol 424 no 1 pp 139ndash1532007

[45] F Zhang ldquoQuaternions and matrices of quaternionsrdquo LinearAlgebra and Its Applications vol 251 pp 21ndash57 1997

[46] R A Horn and C R Johnson Topics in Matrix AnalysisCambridge University Press Cambridge UK 1991

[47] Y X Peng X Y Hu and L Zhang ldquoAn iterationmethod for thesymmetric solutions and the optimal approximation solutionof the matrix equation 119860119883119861 = 119862rdquo Applied Mathematics andComputation vol 160 no 3 pp 763ndash777 2005

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 961568 6 pageshttpdxdoiorg1011552013961568

Research ArticleThe Optimization on Ranks and Inertias of a QuadraticHermitian Matrix Function and Its Applications

Yirong Yao

Department of Mathematics Shanghai University Shanghai 200444 China

Correspondence should be addressed to Yirong Yao yryaostaffshueducn

Received 3 December 2012 Accepted 9 January 2013

Academic Editor Yang Zhang

Copyright copy 2013 Yirong Yao This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We solve optimization problems on the ranks and inertias of the quadratic Hermitian matrix function 119876 minus 119883119875119883lowast subject to a

consistent system of matrix equations 119860119883 = 119862 and 119883119861 = 119863 As applications we derive necessary and sufficient conditions for thesolvability to the systems of matrix equations and matrix inequalities 119860119883 = 119862119883119861 = 119863 and 119883119875119883lowast = (gt lt ge le)119876 in the Lownerpartial ordering to be feasible respectively The findings of this paper widely extend the known results in the literature

1 Introduction

Throughout this paper we denote the complex number fieldby C The notations C119898times119899 and C119898times119898

ℎstand for the sets

of all 119898 times 119899 complex matrices and all 119898 times 119898 complexHermitian matrices respectivelyThe identity matrix with anappropriate size is denoted by 119868 For a complex matrix 119860the symbols 119860lowast and 119903(119860) stand for the conjugate transposeand the rank of119860 respectivelyTheMoore-Penrose inverse of119860 isin C119898times119899 denoted by119860dagger is defined to be the unique solution119883 to the following four matrix equations

(1) 119860119883119860 = 119860 (2) 119883119860119883 = 119883

(3) (119860119883)lowast

= 119860119883 (4) (119883119860)lowast

= 119883119860

(1)

Furthermore 119871119860and 119877

119860stand for the two projectors 119871

119860=

119868 minus 119860dagger

119860 and 119877119860= 119868 minus 119860119860

dagger induced by 119860 respectively It isknown that 119871

119860= 119871lowast

119860and 119877

119860= 119877lowast

119860 For 119860 isin C119898times119898

ℎ its inertia

I119899(119860) = (119894

+(119860) 119894minus(119860) 1198940(119860)) (2)

is the triple consisting of the numbers of the positive nega-tive and zero eigenvalues of 119860 counted with multiplicitiesrespectively It is easy to see that 119894

+(119860) + 119894

minus(119860) = 119903(119860) For

two Hermitian matrices 119860 and 119861 of the same sizes we say119860 gt 119861 (119860 ge 119861) in the Lowner partial ordering if 119860 minus 119861 ispositive (nonnegative) definite

The investigation on maximal and minimal ranks andinertias of linear and quadratic matrix function is activein recent years (see eg [1ndash24]) Tian [21] considered themaximal and minimal ranks and inertias of the Hermitianquadratic matrix function

ℎ (119883) = 119860119883119861119883lowast

119860lowast

+ 119860119883119862 + 119862lowast

119883lowast

119860lowast

+ 119863 (3)

where 119861 and 119863 are Hermitian matrices Moreover Tian [22]investigated the maximal and minimal ranks and inertias ofthe quadratic Hermitian matrix function

119891 (119883) = 119876 minus 119883119875119883lowast (4)

such that 119860119883 = 119862The goal of this paper is to give the maximal andminimal

ranks and inertias of the matrix function (4) subject to theconsistent system of matrix equations

119860119883 = 119862 119883119861 = 119863 (5)

where 119876 isin C119899times119899ℎ

119875 isin C119901times119901

ℎare given complex matrices

As applications we consider the necessary and sufficient

2 Journal of Applied Mathematics

conditions for the solvability to the systems of matrix equa-tions and inequality

119860119883 = 119862 119883119861 = 119863 119883119875119883lowast

= 119876

119860119883 = 119862 119883119861 = 119863 119883119875119883lowast

gt 119876

119860119883 = 119862 119883119861 = 119863 119883119875119883lowast

lt 119876

119860119883 = 119862 119883119861 = 119863 119883119875119883lowast

ge 119876

119860119883 = 119862 119883119861 = 119863 119883119875119883lowast

le 119876

(6)

in the Lowner partial ordering to be feasible respectively

2 The Optimization on Ranks and Inertiasof (4) Subject to (5)

In this section we consider the maximal and minimal ranksand inertias of the quadratic Hermitian matrix function (4)subject to (5) We begin with the following lemmas

Lemma 1 (see [3]) Let 119860 isin C119898times119898ℎ

119861 isin C119898times119901 and 119862 isin C119902times119898

be given and denote

1198751= [

119860 119861

119861lowast

0] 119875

2= [

119860 119862lowast

119862 0]

1198753= [

119860 119861 119862lowast

119861lowast

0 0] 119875

4= [

119860 119861 119862lowast

119862 0 0]

(7)

Then

max119884isinC119901times119902

119903 [119860 minus 119861119884119862 minus (119861119884119862)lowast

]

= min 119903 [119860 119861 119862lowast

] 119903 (1198751) 119903 (119875

2)

min119884isinC119901times119902

119903 [119860 minus 119861119884119862 minus (119861119884119862)lowast

]

= 2119903 [119860 119861 119862lowast

]

+max 119908++ 119908minus 119892++ 119892minus 119908++ 119892minus 119908minus+ 119892+

max119884isinC119901times119902

119894plusmn[119860 minus 119861119884119862 minus (119861119884119862)

lowast

] = min 119894plusmn(1198751) 119894plusmn(1198752)

min119884isinC119901times119902

119894plusmn[119860 minus 119861119884119862 minus (119861119884119862)

lowast

]

= 119903 [119860 119861 119862lowast

] +max 119894plusmn(1198751) minus 119903 (119875

3) 119894plusmn(1198752) minus 119903 (119875

4)

(8)

where

119908plusmn= 119894plusmn(1198751) minus 119903 (119875

3) 119892

plusmn= 119894plusmn(1198752) minus 119903 (119875

4) (9)

Lemma 2 (see [4]) Let 119860 isin C119898times119899 119861 isin C119898times119896 119862 isin C119897times119899 119863 isin

C119898times119901 119864 isin C119902times119899 119876 isin C1198981times119896 and 119875 isin C119897times1198991 be given Then

(1) 119903 (119860) + 119903 (119877119860119861) = 119903 (119861) + 119903 (119877

119861119860) = 119903 [119860 119861]

(2) 119903 (119860) + 119903 (119862119871119860) = 119903 (119862) + 119903 (119860119871

119862) = 119903 [

119860

119862]

(3) 119903 (119861) + 119903 (119862) + 119903 (119877119861119860119871119862) = 119903 [

119860 119861

119862 0]

(4) 119903 (119875) + 119903 (119876) + 119903 [119860 119861119871

119876

119877119875119862 0

] = 119903[

[

119860 119861 0

119862 0 119875

0 119876 0

]

]

(5) 119903 [119877119861119860119871119862119877119861119863

119864119871119862

0] + 119903 (119861) + 119903 (119862) = 119903[

[

119860 119863 119861

119864 0 0

119862 0 0

]

]

(10)

Lemma 3 (see [23]) Let 119860 isin C119898times119898ℎ

119861 isin C119898times119899 119862 isin C119899times119899ℎ

119876 isin C119898times119899 and 119875 isin C119901times119899 be given and 119879 isin C119898times119898 benonsingular Then

(1) 119894plusmn(119879119860119879

lowast

) = 119894plusmn(119860)

(2) 119894plusmn[119860 0

0 119862] = 119894plusmn(119860) + 119894

plusmn(119862)

(3) 119894plusmn[0 119876

119876lowast

0] = 119903 (119876)

(4) 119894plusmn[

119860 119861119871119875

119871119875119861lowast

0] + 119903 (119875) = 119894

plusmn

[

[

119860 119861 0

119861lowast

0 119875lowast

0 119875 0

]

]

(11)

Lemma 4 Let 119860 119862 119861 and 119863 be given Then the followingstatements are equivalent

(1) System (5) is consistent

(2) Let

119903 [119860 119862] = 119903 (119860) [119863

119861] = 119903 (119861) 119860119863 = 119862119861 (12)

In this case the general solution can be written as

119883 = 119860dagger

119862 + 119871119860119863119861dagger

+ 119871119860119881119877119861 (13)

where 119881 is an arbitrary matrix over C with appropriate size

Now we give the fundamental theorem of this paper

Journal of Applied Mathematics 3

Theorem5 Let119891(119883) be as given in (4) and assume that119860119883 =

119862 and 119883119861 = 119863 in (5) is consistent Then

max119860119883=119862119883119861=119863

119903 (119876 minus 119883119875119883lowast

)

= min

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861)

minus 119903 (119875) 2119899 + 119903 (119860119876119860lowast

minus 119862119875119862lowast

)

minus2119903 (119860) 119903 [

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 2119903 (119861)

min119860119883=119862119883119861=119863

119903 (119876 minus 119883119875119883lowast

)

= 2119899 + 2119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 2119903 (119860) minus 2119903 (119861) minus 119903 (119875)

+max 119904++ 119904minus 119905++ 119905minus 119904++ 119905minus 119904minus+ 119905+

max119860119883=119862119883119861=119863

119894plusmn(119876 minus 119883119875119883

lowast

)

= min

119899 + 119894plusmn(119860119876119860

lowast

minus 119862119875119862lowast

)

minus119903 (119860) 119894plusmn

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 119903 (119861)

(14)

min119860119883=119862119883119861=119863

119894plusmn(119876 minus 119883119875119883

lowast

)

= 119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861)

minus 119894plusmn(119875) +max 119904

plusmn 119905plusmn

(15)

where

119904plusmn=minus119899+119903 (119860)minus119894

∓(119875)+119894

plusmn(119860119876119860

lowast

minus119862119875119862lowast

)minus119903 [119862119875 119860119876119860

lowast

119861lowast

119863lowast

119860lowast]

119905plusmn=minus119899+119903 (119860)minus119894

∓(119875)+119894

plusmn

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus[

[

0 119875 119861

119860119876 119862119875 0

119863lowast

119861lowast

0

]

]

(16)

Proof It follows from Lemma 4 that the general solution of(4) can be expressed as

119883 = 1198830+ 119871119860119881119877119861 (17)

where 119881 is an arbitrary matrix over C and 1198830is a special

solution of (5) Then

119876 minus 119883119875119883lowast

= 119876 minus (1198830+ 119871119860119881119877119861) 119875(119883

0+ 119871119860119881119877119861)lowast

(18)

Note that

119903 [119876 minus (1198830+ 119871119860119881119877119861) 119875(119883

0+ 119871119860119881119877119861)lowast

]

= 119903 [119876 (119883

0+ 119871119860119881119877119861) 119875

119875(1198830+ 119871119860119881119877119861)lowast

119875] minus 119903 (119875)

= 119903 [[119876 119883

0119875

119875119883lowast

0119875] + [

119871119860

0]119881 [0 119877

119861119875]

+([119871119860

0]119881 [0 119877

119861119875])

lowast

] minus 119903 (119875)

(19)

119894plusmn[119876 minus (119883

0+ 119871119860119881119877119861) 119875(119883

0+ 119871119860119881119877119861)lowast

]

= 119894plusmn[

119876 (1198830+ 119871119860119881119877119861) 119875

119875(1198830+ 119871119860119881119877119861)lowast

119875] minus 119894plusmn(119875)

= 119894plusmn[[

119876 1198830119875

119875119883lowast

0119875] + [

119871119860

0]119881 [0 119877

119861119875]

+([119871119860

0]119881 [0 119877

119861119875])

lowast

] minus 119894plusmn(119875)

(20)

Let

119902 (119881) = [119876 119883

0119875

119875119883lowast

0119875] + [

119871119860

0]119881 [0 119877

119861119875]

+ ([119871119860

0]119881 [0 119877

119861119875])

lowast

(21)

Applying Lemma 1 to (19) and (20) yields

max119881

119903 [119902 (119881)] = min 119903 (119872) 119903 (1198721) 119903 (119872

2)

min119881

119903 [119902 (119881)]

= 2119903 (119872) +max 119904++ 119904minus 119905++ 119905minus 119904++ 119905minus 119904minus+ 119905+

max119881

119894plusmn[119902 (119881)] = min 119894

plusmn(1198721) 119894plusmn(1198722)

min119881

119894plusmn[119902 (119881)] = 119903 (119872) +max 119904

plusmn 119905plusmn

(22)

where

119872=[119876 119883

0119875 119871119860

0

119875119883lowast

0119875 0 119875119877

119861

] 1198721=[

[

119876 1198830119875 119871119860

119875119883lowast

0119875 0

119871119860

0 0

]

]

1198722=[

[

119876 1198830119875 0

119875119883lowast

0119875 119875119877

119861

0 119877119861119875 0

]

]

1198723=[

[

119876 1198830119875 119871119860

0

119875119883lowast

0119875 0 119875119877

119861

119871119860

0 0 0

]

]

1198724= [

[

119876 1198830119875 119871119860

0

119875119883lowast

0119875 0 119875119877

119861

0 119877119861119875 0 0

]

]

119904plusmn= 119894plusmn(1198721) minus 119903 (119872

3) 119905

plusmn= 119894plusmn(1198722) minus 119903 (119872

4)

(23)

4 Journal of Applied Mathematics

Applying Lemmas 2 and 3 elementary matrix operations andcongruence matrix operations we obtain

119903 (119872) = 119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861)

119903 (1198721) = 2119899 + 119903 (119860119876119860

lowast

minus 119862119875119862lowast

) minus 2119903 (119860) + 119903 (119875)

119894plusmn(1198721) = 119899 + 119894

plusmn(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 (119860) + 119894plusmn(119875)

119903 (1198722) = 119903[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 2119903 (119861) + 119903 (119875)

119894plusmn(1198722) = 119894plusmn

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 119903 (119861) + 119894plusmn(119875)

119903 (1198723) = 2119899 + 119903 (119875) minus 2119903 (119860) minus 119903 (119861) + 119903 [

119862119875 119860119876119860lowast

119861lowast

119863lowast

119860lowast]

119903 (1198724) = 119899 + 119903 (119875) + 119903[

[

0 119875 119861

119860119876 119862119875 0

119863lowast

119861lowast

0

]

]

minus 2119903 (119861) minus 119903 (119860)

(24)

Substituting (24) into (22) we obtain the results

Using immediately Theorem 5 we can easily get thefollowing

Theorem 6 Let 119891(119883) be as given in (4) 119904plusmnand let 119905

plusmnbe as

given in Theorem 5 and assume that 119860119883 = 119862 and 119883119861 = 119863 in(5) are consistent Then we have the following

(a) 119860119883 = 119862 and 119883119861 = 119863 have a common solution suchthat 119876 minus 119883119875119883

lowast

ge 0 if and only if

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894minus(119875) + 119904

minusle 0

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894minus(119875) + 119905

minusle 0

(25)

(b) 119860119883 = 119862 and 119883119861 = 119863 have a common solution suchthat 119876 minus 119883119875119883

lowast

le 0 if and only if

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894+(119875) + 119904

+le 0

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894+(119875) + 119905

+le 0

(26)

(c) 119860119883 = 119862 and 119883119861 = 119863 have a common solution suchthat 119876 minus 119883119875119883

lowast

gt 0 if and only if

119894+(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 (119860) ge 0

119894+

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 119903 (119861) ge 119899

(27)

(d) 119860119883 = 119862 and 119883119861 = 119863 have a common solution suchthat 119876 minus 119883119875119883

lowast

lt 0 if and only if

119894minus(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 (119860) ge 0

119894minus

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 119903 (119861) ge 119899

(28)

(e) All common solutions of 119860119883 = 119862 and 119883119861 = 119863 satisfy119876 minus 119883119875119883

lowast

ge 0 if and only if

119899 + 119894minus(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 (119860) = 0

or 119894minus

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 119903 (119861) = 0

(29)

(f) All common solutions of 119860119883 = 119862 and 119883119861 = 119863 satisfy119876 minus 119883119875119883

lowast

le 0 if and only if

119899 + 119894+(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 (119860) = 0

or 119894+

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 119903 (119861) = 0

(30)

(g) All common solutions of 119860119883 = 119862 and 119883119861 = 119863 satisfy119876 minus 119883119875119883

lowast

gt 0 if and only if

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894+(119875) + 119904

+= 119899

(31)or

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894+(119875) + 119905

+= 119899

(32)

(h) All common solutions of 119860119883 = 119862 and 119883119861 = 119863 satisfy119876 minus 119883119875119883

lowast

lt 0 if and only if

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894minus(119875) + 119904

minus= 119899

(33)or

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894minus(119875) + 119905

minus= 119899

(34)

Journal of Applied Mathematics 5

(i) 119860119883 = 119862 119883119861 = 119863 and 119876 = 119883119875119883lowast have a common

solution if and only if

2119899+2119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus2119903 (119860)minus2119903 (119861)minus119903 (119875)+119904++119904minusle0

2119899+2119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus2119903 (119860)minus2119903 (119861)minus119903 (119875)+119905++119905minusle0

2119899+2119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus2119903 (119860)minus2119903 (119861)minus119903 (119875) + 119904++119905minusle0

2119899+2119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus2119903 (119860)minus2119903 (119861)minus119903 (119875) + 119904minus+119905+le0

(35)

Let 119875 = 119868 in Theorem 5 we get the following corollary

Corollary 7 Let 119876 isin C119899times119899 119860 119861 119862 and119863 be given Assumethat (5) is consistent Denote

1198791= [

119862 119860119876

119861lowast

119863lowast] 119879

2= 119860119876119860

lowast

minus 119862119862lowast

1198793= [

119876 119863

119863lowast

119861lowast

119861] 119879

4= [

119862 119860119876119860lowast

119861lowast

119863lowast

119860lowast]

1198795= [

119862119861 119860119876

119861lowast

119861 119863lowast]

(36)

Then

max119860119883=119862119883119861=119863

119903 (119876 minus 119883119883lowast

)

= min 119899 + 119903 (1198791) minus 119903 (119860) minus 119903 (119861) 2119899 + 119903 (119879

2)

minus2119903 (119860) 119899 + 119903 (1198793) minus 2119903 (119861)

min119860119883=119862119883119861=119863

119903 (119876 minus 119883119883lowast

)

= 2119903 (1198791) +max 119903 (119879

2) minus 2119903 (119879

4) minus119899 + 119903 (119879

3)

minus 2119903 (1198795) 119894+(1198792) + 119894minus(1198793)

minus 119903 (1198794) minus 119903 (119879

5) minus119899 + 119894

minus(1198792)

+119894+(1198793) minus 119903 (119879

4) minus 119903 (119879

5)

max119860119883=119862119883119861=119863

119894+(119876 minus 119883119883

lowast

)

= min 119899 + 119894+(1198792) minus 119903 (119860) 119894

+(1198793) minus 119903 (119861)

max119860119883=119862119883119861=119863

119894minus(119876 minus 119883119883

lowast

)

= min 119899 + 119894minus(1198792) minus 119903 (119860) 119899 + 119894

minus(1198793) minus 119903 (119861)

min119860119883=119862119883119861=119863

119894+(119876 minus 119883119883

lowast

)

= 119903 (1198791) +max 119894

+(1198792) minus 119903 (119879

4) 119894+(1198793) minus 119899 minus 119903 (119879

5)

min119860119883=119862119883119861=119863

119894minus(119876 minus 119883119883

lowast

)

= 119903 (1198791) +max 119894

minus(1198792) minus 119903 (119879

4) 119894minus(1198793) minus 119903 (119879

5)

(37)

Remark 8 Corollary 7 is one of the results in [24]

Let 119861 and 119863 vanish in Theorem 5 then we can obtainthe maximal and minimal ranks and inertias of (4) subjectto 119860119883 = 119862

Corollary 9 Let 119891(119883) be as given in (4) and assume that119860119883 = 119862 is consistent Then

max119860119883=119862

119903 (119876 minus 119883119875119883lowast

)

= min 119899 + 119903 [119860119876 119862119875] minus 119903 (119860) minus 119903 (119861)

2119899 + 119903 (119860119876119860lowast

minus 119862119875119862lowast

) minus 2119903 (119860) 119903 (119876) + 119903 (119875)

min119860119883=119862

119903 (119876 minus 119883119875119883lowast

)

= 2119899 + 2119903 [119860119876 119862119875] minus 2119903 (119860)

+max 119904++ 119904minus 119905++ 119905minus 119904++ 119905minus 119904minus+ 119905+

max119860119883=119862

119894plusmn(119876 minus 119883119875119883

lowast

)

= min 119899 + 119894plusmn(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 (119860) 119894plusmn(119876) + 119894

∓(119875)

min119860119883=119862

119894plusmn(119876 minus 119883119875119883

lowast

)

= 119899 + 119903 [119860119876 119862119875] minus 119903 (119860) + 119894∓(119875) +max 119904

plusmn 119905plusmn

(38)

where

119904plusmn= minus 119899 + 119903 (119860) minus 119894

∓(119875)

+ 119894plusmn(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 [119862119875 119860119876119860lowast

]

119905plusmn= minus119899 + 119903 (119860) + 119894

plusmn(119876) minus 119903 (119875) minus [119860119876 119862119875]

(39)

Remark 10 Corollary 9 is one of the results in [22]

References

[1] D L Chu Y S Hung and H J Woerdeman ldquoInertia and rankcharacterizations of somematrix expressionsrdquo SIAM Journal onMatrix Analysis and Applications vol 31 no 3 pp 1187ndash12262009

[2] Z H He and Q W Wang ldquoSolutions to optimization problemson ranks and inertias of a matrix function with applicationsrdquo

6 Journal of Applied Mathematics

AppliedMathematics andComputation vol 219 no 6 pp 2989ndash3001 2012

[3] Y Liu andY Tian ldquoMax-min problems on the ranks and inertiasof the matrix expressions119860minus119861119883119862plusmn(119861119883119862)lowast with applicationsrdquoJournal of OptimizationTheory and Applications vol 148 no 3pp 593ndash622 2011

[4] G Marsaglia and G P H Styan ldquoEqualities and inequalities forranks of matricesrdquo Linear and Multilinear Algebra vol 2 pp269ndash292 197475

[5] X Zhang Q W Wang and X Liu ldquoInertias and ranks ofsome Hermitian matrix functions with applicationsrdquo CentralEuropean Journal of Mathematics vol 10 no 1 pp 329ndash3512012

[6] Q W Wang Y Zhou and Q Zhang ldquoRanks of the commonsolution to six quaternion matrix equationsrdquo Acta Mathemati-cae Applicatae Sinica vol 27 no 3 pp 443ndash462 2011

[7] Q Zhang and Q W Wang ldquoThe (119875 119876)-(skew)symmetricextremal rank solutions to a system of quaternion matrixequationsrdquo Applied Mathematics and Computation vol 217 no22 pp 9286ndash9296 2011

[8] D Z Lian QWWang and Y Tang ldquoExtreme ranks of a partialbanded block quaternion matrix expression subject to somematrix equationswith applicationsrdquoAlgebraColloquium vol 18no 2 pp 333ndash346 2011

[9] H S Zhang andQWWang ldquoRanks of submatrices in a generalsolution to a quaternion system with applicationsrdquo Bulletin ofthe Korean Mathematical Society vol 48 no 5 pp 969ndash9902011

[10] Q W Wang and J Jiang ldquoExtreme ranks of (skew-)Hermitiansolutions to a quaternion matrix equationrdquo Electronic Journal ofLinear Algebra vol 20 pp 552ndash573 2010

[11] I A Khan Q W Wang and G J Song ldquoMinimal ranks ofsome quaternionmatrix expressions with applicationsrdquoAppliedMathematics and Computation vol 217 no 5 pp 2031ndash20402010

[12] Q Wang S Yu and W Xie ldquoExtreme ranks of real matricesin solution of the quaternion matrix equation 119860119883119861 = 119862 withapplicationsrdquo Algebra Colloquium vol 17 no 2 pp 345ndash3602010

[13] Q W Wang H S Zhang and G J Song ldquoA new solvable con-dition for a pair of generalized Sylvester equationsrdquo ElectronicJournal of Linear Algebra vol 18 pp 289ndash301 2009

[14] Q W Wang S W Yu and Q Zhang ldquoThe real solutions toa system of quaternion matrix equations with applicationsrdquoCommunications in Algebra vol 37 no 6 pp 2060ndash2079 2009

[15] Q W Wang and C K Li ldquoRanks and the least-norm of thegeneral solution to a system of quaternion matrix equationsrdquoLinear Algebra and its Applications vol 430 no 5-6 pp 1626ndash1640 2009

[16] Q W Wang H S Zhang and S W Yu ldquoOn solutions to thequaternionmatrix equation119860119883119861+119862119884119863 = 119864rdquoElectronic Journalof Linear Algebra vol 17 pp 343ndash358 2008

[17] Q W Wang G J Song and X Liu ldquoMaximal and minimalranks of the common solution of some linear matrix equationsover an arbitrary division ring with applicationsrdquo AlgebraColloquium vol 16 no 2 pp 293ndash308 2009

[18] Q W Wang G J Song and C Y Lin ldquoRank equalities relatedto the generalized inverse 119860

(2)

119879119878with applicationsrdquo Applied

Mathematics and Computation vol 205 no 1 pp 370ndash3822008

[19] Q W Wang S W Yu and C Y Lin ldquoExtreme ranks of alinear quaternionmatrix expression subject to triple quaternionmatrix equations with applicationsrdquo Applied Mathematics andComputation vol 195 no 2 pp 733ndash744 2008

[20] Q W Wang G J Song and C Y Lin ldquoExtreme ranks ofthe solution to a consistent system of linear quaternion matrixequations with an applicationrdquo Applied Mathematics and Com-putation vol 189 no 2 pp 1517ndash1532 2007

[21] Y Tian ldquoFormulas for calculating the extremum ranks andinertias of a four-term quadratic matrix-valued function andtheir applicationsrdquo Linear Algebra and its Applications vol 437no 3 pp 835ndash859 2012

[22] Y Tian ldquoSolving optimization problems on ranks and inertiasof some constrained nonlinearmatrix functions via an algebraiclinearization methodrdquo Nonlinear Analysis Theory Methods ampApplications vol 75 no 2 pp 717ndash734 2012

[23] Y Tian ldquoMaximization and minimization of the rank andinertia of the Hermitianmatrix expression119860minus119861119883minus(119861119883)lowast withapplicationsrdquo Linear Algebra and its Applications vol 434 no10 pp 2109ndash2139 2011

[24] Q W Wang X Zhang and Z H He ldquoOn the Hermitianstructures of the solution to a pair of matrix equationsrdquo Linearand Multilinear Algebra vol 61 no 1 pp 73ndash90 2013

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2012 Article ID 471573 19 pagesdoi1011552012471573

Research ArticleConstrained Solutions of a System ofMatrix Equations

Qing-Wen Wang1 and Juan Yu1 2

1 Department of Mathematics Shanghai University 99 Shangda Road Shanghai 200444 China2 Department of Basic Mathematics China University of Petroleum Qingdao 266580 China

Correspondence should be addressed to Qing-Wen Wang wqw858yahoocomcn

Received 26 September 2012 Accepted 7 December 2012

Academic Editor Panayiotis J Psarrakos

Copyright q 2012 Q-W Wang and J Yu This is an open access article distributed under theCreative Commons Attribution License which permits unrestricted use distribution andreproduction in any medium provided the original work is properly cited

We derive the necessary and sufficient conditions of and the expressions for the orthogonalsolutions the symmetric orthogonal solutions and the skew-symmetric orthogonal solutions ofthe system of matrix equations AX = B and XC = D respectively When the matrix equationsare not consistent the least squares symmetric orthogonal solutions and the least squares skew-symmetric orthogonal solutions are respectively given As an auxiliary an algorithm is providedto compute the least squares symmetric orthogonal solutions and meanwhile an example ispresented to show that it is reasonable

1 Introduction

Throughout this paper the following notations will be used Rmtimesn OR

ntimesn SRntimesn and ASR

ntimesn

denote the set of all m times n real matrices the set of all n times n orthogonal matrices the set of alln times n symmetric matrices and the set of all n times n skew-symmetric matrices respectively In isthe identity matrix of order n (middot)T and tr(middot) represent the transpose and the trace of the realmatrix respectively middot stands for the Frobenius norm induced by the inner product Thefollowing two definitions will also be used

Definition 11 (see [1]) A real matrix X isin Rntimesn is said to be a symmetric orthogonal matrix if

XT = X and XTX = In

Definition 12 (see [2]) A real matrix X isin R2mtimes2m is called a skew-symmetric orthogonal

matrix if XT = minusX and XTX = In

The set of all n times n symmetric orthogonal matrices and the set of all 2m times 2m skew-symmetric orthogonal matrices are respectively denoted by SOR

ntimesn and SSOR2mtimes2m Since

2 Journal of Applied Mathematics

the linear matrix equation(s) and its optimal approximation problem have great applicationsin structural design biology control theory and linear optimal control and so forth seefor example [3ndash5] there has been much attention paid to the linear matrix equation(s) Thewell-known system of matrix equations

AX = B XC = D (11)

as one kind of linear matrix equations has been investigated by many authors and a seriesof important and useful results have been obtained For instance the system (11) withunknown matrix X being bisymmetric centrosymmetric bisymmetric nonnegative definiteHermitian and nonnegative definite and (PQ)-(skew) symmetric has been respectivelyinvestigated by Wang et al [6 7] Khatri and Mitra [8] and Zhang and Wang [9] Of courseif the solvability conditions of system (11) are not satisfied we may consider its least squaressolution For example Li et al [10] presented the least squares mirrorsymmetric solutionYuan [11] got the least-squares solution Some results concerning the system (11) can also befound in [12ndash18]

Symmetric orthogonal matrices and skew-symmetric orthogonal matrices play impor-tant roles in numerical analysis and numerical solutions of partial differential equationsPapers [1 2] respectively derived the symmetric orthogonal solution X of the matrixequation XC = D and the skew-symmetric orthogonal solution X of the matrix equationAX = B Motivated by the work mentioned above we in this paper will respectively studythe orthogonal solutions symmetric orthogonal solutions and skew-symmetric orthogonalsolutions of the system (11) Furthermore if the solvability conditions are not satisfiedthe least squares skew-symmetric orthogonal solutions and the least squares symmetricorthogonal solutions of the system (11) will be also given

The remainder of this paper is arranged as follows In Section 2 some lemmas areprovided to give the main results of this paper In Sections 3 4 and 5 the necessary andsufficient conditions of and the expression for the orthogonal the symmetric orthogonaland the skew-symmetric orthogonal solutions of the system (11) are respectively obtainedIn Section 6 the least squares skew-symmetric orthogonal solutions and the least squaressymmetric orthogonal solutions of the system (11) are presented respectively In additionan algorithm is provided to compute the least squares symmetric orthogonal solutions andmeanwhile an example is presented to show that it is reasonable Finally in Section 7 someconcluding remarks are given

2 Preliminaries

In this section we will recall some lemmas and the special C-S decomposition which will beused to get the main results of this paper

Lemma 21 (see [1 Lemmas 1 and 2]) Given C isin R2mtimesn D isin R

2mtimesn The matrix equation YC =D has a solution Y isin OR

2mtimes2m if and only if DTD = CTC Let the singular value decompositions ofC and D be respectively

C = U

(Π 00 0

)V T D = W

(Π 00 0

)V T (21)

Journal of Applied Mathematics 3

where

Π = diag(σ1 σk) gt 0 k = rank(C) = rank(D)

U =(U1 U2

) isin OR2mtimes2m U1 isin R

2mtimesk

W =(W1 W2

) isin OR2mtimes2m W1 isin R

2mtimesk V =(V1 V2

) isin ORntimesn V1 isin R

ntimesk

(22)

Then the orthogonal solutions of YC = D can be described as

Y = W

(Ik 00 P

)UT (23)

where P isin OR(2mminusk)times(2mminusk) is arbitrary

Lemma 22 (see [2 Lemmas 1 and 2]) GivenA isin Rntimesm B isin R

ntimesm The matrix equationAX = Bhas a solution X isin OR

mtimesm if and only if AAT = BBT Let the singular value decompositions of Aand B be respectively

A = U

(Σ 00 0

)V T B = U

(Σ 00 0

)QT (24)

where

Σ = diag(δ1 δl) gt 0 l = rank(A) = rank(B) U =(U1 U2

) isin ORntimesn U1 isin R

ntimesl

V =(V1 V2

) isin ORmtimesm V1 isin R

mtimesl Q =(Q1 Q2

) isin ORmtimesm Q1 isin R

mtimesl(25)

Then the orthogonal solutions of AX = B can be described as

X = V

(Il 00 W

)QT (26)

whereW isin OR(mminusl)times(mminusl) is arbitrary

Lemma 23 (see [2 Theorem 1]) If

X =(X11 X12

X21 X22

)isin OR

2mtimes2m X11 isin ASRktimesk (27)

then the C-S decomposition of X can be expressed as

(D1 00 D2

)T(X11 X12

X21 X22

)(D1 00 R2

)=(Σ11 Σ12

Σ21 Σ22

) (28)

4 Journal of Applied Mathematics

where D1 isin ORktimesk D2 R2 isin OR

(2mminusk)times(2mminusk)

Σ11 =

⎛⎜⎝

I 0 00 C 00 0 0

⎞⎟⎠ Σ12 =

⎛⎝0 0 0

0 S 00 0 I

⎞⎠

Σ21 =

⎛⎝0 0 0

0 S 00 0 I

⎞⎠ Σ22 =

⎛⎜⎝

I 0 00 minusCT 00 0 0

⎞⎟⎠

I = diag(I1 Ir1

) I1 = middot middot middot = Ir1 =

(0 1minus1 0

) C = diag(C1 Cl1)

Ci =(

0 ciminusci 0

) i = 1 l1 S = diag(S1 Sl1)

Si =(si 00 si

) ST

i Si + CTi Ci = I2 i = 1 l1

2r1 + 2l1 = rank(X11) Σ21 = ΣT12 k minus 2r1 = rank(X21)

(29)

Lemma 24 (see [1 Theorem 1]) If

K =(K11 K12

K21 K22

)isin OR

mtimesm K11 isin SRltimesl (210)

then the C-S decomposition of K can be described as

(D1 00 D2

)T(K11 K12

K21 K22

)(D1 00 R2

)=(Π11 Π12

Π21 Π22

) (211)

where D1 isin ORltimesl D2 R2 isin OR

(mminusl)times(mminusl)

Π11 =

⎛⎜⎝

I 0 00 C 00 0 0

⎞⎟⎠ Π12 =

⎛⎝0 0 0

0 S 00 0 I

⎞⎠

Π21 =

⎛⎝0 0 0

0 S 00 0 I

⎞⎠ Π22 =

⎛⎜⎝

I 0 00 minusC 00 0 0

⎞⎟⎠

I = diag(i1 ir) ij = plusmn1 j = 1 r

S = diag(s1prime slprime) C = diag(c1prime clprime)

s2iprime =radic

1 minus c2iprime i = 1prime lprime r + lprime = rank(K11) Π21 = ΠT

12

(212)

Journal of Applied Mathematics 5

Remarks 25 In order to know the C-S decomposition of an orthogonal matrix with ak times k leading (skew-) symmetric submatrix for details one can deeply study the proof ofTheorem 1 in [1] and [2]

Lemma 26 Given A isin Rntimesm B isin R

ntimesm Then the matrix equation AX = B has a solution X isinSOR

mtimesm if and only ifAAT = BBT andABT = BAT When these conditions are satisfied the generalsymmetric orthogonal solutions can be expressed as

X = V

(I2lminusr 0

0 G

)QT (213)

where

Q = JQdiag(ID2) isin ORmtimesm V = JV diag(I R2) isin OR

mtimesm

J =

⎛⎝I 0 0

0 0 I0 I 0

⎞⎠

(214)

and G isin SOR(mminus2l+r)times(mminus2l+r) is arbitrary

Proof The Necessity Assume X isin SORmtimesm is a solution of the matrix equation AX = B then

we have

BBT = AXXTAT = AAT

BAT = AXAT = AXTAT = ABT(215)

The Sufficiency Since the equality AAT = BBT holds then by Lemma 22 the singularvalue decompositions of A and B can be respectively expressed as (24) Moreover thecondition ABT = BAT means

U

(Σ 00 0

)V TQ

(Σ 00 0

)UT = U

(Σ 00 0

)QTV

(Σ 00 0

)UT (216)

which can be written as

V T1 Q1 = QT

1V1 (217)

From Lemma 22 the orthogonal solutions of the matrix equation AX = B can be describedas (26) Now we aim to find that X in (26) is also symmetric Suppose that X is symmetricthen we have

(Il 00 WT

)V TQ = QTV

(Il 00 W

) (218)

6 Journal of Applied Mathematics

together with the partitions of the matrices Q and V in Lemma 22 we get

V T1 Q1 = QT

1V1 (219)

QT1V2W = V T

1 Q2 (220)

QT2 V2W = WTV T

2 Q2 (221)

By (217) we can get (219) Now we aim to find the orthogonal solutions of the systemof matrix equations (220) and (221) Firstly we obtain from (220) that QT

1V2(QT1V2)

T =V T

1 Q2(V T1 Q2)

T then by Lemma 22 (220) has an orthogonal solution W By (217) the l times lleading principal submatrix of the orthogonal matrix V TQ is symmetric Then we have fromLemma 24

V T1 Q2 = D1Π12R

T2 (222)

QT1V2 = D1Π12D

T2 (223)

V T2 Q2 = D2Π22R

T2 (224)

From (220) (222) and (223) the orthogonal solution W of (220) is

W = D2

⎛⎝G 0 0

0 I 00 0 I

⎞⎠RT

2 (225)

where G isin OR(mminus2l+r)times(mminus2l+r) is arbitrary Combining (221) (224) and (225) yields GT =

G that is G is a symmetric orthogonal matrix Denote

V = V diag(ID2) Q = Qdiag(I R2) (226)

then the symmetric orthogonal solutions of the matrix equation AX = B can be expressed as

X = V

⎛⎝I 0 0

0 G 00 0 I

⎞⎠QT (227)

Let the partition matrix J be

J =

⎛⎝I 0 0

0 0 I0 I 0

⎞⎠ (228)

Journal of Applied Mathematics 7

compatible with the block matrix

⎛⎝I 0 0

0 G 00 0 I

⎞⎠ (229)

Put

V = JV Q = JQ (230)

then the symmetric orthogonal solutions of the matrix equation AX = B can be described as(213)

Setting A = CT B = DT and X = YT in [2 Theorem 2] and then by Lemmas 21 and23 we can have the following result

Lemma 27 Given C isin R2mtimesn D isin R

2mtimesn Then the equation has a solution Y isin SSOR2mtimes2m if

and only ifDTD = CTC andDTC = minusCTD When these conditions are satisfied the skew-symmetricorthogonal solutions of the matrix equation YC = D can be described as

Y = W

(I 00 H

)UT (231)

where

W = J primeW diag(IminusD2) isin OR2mtimes2m U = J primeUdiag(I R2) isin OR

2mtimes2m

J prime =

⎛⎝I 0 0

0 0 I0 I 0

⎞⎠

(232)

and H isin SSOR2rtimes2r is arbitrary

3 The Orthogonal Solutions of the System (11)

The following theorems give the orthogonal solutions of the system (11)

Theorem 31 Given A B isin Rntimesm and C D isin R

mtimesn suppose the singular value decompositions ofA and B are respectively as (24) Denote

QTC =(C1

C2

) V TD =

(D1

D2

) (31)

8 Journal of Applied Mathematics

where C1 D1 isin Rltimesn and C2 D2 isin R

(mminusl)timesn Let the singular value decompositions of C2 and D2 berespectively

C2 = U

(Π 00 0

)V T D2 = W

(Π 00 0

)V T (32)

where U W isin OR(mminusl)times(mminusl) V isin OR

ntimesn Π isin Rktimesk is diagonal whose diagonal elements are

nonzero singular values of C2 or D2 Then the system (11) has orthogonal solutions if and only if

AAT = BBT C1 = D1 DT2 D2 = CT

2 C2 (33)

In which case the orthogonal solutions can be expressed as

X = V

(Ik+l 00 Gprime

)QT (34)

where

V = V

(Il 00 W

)isin OR

mtimesm Q = Q

(Il 00 U

)isin OR

mtimesm (35)

and Gprime isin OR(mminuskminusl)times(mminuskminusl) is arbitrary

Proof Let the singular value decompositions of A and B be respectively as (24) Since thematrix equation AX = B has orthogonal solutions if and only if

AAT = BBT (36)

then by Lemma 22 its orthogonal solutions can be expressed as (26) Substituting (26) and(31) into the matrix equation XC = D we have C1 = D1 and WC2 = D2 By Lemma 21 thematrix equation WC2 = D2 has orthogonal solution W if and only if

DT2 D2 = CT

2C2 (37)

Let the singular value decompositions of C2 and D2 be respectively

C2 = U

(Π 00 0

)V T D2 = W

(Π 00 0

)V T (38)

where U W isin OR(mminusl)times(mminusl) V isin OR

ntimesn Π isin Rktimesk is diagonal whose diagonal elements are

nonzero singular values of C2 or D2 Then the orthogonal solutions can be described as

W = W

(Ik 00 Gprime

)UT (39)

Journal of Applied Mathematics 9

where Gprime isin OR(mminuskminusl)times(mminuskminusl) is arbitrary Therefore the common orthogonal solutions of the

system (11) can be expressed as

X = V

(Il 00 W

)QT = V

(Il 00 W

)⎛⎝Il 0 0

0 Ik 00 0 Gprime

⎞⎠(Il 00 UT

)QT = V

(Ik+l 00 Gprime

)QT (310)

where

V = V

(Il 00 W

)isin OR

mtimesm Q = Q

(Il 00 U

)isin OR

mtimesm (311)

and Gprime isin OR(mminuskminusl)times(mminuskminusl) is arbitrary

The following theorem can be shown similarly

Theorem 32 GivenA B isin Rntimesm and C D isin R

mtimesn let the singular value decompositions of C andD be respectively as (21) Partition

AW =(A1 A2

) BU =

(B1 B2

) (312)

where A1 B1 isin Rntimesk A2 B2 isin R

ntimes(mminusk) Assume the singular value decompositions of A2 and B2

are respectively

A2 = U

(Σ 00 0

)V T B2 = U

(Σ 00 0

)QT (313)

where V Q isin OR(mminusk)times(mminusk) U isin OR

ntimesn Σ isin Rlprimetimeslprime is diagonal whose diagonal elements are

nonzero singular values of A2 or B2 Then the system (11) has orthogonal solutions if and onlyif

DTD = CTC A1 = B1 A2AT2 = B2B

T2 (314)

In which case the orthogonal solutions can be expressed as

X = W

(Ik+lprime 0

0 H prime

)UT (315)

where

W = W

(Ik 00 W

)isin OR

mtimesm U = U

(Ik 00 U

)isin OR

mtimesm (316)

and H prime isin OR(mminuskminuslprime)times(mminuskminuslprime) is arbitrary

10 Journal of Applied Mathematics

4 The Symmetric Orthogonal Solutions of the System (11)

We now present the symmetric orthogonal solutions of the system (11)

Theorem 41 Given AB isin Rntimesm CD isin R

mtimesn Let the symmetric orthogonal solutions of thematrix equation AX = B be described as in Lemma 26 Partition

QTC =(Cprime

1Cprime

2

) V TD =

(Dprime

1Dprime

2

) (41)

where Cprime1 D

prime1 isin R

(2lminusr)timesn Cprime2 D

prime2 isin R

(mminus2l+r)timesn Then the system (11) has symmetric orthogonalsolutions if and only if

AAT = BBT ABT = BAT Cprime1 = Dprime

1 DprimeT2 Dprime

2 = CprimeT2 Cprime

2 DprimeT2 Cprime

2 = CprimeT2 Dprime

2 (42)

In which case the solutions can be expressed as

X = V

(I 00 Gprimeprime

)QT (43)

where

V = V

(I2lminusr 0

0 W

)isin OR

mtimesm Q = Q

(I2lminusr 0

0 U

)isin OR

mtimesm (44)

and Gprimeprime isin SOR(mminus2l+rminus2lprime+r prime)times(mminus2l+rminus2lprime+r prime) is arbitrary

Proof From Lemma 26 we obtain that the matrix equation AX = B has symmetricorthogonal solutions if and only if AAT = BBT and ABT = BAT When these conditionsare satisfied the general symmetric orthogonal solutions can be expressed as

X = V

(I2lminusr 0

0 G

)QT (45)

where G isin SOR(mminus2l+r)times(mminus2l+r) is arbitrary Q isin OR

mtimesm V isin ORmtimesm Inserting (41) and (45)

into the matrix equation XC = D we get Cprime1 = Dprime

1 and GCprime2 = Dprime

2 By [1 Theorem 2] thematrix equation GCprime

2 = Dprime2 has a symmetric orthogonal solution if and only if

DprimeT2 Dprime

2 = CprimeT2 Cprime

2 DprimeT2 Cprime

2 = CprimeT2 Dprime

2 (46)

In which case the solutions can be described as

G = W

(I2lprimeminusr prime 0

0 Gprimeprime

)UT (47)

Journal of Applied Mathematics 11

where Gprimeprime isin SOR(mminus2l+rminus2lprime+r prime)times(mminus2l+rminus2lprime+r prime) is arbitrary W isin OR

(mminus2l+r)times(mminus2l+r) and U isinOR

(mminus2l+r)times(mminus2l+r) Hence the system (11) has symmetric orthogonal solutions if and onlyif all equalities in (42) hold In which case the solutions can be expressed as

X = V

(I2lminusr 0

0 G

)QT = V

(I2lminusr 0

0 W

)⎛⎝I2lminusr 0

0(I 00 Gprimeprime

)⎞⎠(I2lminusr 0

0 UT

)QT (48)

that is the expression in (43)

The following theorem can also be obtained by the method used in the proof ofTheorem 41

Theorem 42 Given AB isin Rntimesm CD isin R

mtimesn Let the symmetric orthogonal solutions of thematrix equation XC = D be described as

X = M

(I2kminusr 0

0 G

)NT (49)

where M N isin ORmtimesm G isin SOR

(mminus2k+r)times(mminus2k+r) Partition

AM =(M1 M2

) BN =

(N1 N2

) M1N1 isin R

ntimes(2kminusr) M2N2 isin Rntimes(mminus2k+r) (410)

Then the system (11) has symmetric orthogonal solutions if and only if

DTD = CTC DTC = CTD M1 = N1 M2MT2 = N2N

T2 M2N

T2 = N2M

T2

(411)

In which case the solutions can be expressed as

X = M

(I 00 H primeprime

)NT (412)

where

M = M

(I2kminusr 0

0 W1

)isin OR

mtimesm N = N

(I2kminusr 0

0 U1

)isin OR

mtimesm (413)

and H primeprime isin SOR(mminus2k+rminus2kprime+r prime)times(mminus2k+rminus2kprime+r prime) is arbitrary

12 Journal of Applied Mathematics

5 The Skew-Symmetric Orthogonal Solutions of the System (11)

In this section we show the skew-symmetric orthogonal solutions of the system (11)

Theorem 51 Given AB isin Rntimes2m CD isin R

2mtimesn Suppose the matrix equation AX = B has skew-symmetric orthogonal solutions with the form

X = V1

(I 00 H

)QT

1 (51)

whereH isin SSOR2rtimes2r is arbitrary V1 Q1 isin OR

2mtimes2m Partition

QT1C =

(Q1

Q2

) V T

1 D =(V1

V2

) (52)

where Q1 V1 isin R(2mminus2r)timesn Q2 V2 isin R

2rtimesn Then the system (11) has skew-symmetric orthogonalsolutions if and only if

AAT = BBT ABT = minusBAT Q1 = V1 QT2 Q2 = V T

2 V2 QT2V2 = minusV T

2 Q2 (53)

In which case the solutions can be expressed as

X = V1

(I 00 J prime

)QT

1 (54)

where

V = V1

(I2mminus2r 0

0 W

)isin OR

2mtimes2m Q = Q1

(I2mminus2r 0

0 U

)isin OR

2mtimes2m (55)

and J prime isin SSOR2kprimetimes2kprime

is arbitrary

Proof By [2 Theorem 2] the matrix equation AX = B has the skew-symmetric orthogonalsolutions if and only if AAT = BBT and ABT = minusBAT When these conditions are satisfiedthe general skew-symmetric orthogonal solutions can be expressed as (51) Substituting (51)and (52) into the matrix equation XC = D we get Q1 = V1 and HQ2 = V2 From Lemma 27equation HQ2 = V2 has a skew-symmetric orthogonal solution H if and only if

QT2 Q2 = V T

2 V2 QT2V2 = minusV T

2 Q2 (56)

When these conditions are satisfied the solution can be described as

H = W

(I 00 J prime

)UT (57)

Journal of Applied Mathematics 13

where J prime isin SSOR2kprimetimes2kprime

is arbitrary W U isin OR2rtimes2r Inserting (57) into (51) yields that the

system (11) has skew-symmetric orthogonal solutions if and only if all equalities in (53)hold In which case the solutions can be expressed as (54)

Similarly the following theorem holds

Theorem 52 Given AB isin Rntimes2m CD isin R

2mtimesn Suppose the matrix equation XC = D hasskew-symmetric orthogonal solutions with the form

X = W

(I 00 K

)UT (58)

where K isin SSOR2ptimes2p is arbitrary W U isin OR

2mtimes2m Partition

AW =(W1 W2

) BU =

(U1 U2

) (59)

where W1 U1 isin Rntimes(2mminus2p) W2 U2 isin R

ntimes2p Then the system (11) has skew-symmetric orthogonalsolutions if and only if

DTD = CTC DTC = minusCTD W1 = U1 W2WT2 = U2U

T2 U2W

T2 = minusW2U

T2

(510)

In which case the solutions can be expressed as

X = W1

(I 00 J primeprime

)UT

1 (511)

where

W1 = W

(I2mminus2p 0

0 W1

)isin OR

2mtimes2m U1 =

(I2mminus2p 0

0 UT1

)U isin OR

2mtimes2m (512)

and J primeprime isin SSOR2qtimes2q is arbitrary

6 The Least Squares (Skew-) Symmetric Orthogonal Solutions ofthe System (11)

If the solvability conditions of a system of matrix equations are not satisfied it is natural toconsider its least squares solution In this section we get the least squares (skew-) symmetricorthogonal solutions of the system (11) that is seek X isin SSOR

2mtimes2m(SORntimesn) such that

minXisinSSOR2mtimes2m(SORntimesn)

AX minus B2 + XC minusD2 (61)

14 Journal of Applied Mathematics

With the help of the definition of the Frobenius norm and the properties of the skew-symmetric orthogonal matrix we get that

AX minus B2 + XC minusD2 = A2 + B2 + C2 + D2 minus 2 tr(XT(ATB +DCT

)) (62)

Let

ATB +DCT = K

KT minusK

2= T

(63)

Then it follows from the skew-symmetric matrix X that

tr(XT(ATB +DCT

))= tr(XTK

)= tr(minusXK) = tr

(X

(KT minusK

2

))= tr(XT) (64)

Therefore (61) holds if and only if (64) reaches its maximum Now we pay our attention tofind the maximum value of (64) Assume the eigenvalue decomposition of T is

T = E

(Λ 00 0

)ET (65)

with

Λ = diag(Λ1 Λl) Λi =(

0 αi

minusαi 0

) αi gt 0 i = 1 l 2l = rank(T) (66)

Denote

ETXE =(X11 X12

X21 X22

) (67)

partitioned according to

(Λ 00 0

) (68)

then (64) has the following form

tr(XT) = tr(ETXE

(Λ 00 0

))= tr(X11Λ) (69)

Journal of Applied Mathematics 15

Thus by

X11 = I = diag(I1 Il

) (610)

where

Ii =(

0 minus11 0

) i = 1 l (611)

Equation (69) gets its maximum Since ETXE is skew-symmetric it follows from

X = E

(I 00 G

)ET (612)

where G isin SSOR(2mminus2l)times(2mminus2l) is arbitrary that (61) obtains its minimum Hence we have the

following theorem

Theorem 61 Given AB isin Rntimes2m and CD isin R

2mtimesn denote

ATB +DCT = KKT minusK

2= T (613)

and let the spectral decomposition of T be (65) Then the least squares skew-symmetric orthogonalsolutions of the system (11) can be expressed as (612)

If X in (61) is a symmetric orthogonal matrix then by the definition of the Frobeniusnorm and the properties of the symmetric orthogonal matrix (62) holds Let

ATB +DCT = HHT +H

2= N (614)

Then we get that

minXisinSORntimesn

AX minus B2 + XC minusD2 = minXisinSORntimesn

[A2 + B2 + C2 + D2 minus 2 tr(XN)

] (615)

Thus (615) reaches its minimum if and only if tr(XN) obtains its maximum Now we focuson finding the maximum value of tr(XN) Let the spectral decomposition of the symmetricmatrix N be

N = M

(Σ 00 0

)MT (616)

16 Journal of Applied Mathematics

where

Σ =(Σ+ 00 Σminus

) Σ+ = diag(λ1 λs)

λ1 ge λ2 ge middot middot middot ge λs gt 0 Σminus = diag(λs+1 λt)

λt le λtminus1 le middot middot middot le λs+1 lt 0 t = rank(N)

(617)

Denote

MTXM =

(X11 X12

X21 X22

)(618)

being compatible with

(Σ 00 0

) (619)

Then

tr(XN) = tr(MTXM

(Σ 00 0

))= tr

((X11 X12

X21 X22

)(Σ 00 0

))= tr(X11Σ

) (620)

Therefore it follows from

X11 = I =(Is 00 minusItminuss

)(621)

that (620) reaches its maximum Since MTXM is a symmetric orthogonal matrix then whenX has the form

X = M

(I 00 L

)MT (622)

where L isin SOR(nminust)times(nminust) is arbitrary (615) gets its minimum Thus we obtain the following

theorem

Theorem 62 Given AB isin Rntimesm CD isin R

mtimesn denote

ATB +DCT = HHT +H

2= N (623)

and let the eigenvalue decomposition of N be (616) Then the least squares symmetric orthogonalsolutions of the system (11) can be described as (622)

Journal of Applied Mathematics 17

Algorithm 63 Consider the following

Step 1 Input AB isin Rntimesm and CD isin R

mtimesn

Step 2 Compute

ATB +DCT = H

N =HT +H

2

(624)

Step 3 Compute the spectral decomposition of N with the form (616)

Step 4 Compute the least squares symmetric orthogonal solutions of (11) according to(622)

Example 64 Assume

A =

⎛⎜⎜⎜⎜⎜⎝

122 84 minus56 63 94 107118 29 85 69 96 minus78106 23 115 78 67 8936 78 49 119 94 5945 67 78 31 56 116

⎞⎟⎟⎟⎟⎟⎠

B =

⎛⎜⎜⎜⎜⎜⎝

85 94 36 78 63 4727 36 79 94 56 7837 67 86 98 34 29minus43 62 57 74 54 9529 39 minus52 63 78 46

⎞⎟⎟⎟⎟⎟⎠

C =

⎛⎜⎜⎜⎜⎜⎜⎜⎝

54 64 37 56 9736 42 78 minus63 7867 35 minus46 29 28minus27 72 108 37 3819 39 82 56 11289 78 94 79 56

⎞⎟⎟⎟⎟⎟⎟⎟⎠

D =

⎛⎜⎜⎜⎜⎜⎜⎜⎝

79 95 54 28 8687 26 67 84 8157 39 minus29 52 1948 58 18 minus72 5828 79 45 67 9695 41 34 98 39

⎞⎟⎟⎟⎟⎟⎟⎟⎠

(625)

It can be verified that the given matrices ABC and D do not satisfy the solvabilityconditions in Theorem 41 or Theorem 42 So we intend to derive the least squares symmetricorthogonal solutions of the system (11) By Algorithm 63 we have the following results

(1) the least squares symmetric orthogonal solution

X =

⎛⎜⎜⎜⎜⎜⎜⎜⎝

001200 006621 minus024978 027047 091302 minus016218006621 065601 022702 minus055769 024587 037715minus024978 022702 080661 040254 004066 minus026784027047 minus055769 040254 006853 023799 062645091302 024587 004066 023799 minus012142 minus018135minus016218 037715 minus026784 062645 minus0181354 057825

⎞⎟⎟⎟⎟⎟⎟⎟⎠

(626)

(2)

minXisinSORmtimesm

AX minus B2 + XC minusD2 = 198366

∥∥∥XTX minus I6

∥∥∥ = 284882 times 10minus15∥∥∥XT minusX

∥∥∥ = 000000(627)

18 Journal of Applied Mathematics

Remark 65 (1) There exists a unique symmetric orthogonal solution such that (61) holds ifand only if the matrix

N =HT +H

2 (628)

where

H = ATB +DCT (629)

is invertible Example 64 just illustrates it(2) The algorithm about computing the least squares skew-symmetric orthogonal

solutions of the system (11) can be shown similarly we omit it here

7 Conclusions

This paper is devoted to giving the solvability conditions of and the expressions ofthe orthogonal solutions the symmetric orthogonal solutions and the skew-symmetricorthogonal solutions to the system (11) respectively and meanwhile obtaining the leastsquares symmetric orthogonal and skew-symmetric orthogonal solutions of the system (11)In addition an algorithm and an example have been provided to compute its least squaressymmetric orthogonal solutions

Acknowledgments

This research was supported by the grants from Innovation Program of Shanghai MunicipalEducation Commission (13ZZ080) the National Natural Science Foundation of China(11171205) the Natural Science Foundation of Shanghai (11ZR1412500) the DisciplineProject at the corresponding level of Shanghai (A 13-0101-12-005) and Shanghai LeadingAcademic Discipline Project (J50101)

References

[1] C J Meng and X Y Hu ldquoThe inverse problem of symmetric orthogonal matrices and its optimalapproximationrdquo Mathematica Numerica Sinica vol 28 pp 269ndash280 2006 (Chinese)

[2] C J Meng X Y Hu and L Zhang ldquoThe skew-symmetric orthogonal solutions of the matrix equationAX = Brdquo Linear Algebra and Its Applications vol 402 no 1ndash3 pp 303ndash318 2005

[3] D L Chu H C Chan and D W C Ho ldquoRegularization of singular systems by derivative andproportional output feedbackrdquo SIAM Journal on Matrix Analysis and Applications vol 19 no 1 pp21ndash38 1998

[4] K T Joseph ldquoInverse eigenvalue problem in structural designrdquo AIAA Journal vol 30 no 12 pp2890ndash2896 1992

[5] A Jameson and E Kreinder ldquoInverse problem of linear optimal controlrdquo SIAM Journal on Controlvol 11 no 1 pp 1ndash19 1973

[6] Q W Wang ldquoBisymmetric and centrosymmetric solutions to systems of real quaternion matrixequationsrdquo Computers and Mathematics with Applications vol 49 no 5-6 pp 641ndash650 2005

[7] Q W Wang X Liu and S W Yu ldquoThe common bisymmetric nonnegative definite solutions withextreme ranks and inertias to a pair of matrix equationsrdquo Applied Mathematics and Computation vol218 pp 2761ndash2771 2011

Journal of Applied Mathematics 19

[8] C G Khatri and S K Mitra ldquoHermitian and nonnegative definite solutions of linear matrixequationsrdquo SIAM Journal on Applied Mathematics vol 31 pp 578ndash585 1976

[9] Q Zhang and Q W Wang ldquoThe (PQ)-(skew)symmetric extremal rank solutions to a system ofquaternion matrix equationsrdquo Applied Mathematics and Computation vol 217 no 22 pp 9286ndash92962011

[10] F L Li X Y Hu and L Zhang ldquoLeast-squares mirrorsymmetric solution for matrix equations (AX =BXC = D)rdquo Numerical Mathematics vol 15 pp 217ndash226 2006

[11] Y X Yuan ldquoLeast-squares solutions to the matrix equations AX = B and XC = Drdquo AppliedMathematics and Computation vol 216 no 10 pp 3120ndash3125 2010

[12] A Dajic and J J Koliha ldquoPositive solutions to the equations AX = C and XB = D for Hilbert spaceoperatorsrdquo Journal of Mathematical Analysis and Applications vol 333 no 2 pp 567ndash576 2007

[13] F L Li X Y Hu and L Zhang ldquoThe generalized reflexive solution for a class of matrix equations(AX = BXC = D)rdquo Acta Mathematica Scientia vol 28 no 1 pp 185ndash193 2008

[14] S Kumar Mitra ldquoThe matrix equations AX = CXB = Drdquo Linear Algebra and Its Applications vol 59pp 171ndash181 1984

[15] Y Qiu Z Zhang and J Lu ldquoThe matrix equations AX = BXC = D with PX = sXP constraintrdquoApplied Mathematics and Computation vol 189 no 2 pp 1428ndash1434 2007

[16] Y Y Qiu and A D Wang ldquoLeast squares solutions to the equations AX = BXC = B with someconstraintsrdquo Applied Mathematics and Computation vol 204 no 2 pp 872ndash880 2008

[17] Q W Wang X Zhang and Z H He ldquoOn the Hermitian structures of the solution to a pair of matrixequationsrdquo Linear Multilinear Algebra vol 61 no 1 pp 73ndash90 2013

[18] Q X Xu ldquoCommon Hermitian and positive solutions to the adjointable operator equations AX =CXB = Drdquo Linear Algebra and Its Applications vol 429 no 1 pp 1ndash11 2008

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2012 Article ID 398085 14 pagesdoi1011552012398085

Research ArticleOn the Hermitian R-Conjugate Solution ofa System of Matrix Equations

Chang-Zhou Dong1 Qing-Wen Wang2 and Yu-Ping Zhang3

1 School of Mathematics and Science Shijiazhuang University of Economics ShijiazhuangHebei 050031 China

2 Department of Mathematics Shanghai University Shanghai Shanghai 200444 China3 Department of Mathematics Ordnance Engineering College Shijiazhuang Hebei 050003 China

Correspondence should be addressed to Qing-Wen Wang wqw858yahoocomcn

Received 3 October 2012 Accepted 26 November 2012

Academic Editor Yang Zhang

Copyright q 2012 Chang-Zhou Dong et al This is an open access article distributed under theCreative Commons Attribution License which permits unrestricted use distribution andreproduction in any medium provided the original work is properly cited

Let R be an n by n nontrivial real symmetric involution matrix that is R = Rminus1 = RT = In An n times n

complex matrix A is termed R-conjugate if A = RAR where A denotes the conjugate of A Wegive necessary and sufficient conditions for the existence of the Hermitian R-conjugate solutionto the system of complex matrix equations AX = C andXB = D and present an expression ofthe Hermitian R-conjugate solution to this system when the solvability conditions are satisfiedIn addition the solution to an optimal approximation problem is obtained Furthermore the leastsquares Hermitian R-conjugate solution with the least norm to this system mentioned above isconsidered The representation of such solution is also derived Finally an algorithm and numericalexamples are given

1 Introduction

Throughout we denote the complex m times n matrix space by Cmtimesn the real m times n matrix space

by Rmtimesn and the set of all matrices in R

mtimesn with rank r by Rmtimesnr The symbols IAAT Alowast Adagger

and A stand for the identity matrix with the appropriate size the conjugate the transposethe conjugate transpose the Moore-Penrose generalized inverse and the Frobenius norm ofA isin C

mtimesn respectively We use Vn to denote the ntimes n backward matrix having the elements 1along the southwest diagonal and with the remaining elements being zeros

Recall that an n times n complex matrix A is centrohermitian if A = VnAVn Centro-hermitian matrices and related matrices such as k-Hermitian matrices Hermitian Toeplitzmatrices and generalized centrohermitian matrices appear in digital signal processing andothers areas (see [1ndash4]) As a generalization of a centrohermitian matrix and related matrices

2 Journal of Applied Mathematics

Trench [5] gave the definition of R-conjugate matrix A matrix A isin Cntimesn is R-conjugate if

A = RAR where R is a nontrivial real symmetric involution matrix that is R = Rminus1 = RT andR= In At the same time Trench studied the linear equation Az = w for R-conjugate matricesin [5] where zw are known column vectors

Investigating the matrix equation

AX = B (11)

with the unknown matrix X being symmetric reflexive Hermitian-generalized Hamiltonianand repositive definite is a very active research topic [6ndash14] As a generalization of (11) theclassical system of matrix equations

AX = C XB = D (12)

has attracted many authorrsquos attention For instance [15] gave the necessary and sufficientconditions for the consistency of (12) [16 17] derived an expression for the general solutionby using singular value decomposition of a matrix and generalized inverses of matricesrespectively Moreover many results have been obtained about the system (12) with variousconstraints such as bisymmetric Hermitian positive semidefinite reflexive and generalizedreflexive solutions (see [18ndash28]) To our knowledge so far there has been little investigationof the Hermitian R-conjugate solution to (12)

Motivated by the work mentioned above we investigate Hermitian R-conjugatesolutions to (12) We also consider the optimal approximation problem

∥∥∥X minus E∥∥∥ = min

XisinSX

X minus E (13)

where E is a given matrix in Cntimesn and SX the set of all Hermitian R-conjugate solutions to

(12) In many cases the system (12) has not Hermitian R-conjugate solution Hence we needto further study its least squares solution which can be described as follows Let RHC

ntimesn

denote the set of all Hermitian R-conjugate matrices in Cntimesn

SL =X | min

XisinRHCntimesn

(AX minus C2 + XB minusD2

) (14)

Find X isin Cntimesn such that

∥∥∥X∥∥∥ = minXisinSL

X (15)

In Section 2 we present necessary and sufficient conditions for the existence of theHermitian R-conjugate solution to (12) and give an expression of this solution when thesolvability conditions are met In Section 3 we derive an optimal approximation solution to(13) In Section 4 we provide the least squares Hermitian R-conjugate solution to (15) InSection 5 we give an algorithm and a numerical example to illustrate our results

Journal of Applied Mathematics 3

2 R-Conjugate Hermitian Solution to (12)

In this section we establish the solvability conditions and the general expression for theHermitian R-conjugate solution to (12)

We denote RCntimesn and RHC

ntimesn the set of all R-conjugate matrices and Hermitian R-conjugate matrices respectively that is

RCntimesn =

A | A = RAR

HRCntimesn =

A | A = RARA = Alowast

(21)

where R is n times n nontrivial real symmetric involution matrixChang et al in [29] mentioned that for nontrivial symmetric involution matrix R isin

Rntimesn there exist positive integer r and n times n real orthogonal matrix [P Q] such that

R =[P Q

]⎡⎣Ir 0

0 minusInminusr

⎤⎦⎡⎣P

T

QT

⎤⎦ (22)

where P isin Rntimesr Q isin R

ntimes(nminusr) By (22)

RP = P RQ = minusQ PTP = Ir QTQ = Inminusr PTQ = 0 QTP = 0 (23)

Throughout this paper we always assume that the nontrivial symmetric involutionmatrix R is fixed which is given by (22) and (23) Now we give a criterion of judging amatrix is R-conjugate Hermitian matrix

Theorem 21 A matrix K isin HRCntimesn if and only if there exists a symmetric matrix H isin R

ntimesn suchthat K = ΓHΓlowast where

Γ =[P iQ

] (24)

with PQ being the same as (22)

Proof If K isin HRCntimesn then K = RKR By (22)

K = RKR =[P Q

]⎡⎣Ir 0

0 minusInminusr

⎤⎦⎡⎣P

T

QT

⎤⎦K[P Q

]⎡⎣Ir 0

0 minusInminusr

⎤⎦⎡⎣P

T

QT

⎤⎦ (25)

4 Journal of Applied Mathematics

which is equivalent to

⎡⎣P

T

QT

⎤⎦K[P Q

]

=

⎡⎣Ir 0

0 minusInminusr

⎤⎦⎡⎣P

T

QT

⎤⎦K[P Q

]⎡⎣Ir 0

0 minusInminusr

⎤⎦

(26)

Suppose that

⎡⎣P

T

QT

⎤⎦K[P Q

]=

⎡⎣K11 K12

K21 K22

⎤⎦ (27)

Substituting (27) into (26) we obtain

⎡⎣K11 K12

K21 K22

⎤⎦ =

⎡⎣Ir 0

0 minusInminusr

⎤⎦⎡⎣K11 K12

K21 K22

⎤⎦⎡⎣Ir 0

0 minusInminusr

⎤⎦ =

⎡⎣ K11 minusK12

minusK21 K22

⎤⎦ (28)

Hence K11 = K11 K12 = minusK12 K21 = minusK21 K22 = K22 that is K11 iK12 iK21 K22 are realmatrices If we denote M = iK12 N = minusiK21 then by (27)

K =[P Q

]⎡⎣K11 K12

K21 K22

⎤⎦⎡⎣P

T

QT

⎤⎦ =

[P iQ

]⎡⎣K11 M

N K22

⎤⎦⎡⎣ PT

minusiQT

⎤⎦ (29)

Let Γ = [P iQ] and

H =

⎡⎣K11 M

N K22

⎤⎦ (210)

Then K can be expressed as ΓHΓlowast where Γ is unitary matrix and H is a real matrix ByK = Klowast

ΓHTΓlowast = Klowast = K = ΓHΓlowast (211)

we obtain H = HT

Journal of Applied Mathematics 5

Conversely if there exists a symmetric matrix H isin Rntimesn such that K = ΓHΓlowast then it

follows from (23) that

RKR = RΓHΓlowastR = R[P iQ

]H

⎡⎣ PT

minusiQT

⎤⎦R =

[P minusiQ]H

⎡⎣ PT

iQT

⎤⎦ = ΓHΓlowast = K

Klowast = ΓHTΓlowast = ΓHΓlowast = K

(212)

that is K isin HRCntimesn

Theorem 21 implies that an arbitrary complex Hermitian R-conjugate matrix is equiv-alent to a real symmetric matrix

Lemma 22 For any matrix A isin Cmtimesn A = A1 + iA2 where

A1 =A +A

2 A2 =

A minusA

2i (213)

Proof For any matrix A isin Cmtimesn it is obvious that A = A1 + iA2 where A1 A2 are defined

as (213) Now we prove that the decomposition A = A1 + iA2 is unique If there exist B1 B2

such that A = B1 + iB2 then

A1 minus B1 + i(B2 minusA2) = 0 (214)

It follows from A1 A2 B1 andB2 are real matrix that

A1 = B1 A2 = B2 (215)

Hence A = A1 + iA2 holds where A1 A2 are defined as (213)

By Theorem 21 for X isin HRCntimesn we may assume that

X = ΓYΓlowast (216)

where Γ is defined as (24) and Y isin Rntimesn is a symmetric matrix

Suppose that AΓ = A1 + iA2 isin Cmtimesn CΓ = C1 + iC2 isin C

mtimesn ΓlowastB = B1 + iB2 isin Cntimesl and

ΓlowastD = D1 + iD2 isin Cntimesl where

A1 =AΓ +AΓ

2 A2 =

AΓ minusAΓ2i

C1 =CΓ + CΓ

2 C2 =

CΓ minus CΓ2i

B1 =ΓlowastB + ΓlowastB

2 B2 =

ΓlowastB minus ΓlowastB2i

D1 =ΓlowastD + ΓlowastD

2 D2 =

ΓlowastD minus ΓlowastD2i

(217)

6 Journal of Applied Mathematics

Then system (12) can be reduced into

(A1 + iA2)Y = C1 + iC2 Y (B1 + iB2) = D1 + iD2 (218)

which implies that

⎡⎣A1

A2

⎤⎦Y =

⎡⎣C1

C2

⎤⎦ Y

[B1 B2

]=[D1 D2

] (219)

Let

F =

⎡⎣A1

A2

⎤⎦ G =

⎡⎣C1

C2

⎤⎦ K =

[B1 B2

]

L =[D1 D2

] M =

⎡⎣ F

KT

⎤⎦ N =

⎡⎣G

LT

⎤⎦

(220)

Then system (12) has a solution X in HRCntimesn if and only if the real system

MY = N (221)

has a symmetric solution Y in Rntimesn

Lemma 23 (Theorem 1 in [7]) Let A isin Rmtimesn The SVD of matrix A is as follows

A = U

⎡⎣Σ 0

0 0

⎤⎦V T (222)

where U = [U1 U2] isin Rmtimesm and V = [V1 V2] isin R

ntimesn are orthogonal matrices Σ = diag(σ1 σr) σi gt 0 (i = 1 r) r = rank(A) U1 isin R

mtimesr V1 isin Rntimesr Then (11) has a symmetric

solution if and only if

ABT = BAT UT2 B = 0 (223)

In that case it has the general solution

X = V1Σminus1UT1 B + V2V

T2 B

TU1Σminus1V T1 + V2GVT

2 (224)

where G is an arbitrary (n minus r) times (n minus r) symmetric matrix

By Lemma 23 we have the following theorem

Journal of Applied Mathematics 7

Theorem 24 Given A isin Cmtimesn C isin C

mtimesn B isin Cntimesl and D isin C

ntimesl Let A1 A2 C1C2B1 B2 D1 D2 F G K L M andN be defined in (217) (220) respectively Assume thatthe SVD of M isin R

(2m+2l)timesn is as follows

M = U

⎡⎣M1 0

0 0

⎤⎦V T (225)

where U = [U1 U2] isin R(2m+2l)times(2m+2l) and V = [V1 V2] isin R

ntimesn are orthogonal matrices M1 =diag(σ1 σr) σi gt 0 (i = 1 r) r = rank(M) U1 isin R

(2m+2l)timesr V1 isin Rntimesr Then system

(12) has a solution in HRCntimesn if and only if

MNT = NMT UT2 N = 0 (226)

In that case it has the general solution

X = Γ(V1M

minus11 UT

1 N + V2VT2 N

TU1Mminus11 V T

1 + V2GVT2

)Γlowast (227)

where G is an arbitrary (n minus r) times (n minus r) symmetric matrix

3 The Solution of Optimal Approximation Problem (13)

When the set SX of all Hermitian R-conjugate solution to (12) is nonempty it is easy to verifySX is a closed set Therefore the optimal approximation problem (13) has a unique solutionby [30]

Theorem 31 GivenA isin Cmtimesn C isin C

mtimesn B isin Cntimesl D isin C

ntimesl E isin Cntimesn and E1 = (12)(ΓlowastEΓ+

ΓlowastEΓ) Assume SX is nonempty then the optimal approximation problem (13) has a unique solutionX and

X = Γ(V1M

minus11 UT

1 N + V2VT2 N

TU1Mminus11 V T

1 + V2VT2 E1V2V

T2

)Γlowast (31)

Proof Since SX is nonempty X isin SX has the form of (227) By Lemma 22 ΓlowastEΓ can be writtenas

ΓlowastEΓ = E1 + iE2 (32)

where

E1 =12

(ΓlowastEΓ + ΓlowastEΓ

) E2 =

12i

(ΓlowastEΓ minus ΓlowastEΓ

) (33)

8 Journal of Applied Mathematics

According to (32) and the unitary invariance of Frobenius norm

X minus E =∥∥∥Γ(V1M

minus1UT1N + V2V

T2 N

TU1Mminus1V T

1 + V2GVT2

)Γlowast minus E

∥∥∥=∥∥∥(E1 minus V1M

minus1UT1N minus V2V

T2 N

TU1Mminus1V T

1 minus V2GVT2

)+ iE2

∥∥∥(34)

We get

X minus E2 =∥∥∥E1 minus V1M

minus1UT1 N minus V2V

T2 N

TU1Mminus1V T

1 minus V2GVT2

∥∥∥2+ E22 (35)

Then minXisinSXX minus E is consistent if and only if there exists G isin R(nminusr)times(nminusr) such that

min∥∥∥E1 minus V1M

minus1UT1 N minus V2V

T2 N

TU1Mminus1V T

1 minus V2GVT2

∥∥∥ (36)

For the orthogonal matrix V

∥∥∥E1 minus V1Mminus1UT

1N minus V2VT2 N

TU1Mminus1V T

1 minus V2GVT2

∥∥∥2

=∥∥∥V T (E1 minus V1M

minus1UT1 N minus V2V

T2 N

TU1Mminus1V T

1 minus V2GVT2 )V

∥∥∥2

=∥∥∥V T

1

(E1 minus V1M

minus1UT1 N)V1

∥∥∥2+∥∥∥V T

1 (E1 minus V1Mminus1UT

1 N)V2

∥∥∥2

+∥∥∥V T

2

(E1 minus V2V

T2 N

TU1Mminus1V T

1

)V1

∥∥∥2+∥∥∥V T

2

(E1 minus V2GVT

2

)V2

∥∥∥

(37)

Therefore

min∥∥∥E1 minus V1M

minus1UT1 N minus V2V

T2 N

TU1Mminus1V T

1 minus V2GVT2

∥∥∥ (38)

is equivalent to

G = V T2 E1V2 (39)

Substituting (39) into (227) we obtain (31)

4 The Solution of Problem (15)

In this section we give the explicit expression of the solution to (15)

Theorem 41 Given A isin Cmtimesn C isin C

mtimesn B isin Cntimesl and D isin C

ntimesl Let A1 A2 C1 C2B1 B2 D1 D2 F G K L M and N be defined in (217) (220) respectively Assume that the

Journal of Applied Mathematics 9

SVD of M isin R(2m+2l)timesn is as (225) and system (12) has not a solution in HRC

ntimesn Then X isin SL

can be expressed as

X = ΓV

⎡⎣Mminus1

1 UT1NV1 Mminus1

1 UT1NV2

V T2 N

TU1Mminus11 Y22

⎤⎦V TΓlowast (41)

where Y22 isin R(nminusr)times(nminusr) is an arbitrary symmetric matrix

Proof It yields from (217)ndash(221) and (225) that

AX minus C2 + XB minusD2 = MY minusN2

=

∥∥∥∥∥∥U⎡⎣M1 0

0 0

⎤⎦V TY minusN

∥∥∥∥∥∥2

=

∥∥∥∥∥∥⎡⎣M1 0

0 0

⎤⎦V TYV minusUTNV

∥∥∥∥∥∥2

(42)

Assume that

V TYV =

⎡⎣Y11 Y12

Y21 Y22

⎤⎦ Y11 isin R

rtimesr Y22 isin R(nminusr)times(nminusr) (43)

Then we have

AX minus C2 + XB minusD2

=∥∥∥M1Y11 minusUT

1 NV1

∥∥∥2+∥∥∥M1Y12 minusUT

1NV2

∥∥∥2

+∥∥∥UT

2NV1

∥∥∥2+∥∥∥UT

2 NV2

∥∥∥2

(44)

Hence

min(AX minus C2 + XB minusD2

)(45)

is solvable if and only if there exist Y11 Y12 such that

∥∥∥M1Y11 minusUT1 NV1

∥∥∥2= min

∥∥∥M1Y12 minusUT1NV2

∥∥∥2= min

(46)

10 Journal of Applied Mathematics

It follows from (46) that

Y11 = Mminus11 UT

1 NV1

Y12 = Mminus11 UT

1 NV2(47)

Substituting (47) into (43) and then into (216) we can get that the form of elements in SL is(41)

Theorem 42 Assume that the notations and conditions are the same as Theorem 41 Then

∥∥∥X∥∥∥ = minXisinSL

X (48)

if and only if

X = ΓV

⎡⎣Mminus1

1 UT1NV1 Mminus1

1 UT1NV2

V T2 N

TU1Mminus11 0

⎤⎦V TΓlowast (49)

Proof In Theorem 41 it implies from (41) that minXisinSLX is equivalent to X has theexpression (41) with Y22 = 0 Hence (49) holds

5 An Algorithm and Numerical Example

Base on the main results of this paper we in this section propose an algorithm for finding thesolution of the approximation problem (13) and the least squares problem with least norm(15) All the tests are performed by MATLAB 65 which has a machine precision of around10minus16

Algorithm 51 (1) Input A isin Cmtimesn C isin C

mtimesn B isin Cntimesl D isin C

ntimesl(2) Compute A1 A2 C1 C2 B1 B2 D1 D2 F G K L M andN by (217) and

(220)(3) Compute the singular value decomposition of M with the form of (225)(4) If (226) holds then input E isin C

ntimesn and compute the solution X of problem (13)according (31) else compute the solution X to problem (15) by (49)

To show our algorithm is feasible we give two numerical example Let an nontrivialsymmetric involution be

R =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 0 0 0

0 minus1 0 0

0 0 1 0

0 0 0 minus1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦ (51)

Journal of Applied Mathematics 11

We obtain [P Q] in (22) by using the spectral decomposition of R then by (24)

Γ =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 0 0 0

0 0 i 0

0 1 0 0

0 0 0 i

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦ (52)

Example 52 Suppose A isin C2times4 C isin C

2times4 B isin C4times3 D isin C

4times3 and

A =

⎡⎣333 minus 5987i 45i 721 minusi

0 minus066i 7694 1123i

⎤⎦

C =

⎡⎣02679 minus 00934i 00012 + 40762i minus00777 minus 01718i minus12801i

02207 minus01197i 00877 07058i

⎤⎦

B =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

4 + 12i 2369i 4256 minus 5111i

4i 466i 821 minus 5i

0 483i 56 + i

222i minus4666 7i

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

D =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

00616 + 01872i minus00009 + 01756i 16746 minus 00494i

00024 + 02704i 01775 + 04194i 07359 minus 06189i

minus00548 + 03444i 00093 minus 03075i minus04731 minus 01636i

00337i 01209 minus 01864i minus02484 minus 38817i

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

(53)

We can verify that (226) holds Hence system (12) has an Hermitian R-conjugate solutionGiven

E =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

735i 8389i 99256 minus 651i minus46i

155 456i 771 minus 75i i

5i 0 minus4556i minus799

422i 0 51i 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦ (54)

12 Journal of Applied Mathematics

Applying Algorithm 51 we obtain the following

X =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

15597 00194i 28705 00002i

minus00194i 90001 02005i minus39997

28705 minus02005i minus00452 79993i

minus00002i minus39997 minus79993i 56846

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦ (55)

Example 52 illustrates that we can solve the optimal approximation problem withAlgorithm 51 when system (12) have Hermitian R-conjugate solutions

Example 53 Let A B andC be the same as Example 52 and let D in Example 52 be changedinto

D =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

00616 + 01872i minus00009 + 01756i 16746 + 00494i

00024 + 02704i 01775 + 04194i 07359 minus 06189i

minus00548 + 03444i 00093 minus 03075i minus04731 minus 01636i

00337i 01209 minus 01864i minus02484 minus 38817i

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦ (56)

We can verify that (226) does not hold By Algorithm 51 we get

X =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

052 22417i 04914 03991i

minus22417i 86634 01921i minus28232

04914 minus01921i 01406 13154i

minus03991i minus28232 minus13154i 63974

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦ (57)

Example 53 demonstrates that we can get the least squares solution with Algo-rithm 51 when system (12) has not Hermitian R-conjugate solutions

Acknowledgments

This research was supported by the Grants from the Key Project of Scientific ResearchInnovation Foundation of Shanghai Municipal Education Commission (13ZZ080) theNational Natural Science Foundation of China (11171205) the Natural Science Foundationof Shanghai (11ZR1412500) the PhD Programs Foundation of Ministry of Education ofChina (20093108110001) the Discipline Project at the corresponding level of Shanghai (A13-0101-12-005) Shanghai Leading Academic Discipline Project (J50101) the Natural ScienceFoundation of Hebei province (A2012403013) and the Natural Science Foundation of Hebeiprovince (A2012205028) The authors are grateful to the anonymous referees for their helpfulcomments and constructive suggestions

Journal of Applied Mathematics 13

References

[1] R D Hill R G Bates and S R Waters ldquoOn centro-Hermitian matricesrdquo SIAM Journal on MatrixAnalysis and Applications vol 11 no 1 pp 128ndash133 1990

[2] R Kouassi P Gouton and M Paindavoine ldquoApproximation of the Karhunen-Loeve tranformationand Its application to colour imagesrdquo Signal Processing Image Communication vol 16 pp 541ndash5512001

[3] A Lee ldquoCentro-Hermitian and skew-centro-Hermitian matricesrdquo Linear Algebra and Its Applicationsvol 29 pp 205ndash210 1980

[4] Z-Y Liu H-D Cao and H-J Chen ldquoA note on computing matrix-vector products with generalizedcentrosymmetric (centrohermitian) matricesrdquo Applied Mathematics and Computation vol 169 no 2pp 1332ndash1345 2005

[5] W F Trench ldquoCharacterization and properties of matrices with generalized symmetry or skewsymmetryrdquo Linear Algebra and Its Applications vol 377 pp 207ndash218 2004

[6] K-W E Chu ldquoSymmetric solutions of linear matrix equations by matrix decompositionsrdquo LinearAlgebra and Its Applications vol 119 pp 35ndash50 1989

[7] H Dai ldquoOn the symmetric solutions of linear matrix equationsrdquo Linear Algebra and Its Applicationsvol 131 pp 1ndash7 1990

[8] F J H Don ldquoOn the symmetric solutions of a linear matrix equationrdquo Linear Algebra and ItsApplications vol 93 pp 1ndash7 1987

[9] I Kyrchei ldquoExplicit representation formulas for the minimum norm least squares solutions of somequaternion matrix equationsrdquo Linear Algebra and Its Applications vol 438 no 1 pp 136ndash152 2013

[10] Y Li Y Gao and W Guo ldquoA Hermitian least squares solution of the matrix equation AXB = Csubject to inequality restrictionsrdquo Computers amp Mathematics with Applications vol 64 no 6 pp 1752ndash1760 2012

[11] Z-Y Peng and X-Y Hu ldquoThe reflexive and anti-reflexive solutions of the matrix equation AX = BrdquoLinear Algebra and Its Applications vol 375 pp 147ndash155 2003

[12] L Wu ldquoThe re-positive definite solutions to the matrix inverse problem AX = Brdquo Linear Algebra andIts Applications vol 174 pp 145ndash151 1992

[13] S F Yuan Q W Wang and X Zhang ldquoLeast-squares problem for the quaternion matrix equationAXB + CYD = E over different constrained matricesrdquo Article ID 722626 International Journal ofComputer Mathematics In press

[14] Z-Z Zhang X-Y Hu and L Zhang ldquoOn the Hermitian-generalized Hamiltonian solutions of linearmatrix equationsrdquo SIAM Journal on Matrix Analysis and Applications vol 27 no 1 pp 294ndash303 2005

[15] F Cecioni ldquoSopra operazioni algebricherdquo Annali della Scuola Normale Superiore di Pisa vol 11 pp17ndash20 1910

[16] K-W E Chu ldquoSingular value and generalized singular value decompositions and the solution oflinear matrix equationsrdquo Linear Algebra and Its Applications vol 88-89 pp 83ndash98 1987

[17] S K Mitra ldquoA pair of simultaneous linear matrix equations A1XB1 = C1 A2XB2 = C2 and a matrixprogramming problemrdquo Linear Algebra and Its Applications vol 131 pp 107ndash123 1990

[18] H-X Chang and Q-W Wang ldquoReflexive solution to a system of matrix equationsrdquo Journal of ShanghaiUniversity vol 11 no 4 pp 355ndash358 2007

[19] A Dajic and J J Koliha ldquoEquations ax = c and xb = d in rings and rings with involution withapplications to Hilbert space operatorsrdquo Linear Algebra and Its Applications vol 429 no 7 pp 1779ndash1809 2008

[20] A Dajic and J J Koliha ldquoPositive solutions to the equations AX = C and XB = D for Hilbert spaceoperatorsrdquo Journal of Mathematical Analysis and Applications vol 333 no 2 pp 567ndash576 2007

[21] C-Z Dong Q-W Wang and Y-P Zhang ldquoThe common positive solution to adjointable operatorequations with an applicationrdquo Journal of Mathematical Analysis and Applications vol 396 no 2 pp670ndash679 2012

[22] X Fang J Yu and H Yao ldquoSolutions to operator equations on Hilbert Clowast-modulesrdquo Linear Algebraand Its Applications vol 431 no 11 pp 2142ndash2153 2009

[23] C G Khatri and S K Mitra ldquoHermitian and nonnegative definite solutions of linear matrixequationsrdquo SIAM Journal on Applied Mathematics vol 31 no 4 pp 579ndash585 1976

[24] I Kyrchei ldquoAnalogs of Cramerrsquos rule for the minimum norm least squares solutions of some matrixequationsrdquo Applied Mathematics and Computation vol 218 no 11 pp 6375ndash6384 2012

[25] F L Li X Y Hu and L Zhang ldquoThe generalized reflexive solution for a class of matrix equations(AX = CXB = D)rdquo Acta Mathematica Scientia vol 28B no 1 pp 185ndash193 2008

14 Journal of Applied Mathematics

[26] Q-W Wang and C-Z Dong ldquoPositive solutions to a system of adjointable operator equations overHilbert Clowast-modulesrdquo Linear Algebra and Its Applications vol 433 no 7 pp 1481ndash1489 2010

[27] Q Xu ldquoCommon Hermitian and positive solutions to the adjointable operator equations AX = CXB = Drdquo Linear Algebra and Its Applications vol 429 no 1 pp 1ndash11 2008

[28] F Zhang Y Li and J Zhao ldquoCommon Hermitian least squares solutions of matrix equations A1XAlowast1 =

B1 and A2XAlowast2 = B2 subject to inequality restrictionsrdquo Computers ampMathematics with Applications vol

62 no 6 pp 2424ndash2433 2011[29] H-X Chang Q-W Wang and G-J Song ldquo(R S)-conjugate solution to a pair of linear matrix

equationsrdquo Applied Mathematics and Computation vol 217 no 1 pp 73ndash82 2010[30] E W Cheney Introduction to Approximation Theory McGraw-Hill New York NY USA 1966

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2012 Article ID 784620 13 pagesdoi1011552012784620

Research ArticlePerturbation Analysis for the Matrix EquationX minussumm

i=1 Alowasti XAi +

sumnj=1 B

lowastj XBj = I

Xue-Feng Duan1 2 and Qing-Wen Wang3

1 School of Mathematics and Computational Science Guilin University of Electronic TechnologyGuilin 541004 China

2 Research Center on Data Science and Social Computing Guilin University of Electronic TechnologyGuilin 541004 China

3 Department of Mathematics Shanghai University Shanghai 200444 China

Correspondence should be addressed to Qing-Wen Wang wqw858yahoocomcn

Received 18 September 2012 Accepted 19 November 2012

Academic Editor Yang Zhang

Copyright q 2012 X-F Duan and Q-W Wang This is an open access article distributed underthe Creative Commons Attribution License which permits unrestricted use distribution andreproduction in any medium provided the original work is properly cited

We consider the perturbation analysis of the matrix equation X minussummi=1 A

lowasti XAi +

sumnj=1 B

lowastj XBj = I

Based on the matrix differentiation we first give a precise perturbation bound for the positivedefinite solution A numerical example is presented to illustrate the sharpness of the perturbationbound

1 Introduction

In this paper we consider the matrix equation

X minusmsumi=1

Alowasti XAi +

nsumj=1

Blowastj XBj = I (11)

where A1 A2 Am B1 B2 Bn are n times n complex matrices I is an n times n identity matrixm n are nonnegative integers and the positive definite solution X is practical interestHere Alowast

i and Blowasti denote the conjugate transpose of the matrices Ai and Bi respectively

Equation (11) arises in solving some nonlinear matrix equations with Newton method Seefor example the nonlinear matrix equation which appears in Sakhnovich [1] Solving these

2 Journal of Applied Mathematics

nonlinear matrix equations gives rise to (11) On the other hand (11) is the general case ofthe generalized Lyapunov equation

MYSlowast + SYMlowast +tsum

k=1

NkYNlowastk + CClowast = 0 (12)

whose positive definite solution is the controllability Gramian of the bilinear control system(see [2 3] for more details)

Mx(t) = Sx(t) +tsum

k=1

Nkx(t)uk(t) + Cu(t) (13)

Set

E =1radic2(M minus S + I) F =

1radic2(M + S + I) G = S minus I Q = CClowast

X = Qminus12YQminus12 Ai = Qminus12NiQ12 i = 1 2 t

At+1 = Qminus12FQ12 At+2 = Qminus12GQ12

B1 = Qminus12EQ12 B2 = Qminus12CQ12

(14)

then (12) can be equivalently written as (11) with m = t + 2 and n = 2Some special cases of (11) have been studied Based on the kronecker product and

fixed point theorem in partially ordered sets Reurings [4] and Ran and Reurings [5 6] gavesome sufficient conditions for the existence of a unique positive definite solution of the linearmatrix equations X minussumm

i=1 Alowasti XAi = I and X +

sumnj=1 B

lowasti XBi = I And the expressions for these

unique positive definite solutions were also derived under some constraint conditions Forthe general linear matrix equation (11) Reurings [[4] Page 61] pointed out that it is hard tofind sufficient conditions for the existence of a positive definite solution because the map

G(X) = I +msumi=1

Alowasti XAi minus

nsumj=1

Blowastj XBj (15)

is not monotone and does not map the set of n times n positive definite matrices into itselfRecently Berzig [7] overcame these difficulties by making use of Bhaskar-Lakshmikanthamcoupled fixed point theorem and gave a sufficient condition for (11) existing a uniquepositive definite solution An iterative method was constructed to compute the uniquepositive definite solution and the error estimation was given too

Recently the matrix equations of the form (11) have been studied by many authors(see [8ndash14]) Some numerical methods for solving the well-known Lyapunov equationX + ATXA = Q such as Bartels-Stewart method and Hessenberg-Schur method have beenproposed in [12] Based on the fixed point theorem the sufficient and necessary conditions forthe existence of a positive definite solution of the matrix equation Xs plusmnAlowastXminustA = Q s t isin Nhave been given in [8 9] The fixed point iterative method and inversion-free iterative method

Journal of Applied Mathematics 3

were developed for solving the matrix equations X plusmn AlowastXminusαA = Q α gt 0 in [13 14] Bymaking use of the fixed point theorem of mixed monotone operator the matrix equationX minussumm

i=1 Alowasti X

δiAi = Q 0 lt |δi| lt 1 was studied in [10] and derived a sufficient condition forthe existence of a positive definite solution Assume that F maps positive definite matriceseither into positive definite matrices or into negative definite matrices the general nonlinearmatrix equation X +AlowastF(X)A = Q was studied in [11] and the fixed point iterative methodwas constructed to compute the positive definite solution under some additional conditions

Motivated by the works and applications in [2ndash6] we continue to study the matrixequation (11) Based on a new mathematical tool (ie the matrix differentiation) we firstlygive a differential bound for the unique positive definite solution of (11) and then use it toderive a precise perturbation bound for the unique positive definite solution A numericalexample is used to show that the perturbation bound is very sharp

Throughout this paper we write B gt 0 (B ge 0) if the matrix B is positive definite(semidefinite) If B minus C is positive definite (semidefinite) then we write B gt C (B ge C) Ifa positive definite matrix X satisfies B le X le C we denote that X isin [BC] The symbolsλ1(B) and λn(B) denote the maximal and minimal eigenvalues of an n times n Hermitian matrixB respectively The symbol Hntimesn stands for the set of n times n Hermitian matrices The symbolB denotes the spectral norm of the matrix B

2 Perturbation Analysis for the Matrix Equation (11)

Based on the matrix differentiation we firstly give a differential bound for the unique positivedefinite solution XU of (11) and then use it to derive a precise perturbation bound for XU inthis section

Definition 21 ([15] Definition 36) Let F = (fij)mtimesn then the matrix differentiation of F isdF = (dfij)mtimesn For example let

F =(

s + t s2 minus 2t2s + t3 t2

) (21)

Then

dF =(

ds + dt 2sds minus 2dt2ds + 3t2dt 2tdt

) (22)

Lemma 22 ([15] Theorem 32) The matrix differentiation has the following properties

(1) d(F1 plusmn F2) = dF1 plusmn dF2

(2) d(kF) = k(dF) where k is a complex number

(3) d(Flowast) = (dF)lowast

(4) d(F1F2F3) = (dF1)F2F3 + F1(dF2)F3 + F1F2(dF3)

(5) dFminus1 = minusFminus1(dF)Fminus1

(6) dF = 0 where F is a constant matrix

4 Journal of Applied Mathematics

Lemma 23 ([7] Theorem 31) If

msumi=1

Alowasti Ai lt

12I

nsumj=1

Blowastj Bj lt

12I (23)

then (11) has a unique positive definite solution XU and

XU isin⎡⎣I minus 2

nsumj=1

Blowastj Bj I + 2

msumi=1

Alowasti Ai

⎤⎦ (24)

Theorem 24 If

msumi=1

Ai2 lt12

nsumj=1

∥∥Bj

∥∥2lt

12 (25)

then (11) has a unique positive definite solution XU and it satisfies

dXU le2(

1 + 2summ

i=1 Ai2)[summ

i=1(AidAi) +sumn

j=1(∥∥Bj

∥∥∥∥dBj

∥∥)]

1 minussummi=1 Ai2 minussumn

j=1

∥∥Bj

∥∥2 (26)

Proof Since

λ1(Alowast

i Ai

) le ∥∥Alowasti Ai

∥∥ le Ai2 i = 1 2 m

λ1

(Blowastj Bj

)le∥∥∥Blowast

j Bj

∥∥∥ le ∥∥Bj

∥∥2 j = 1 2 n

(27)

then

Alowasti Ai le λ1

(Alowast

i Ai

)I le ∥∥Alowast

i Ai

∥∥I le Ai2I i = 1 2 m

Blowastj Bj le λ1

(Blowastj Bj

)I le∥∥∥Blowast

j Bj

∥∥∥ le ∥∥Bj

∥∥2I j = 1 2 n

(28)

consequently

msumi=1

Alowasti Ai le

msumi=1

Ai2I

nsumj=1

Blowastj Bj le

nsumj=1

∥∥Bj

∥∥2I

(29)

Journal of Applied Mathematics 5

Combining (25)ndash(29) we have

msumi=1

Alowasti Ai le

msumi=1

Ai2I lt12I

nsumj=1

Blowastj Bj le

nsumj=1

∥∥Bj

∥∥2I lt

12I

(210)

Then by Lemma 23 we obtain that (11) has a unique positive definite solution XU whichsatisfies

XU isin⎡⎣I minus 2

nsumj=1

Blowastj Bj I + 2

msumi=1

Alowasti Ai

⎤⎦ (211)

Noting that XU is the unique positive definite solution of (11) then

XU minusmsumi=1

Alowasti XUAi +

nsumj=1

Blowastj XUBj = I (212)

It is known that the elements of XU are differentiable functions of the elements of Ai and BiDifferentiating (212) and by Lemma 22 we have

dXU minusmsumi=1

[(dAlowast

i

)XUAi +Alowast

i (dXU)Ai +Alowasti XU(dAi)

]

+nsumj=1

[(dBlowast

j

)XUBj + Blowast

j (dXU)Bj + Blowastj XU

(dBj

)]= 0

(213)

which implies that

dXU minusmsumi=1

Alowasti (dXU)Ai +

nsumj=1

Blowastj (dXU)Bj

=msumi=1

(dAlowast

i

)XUAi +

msumi=1

Alowasti XU(dAi) minus

nsumj=1

(dBlowast

j

)XUBj minus

nsumj=1

Blowastj XU

(dBj

)

(214)

6 Journal of Applied Mathematics

By taking spectral norm for both sides of (214) we obtain that

∥∥∥∥∥∥dXU minusmsumi=1

Alowasti (dXU)Ai +

nsumj=1

Blowastj (dXU)Bj

∥∥∥∥∥∥

=

∥∥∥∥∥∥msumi=1

(dAlowast

i

)XUAi +

msumi=1

Alowasti XU(dAi) minus

nsumj=1

(dBlowast

j

)XUBj minus

nsumj=1

Blowastj XU

(dBj

)∥∥∥∥∥∥

le∥∥∥∥∥

msumi=1

(dAlowast

i

)XUAi

∥∥∥∥∥ +∥∥∥∥∥

msumi=1

Alowasti XU(dAi)

∥∥∥∥∥ +∥∥∥∥∥∥

nsumj=1

(dBlowast

j

)XUBj

∥∥∥∥∥∥ +∥∥∥∥∥∥

nsumj=1

Blowastj XU

(dBj

)∥∥∥∥∥∥

lemsumi=1

∥∥(dAlowasti

)XUAi

∥∥ + msumi=1

∥∥Alowasti XU(dAi)

∥∥ + nsumj=1

∥∥∥(dBlowastj

)XUBj

∥∥∥ +nsumj=1

∥∥∥Blowastj XU

(dBj

)∥∥∥

lemsumi=1

∥∥dAlowasti

∥∥XUAi +msumi=1

∥∥Alowasti

∥∥XUdAi +nsumj=1

∥∥∥dBlowastj

∥∥∥XU∥∥Bj

∥∥ + nsumj=1

∥∥∥Blowastj

∥∥∥XU∥∥dBj

∥∥

= 2msumi=1

(AiXUdAi) + 2nsumj=1

(∥∥Bj

∥∥XU∥∥dBj

∥∥)

= 2

⎡⎣ msum

i=1

(AidAi) +nsumj=1

(∥∥Bj

∥∥∥∥dBj

∥∥)⎤⎦XU

(215)

and noting (211) we obtain that

XU le∥∥∥∥∥I + 2

msumi=1

Alowasti Ai

∥∥∥∥∥ le 1 + 2msumi=1

Ai2 (216)

Then

∥∥∥∥∥∥dXU minusmsumi=1

Alowasti (dXU)Ai +

nsumj=1

Blowastj (dXU)Bj

∥∥∥∥∥∥

le 2

⎡⎣ msum

i=1

(AidAi) +nsumj=1

(∥∥Bj

∥∥∥∥dBj

∥∥)⎤⎦XU

le 2

(1 + 2

msumi=1

Ai2

)⎡⎣ msum

i=1

(AidAi) +nsumj=1

(∥∥Bj

∥∥∥∥dBj

∥∥)⎤⎦

(217)

Journal of Applied Mathematics 7∥∥∥∥∥∥dXU minus

msumi=1

Alowasti (dXU)Ai +

nsumj=1

Blowastj (dXU)Bj

∥∥∥∥∥∥

ge dXU minus∥∥∥∥∥

msumi=1

Alowasti (dXU)Ai

∥∥∥∥∥ minus∥∥∥∥∥∥

nsumj=1

Blowastj (dXU)Bj

∥∥∥∥∥∥

ge dXU minusmsumi=1

∥∥Alowasti (dXU)Ai

∥∥ minus nsumj=1

∥∥∥Blowastj (dXU)Bj

∥∥∥

ge dXU minusmsumi=1

∥∥Alowasti

∥∥dXUAi minusnsumj=1

∥∥∥Blowastj

∥∥∥dXU∥∥Bj

∥∥

=

⎛⎝1 minus

msumi=1

Ai2 minusnsumj=1

∥∥Bj

∥∥2

⎞⎠dXU

(218)

Due to (25) we have

1 minusmsumi=1

Ai2 minusnsumj=1

∥∥Bj

∥∥2gt 0 (219)

Combining (217) (218) and noting (219) we have

⎛⎝1 minus

msumi=1

Ai2 minusnsumj=1

∥∥Bj

∥∥2

⎞⎠dXU

le∥∥∥∥∥∥dXU minus

msumi=1

Alowasti (dXU)Ai +

nsumj=1

Blowastj (dXU)Bj

∥∥∥∥∥∥

le 2

(1 + 2

msumi=1

Ai2

)⎡⎣ msum

i=1

(AidAi) +nsumj=1

(∥∥Bj

∥∥∥∥dBj

∥∥)⎤⎦

(220)

which implies that

dXU le2(

1 + 2summ

i=1 Ai2)[summ

i=1(AidAi) +sumn

j=1(∥∥Bj

∥∥∥∥dBj

∥∥)]

1 minussummi=1 Ai2 minussumn

j=1

∥∥Bj

∥∥2 (221)

8 Journal of Applied Mathematics

Theorem 25 Let A1 A2 Am B1 B2 Bn be perturbed matrices of A1 A2 AmB1 B2 Bn in (11) and Δi = Ai minusAi i = 1 2 m Δj = Bj minus Bj j = 1 2 n If

msumi=1

Ai2 lt12

nsumj=1

∥∥Bj

∥∥2lt

12 (222)

2msumi=1

(AiΔi) +msumi=1

Ai2 lt12minus

msumi=1

Ai2 (223)

2nsumj=1

(∥∥Bj

∥∥∥∥Δj

∥∥) + nsumj=1

∥∥Δj

∥∥2lt

12minus

nsumj=1

∥∥Bj

∥∥2 (224)

then (11) and its perturbed equation

X minusmsumi=1

Alowasti XAi +

nsumj=1

Blowastj XBj = I (225)

have unique positive definite solutions XU and XU respectively which satisfy

∥∥∥XU minusXU

∥∥∥ le Serr (226)

where

Serr =2[1 + 2

summi=1 (Ai + Δi)2

][summi=1 (Ai + Δi)2Δi +

sumnj=1(∥∥Bj

∥∥ + ∥∥Δj

∥∥)2∥∥Δj

∥∥]

1 minussummi=1 (Ai + Δi)2 minussumn

j=1(∥∥Bj

∥∥ + ∥∥Δj

∥∥)2

(227)

Proof By (222) and Theorem 24 we know that (11) has a unique positive definite solutionXU And by (223) we have

msumi=1

∥∥∥Ai

∥∥∥2=

msumi=1

Ai + Δi2 lemsumi=1

(Ai + Δi)2

=msumi=1

(Ai2 + 2AiΔi + Δi2

)

=msumi=1

Ai2 + 2msumi=1

(AiΔi) +msumi=1

Δi2

ltmsumi=1

Ai2 +12minus

msumi=1

Ai2 =12

(228)

Journal of Applied Mathematics 9

similarly by (224) we have

nsumj=1

∥∥∥Bj

∥∥∥2lt

12 (229)

By (228) (229) and Theorem 24 we obtain that the perturbed equation (225) has a uniquepositive definite solution XU

Set

Ai(t) = Ai + tΔi Bj(t) = Bj + tΔj t isin [0 1] (230)

then by (223) we have

msumi=1

Ai(t)2 =msumi=1

Ai + tΔi2 lemsumi=1

(Ai + tΔi)2

=msumi=1

(Ai2 + 2tAiΔi + t2Δi2

)

lemsumi=1

(Ai2 + 2AiΔi + Δi2

)

=msumi=1

Ai2 + 2msumi=1

(AiΔi) +msumi=1

Δi2

ltmsumi=1

Ai2 +12minus

msumi=1

Ai2 =12

(231)

similarly by (224) we have

nsumj=1

∥∥Bj(t)∥∥2 =

msumi=1

∥∥Bj + tΔj

∥∥2lt

12 (232)

Therefore by (231) (232) and Theorem 24 we derive that for arbitrary t isin [0 1] thematrix equation

X minusmsumi=1

Alowasti (t)XAi(t) +

nsumj=1

Blowastj (t)XBj(t) = I (233)

has a unique positive definite solution XU(t) especially

XU(0) = XU XU(1) = XU (234)

10 Journal of Applied Mathematics

From Theorem 24 it follows that

∥∥∥XU minusXU

∥∥∥ = XU(1) minusXU(0) =

∥∥∥∥∥int1

0dXU(t)

∥∥∥∥∥ leint1

0dXU(t)

leint1

0

2(

1 + 2summ

i=1 Ai(t)2)[summ

i=1(Ai(t)dAi(t)) +sumn

j=1(∥∥Bj(t)

∥∥∥∥dBj(t)∥∥)]

1 minussummi=1 Ai(t)2 minussumn

j=1

∥∥Bj(t)∥∥2

leint1

0

2(

1 + 2summ

i=1 Ai(t)2)[summ

i=1(Ai(t)Δidt) +sumn

j=1(∥∥Bj(t)

∥∥∥∥Δj

∥∥dt)]

1 minussummi=1 Ai(t)2 minussumn

j=1

∥∥Bj(t)∥∥2

=int1

0

2(

1 + 2summ

i=1 Ai(t)2)[summ

i=1(Ai(t)Δi) +sumn

j=1(∥∥Bj(t)

∥∥∥∥Δj

∥∥)]

1 minussummi=1 Ai(t)2 minussumn

j=1

∥∥Bj(t)∥∥2

dt

(235)

Noting that

Ai(t) = Ai + tΔi le Ai + tΔi i = 1 2 m∥∥Bj(t)

∥∥ =∥∥Bj + tΔi

∥∥ le ∥∥Bj

∥∥ + tΔi j = 1 2 n(236)

and combining Mean Value Theorem of Integration we have

∥∥∥XU minusXU

∥∥∥

leint1

0

2(

1 + 2summ

i=1 Ai(t)2)[summ

i=1(Ai(t)Δi) +sumn

j=1(∥∥Bj(t)

∥∥∥∥Δj

∥∥)]

1 minussummi=1 Ai(t)2 minussumn

j=1

∥∥Bj(t)∥∥2

dt

leint1

0

2[1 + 2

summi=1 (Ai + tΔi)2

][summi=1 (Ai + tΔi)2Δi +

sumnj=1(∥∥Bj

∥∥ + t∥∥Δj

∥∥)2∥∥Δj

∥∥]

1 minussummi=1 (Ai + tΔi)2 minussumn

j=1(∥∥Bj

∥∥ + t∥∥Δj

∥∥)2dt

=2[1 + 2

summi=1 (Ai + ξΔi)2

][summi=1 (Ai + ξΔi)2Δi +

sumnj=1(∥∥Bj

∥∥ + ξ∥∥Δj

∥∥)2∥∥Δj

∥∥]

1 minussummi=1 (Ai + ξΔi)2 minussumn

j=1(∥∥Bj

∥∥ + ξ∥∥Δj

∥∥)2

times (1 minus 0) ξ isin [0 1]

le2[1 + 2

summi=1 (Ai + Δi)2

][summi=1 (Ai + Δi)2Δi +

sumnj=1(∥∥Bj

∥∥ + ∥∥Δj

∥∥)2∥∥Δj

∥∥]

1 minussummi=1 (Ai + Δi)2 minussumn

j=1(∥∥Bj

∥∥ + ∥∥Δj

∥∥)2= Serr

(237)

Journal of Applied Mathematics 11

3 Numerical Experiments

In this section we use a numerical example to confirm the correctness of Theorem 25 and theprecision of the perturbation bound for the unique positive definite solution XU of (11)

Example 31 Consider the symmetric linear matrix equation

X minusAlowast1XA1 minusAlowast

2XA2 + Blowast1XB1 + Blowast

2XB2 = I (31)

and its perturbed equation

X minus Alowast1XA1 minus Alowast

2XA2 + Blowast1XB1 + Blowast

2XB2 = I (32)

where

A1 =

⎛⎝ 002 minus010 minus002

008 minus010 002minus006 minus012 014

⎞⎠ A2 =

⎛⎝ 008 minus010 minus002

008 minus010 002minus006 minus012 014

⎞⎠

B1 =

⎛⎝ 047 002 004

minus010 036 minus002minus004 001 047

⎞⎠ B2 =

⎛⎝010 010 005

015 0275 0075005 005 0175

⎞⎠

A1 = A1 +

⎛⎝ 05 01 minus02

minus04 02 06minus02 01 minus01

⎞⎠ times 10minusj A2 = A2 +

⎛⎝minus04 01 minus02

05 07 minus1311 09 06

⎞⎠ times 10minusj

B1 = B1 +

⎛⎝ 08 02 005

minus02 012 014minus025 minus02 026

⎞⎠ times 10minusj B2 = B2 +

⎛⎝ 02 02 01

minus03 015 minus01501 minus01 025

⎞⎠ times 10minusj j isin N

(33)

It is easy to verify that the conditions (222)ndash(224) are satisfied then (31) and itsperturbed equation (32) have unique positive definite solutions XU and XU respectivelyFrom Berzig [7] it follows that the sequences Xk and Yk generated by the iterative method

X0 = 0 Y0 = 2I

Xk+1 = I +Alowast1XkA1 +Alowast

2XkA2 minus Blowast1YkB1 minus Blowast

2YkB2

Yk+1 = I +Alowast1YkA1 +Alowast

2YkA2 minus Blowast1XkB1 minus Blowast

2XkB2 k = 0 1 2

(34)

both converge to XU Choose τ = 10 times 10minus15 as the termination scalar that is

R(X) =∥∥X minusAlowast

1XA1 minusAlowast2XA2 + Blowast

1XB1 + Blowast2XB2 minus I

∥∥ le τ = 10 times 10minus15 (35)

By using the iterative method (34) we can get the computed solution Xk of (31) SinceR(Xk) lt 10 times 10minus15 then the computed solution Xk has a very high precision For simplicity

12 Journal of Applied Mathematics

Table 1 Numerical results for the different values of j

j 2 3 4 5 6

XU minusXUXU 213 times 10minus4 616 times 10minus6 523 times 10minus8 437 times 10minus10 845 times 10minus12

SerrXU 342 times 10minus4 713 times 10minus6 663 times 10minus8 512 times 10minus10 977 times 10minus12

we write the computed solution as the unique positive definite solution XU Similarly we canalso get the unique positive definite solution XU of the perturbed equation (32)

Some numerical results on the perturbation bounds for the unique positive definitesolution XU are listed in Table 1

From Table 1 we see that Theorem 25 gives a precise perturbation bound for theunique positive definite solution of (31)

4 Conclusion

In this paper we study the matrix equation (11) which arises in solving some nonlinearmatrix equations and the bilinear control system A new method of perturbation analysisis developed for the matrix equation (11) By making use of the matrix differentiationand its elegant properties we derive a precise perturbation bound for the unique positivedefinite solution of (11) A numerical example is presented to illustrate the sharpness of theperturbation bound

Acknowledgments

This research was supported by the National Natural Science Foundation of China(11101100 11171205 11261014 11226323) the Natural Science Foundation of GuangxiProvince (2012GXNSFBA053006 2011GXNSFA018138) the Key Project of Scientific ResearchInnovation Foundation of Shanghai Municipal Education Commission (13ZZ080) theNatural Science Foundation of Shanghai (11ZR1412500) the PhD Programs Foundation ofMinistry of Education of China (20093108110001) the Discipline Project at the correspondinglevel of Shanghai (A 13-0101-12-005) and Shanghai Leading Academic Discipline Project(J50101) The authors wish to thank Professor Y Zhang and the anonymous referees forproviding very useful suggestions for improving this paper The authors also thank DrXiaoshan Chen for discussing the properties of matrix differentiation

References

[1] L A Sakhnovich Interpolation Theory and Its Applications Mathematics and Its Applications KluwerAcademic Publishers Dordrecht The Netherlands 1997

[2] T Damm ldquoDirect methods and ADI-preconditioned Krylov subspace methods for generalizedLyapunov equationsrdquo Numerical Linear Algebra with Applications vol 15 no 9 pp 853ndash871 2008

[3] L Zhang and J Lam ldquoOn H2 model reduction of bilinear systemsrdquo Automatica vol 38 no 2 pp205ndash216 2002

[4] M C B Reurings Symmetric matrix equations [PhD thesis] Vrije Universiteit Amsterdam TheNetherlands 2003

[5] A C M Ran and M C B Reurings ldquoThe symmetric linear matrix equationrdquo Electronic Journal ofLinear Algebra vol 9 pp 93ndash107 2002

Journal of Applied Mathematics 13

[6] A C M Ran and M C B Reurings ldquoA fixed point theorem in partially ordered sets and someapplications to matrix equationsrdquo Proceedings of the American Mathematical Society vol 132 no 5 pp1435ndash1443 2004

[7] M Berzig ldquoSolving a class of matrix equation via Bhaskar-Lakshmikantham coupled fixed pointtheoremrdquo Applied Mathematics Letters vol 25 no 11 pp 1638ndash1643 2012

[8] J Cai and G Chen ldquoSome investigation on Hermitian positive definite solutions of the matrixequation Xs +AlowastXminustA = Qrdquo Linear Algebra and Its Applications vol 430 no 8-9 pp 2448ndash2456 2009

[9] X Duan and A Liao ldquoOn the existence of Hermitian positive definite solutions of the matrix equationXs +AlowastXminustA = Qrdquo Linear Algebra and its Applications vol 429 no 4 pp 673ndash687 2008

[10] X Duan A Liao and B Tang ldquoOn the nonlinear matrix equation X minus summi=1 A

lowasti X

δiAi = Qrdquo LinearAlgebra and its Applications vol 429 no 1 pp 110ndash121 2008

[11] S M El-Sayed and A C M Ran ldquoOn an iteration method for solving a class of nonlinear matrixequationsrdquo SIAM Journal on Matrix Analysis and Applications vol 23 no 3 pp 632ndash645 2001

[12] Z Gajic and M T J Qureshi Lyapunov Matrix Equation in System Stability and Control Academic PressLondon UK 1995

[13] V I Hasanov ldquoNotes on two perturbation estimates of the extreme solutions to the equations X plusmnAlowastXminus1A = Qrdquo Applied Mathematics and Computation vol 216 no 5 pp 1355ndash1362 2010

[14] Z-Y Peng S M El-Sayed and X-l Zhang ldquoIterative methods for the extremal positive definitesolution of the matrix equation X + AlowastXminusαA = Qrdquo Journal of Computational and Applied Mathematicsvol 200 no 2 pp 520ndash527 2007

[15] B R Fang J D Zhou and Y M Li Matrix Theory Tsinghua University Press Beijing China 2006

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2012 Article ID 483624 11 pagesdoi1011552012483624

Research ArticleRefinements of Kantorovich Inequality forHermitian Matrices

Feixiang Chen

School of Mathematics and Statistics Chongqing Three Gorges University WanzhouChongqing 404000 China

Correspondence should be addressed to Feixiang Chen cfx2002126com

Received 28 August 2012 Accepted 5 November 2012

Academic Editor K C Sivakumar

Copyright q 2012 Feixiang Chen This is an open access article distributed under the CreativeCommons Attribution License which permits unrestricted use distribution and reproduction inany medium provided the original work is properly cited

Some new Kantorovich-type inequalities for Hermitian matrix are proposed in this paper Weconsider what happens to these inequalities when the positive definite matrix is allowed to beinvertible and provides refinements of the classical results

1 Introduction and Preliminaries

We first state the well-known Kantorovich inequality for a positive definite Hermite matrix(see [1 2]) let A isin Mn be a positive definite Hermitian matrix with real eigenvalues λ1 leλ2 le middot middot middot le λn Then

1 le xlowastAxxlowastAminus1x le (λ1 + λn)2

4λ1λn (11)

for any x isin Cn x = 1 where Alowast denotes the conjugate transpose of matrix A A matrix

A isin Mn is Hermitian if A = Alowast An equivalent form of this result is incorporated in

0 le xlowastAxxlowastAminus1x minus 1 le (λn minus λ1)2

4λ1λn (12)

for any x isin Cn x = 1

Attributed to Kantorovich the inequality has built up a considerable literatureThis typically comprises generalizations Examples are [3ndash5] for matrix versions Operator

2 Journal of Applied Mathematics

versions are developed in [6 7] Multivariate versions have been useful in statistics to assessthe robustness of least squares see [8 9] and the references therein

Due to the important applications of the original Kantorovich inequality for matrices[10] in Statistics [8 11 12] and Numerical Analysis [13 14] any new inequality of this typewill have a flow of consequences in the areas of applications

Motivated by the interest in both pure and applied mathematics outlined abovewe establish in this paper some improvements of Kantorovich inequalities The classicalKantorovich-type inequalities are modified to apply not only to positive definite but also toinvertible Hermitian matrices As natural tools in deriving the new results the recent Gruss-type inequalities for vectors in inner product in [6 15ndash19] are utilized

To simplify the proof we first introduce some lemmas

2 Lemmas

Let B be a Hermitian matrix with real eigenvalues μ1 le μ2 le middot middot middot le μn if A minus B is positivesemidefinite we write

A ge B (21)

that is λi ge μi i = 1 2 n On Cn we have the standard inner product defined by 〈x y〉 =sumn

i=1 xiyi where x = (x1 xn)lowast isin C

n and y = (y1 yn)lowast isin C

n

Lemma 21 Let a b c and d be real numbers then one has the following inequality

(a2 minus b2

)(c2 minus d2

)le (ac minus bd)2 (22)

Lemma 22 Let A and B be Hermitian matrices if AB = BA then

AB le (A + B)2

4 (23)

Lemma 23 Let A ge 0 B ge 0 if AB = BA then

AB ge 0 (24)

3 Some Results

The following lemmas can be obtained from [16ndash19] by replacing Hilbert space (H 〈middot middot〉) withinner product spaces C

n so we omit the details

Lemma 31 Let u v and e be vectors in Cn and e = 1 If α β δ and γ are real or complex

numbers such that

Relangβe minus u u minus αe

rang ge 0 Relangδe minus v v minus γe

rang ge 0 (31)

Journal of Applied Mathematics 3

then

|〈u v〉 minus 〈u e〉〈e v〉| le 14(β minus α

)(δ minus γ

) minus [Re

langβe minus u u minus αe

rangRe

langδe minus v v minus γe

rang]12 (32)

Lemma 32 With the assumptions in Lemma 31 one has

|〈u v〉 minus 〈u e〉〈e v〉| le 14(β minus α

)(γ minus δ

) minus∣∣∣∣〈u e〉 minus α + β

2

∣∣∣∣∣∣∣∣〈v e〉 minus γ + δ

2

∣∣∣∣ (33)

Lemma 33 With the assumptions in Lemma 31 if Re(βα) ge 0 Re(δγ) ge 0 one has

|〈u v〉 minus 〈u e〉〈e v〉| le∣∣β minus α

∣∣∣∣δ minus γ∣∣

4[Re

(βα

)Re

(δγ

)]12|〈u e〉〈e v〉| (34)

Lemma 34 With the assumptions in Lemma 33 one has

|〈u v〉 minus 〈u e〉〈e v〉|

le(∣∣α + β

∣∣ minus 2[Re

(βα

)]12)(∣∣δ + γ

∣∣ minus 2[Re

(δγ

)]12)12

[|〈u e〉〈e v〉|]12(35)

4 New Kantorovich Inequalities for Hermitian Matrices

For a Hermitian matrix A as in [6] we define the following transform

C(A) = (λnI minusA)(A minus λ1I) (41)

When A is invertible if λ1λn gt 0 then

C(Aminus1

)=(

1λ1

I minusAminus1)(

Aminus1 minus 1λn

I

) (42)

Otherwise λ1λn lt 0 then

C(Aminus1

)=(

1λk+1

I minusAminus1)(

Aminus1 minus 1λk

I

) (43)

where

λ1 le middot middot middot le λk lt 0 lt λk+1 le middot middot middot le λn (44)

From Lemma 23 we can conclude that C(A) ge 0 and C(Aminus1) ge 0

4 Journal of Applied Mathematics

For two Hermitian matrices A and B and x isin Cn x = 1 we define the following

functional

G(ABx) = 〈AxBx〉 minus 〈Ax x〉〈x Bx〉 (45)

When A = B we denote

G(Ax) = Ax2 minus 〈Ax x〉2 (46)

for x isin Cn x = 1

Lemma 41 With notations above and for x isin Cn x = 1 then

0 le 〈C(A)x x〉 le (λn minus λ1)2

4 (47)

if λnλ1 gt 0

0 lelangC(Aminus1

)x x

rangle (λn minus λ1)

2

4(λnλ1)2 (48)

if λnλ1 lt 0

0 lelangC(Aminus1

)x x

rangle (λk+1 minus λk)

2

4(λk+1λk)2 (49)

Proof From C(A) ge 0 then

〈C(A)x x〉 ge 0 (410)

While from Lemma 22 we can get

C(A) = (λnI minusA)(A minus λ1I) le (λn minus λ1)2

4I (411)

Then 〈C(A)x x〉 le (λn minus λ1)24 is straightforward The proof for C(Aminus1) is similar

Lemma 42 With notations above and for x isin Cn x = 1 then

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le G(Ax)G

(Aminus1x

) (412)

Journal of Applied Mathematics 5

Proof Thus

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2

=∣∣∣xlowast

((xlowastAminus1x

)I minusAminus1

)((xlowastAx)I minusA)x

∣∣∣2

le∥∥∥((xlowastAminus1x

)I minusAminus1

)x∥∥∥2

((xlowastAx)I minusA)x2

(413)

while

((xlowastAx)I minusA)x2 = xlowast((xlowastAx)2I minus 2(xlowastAx)A +A2

)x

= xlowastA2x minus (xlowastAx)2

= Ax2 minus 〈Ax x〉2

= G(Ax)

(414)

Similarly we can get ((xlowastAminus1x)I minusAminus1)x2 = G(Aminus1x) then we complete the proof

Theorem 43 Let A B be two Hermitian matrices and C(A) ge 0 C(B) ge 0 are defined as abovethen

|G(ABx)| le 14(λn minus λ1)

(μn minus μ1

) minus [〈C(A)x x〉〈C(B)x x〉]12

|G(ABx)| le 14(λn minus λ1)

(μn minus μ1

) minus∣∣∣∣lang(

A minus λ1 + λn2

)x x

rang∣∣∣∣∣∣∣∣lang(

B minus μ1 + μn

2

)x x

rang∣∣∣∣(415)

for any x isin Cn x = 1

If λ1λn gt 0 μ1μn gt 0 then

|G(ABx)| le (λn minus λ1)(μn minus μ1

)4[(λnλ1)

(μnμ1

)]12|〈Ax x〉〈Bx x〉|

|G(ABx)| le(

|λ1 + λn| minus 2(λ1λn)12

)(∣∣μ1 + μn

∣∣ minus 2(μ1μn

)12)12

[|〈Ax x〉〈Bx x〉|]12

(416)

for any x isin Cn x = 1

Proof The proof follows by Lemmas 31 32 33 and 34 on choosing u = Ax v = Bx ande = x β = λn α = λ1 δ = μn and γ = μ1 x isin C

n x = 1 respectively

6 Journal of Applied Mathematics

Corollary 44 Let A be a Hermitian matrices and C(A) ge 0 is defined as above then

|G(Ax)| le 14(λn minus λ1)

2 minus 〈C(A)x x〉 (417)

|G(Ax)| le 14(λn minus λ1)

2 minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣2

(418)

for any x isin Cn x = 1

If λ1λn gt 0 then

|G(Ax)| le (λn minus λ1)2

4(λnλ1)|〈Ax x〉|2 (419)

|G(Ax)| le(|λ1 + λn| minus 2

radicλ1λn

)|〈Ax x〉| (420)

for any x isin Cn x = 1

Proof The proof follows by Theorem 43 on choosing A = B respectively

Corollary 45 Let A be a Hermitian matrices and C(Aminus1) ge 0 is defined as above then one has thefollowing

If λ1λn gt 0 then

∣∣∣G(Aminus1x

)∣∣∣ le (λn minus λ1)2

4(λnλ1)2

minuslangC(Aminus1

)x x

rang (421)

∣∣∣G(Aminus1x

)∣∣∣ le (λn minus λ1)2

4(λnλ1)2

minus∣∣∣∣lang(

Aminus1 minus λ1 + λn2λnλ1

I

)x x

rang∣∣∣∣2

(422)

∣∣∣G(Aminus1x

)∣∣∣ le (λn minus λ1)2

4λnλ1

∣∣∣langAminus1x xrang∣∣∣2

(423)

∣∣∣G(Aminus1x

)∣∣∣ le(

|λ1 + λn|λ1λn

minus 21radicλ1λn

)∣∣∣langAminus1x xrang∣∣∣ (424)

for any x isin Cn x = 1

If λ1λn lt 0 then

∣∣∣G(Aminus1x

)∣∣∣ le (λk minus λk+1)2

4(λkλk+1)2

minuslangC(Aminus1

)x x

rang (425)

Journal of Applied Mathematics 7

where C(Aminus1) = ((1λk+1)I minusAminus1)(Aminus1 minus (1λk)I) and

∣∣∣G(Aminus1x

)∣∣∣ le (λk minus λk+1)2

4(λkλk+1)2

minus∣∣∣∣lang(

Aminus1 minus λk+1 + λk2λkλk+1

I

)x x

rang∣∣∣∣2

(426)

for any x isin Cn x = 1

Proof The proof follows by Corollary 44 by replacing A with Aminus1 respectively

Theorem 46 LetA be an ntimesn invertible Hermitian matrix with real eigenvalues λ1 le λ2 le middot middot middot le λnthen one has the following

If λ1λn gt 0 then

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le (λn minus λ1)

2

4(λnλ1)minusradic〈C(A)x x〉langC(Aminus1

)x x

rang (427)

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le (λn minus λ1)

2

4(λnλ1)minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣∣∣∣∣lang(

Aminus1 minus λ1 + λn2λnλ1

I

)x x

rang∣∣∣∣(428)

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le (λn minus λ1)

2

4(λnλ1)2 〈Ax x〉

langAminus1x x

rang (429)

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le

(radic|λn| minus

radic|λ1|

)2

radicλnλ1

radic〈Ax x〉〈Aminus1x x〉 (430)

for any x isin Cn x = 1

If λ1λn lt 0 then

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le (λn minus λ1)(λk+1 minus λk)

4|λkλk+1| minusradic〈C(A)x x〉langC(Aminus1

)x x

rang (431)

where C(Aminus1) = ((1λk+1)I minusAminus1)(Aminus1 minus (1λk)I)

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le (λn minus λ1)(λk+1 minus λk)

4|λkλk+1|

minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣∣∣∣∣lang(

Aminus1 minus λk + λk+1

2λ1λk+1I

)x x

rang∣∣∣∣(432)

Proof Considering

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le G(Ax)G

(Aminus1x

) (433)

8 Journal of Applied Mathematics

When λ1λn gt 0 from (417) and (421) we get

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le

14(λn minus λ1)

2 minus 〈C(A)x x〉

(λn minus λ1)2

4(λnλ1)2

minuslangC(Aminus1

)x x

rang (434)

From (a2 minus b2)(c2 minus d2) le (ac minus bd)2 we have

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le

(λn minus λ1)

2

4(λnλ1)minusradic[〈C(A)x x〉langC(Aminus1

)x x

rang]2

(435)

then the conclusion (427) holdsSimilarly from (418) and (422) we get

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le

14(λn minus λ1)

2 minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣2

times(λn minus λ1)

2

4(λnλ1)2

minus∣∣∣∣lang(

Aminus1 minus λ1 + λn2λnλ1

I

)x x

rang∣∣∣∣2

le(λn minus λ1)

2

4(λnλ1)minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣∣∣∣∣lang(

Aminus1 minus λ1 + λn2λnλ1

I

)x x

rang∣∣∣∣2

(436)

then the conclusion (428) holdsFrom (419) and (423) we get

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le (λn minus λ1)

2

4(λnλ1)|〈Ax x〉|2 (λn minus λ1)

2

4(λnλ1)

∣∣∣langAminus1x xrang∣∣∣2

(437)

then the conclusion (429) holdsFrom (420) and (424) we get

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le

(|λ1 + λn| minus 2

radicλ1λn

)|〈Ax x〉|

(|λ1 + λn|λ1λn

minus 21radicλ1λn

)∣∣∣langAminus1x xrang∣∣∣(438)

then the conclusion (430) holdsWhen λ1λn lt 0 from (417) and (425) we get

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2le

14(λn minus λ1)

2 minus 〈C(A)x x〉

(λk minus λk+1)2

4(λkλk+1)2

minuslangC(Aminus1

)x x

rang (439)

then the conclusion (431) holds

Journal of Applied Mathematics 9

From (418) and (426) we get

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le

14(λn minus λ1)

2 minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣2

times(λk minus λk+1)

2

4(λkλk+1)2

minus∣∣∣∣lang(

Aminus1 minus λk+1 + λk2λkλk+1

I

)x x

rang∣∣∣∣2

le(λn minus λ1)(λk+1 minus λk)

4|λkλk+1|

minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣∣∣∣∣lang(

Aminus1 minus λk + λk+1

2λ1λk+1I

)x x

rang∣∣∣∣2

(440)

then the conclusion (432) holds

Corollary 47 With the notations above for any x isin Cn x = 1 one lets

|G1(Ax)| = 14(λn minus λ1)

2 minus 〈C(A)x x〉

|G2(Ax)| = 14(λn minus λ1)

2 minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣2

(441)

If λ1λn gt 0 one lets

|G3(Ax)| = (λn minus λ1)2

4(λnλ1)|〈Ax x〉|2

|G4(Ax)| =(|λ1 + λn| minus 2

radicλ1λn

)|〈Ax x〉|

(442)

If λ1λn gt 0

∣∣∣G1(Aminus1x

)∣∣∣ = (λn minus λ1)2

4(λnλ1)2

minuslangC(Aminus1

)x x

rang

∣∣∣G2(Aminus1x

)∣∣∣ = (λn minus λ1)2

4(λnλ1)2

minus∣∣∣∣lang(

Aminus1 minus λ1 + λn2λnλ1

I

)x x

rang∣∣∣∣2

∣∣∣G3(Aminus1x

)∣∣∣ = (λn minus λ1)2

4λnλ1

∣∣∣langAminus1x xrang∣∣∣2

∣∣∣G4(Aminus1x

)∣∣∣ =(

|λ1 + λn|λ1λn

minus 21radicλ1λn

)∣∣∣langAminus1x xrang∣∣∣

(443)

10 Journal of Applied Mathematics

If λ1λn lt 0 then

∣∣∣G5(Aminus1x

)∣∣∣ = (λk minus λk+1)2

4(λkλk+1)2

minuslangC(Aminus1

)x x

rang (444)

where C(Aminus1) = ((1λk+1)I minusAminus1)(Aminus1 minus (1λk)I) and

∣∣∣G6(Aminus1x

)∣∣∣ = (λk minus λk+1)2

4(λkλk+1)2

minus∣∣∣∣lang(

Aminus1 minus λk+1 + λk2λkλk+1

I

)x x

rang∣∣∣∣2

(445)

Then one has the followingIf λ1λn gt 0

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le

radicGlowast(Ax)Glowast(Aminus1x

) (446)

where

Glowast(Ax) = minG1(Ax) G2(Ax) G3(Ax) G4(Ax)

Glowast(Ax) = minG1

(Aminus11x

) G2

(Aminus1x

) G3

(Aminus1x

) G4

(Aminus1x

)

(447)

If λ1λn lt 0

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le

radicG(Ax)Gbull(Aminus1x

) (448)

where

G(Ax) = minG1(Ax) G2(Ax)

Gbull(Ax) = minG5

(Aminus1x

) G6

(Aminus1x

)

(449)

Proof The proof follows from that the conclusions in Corollaries 44 and 45 are independent

Remark 48 It is easy to see that if λ1 gt 0 λn gt 0 our result coincides with the inequalityof operator versions in [6] So we conclude that our results give an improvement of theKantorovich inequality [6] that applies to all invertible Hermite matrices

5 Conclusion

In this paper we introduce some new Kantorovich-type inequalities for the invertibleHermitian matrices Inequalities (427) and (431) are the same as [4] but our proof is simpleIn Theorem 46 if λ1 gt 0 λn gt 0 the results are similar to the well-known Kantorovich-typeinequalities for operators in [6] Moreover for any invertible Hermitian matrix there exists asimilar inequality

Journal of Applied Mathematics 11

Acknowledgments

The authors are grateful to the associate editor and two anonymous referees whose commentshave helped to improve the final version of the present work This work was supported by theNatural Science Foundation of Chongqing Municipal Education Commission (no KJ091104)

References

[1] R A Horn and C R Johnson Matrix Analysis Cambridge University Press Cambridge UK 1985[2] A S Householder The Theory of Matrices in Numerical Analysis Blaisdell New York NY USA 1964[3] S Liu and H Neudecker ldquoSeveral matrix Kantorovich-type inequalitiesrdquo Journal of Mathematical

Analysis and Applications vol 197 no 1 pp 23ndash26 1996[4] A W Marshall and I Olkin ldquoMatrix versions of the Cauchy and Kantorovich inequalitiesrdquo

Aequationes Mathematicae vol 40 no 1 pp 89ndash93 1990[5] J K Baksalary and S Puntanen ldquoGeneralized matrix versions of the Cauchy-Schwarz and

Kantorovich inequalitiesrdquo Aequationes Mathematicae vol 41 no 1 pp 103ndash110 1991[6] S S Dragomir ldquoNew inequalities of the Kantorovich type for bounded linear operators in Hilbert

spacesrdquo Linear Algebra and its Applications vol 428 no 11-12 pp 2750ndash2760 2008[7] P G Spain ldquoOperator versions of the Kantorovich inequalityrdquo Proceedings of the American

Mathematical Society vol 124 no 9 pp 2813ndash2819 1996[8] S Liu and H Neudecker ldquoKantorovich inequalities and efficiency comparisons for several classes of

estimators in linear modelsrdquo Statistica Neerlandica vol 51 no 3 pp 345ndash355 1997[9] P Bloomfield and G S Watson ldquoThe inefficiency of least squaresrdquo Biometrika vol 62 pp 121ndash128

1975[10] L V Kantorovic ldquoFunctional analysis and applied mathematicsrdquo Uspekhi Matematicheskikh Nauk vol

3 no 6(28) pp 89ndash185 1948 (Russian)[11] C G Khatri and C R Rao ldquoSome extensions of the Kantorovich inequality and statistical

applicationsrdquo Journal of Multivariate Analysis vol 11 no 4 pp 498ndash505 1981[12] C T Lin ldquoExtrema of quadratic forms and statistical applicationsrdquo Communications in Statistics A

vol 13 no 12 pp 1517ndash1520 1984[13] A Galantai ldquoA study of Auchmutyrsquos error estimaterdquo Computers amp Mathematics with Applications vol

42 no 8-9 pp 1093ndash1102 2001[14] P D Robinson and A J Wathen ldquoVariational bounds on the entries of the inverse of a matrixrdquo IMA

Journal of Numerical Analysis vol 12 no 4 pp 463ndash486 1992[15] S S Dragomir ldquoA generalization of Grussrsquos inequality in inner product spaces and applicationsrdquo

Journal of Mathematical Analysis and Applications vol 237 no 1 pp 74ndash82 1999[16] S S Dragomir ldquoSome Gruss type inequalities in inner product spacesrdquo Journal of Inequalities in Pure

and Applied Mathematics vol 4 no 2 article 42 2003[17] S S Dragomir ldquoOn Bessel and Gruss inequalities for orthonormal families in inner product spacesrdquo

Bulletin of the Australian Mathematical Society vol 69 no 2 pp 327ndash340 2004[18] S S Dragomir ldquoReverses of Schwarz triangle and Bessel inequalities in inner product spacesrdquo Journal

of Inequalities in Pure and Applied Mathematics vol 5 no 3 article 76 2004[19] S S Dragomir ldquoReverses of the Schwarz inequality generalising a Klamkin-McLenaghan resultrdquo

Bulletin of the Australian Mathematical Society vol 73 no 1 pp 69ndash78 2006[20] Z Liu L Lu and K Wang ldquoOn several matrix Kantorovich-type inequalitiesrdquo Journal of Inequalities

and Applications vol 2010 Article ID 571629 5 pages 2010

Copyright copy 2013 Hindawi Publishing Corporation All rights reserved

This is a special issue published in ldquoJournal of Applied Mathematicsrdquo All articles are open access articles distributed under the CreativeCommons Attribution License which permits unrestricted use distribution and reproduction in any medium provided the originalwork is properly cited

Editorial Board

Saeid Abbasbandy IranMina B Abd-El-Malek EgyptMohamed A Abdou EgyptSubhas Abel IndiaMostafa Adimy FranceCarlos J S Alves PortugalMohamad Alwash USAIgor Andrianov GermanySabri Arik TurkeyFrancis TK Au Hong KongOlivier Bahn CanadaRoberto Barrio SpainAlfredo Bellen ItalyJafar Biazar IranHester Bijl The NetherlandsAnjan Biswas Saudi ArabiaStephane PA Bordas USAJames Robert Buchanan USAAlberto Cabada SpainXiao Chuan Cai USAJinde Cao ChinaAlexandre Carvalho BrazilSong Cen ChinaQianshun S Chang ChinaTai-Ping Chang TaiwanShih-sen Chang ChinaRushan Chen ChinaXinfu Chen USAKe Chen UKEric Cheng Hong KongFrancisco Chiclana UKJen-Tzung Chien TaiwanC S Chien TaiwanHan H Choi Republic of KoreaTin-Tai Chow ChinaM S H Chowdhury MalaysiaCarlos Conca ChileVitor Costa PortugalLivija Cveticanin SerbiaEric de Sturler USAOrazio Descalzi ChileKai Diethelm GermanyVit Dolejsi Czech RepublicBo-Qing Dong ChinaMagdy A Ezzat Egypt

Meng Fan ChinaYa Ping Fang ChinaAntonio J M Ferreira PortugalMichel Fliess FranceM A Fontelos SpainHuijun Gao ChinaBernard J Geurts The NetherlandsJamshid Ghaboussi USAPablo Gonzalez-Vera SpainLaurent Gosse ItalyK S Govinder South AfricaJose L Gracia SpainYuantong Gu AustraliaZhihong GUAN ChinaNicola Guglielmi ItalyFrederico G Guimaraes BrazilVijay Gupta IndiaBo Han ChinaMaoan Han ChinaPierre Hansen CanadaFerenc Hartung HungaryXiaoqiao He Hong KongLuis Javier Herrera SpainJ Hoenderkamp The NetherlandsYing Hu FranceNing Hu JapanZhilong L Huang ChinaKazufumi Ito USATakeshi Iwamoto JapanGeorge Jaiani GeorgiaZhongxiao Jia ChinaTarun Kant IndiaIdo Kanter IsraelAbdul Hamid Kara South AfricaHamid Reza Karimi NorwayJae-Wook Kim UKJong Hae Kim Republic of KoreaKazutake Komori JapanFanrong Kong USAVadim Krysko RussiaJin L Kuang SingaporeMiroslaw Lachowicz PolandHak-Keung Lam UKTak-Wah Lam Hong KongPGL Leach Cyprus

Yongkun Li ChinaWan-Tong Li ChinaJ Liang ChinaChing-Jong Liao TaiwanChong Lin ChinaMingzhu Liu ChinaChein-Shan Liu TaiwanKang Liu USAYansheng Liu ChinaFawang Liu AustraliaShutian Liu ChinaZhijun Liu ChinaJulian Lopez-Gomez SpainShiping Lu ChinaGert Lube GermanyNazim Idrisoglu Mahmudov TurkeyOluwole Daniel Makinde South AfricaFrancisco J Marcellan SpainGuiomar Martın-Herran SpainNicola Mastronardi ItalyMichael McAleer The NetherlandsStephane Metens FranceMichael Meylan AustraliaAlain Miranville FranceRam N Mohapatra USAJaime E Munoz Rivera BrazilJavier Murillo SpainRoberto Natalini ItalySrinivasan Natesan IndiaJiri Nedoma Czech RepublicJianlei Niu Hong KongRoger Ohayon FranceJavier Oliver SpainDonal OrsquoRegan IrelandMartin Ostoja-Starzewski USATurgut Ozis TurkeyClaudio Padra ArgentinaReinaldo Martinez Palhares BrazilFrancesco Pellicano ItalyJuan Manuel Pena SpainRicardo Perera SpainMalgorzata Peszynska USAJames F Peters CanadaMark A Petersen South AfricaMiodrag Petkovic Serbia

Vu Ngoc Phat VietnamAndrew Pickering SpainHector Pomares SpainMaurizio Porfiri USAMario Primicerio ItalyMorteza Rafei The NetherlandsRoberto Reno ItalyJacek Rokicki PolandDirk Roose BelgiumCarla Roque PortugalDebasish Roy IndiaSamir H Saker EgyptMarcelo A Savi BrazilWolfgang Schmidt GermanyEckart Schnack GermanyMehmet Sezer TurkeyNaseer Shahzad Saudi ArabiaFatemeh Shakeri IranJian Hua Shen ChinaHui-Shen Shen ChinaFernando Simoes PortugalTheodore E Simos Greece

Abdel-Maksoud A Soliman EgyptXinyu Song ChinaQiankun Song ChinaYuri N Sotskov BelarusPeter J C Spreij The NetherlandsNiclas Stromberg SwedenRay KL Su Hong KongJitao Sun ChinaWenyu Sun ChinaXianHua Tang ChinaAlexander Timokha NorwayMariano Torrisi ItalyJung-Fa Tsai TaiwanCh Tsitouras GreeceKuppalapalle Vajravelu USAAlvaro Valencia ChileErik Van Vleck USAEzio Venturino ItalyJesus Vigo-Aguiar SpainMichael N Vrahatis GreeceBaolin Wang ChinaMingxin Wang China

Qing-Wen Wang ChinaGuangchen Wang ChinaJunjie Wei ChinaLi Weili ChinaMartin Weiser GermanyFrank Werner GermanyShanhe Wu ChinaDongmei Xiao ChinaGongnan Xie ChinaYuesheng Xu USASuh-Yuh Yang TaiwanBo Yu ChinaJinyun Yuan BrazilAlejandro Zarzo SpainGuisheng Zhai JapanJianming Zhan ChinaZhihua Zhang ChinaJingxin Zhang AustraliaShan Zhao USAChongbin Zhao AustraliaRenat Zhdanov USAHongping Zhu China

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On Nonnegative Moore-Penrose Inverses of Perturbed Matrices Shani Jose and K C SivakumarVolume 2013 Article ID 680975 7 pages

A New Construction of Multisender Authentication Codes from Polynomials over Finite FieldsXiuli WangVolume 2013 Article ID 320392 7 pages

Geometrical and Spectral Properties of the Orthogonal Projections of the Identity Luis GonzalezAntonio Suarez and Dolores GarcıaVolume 2013 Article ID 435730 8 pages

A Parameterized Splitting Preconditioner for Generalized Saddle Point ProblemsWei-Hua Luo and Ting-Zhu HuangVolume 2013 Article ID 489295 6 pages

Solving Optimization Problems on Hermitian Matrix Functions with ApplicationsXiang Zhang and Shu-Wen XiangVolume 2013 Article ID 593549 11 pages

Left and Right Inverse Eigenpairs Problem for κ-Hermitian Matrices Fan-Liang Li Xi-Yan Huand Lei ZhangVolume 2013 Article ID 230408 6 pages

Completing a 2times 2 Block Matrix of Real Quaternions with a Partial Specified InverseYong Lin and Qing-Wen WangVolume 2013 Article ID 271978 5 pages

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Contents

On the Hermitian R-Conjugate Solution of a System of Matrix Equations Chang-Zhou DongQing-Wen Wang and Yu-Ping ZhangVolume 2012 Article ID 398085 14 pages

Perturbation Analysis for the Matrix Equation X minussummi=1 A

lowasti XAi +

sumnj=1 B

lowastj XBj = I

Xue-Feng Duan and Qing-Wen WangVolume 2012 Article ID 784620 13 pages

Refinements of Kantorovich Inequality for Hermitian Matrices Feixiang ChenVolume 2012 Article ID 483624 11 pages

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 156136 3 pageshttpdxdoiorg1011552013156136

EditorialAdvances in Matrices Finite and Infinite with Applications

P N Shivakumar1 Panayiotis Psarrakos2 K C Sivakumar3 and Yang Zhang1

1 Department of Mathematics University of Manitoba Winnipeg MB Canada R3T 2N22Department of Mathematics School of Applied Mathematical and Physical SciencesNational Technical Univeristy of Athens Zografou Campus 15780 Athens Greece

3 Department of Mathematics Indian Institute of Technology Madras Chennai 600 036 India

Correspondence should be addressed to P N Shivakumar shivakuccumanitobaca

Received 13 June 2013 Accepted 13 June 2013

Copyright copy 2013 P N Shivakumar et al This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

In the mathematical modeling of many problems includingbut not limited to physical sciences biological sciencesengineering image processing computer graphics medicalimaging and even social sciences lately solution methodsmost frequently involve matrix equations In conjunctionwith powerful computing machines complex numericalcalculations employing matrix methods could be performedsometimes even in real time It is well known that infinitematrices arise more naturally than finite matrices and havea colorful history in development from sequences seriesand quadratic forms Modern viewpoint considers infinitematrices more as operators defined between certain spe-cific infinite-dimensional normed spaces Banach spacesor Hilbert spaces Giant and rapid advancements in thetheory and applications of finite matrices have been madeover the past several decades These developments havebeen well documented in many books monographs andresearch journals The present topics of research includediagonally dominantmatrices theirmany extensions inverseof matrices their recursive computation singular matricesgeneralized inverses inverse positive matrices with specificemphasis on their applications to economics like the Leon-tief input-output models and use of finite difference andfinite element methods in the solution of partial differentialequations perturbationmethods and eigenvalue problems ofinterest and importance to numerical analysts statisticiansphysical scientists and engineers to name a few

The aim of this special issue is to announce certainrecent advancements in matrices finite and infinite andtheir applications For a review of infinite matrices andapplications see P N Shivakumar and K C Sivakumar

(Linear Algebra Appl 2009) Specifically an example of anapplication can be found in the classical problem ldquoshapeof a drumrdquo in P N Shivakumar W Yan and Y Zhang(WSES transactions in Mathematics 2011) The focal themesof this special issue are inverse positivematrices including119872-matrices applications of operator theorymatrix perturbationtheory and pseudospectra matrix functions and generalizedeigenvalue problems and inverse problems including scat-tering and matrices over quaternions In the following wepresent a brief overview of a few of the central topics of thisspecial issue

For119872-matrices let us start by mentioning the most citedbooks byR Berman andR J Plemmons (SIAM 1994) and byR A Horn and C R Johnson (Cambridge 1994) as excellentsources One of the most important properties of invertible119872-matrices is that their inverses are nonnegative There areseveral works that have considered generalizations of someof the important properties of 119872-matrices We only pointout a few of them here The article by D Mishra and K CSivakumar (Oper Matrices 2012) considers generalizationsof inverse nonnegativity to the Moore-Penrose inverse whilemore general classes of matrices are the objects of discussionin D Mishra and K C Sivakumar (Linear Algebra Appl2012) and A N Sushama K Premakumari and K CSivakumar (Electron J Linear Algebra 2012)

A brief description of the papers in the issue is as followsIn the work of Z Zhang the problem of solving cer-

tain optimization problems on Hermitian matrix functionsis studied Specifically the author considers the extremalinertias and ranks of a given function 119891(119883 119884) C119898times119899 rarr

C119898times119899 (which is defined in terms of matrices 119860 119861 119862 and 119863

2 Journal of Applied Mathematics

inC119898times119899) subject to119883 119884 satisfying certain matrix equationsAs applications the author presents necessary and sufficientconditions for 119891 to be positive definite positive semidefiniteand so forth As another application the author presents acharacterization for 119891(119883 119884) = 0 again subject to 119883 119884satisfying certain matrix equations

The task of determining parametrized splitting precon-ditioners for certain generalized saddle point problems isundertaken byW-H Luo and T-Z Huang in their workThewell-known and very widely applicable Sherman-Morrison-Woodbury formula for the sum of matrices is used indetermining a preconditioner for linear equations wherethe coefficient matrix may be singular and nonsymmetricIllustrations of the procedure are provided for Stokes andOseen problems

The work reported by A Gonzalez A Suarez and DGarcia deals with the problem of estimating the Frobeniuscondition of the matrix 119860119873 given a matrix 119860 isin R119899times119899 where119873 satisfies the condition that 119873 = argmin

119883isin119878119868 minus 119860119883

119865

Here 119878 is a certain subspace of R119899 and sdot 119865denotes the

Frobenius norm The authors derive certain spectral andgeometrical properties of the preconditioner119873 as above

Let 119861 119862 119863 and 119864 be (not necessarily square) matriceswith quaternion entries so that the matrix 119872 = (

119860 119861

119862 119863)

is square Y Lin and Q-W Wang present necessary andsufficient conditions so that the top left block 119860 of119872 existssatisfying the conditions that119872 is nonsingular and thematrix119864 is the top left block of119872minus1 A general expression for suchan 119860 when it exists is obtained A numerical illustration ispresented

For a complex hermitian 119860 of order 119899 with eigenvalues1205821le 1205822le sdot sdot sdot le 120582

119899 the celebrated Kantorovich inequality is

1 le 119909lowast

119860119909119909lowast

119860minus1

119909 le (120582119899+1205821)2

41205821120582119899 for all unit vectors 119909 isin

C119899 Equivalently 0 le 119909lowast119860119909119909lowast119860minus1119909minus1 le (120582119899minus1205821)2

41205821120582119899 for

all unit vectors 119909 isin C119899 F Chen in his work on refinements ofKantorovich inequality for hermitian matrices obtains somerefinements of the second inequality where the proofs useonly elementary ideas (httpwwwmathntuagrsimppsarr)

It is pertinent to point out that among other things aninteresting generalization of 119872-matrices is obtained by SJose and K C Sivakumar The authors consider the prob-lem of nonnegativity of the Moore-Penrose inverse of aperturbation of the form 119860minus119883119866119884

119879 given that 119860dagger ge 0 Usinga generalized version of the Sherman-Morrison-Woodburyformula conditions for (119860 minus 119883119866119884

119879

)dagger to be nonnegative

are derived Applications of the results are presented brieflyIterative versions of the results are also studied

It is well known that the concept of quaternions wasintroduced by Hamilton in 1843 and has played an importantrole in contemporary mathematics such as noncommuta-tive algebra analysis and topology Nowadays quaternionmatrices are widely and heavily used in computer sciencequantum physics altitude control robotics signal and colourimage processing and so on Many questions can be reducedto quaternion matrices and solving quaternion matrix equa-tionsThis special issue has provided an excellent opportunityfor researchers to report their recent results

Let H119899times119899 denote the space of all 119899 times 119899 matrices whoseentries are quaternions Let 119876 isin H119899times119899 be Hermitian and self-invertible 119883 isin H119899times119899 is called reflexive (relative to 119876) if 119883 =

119876119883119876 The authors F-L Li X-Y Hu and L Zhang presentan efficient algorithm for finding the reflexive solution of thequaternion matrix equation 119860119883119861 + 119862119883

lowast

119863 = 119865 They alsoshow that given an appropriate initialmatrix their procedureyields the least reflexive solutionwith respect to the Frobeniusnorm

In the work on the ranks of a constrained hermitianmatrix expression S W Yu gives formulas for the maximaland minimal ranks of the quaternion Hermitian matrixexpression 119862

4minus 1198604119883119860lowast

4 where 119883 is a Hermitian solution

to quaternion matrix equations 1198601119883 = 119862

1 1198831198611= 1198622 and

1198603119883119860lowast

3= 1198623 and also applies this result to some special

system of quaternion matrix equationsY Yao reports his findings on the optimization on ranks

and inertias of a quadratic hermitian matrix function Hepresents one solution for optimization problems on the ranksand inertias of the quadratic Hermitian matrix function 119876 minus119883119875119883lowast subject to a consistent system of matrix equations

119860119883 = 119862 and 119883119861 = 119863 As applications he derives necessaryand sufficient conditions for the solvability to the systems ofmatrix equations and matrix inequalities 119860119883 = 119862 119883119861 = 119863and119883119875119883lowast = (gt lt ge le)119876

Let 119877 be a nontrivial real symmetric involution matrixA complex matrix 119860 is called 119877-conjugate if 119860 = 119877119860119877In the work on the hermitian 119877-conjugate solution of asystem of matrix equations C-Z Dong Q-W Wang andY-P Zhang present necessary and sufficient conditions forthe existence of the hermitian 119877-conjugate solution to thesystem of complex matrix equations 119860119883 = 119862 and 119883119861 = 119863An expression of the Hermitian 119877-conjugate solution is alsopresented

A perturbation analysis for the matrix equation 119883 minus

sum119898

119894=1119860lowast

119894119883119860119894+ sum119899

119895=1119861lowast

119895119883119861119895= 119868 is undertaken by X-F Duan

and Q-W Wang They give a precise perturbation bound fora positive definite solution A numerical example is presentedto illustrate the sharpness of the perturbation bound

Linear matrix equations and their optimal approxima-tion problems have great applications in structural designbiology statistics control theory and linear optimal controland as a consequence they have attracted the attention ofseveral researchers for many decades In the work of Q-WWang and J Yu necessary and sufficient conditions of and theexpressions for orthogonal solutions symmetric orthogonalsolutions and skew-symmetric orthogonal solutions of thesystem of matrix equations 119860119883 = 119861 and 119883119862 = 119863 arederived When the solvability conditions are not satisfied theleast squares symmetric orthogonal solutions and the leastsquares skew-symmetric orthogonal solutions are obtainedA methodology is provided to compute the least squaressymmetric orthogonal solutions and an illustrative exampleis given to illustrate the proposed algorithm

Many real-world engineering systems are too complex tobe defined in precise terms and in many matrix equationssome or all of the system parameters are vague or impreciseThus solving a fuzzy matrix equation becomes important In

Journal of Applied Mathematics 3

the work reported by X Guo and D Shang the fuzzy matrixequation 119860 otimes 119883 otimes 119861 = 119862 in which 119860 119861 and 119862 are 119898 times 119898119899 times 119899 and 119898 times 119899 nonnegative LR fuzzy numbers matricesrespectively is investigated This fuzzy matrix system whichhas a wide use in control theory and control engineering isextended into three crisp systems of linear matrix equationsaccording to the arithmetic operations of LR fuzzy numbersBased on the pseudoinversematrix a computingmodel to thepositive fully fuzzy linear matrix equation is provided andthe fuzzy approximate solution of original fuzzy systems isobtained by solving the crisp linear matrix systems In addi-tion the existence condition of nonnegative fuzzy solutionis discussed and numerical examples are given to illustratethe proposed method Overall the technique of LR fuzzynumbers and their operations appears to be a strong tool ininvestigating fully fuzzy matrix equations

Left and right inverse eigenpairs problem arises in anatural way when studying spectral perturbations ofmatricesand relative recursive problems F-L Li X-Y Hu and LZhang consider in their work left and right inverse eigenpairsproblem for 119896-hermitian matrices and its optimal approx-imate problem The class of 119896-hermitian matrices containshermitian and perhermitian matrices and has numerousapplications in engineering and statistics Based on thespecial properties of 119896-hermitian matrices and their left andright eigenpairs an equivalent problem is obtained Thencombining a new inner product of matrices necessary andsufficient conditions for the solvability of the problem aregiven and its general solutions are derived Furthermorethe optimal approximate solution a calculation procedure tocompute the unique optimal approximation and numericalexperiments are provided

Finally let us point out a work on algebraic coding theoryRecall that a communication system in which a receivercan verify the authenticity of a message sent by a group ofsenders is referred to as a multisender authentication codeIn the work reported by X Wang the author constructsmulti-sender authentication codes using polynomials overfinite fields Here certain important parameters and theprobabilities of deceptions of these codes are also computed

Acknowledgments

We wish to thank all the authors for their contributions andthe reviewers for their valuable comments in bringing thespecial issue to fruition

P N ShivakumarPanayiotis Psarrakos

K C SivakumarYang Zhang

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 680975 7 pageshttpdxdoiorg1011552013680975

Research ArticleOn Nonnegative Moore-Penrose Inverses of Perturbed Matrices

Shani Jose and K C Sivakumar

Department of Mathematics Indian Institute of Technology Madras Chennai Tamil Nadu 600036 India

Correspondence should be addressed to K C Sivakumar kcskumariitmacin

Received 22 April 2013 Accepted 7 May 2013

Academic Editor Yang Zhang

Copyright copy 2013 S Jose and K C Sivakumar This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

Nonnegativity of theMoore-Penrose inverse of a perturbation of the form119860minus119883119866119884119879 is consideredwhen119860dagger ge 0 Using a generalizedversion of the Sherman-Morrison-Woodbury formula conditions for (119860 minus 119883119866119884119879)dagger to be nonnegative are derived Applications ofthe results are presented briefly Iterative versions of the results are also studied

1 Introduction

We consider the problem of characterizing nonnegativityof the Moore-Penrose inverse for matrix perturbations ofthe type 119860 minus 119883119866119884

119879 when the Moore-Penrose inverse of119860 is nonnegative Here we say that a matrix 119861 = (119887

119894119895) is

nonnegative and denote it by 119861 ge 0 if 119887119894119895ge 0 forall119894 119895 This

problemwasmotivated by the results in [1] where the authorsconsider an 119872-matrix 119860 and find sufficient conditions forthe perturbed matrix (119860 minus 119883119884119879) to be an119872-matrix Let usrecall that a matrix 119861 = (119887

119894119895) is said to 119885-matrix if 119887

119894119895le 0

for all 119894 119895 119894 = 119895 An119872-matrix is a nonsingular 119885-matrix withnonnegative inverse The authors in [1] use the well-knownSherman-Morrison-Woodbury (SMW) formula as one of theimportant tools to prove their main resultThe SMW formulagives an expression for the inverse of (119860minus119883119884119879) in terms of theinverse of119860 when it exists When119860 is nonsingular119860minus119883119884119879

is nonsingular if and only 119868 minus 119884119879119860minus1119883 is nonsingular In thatcase

(119860 minus 119883119884119879

)minus1

= 119860minus1

minus 119860minus1

119883(119868 minus 119884119879

119860minus1

119883)minus1

119884119879

119860minus1

(1)

Themain objective of the present work is to study certainstructured perturbations 119860 minus 119883119884119879 of matrices 119860 such thattheMoore-Penrose inverse of the perturbation is nonnegativewhenever the Moore-Penrose inverse of 119860 is nonnegativeClearly this class of matrices includes the class of matricesthat have nonnegative inverses especially119872-matrices In ourapproach extensions of SMW formula for singular matrices

play a crucial role Let us mention that this problem has beenstudied in the literature (See for instance [2] formatrices and[3] for operators over Hilbert spaces) We refer the reader tothe references in the latter for other recent extensions

In this paper first we present alternative proofs ofgeneralizations of the SMW formula for the cases of theMoore-Penrose inverse (Theorem 5) and the group inverse(Theorem 6) in Section 3 In Section 4 we characterize thenonnegativity of (119860 minus 119883119866119884119879)dagger This is done in Theorem 9and is one of the main results of the present work As aconsequence we present a result for119872-matrices which seemsnew We present a couple of applications of the main resultin Theorems 13 and 15 In the concluding section we studyiterative versions of the results of the second section Weprove two characterizations for (119860minus119883119884119879)dagger to be nonnegativein Theorems 18 and 21

Before concluding this introductory section let us givea motivation for the work that we have undertaken hereIt is a well-documented fact that 119872-matrices arise quiteoften in solving sparse systems of linear equations Anextensive theory of119872-matrices has been developed relativeto their role in numerical analysis involving the notion ofsplitting in iterativemethods and discretization of differentialequations in the mathematical modeling of an economyoptimization and Markov chains [4 5] Specifically theinspiration for the present study comes from the work of [1]where the authors consider a system of linear inequalitiesarising out of a problem in third generation wireless commu-nication systemsThematrix defining the inequalities there is

2 Journal of Applied Mathematics

an 119872-matrix In the likelihood that the matrix of thisproblem is singular (due to truncation or round-off errors)the earlier method becomes inapplicable Our endeavour isto extend the applicability of these results to more generalmatrices for instance matrices with nonnegative Moore-Penrose inverses Finally as mentioned earlier sincematriceswith nonnegative generalized inverses include in particular119872-matrices it is apparent that our results are expected toenlarge the applicability of themethods presently available for119872-matrices even in a very general framework including thespecific problem mentioned above

2 Preliminaries

Let R R119899 and R119898times119899 denote the set of all real numbers the119899-dimensional real Euclidean space and the set of all 119898 times 119899matrices over R For 119860 isin R119898times119899 let 119877(119860) 119873(119860) 119877(119860)perp and119860119879 denote the range space the null space the orthogonal

complement of the range space and the transpose of thematrix 119860 respectively For x = (119909

1 1199092 119909

119899)119879

isin R119899 wesay that x is nonnegative that is x ge 0 if and only if x

119894ge 0

for all 119894 = 1 2 119899 As mentioned earlier for a matrix 119861 weuse 119861 ge 0 to denote that all the entries of 119861 are nonnegativeAlso we write 119860 le 119861 if 119861 minus 119860 ge 0

Let 120588(119860) denote the spectral radius of the matrix 119860 If120588(119860) lt 1 for 119860 isin R119899times119899 then 119868 minus 119860 is invertible Thenext result gives a necessary and sufficient condition for thenonnegativity of (119868 minus 119860)minus1 This will be one of the results thatwill be used in proving the first main result

Lemma 1 (see [5 Lemma 21 Chapter 6]) Let 119860 isin R119899times119899 benonnegative Then 120588(119860) lt 1 if and only if (119868 minus 119860)minus1 exists and

(119868 minus 119860)minus1

=

infin

sum

119896=0

119860119896

ge 0 (2)

More generally matrices having nonnegative inversesare characterized using a property called monotonicity Thenotion of monotonicity was introduced by Collatz [6] A real119899 times 119899matrix 119860 is called monotone if 119860x ge 0 rArr x ge 0 It wasproved by Collatz [6] that 119860 is monotone if and only if 119860minus1exists and 119860minus1 ge 0

One of the frequently used tools in studying monotonematrices is the notion of a regular splitting We only refer thereader to the book [5] for more details on the relationshipbetween these concepts

The notion of monotonicity has been extended in a vari-ety of ways to singular matrices using generalized inversesFirst let us briefly review the notion of two importantgeneralized inverses

For 119860 isin R119898times119899 the Moore-Penrose inverse is the unique119885 isin R119899times119898 satisfying the Penrose equations 119860119885119860 = 119860119885119860119885 = 119885 (119860119885)119879 = 119860119885 and (119885119860)119879 = 119885119860 The uniqueMoore-Penrose inverse of119860 is denoted by119860dagger and it coincideswith 119860minus1 when 119860 is invertible

The following theorem by Desoer and Whalen whichis used in the sequel gives an equivalent definition for theMoore-Penrose inverse Let us mention that this result wasproved for operators between Hilbert spaces

Theorem 2 (see [7]) Let 119860 isin R119898times119899 Then 119860dagger isin R119899times119898 is theunique matrix119883 isin R119899times119898 satisfying

(i) 119885119860119909 = 119909 forall119909 isin 119877(119860119879)(ii) 119885119910 = 0 forall119910 isin 119873(119860119879)

Now for119860 isin R119899times119899 any119885 isin R119899times119899 satisfying the equations119860119885119860 = 119860 119885119860119885 = 119885 and 119860119885 = 119885119860 is called the groupinverse of 119860 The group inverse does not exist for everymatrix But whenever it exists it is unique A necessary andsufficient condition for the existence of the group inverse of119860 is that the index of119860 is 1 where the index of a matrix is thesmallest positive integer 119896 such that rank (119860119896+1) = rank (119860119896)

Some of the well-known properties of theMoore-Penroseinverse and the group inverse are given as follows 119877(119860119879) =119877(119860dagger

)119873(119860119879) = 119873(119860dagger) 119860dagger119860 = 119875119877(119860119879) and 119860119860dagger = 119875

119877(119860) In

particular x isin 119877(119860119879) if and only if x = 119860dagger119860x Also 119877(119860) =119877(119860

) 119873(119860) = 119873(119860

) and 119860119860 = 119860119860

= 119875119877(119860)119873(119860)

Herefor complementary subspaces 119871 and119872 of R119896 119875

119871119872denotes

the projection ofR119896 onto 119871 along119872 119875119871denotes 119875

119871119872if119872 =

119871perp For details we refer the reader to the book [8]Inmatrix analysis a decomposition (splitting) of amatrix

is considered in order to study the convergence of iterativeschemes that are used in the solution of linear systems of alge-braic equations As mentioned earlier regular splittings areuseful in characterizing matrices with nonnegative inverseswhereas proper splittings are used for studying singularsystems of linear equations Let us next recall this notionFor a matrix 119860 isin R119898times119899 a decomposition 119860 = 119880 minus 119881 iscalled a proper splitting [9] if 119877(119860) = 119877(119880) and 119873(119860) =119873(119880) It is rather well-known that a proper splitting existsfor every matrix and that it can be obtained using a full-rank factorization of the matrix For details we refer to [10]Certain properties of a proper splitting are collected in thenext result

Theorem 3 (see [9 Theorem 1]) Let 119860 = 119880 minus 119881 be a propersplitting of 119860 isin R119898times119899 Then

(a) 119860 = 119880(119868 minus 119880dagger119881)(b) 119868 minus 119880dagger119881 is nonsingular and(c) 119860dagger = (119868 minus 119880dagger119881)minus1119880dagger

The following result by Berman and Plemmons [9] givesa characterization for 119860dagger to be nonnegative when 119860 has aproper splitting This result will be used in proving our firstmain result

Theorem 4 (see [9 Corollary 4]) Let 119860 = 119880 minus 119881 be a propersplitting of 119860 isin R119898times119899 where 119880dagger ge 0 and 119880dagger119881 ge 0 Then119860dagger

ge 0 if and only if 120588(119880dagger119881) lt 1

3 Extensions of the SMW Formula forGeneralized Inverses

The primary objects of consideration in this paper are gen-eralized inverses of perturbations of certain types of a matrix

Journal of Applied Mathematics 3

119860 Naturally extensions of the SMW formula for generalizedinverses are relevant in the proofs Inwhat follows we presenttwo generalizations of the SMW formula for matrices Wewould like to emphasize that our proofs also carry overto infinite dimensional spaces (the proof of the first resultis verbatim and the proof of the second result with slightmodifications applicable to range spaces instead of ranks ofthe operators concerned) However we are confining ourattention to the case of matrices Let us also add that theseresults have been proved in [3] for operators over infinitedimensional spaces We have chosen to include these resultshere as our proofs are different from the ones in [3] and thatour intention is to provide a self-contained treatment

Theorem 5 (see [3 Theorem 21]) Let 119860 isin R119898times119899 119883 isin R119898times119896and 119884 isin R119899times119896 be such that

119877 (119883) sube 119877 (119860) 119877 (119884) sube 119877 (119860119879

) (3)

Let 119866 isin R119896times119896 and Ω = 119860 minus 119883119866119884119879 If

119877 (119883119879

) sube 119877 ((119866dagger

minus 119884119879

119860dagger

119883)119879

)

119877 (119884119879

) sube 119877 (119866dagger

minus 119884119879

119860dagger

119883) 119877 (119883119879

) sube 119877 (119866)

119877 (119884119879

) sube 119877 (119866119879

)

(4)

then

Ωdagger

= 119860dagger

+ 119860dagger

119883(119866dagger

minus 119884119879

119860dagger

119883)dagger

119884119879

119860dagger

(5)

Proof Set 119885 = 119860dagger + 119860dagger119883119867dagger119884119879119860dagger where119867 = 119866dagger

minus 119884119879

119860dagger

119883From the conditions (3) and (4) it follows that 119860119860dagger119883 = 119883119860dagger

119860119884 = 119884119867dagger119867119883119879 = 119883119879119867119867dagger119884119879 = 119884119879119866119866dagger119883119879 = 119883119879 and119866dagger

119866119884119879

= 119884119879

Now

119860dagger

119883119867dagger

119884119879

minus 119860dagger

119883119867dagger

119884119879

119860dagger

119883119866119884119879

= 119860dagger

119883119867dagger

(119866dagger

119866119884119879

minus 119884119879

119860dagger

119883119866119884119879

)

= 119860dagger

119883119867dagger

119867119866119884119879

= 119860dagger

119883119866119884119879

(6)

Thus 119885Ω = 119860dagger

119860 Since 119877(Ω119879) sube 119877(119860119879

) it follows that119885Ω(x) = 119909 forallx isin 119877(Ω119879)

Let y isin 119873(Ω119879

) Then 119860119879y minus 119884119866119879119883119879y = 0 so that119883119879

119860dagger119879

(119860119879

minus 119884119866119879

119883119879

)y = 0 Substituting 119883119879 = 119866119866dagger119883119879 andsimplifying it we get119867119879119866119879119883119879y = 0 Also119860119879y = 119884119866119879119883119879y =119884119867119867dagger

119866119879

119883119879y = 0 and so 119860daggery = 0 Thus 119885y = 0 for

y isin 119873(Ω119879) Hence by Theorem 2 119885 = Ωdagger

The result for the group inverse follows

Theorem 6 Let 119860 isin R119899times119899 be such that 119860 exists Let 119883119884 isin R119899times119896 and 119866 be nonsingular Assume that 119877(119883) sube 119877(119860)119877(119884) sube 119877(119860

119879

) and 119866minus1 minus 119884119879119860119883 is nonsingular Suppose that

rank (119860 minus 119883119866119884119879) = rank (119860) Then (119860 minus 119883119866119884119879) exists andthe following formula holds

(119860 minus 119883119866119884119879

)= 119860

+ 119860

119883(119866minus1

minus 119884119879

119860119883)minus1

119884119879

119860 (7)

Conversely if (119860minus119883119866119884119879) exists then the formula above holdsand we have rank (119860 minus 119883119866119884119879) = rank (119860)

Proof Since 119860 exists 119877(119860) and 119873(119860) are complementarysubspaces of R119899times119899

Suppose that rank (119860 minus 119883119866119884119879) = rank (119860) As 119877(119883) sube119877(119860) it follows that 119877(119860 minus 119883119866119884119879) sube 119877(119860) Thus 119877(119860 minus119883119866119884119879

) = 119877(119860) By the rank-nullity theorem the nullity ofboth 119860 minus 119883119866119884119879 and 119860 are the same Again since 119877(119884) sube119877(119860119879

) it follows that 119873(119860 minus 119883119866119884119879

) = 119873(119860) Thus119877(119860 minus 119883119866119884

119879

) and 119873(119860 minus 119883119866119884119879) are complementary sub-spaces This guarantees the existence of the group inverse of119860 minus 119883119866119884

119879Conversely suppose that (119860 minus 119883119866119884119879) exists It can be

verified by direct computation that 119885 = 119860+ 119860

119883(119866minus1

minus

119884119879

119860119883)minus1

119884119879

119860 is the group inverse of 119860 minus 119883119866119884119879 Also we

have (119860minus119883119866119884119879)(119860minus119883119866119884119879) = 119860119860 so that119877(119860minus119883119866119884119879) =119877(119860) and hence the rank (119860 minus 119883119866119884119879) = rank (119860)

We conclude this section with a fairly old result [2] as aconsequence of Theorem 5

Theorem 7 (see [2 Theorem 15]) Let 119860 isin R119898times119899 of rank 119903119883 isin R119898times119903 and 119884 isin R119899times119903 Let119866 be an 119903times119903 nonsingular matrixAssume that 119877(119883) sube 119877(119860) 119877(119884) sube 119877(119860119879) and 119866minus1 minus 119884119879119860dagger119883is nonsingular Let Ω = 119860 minus 119883119866119884119879 Then

(119860 minus 119883119866119884119879

)dagger

= 119860dagger

+ 119860dagger

119883(119866minus1

minus 119884119879

119860dagger

119883)minus1

119884119879

119860dagger

(8)

4 Nonnegativity of (119860minus119883119866119884119879)dagger

In this section we consider perturbations of the form 119860 minus

119883119866119884119879 and derive characterizations for (119860 minus 119883119866119884119879)dagger to be

nonnegative when 119860dagger ge 0 119883 ge 0 119884 ge 0 and 119866 ge 0 In orderto motivate the first main result of this paper let us recall thefollowing well known characterization of119872-matrices [5]

Theorem 8 Let 119860 be a 119885-matrix with the representation 119860 =119904119868 minus 119861 where 119861 ge 0 and 119904 ge 0 Then the following statementsare equivalent

(a) 119860minus1 exists and 119860minus1 ge 0

(b) There exists 119909 gt 0 such that 119860119909 gt 0

(c) 120588(119861) lt 119904

Let us prove the first result of this article This extendsTheorem 8 to singular matrices We will be interested inextensions of conditions (a) and (c) only

Theorem 9 Let 119860 = 119880 minus 119881 be a proper splitting of 119860 isin R119898times119899

with 119880dagger ge 0 119880dagger119881 ge 0 and 120588(119880dagger119881) lt 1 Let 119866 isin R119896times119896 benonsingular and nonnegative 119883 isin R119898times119896 and 119884 isin R119899times119896 be

4 Journal of Applied Mathematics

nonnegative such that 119877(119883) sube 119877(119860) 119877(119884) sube 119877(119860119879) and119866minus1 minus119884119879

119860dagger

119883 is nonsingular LetΩ = 119860minus119883119866119884119879Then the followingare equivalent

(a) Ωdagger ge 0(b) (119866minus1 minus 119884119879119860dagger119883)minus1 ge 0(c) 120588(119880dagger119882) lt 1 where119882 = 119881 + 119883119866119884

119879

Proof First we observe that since 119877(119883) sube 119877(119860) 119877(119884) sube119877(119860119879

) and 119866minus1 minus 119884119879119860dagger119883 is nonsingular by Theorem 7 wehave

Ωdagger

= 119860dagger

+ 119860dagger

119883(119866minus1

minus 119884119879

119860dagger

119883)minus1

119884119879

119860dagger

(9)

We thus haveΩΩdagger = 119860119860dagger andΩdaggerΩ = 119860dagger119860 Therefore

119877 (Ω) = 119877 (119860) 119873 (Ω) = 119873 (119860) (10)

Note that the first statement also implies that 119860dagger ge 0 byTheorem 4

(a) rArr (b) By taking 119864 = 119860 and 119865 = 119883119866119884119879 we get

Ω = 119864 minus 119865 as a proper splitting for Ω such that 119864dagger = 119860dagger ge 0and 119864dagger119865 = 119860dagger119883119866119884119879 ge 0 (since 119883119866119884119879 ge 0) Since Ωdagger ge 0by Theorem 4 we have 120588(119860dagger119883119866119884119879) = 120588(119864

dagger

119865) lt 1 Thisimplies that 120588(119884119879119860dagger119883119866) lt 1 We also have 119884119879119860dagger119883119866 ge 0Thus by Lemma 1 (119868minus119884119879119860dagger119883119866)minus1 exists and is nonnegativeBut we have 119868 minus 119884119879119860dagger119883119866 = (119866

minus1

minus 119884119879

119860dagger

119883)119866 Now 0 le(119868 minus 119884

119879

119860dagger

119883119866)minus1

= 119866minus1

(119866minus1

minus 119884119879

119860dagger

119883)minus1 This implies that

(119866minus1

minus 119884119879

119860dagger

119883)minus1

ge 0 since 119866 ge 0 This proves (b)(b) rArr (c) We have119880minus119882 = 119880minus119881minus119883119866119884

119879

= 119860minus119883119884119879

=

Ω Also 119877(Ω) = 119877(119860) = 119877(119880) and 119873(Ω) = 119873(119860) = 119873(119880)So Ω = 119880 minus 119882 is a proper splitting Also 119880dagger ge 0 and119880dagger

119882 = 119880dagger

(119881+119883119866119884119879

) ge 119880dagger

119881 ge 0 Since (119866minus1minus119884119879119860dagger119883)minus1 ge0 it follows from (9) that Ωdagger ge 0 (c) now follows fromTheorem 4

(c) rArr (a) Since119860 = 119880minus119881 we haveΩ = 119880minus119881minus119883119866119884119879 =119880minus119882 Also we have119880dagger ge 0119880dagger119881 ge 0 Thus119880dagger119882 ge 0 since119883119866119884119879

ge 0 Now by Theorem 4 we are done if the splittingΩ = 119880 minus119882 is a proper splitting Since 119860 = 119880 minus 119881 is a propersplitting we have 119877(119880) = 119877(119860) and 119873(119880) = 119873(119860) Nowfrom the conditions in (10) we get that 119877(119880) = 119877(Ω) and119873(119880) = 119873(Ω) Hence Ω = 119880 minus 119882 is a proper splitting andthis completes the proof

The following result is a special case of Theorem 9

Theorem 10 Let119860 = 119880minus119881 be a proper splitting of119860 isin R119898times119899

with 119880dagger ge 0 119880dagger119881 ge 0 and 120588(119880dagger119881) lt 1 Let 119883 isin R119898times119896 and119884 isin R119899times119896 be nonnegative such that 119877(119883) sube 119877(119860) 119877(119884) sube119877(119860119879

) and 119868minus119884119879119860dagger119883 is nonsingular LetΩ = 119860minus119883119884119879 Thenthe following are equivalent

(a) Ωdagger ge 0(b) (119868 minus 119884119879119860dagger119883)minus1 ge 0(c) 120588(119880dagger119882) lt 1 where119882 = 119881 + 119883119884

119879

The following consequence of Theorem 10 appears to benew This gives two characterizations for a perturbed 119872-matrix to be an119872-matrix

Corollary 11 Let119860 = 119904119868minus119861where119861 ge 0 and 120588(119861) lt 119904 (ie119860is an119872-matrix) Let 119883 isin R119898times119896 and 119884 isin R119899times119896 be nonnegativesuch that 119868 minus 119884119879119860dagger119883 is nonsingular Let Ω = 119860 minus 119883119884119879 Thenthe following are equivalent

(a) Ωminus1 ge 0(b) (119868 minus 119884119879119860dagger119883)minus1 ge 0(c) 120588(119861(119868 + 119883119884119879)) lt 119904

Proof From the proof ofTheorem 10 since 119860 is nonsingularit follows that ΩΩdagger = 119868 and ΩdaggerΩ = 119868 This shows that Ωis invertible The rest of the proof is omitted as it is an easyconsequence of the previous result

In the rest of this section we discuss two applicationsof Theorem 10 First we characterize the least element ina polyhedral set defined by a perturbed matrix Next weconsider the following Suppose that the ldquoendpointsrdquo of aninterval matrix satisfy a certain positivity property Then allmatrices of a particular subset of the interval also satisfythat positivity condition The problem now is if that we aregiven a specific structured perturbation of these endpointswhat conditions guarantee that the positivity property for thecorresponding subset remains valid

The first result is motivated by Theorem 12 below Let usrecall that with respect to the usual order an element xlowast isinX sube R119899 is called a least element ofX if it satisfies xlowast le x forall x isin X Note that a nonempty set may not have the leastelement and if it exists then it is unique In this connectionthe following result is known

Theorem 12 (see [11 Theorem 32]) For 119860 isin R119898times119899 and b isinR119898 let

Xb = x isin R119899

119860x + y ge b 119875119873(119860)

x = 0

119875119877(119860)

y = 0 119891119900119903 119904119900119898119890 y isin R119898 (11)

Then a vector xlowast is the least element ofXb if and only if xlowast =119860daggerb isin Xb with 119860dagger ge 0

Now we obtain the nonnegative least element of apolyhedral set defined by a perturbed matrix This is animmediate application of Theorem 10

Theorem 13 Let 119860 isin R119898times119899 be such that 119860dagger ge 0 Let 119883 isin

R119898times119896 and let 119884 isin R119899times119896 be nonnegative such that 119877(119883) sube119877(119860) 119877(119884) sube 119877(119860119879) and 119868 minus 119884119879119860dagger119883 is nonsingular Supposethat (119868 minus 119884119879119860dagger119883)minus1 ge 0 For b isin R119898 119887 ge 0 let

Sb = x isin R119899

Ωx + y ge b 119875119873(Ω)

x = 0

119875119877(Ω)

y = 0 119891119900119903 119904119900119898119890 y isin R119898 (12)

where Ω = 119860 minus 119883119884119879 Then xlowast = (119860 minus 119883119884119879)daggerb is the least

element of S119887

Proof From the assumptions using Theorem 10 it fol-lows that Ωdagger ge 0 The conclusion now follows fromTheorem 12

Journal of Applied Mathematics 5

To state and prove the result for interval matrices let usfirst recall the notion of interval matrices For119860 119861 isin R119898 times 119899an interval (matrix) 119869 = [119860 119861] is defined as 119869 = [119860 119861] = 119862 119860 le 119862 le 119861 The interval 119869 = [119860 119861] is said to be range-kernelregular if 119877(119860) = 119877(119861) and 119873(119860) = 119873(119861) The followingresult [12] provides necessary and sufficient conditions for119862dagger

ge 0 for 119862 isin 119870 where 119870 = 119862 isin 119869 119877(119862) = 119877(119860) =

119877(119861)119873(119862) = 119873(119860) = 119873(119861)

Theorem 14 Let 119869 = [119860 119861] be range-kernel regular Then thefollowing are equivalent

(a) 119862dagger ge 0 whenever 119862 isin 119870(b) 119860dagger ge 0 and 119861dagger ge 0

In such a case we have 119862dagger = suminfin119895=0(119861dagger

(119861 minus 119862))119895

119861dagger

Now we present a result for the perturbation

Theorem 15 Let 119869 = [119860 119861] be range-kernel regular with119860dagger ge0 and 119861dagger ge 0 Let 119883

1 1198832 1198841 and 119884

2be nonnegative matrices

such that 119877(1198831) sube 119877(119860) 119877(119884

1) sube 119877(119860

119879

) 119877(1198832) sube 119877(119861) and

119877(1198842) sube 119877(119861

119879

) Suppose that 119868minus1198841198791119860dagger

1198831and 119868minus119884119879

2119860dagger

1198832are

nonsingular with nonnegative inverses Suppose further that1198832119884119879

2le 1198831119884119879

1 Let 119869 = [119864 119865] where 119864 = 119860 minus 119883

1119884119879

1and

119865 = 119861 minus 1198832119884119879

2 Finally let = 119885 isin 119869 119877(119885) = 119877(119864) =

119877(119865) 119886119899119889 119873(119885) = 119873(119864) = 119873(119865) Then

(a) 119864dagger ge 0 and 119865dagger ge 0(b) 119885dagger ge 0 whenever 119885 isin

In that case 119885dagger = suminfin119895=0(119861dagger

119861 minus 119865dagger

119885)119895

119865dagger

Proof It follows fromTheorem 7 that

119864dagger

= 119860dagger

+ 119860dagger

1198831(119868 minus 119884

119879

1119860dagger

1198831)minus1

119884119879

1119860dagger

119865dagger

= 119861dagger

+ 119861dagger

1198832(119868 minus 119884

119879

2119861dagger

1198832)minus1

119884119879

2119861dagger

(13)

Also we have 119864119864dagger = 119860119860dagger 119864dagger119864 = 119860

dagger

119860 119865dagger119865 = 119861dagger

119861and 119865119865dagger = 119861119861

dagger Hence 119877(119864) = 119877(119860) 119873(119864) = 119873(119860)119877(119865) = 119877(119861) and 119873(119865) = 119873(119861) This implies that theinterval 119869 is range-kernel regular Now since 119864 and 119865 satisfythe conditions of Theorem 10 we have 119864dagger ge 0 and 119865dagger ge 0proving (a) Hence by Theorem 14 119885dagger ge 0 whenever 119885 isin Again by Theorem 14 we have 119885dagger = suminfin

119895=0(119865dagger

(119865 minus 119885))119895

119865dagger

=

suminfin

119895=0(119861dagger

119861 minus 119865dagger

119885)119895

119865dagger

5 Iterations That Preserve Nonnegativity ofthe Moore-Penrose Inverse

In this section we present results that typically provideconditions for iteratively defined matrices to have nonneg-ative Moore-Penrose inverses given that the matrices thatwe start with have this property We start with the followingresult about the rank-one perturbation case which is a directconsequence of Theorem 10

Theorem 16 Let 119860 isin R119899times119899 be such that 119860dagger ge 0 Let x isin R119898

and y isin R119899 be nonnegative vectors such that x isin 119877(119860) y isin119877(119860119879

) and 1 minus y119879119860daggerx = 0 Then (119860 minus xy119879)dagger ge 0 if and only ify119879119860daggerx lt 1

Example 17 Let us consider 119860 = (13) (1 minus1 1

minus1 4 minus1

1 minus1 1

) It can be

verified that 119860 can be written in the form 119860 = (1198682

119890119879

1

) (1198682+

e1119890119879

1)minus1

119862minus1

(1198682+ e1119890119879

1)minus1

(1198682e1) where e

1= (1 0)

119879 and 119862 isthe 2 times 2 circulant matrix generated by the row (1 12) Wehave 119860dagger = (12) ( 2 1 21 2 1

2 1 2

) ge 0 Also the decomposition 119860 =

119880minus119881 where119880 = (13) ( 1 minus1 2minus1 4 minus2

1 minus1 2

) and119881 = (13) ( 0 0 10 2 minus10 0 1

) isa proper splitting of 119860 with 119880dagger ge 0 119880dagger119881 ge 0 and 120588(119880dagger119881) lt1

Let x = y = (13 16 13)119879 Then x and let y are

nonnegative x isin 119877(119860) and y isin 119877(119860119879) We have 1 minus y119879119860daggerx =512 gt 0 and 120588(119880dagger119882) lt 1 for119882 = 119881 + xy119879 Also it can beseen that (119860 minus xy119879)dagger = (120) ( 47 28 4728 32 28

47 28 47

) ge 0 This illustratesTheorem 10

Let x1 x2 x

119896isin R119898 and y

1 y2 y

119896isin R119899 be nonneg-

ative Denote 119883119894= (x1 x2 x

119894) and 119884

119894= (y1 y2 y

119894) for

119894 = 1 2 119896 Then119860minus119883119896119884119879

119896= 119860minussum

119896

119894=1x119894y119879119894 The following

theorem is obtained by a recurring application of the rank-one result of Theorem 16

Theorem 18 Let119860 isin R119898times119899 and let119883119894 119884119894be as above Further

suppose that x119894isin 119877(119860 minus 119883

119894minus1119884119879

119894minus1) y119894isin 119877(119860 minus 119883

119894minus1119884119879

119894minus1)119879 and

1 minus y119879119894(119860 minus 119883

119894minus1119884119879

119894minus1)daggerx119894be nonzero Let (119860 minus 119883

119894minus1119884119879

119894minus1)dagger

ge 0for all 119894 = 1 2 119896 Then (119860 minus 119883

119896119884119879

119896)dagger

ge 0 if and only ify119879119894(119860 minus 119883

119894minus1119884119879

119894minus1)daggerx119894lt 1 where 119860 minus 119883

0119884119879

0is taken as 119860

Proof Set 119861119894= 119860 minus 119883

119894119884119879

119894for 119894 = 0 1 119896 where 119861

0is

identified as 119860 The conditions in the theorem can be writtenas

x119894isin 119877 (119861

119894minus1) y

119894isin 119877 (119861

119879

119894minus1)

1 minus y119879119894119861dagger

119894minus1x119894= 0

forall119894 = 1 2 119896

(14)

Also we have 119861dagger119894ge 0 for all 119894 = 0 1 119896 minus 1

Now assume that 119861dagger119896ge 0 Then by Theorem 16 and the

conditions in (14) for 119894 = 119896 we have y119896119861dagger

119896minus1x119896lt 1 Also

since by assumption 119861dagger119894ge 0 for all 119894 = 0 1 119896 minus 1 we have

y119879119894119861dagger

119894minus1x119894lt 1 for all 119894 = 1 2 119896

The converse part can be proved iteratively Then condi-tion y119879

11198610

daggerx1lt 1 and the conditions in (14) for 119894 = 1 imply

that 1198611

dagger

ge 0 Repeating the argument for 119894 = 2 to 119896 proves theresult

The following result is an extension of Lemma 23 in [1]which is in turn obtained as a corollaryThis corollary will beused in proving another characterization for (119860minus119883

119896119884119879

119896)dagger

ge 0

6 Journal of Applied Mathematics

Theorem 19 Let 119860 isin R119899times119899 bc isin R119899 be such that minusb ge 0minusc ge 0 and 120572(isin R) gt 0 Further suppose that b isin 119877(119860) andc isin 119877(119860119879) Let 119860 = ( 119860 b

c119879 120572 ) isin R(119899+1)times(119899+1) Then 119860dagger ge 0 if andonly if 119860dagger ge 0 and c119879119860daggerb lt 120572

Proof Let us first observe that 119860119860daggerb = b and c119879119860dagger119860 = c119879Set

119883 =(

119860dagger

+119860daggerbc119879119860dagger

120572 minus c119879119860daggerbminus119860daggerb

120572 minus c119879119860daggerbminus

c119879119860dagger

120572 minus c119879119860daggerb1

120572 minus c119879119860daggerb) (15)

It then follows that 119860119883 = ( 119860119860dagger 00119879 1 ) and 119883119860 = (119860dagger119860 0

0119879 1 ) Usingthese two equations it can easily be shown that119883 = 119860dagger

Suppose that 119860dagger ge 0 and 120573 = 120572 minus c119879119860daggerb gt 0 Then119860dagger

ge 0 Conversely suppose that 119860dagger ge 0 Then we musthave 120573 gt 0 Let e

1 e2 e

119899+1 denote the standard basis

of R119899+1 Then for 119894 = 1 2 119899 we have 0 le 119860dagger

(e119894) =

(119860daggere119894+(119860daggerbc119879119860daggere

119894120573) minus(c119879119860daggere

119894120573))119879 Since c le 0 it follows

that 119860daggere119894ge 0 for 119894 = 1 2 119899 Thus 119860dagger ge 0

Corollary 20 (see [1 Lemma 23]) Let 119860 isin R119899times119899 benonsingular Let b c isin R119899 be such that minusb ge 0 minusc ge 0

and 120572(isin R) gt 0 Let 119860 = (119860 bc119879 120572 ) Then 119860minus1 isin R(119899+1)times(119899+1)

is nonsingular with 119860minus1 ge 0 if and only if 119860minus1 ge 0 andc119879119860minus1b lt 120572

Now we obtain another necessary and sufficient condi-tion for (119860 minus 119883

119896119884119879

119896)dagger to be nonnegative

Theorem 21 Let 119860 isin R119898times119899 be such that 119860dagger ge 0 Let x119894 y119894

be nonnegative vectors inR119899 andR119898 respectively for every 119894 =1 119896 such that

x119894isin 119877 (119860 minus 119883

119894minus1119884119879

119894minus1) y

119894isin 119877(119860 minus 119883

119894minus1119884119879

119894minus1)119879

(16)

where 119883119894= (x1 x

119894) 119884119894= (y1 y

119894) and 119860 minus 119883

0119884119879

0is

taken as119860 If119867119894= 119868119894minus119884119879

119894119860dagger

119883119894is nonsingular and 1minus y119879

119894119860daggerx119894

is positive for all 119894 = 1 2 119896 then (119860 minus 119883119896119884119879

119896)dagger

ge 0 if andonly if

y119879119894119860daggerx119894lt 1 minus y119879

119894119860dagger

119883119894minus1119867minus1

119894minus1119884119879

119894minus1119860daggerx119894 forall119894 = 1 119896 (17)

Proof The range conditions in (16) imply that 119877(119883119896) sube 119877(119860)

and 119877(119884119896) sube 119877(119860

119879

) Also from the assumptions it followsthat 119867

119896is nonsingular By Theorem 10 (119860 minus 119883

119896119884119879

119896)dagger

ge 0 ifand only if119867minus1

119896ge 0 Now

119867119896= (

119867119896minus1

minus119884119879

119896minus1119860daggerx119896

minusy119879119896119860dagger

119883119896minus1

1 minus y119879119896119860daggerx119896

) (18)

Since 1minusy119879119896119860daggerx119896gt 0 usingCorollary 20 it follows that119867minus1

119896ge

0 if and only if y119879119896119860dagger

119883119896minus1119867minus1

119896minus1119884119879

119896minus1119860daggerx119896lt 1 minus y119879

119896119860daggerx119896and

119867minus1

119896minus1ge 0

Now applying the above argument to the matrix 119867119896minus1

we have that 119867minus1

119896minus1ge 0 holds if and only if 119867minus1

119896minus2ge 0

holds and y119879119896minus1119860dagger

119883119896minus2(119868119896minus2

minus 119884119879

119896minus2119860dagger

119883119896minus2)minus1

119884119879

119896minus2119860daggerx119896minus1 lt

1minus y119879119896minus1119860daggerx119896minus1 Continuing the above argument we get (119860minus

119883119896y119879119896)dagger

ge 0 if and only if y119879119894119860dagger

119883119894minus1119867minus1

119894minus1119884119879

119894minus1119860daggerx119894lt 1minusy119879

119894119860daggerx119894

forall119894 = 1 119896 This is condition (17)

We conclude the paper by considering an extension ofExample 17

Example 22 For a fixed 119899 let 119862 be the circulant matrixgenerated by the row vector (1 (12) (13) (1(119899 minus 1)))Consider

119860 = (119868

119890119879

1

) (119868 + e1119890119879

1)minus1

119862minus1

(119868 + e1119890119879

1)minus1

(119868 e1) (19)

where 119868 is the identity matrix of order 119899 minus 1 e1is (119899 minus 1) times 1

vector with 1 as the first entry and 0 elsewhere Then 119860 isin

R119899times119899 119860dagger is the 119899 times 119899 nonnegative Toeplitz matrix

119860dagger

=

((((((

(

11

119899 minus 1

1

119899 minus 2

1

21

1

21

1

119899 minus 1

1

3

1

2

1

119899 minus 1

1

119899 minus 2

1

119899 minus 3 1

1

119899 minus 1

11

119899 minus 1

1

119899 minus 2

1

21

))))))

)

(20)

Let 119909119894be the first row of 119861dagger

119894minus1written as a column vector

multiplied by 1119899119894 and let 119910119894be the first column of 119861dagger

119894minus1

multiplied by 1119899119894 where 119861119894= 119860 minus 119883

119894119884119879

119894 with 119861

0being

identified with 119860 119894 = 0 2 119896 We then have x119894ge 0

y119894ge 0 119909

1isin 119877(119860) and 119910

1isin 119877(119860

119879

) Now if 119861dagger119894minus1

ge 0 foreach iteration then the vectors 119909

119894and 119910

119894will be nonnegative

Hence it is enough to check that 1 minus 119910119879119894119861dagger

119894minus1119909119894gt 0 in order

to get 119861dagger119894ge 0 Experimentally it has been observed that for

119899 gt 3 the condition 1 minus 119910119879119894119861dagger

119894minus1119909119894gt 0 holds true for any

iteration However for 119899 = 3 it is observed that 119861dagger1ge 0 119861dagger

2ge

0 1 minus 1199101198791119860dagger

1199091gt 0 and 1 minus 119910dagger

2119861dagger

11199092gt 0 But 1 minus 119910119879

3119861dagger

21199093lt 0

and in that case we observe that 119861dagger3is not nonnegative

References

[1] J Ding W Pye and L Zhao ldquoSome results on structured119872-matrices with an application to wireless communicationsrdquoLinear Algebra and its Applications vol 416 no 2-3 pp 608ndash614 2006

[2] S K Mitra and P Bhimasankaram ldquoGeneralized inverses ofpartitionedmatrices and recalculation of least squares estimatesfor data ormodel changesrdquo Sankhya A vol 33 pp 395ndash410 1971

[3] C Y Deng ldquoA generalization of the Sherman-Morrison-Woodbury formulardquo Applied Mathematics Letters vol 24 no9 pp 1561ndash1564 2011

[4] A Berman M Neumann and R J SternNonnegative Matricesin Dynamical Systems JohnWiley amp Sons New York NY USA1989

Journal of Applied Mathematics 7

[5] A Berman and R J Plemmons Nonnegative Matrices inthe Mathematical Sciences Society for Industrial and AppliedMathematics (SIAM) 1994

[6] L Collatz Functional Analysis and Numerical MathematicsAcademic Press New York NY USA 1966

[7] C A Desoer and B H Whalen ldquoA note on pseudoinversesrdquoSociety for Industrial and Applied Mathematics vol 11 pp 442ndash447 1963

[8] A Ben-Israel and T N E Greville Generalized InversesTheoryand Applications Springer New York NY USA 2003

[9] A Berman and R J Plemmons ldquoCones and iterative methodsfor best least squares solutions of linear systemsrdquo SIAM Journalon Numerical Analysis vol 11 pp 145ndash154 1974

[10] D Mishra and K C Sivakumar ldquoOn splittings of matrices andnonnegative generalized inversesrdquo Operators and Matrices vol6 no 1 pp 85ndash95 2012

[11] D Mishra and K C Sivakumar ldquoNonnegative generalizedinverses and least elements of polyhedral setsrdquo Linear Algebraand its Applications vol 434 no 12 pp 2448ndash2455 2011

[12] M R Kannan and K C Sivakumar ldquoMoore-Penrose inversepositivity of interval matricesrdquo Linear Algebra and its Applica-tions vol 436 no 3 pp 571ndash578 2012

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 320392 7 pageshttpdxdoiorg1011552013320392

Research ArticleA New Construction of Multisender Authentication Codes fromPolynomials over Finite Fields

Xiuli Wang

College of Science Civil Aviation University of China Tianjin 300300 China

Correspondence should be addressed to Xiuli Wang xlwangcauceducn

Received 3 February 2013 Accepted 7 April 2013

Academic Editor Yang Zhang

Copyright copy 2013 Xiuli Wang This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Multisender authentication codes allow a group of senders to construct an authenticated message for a receiver such that thereceiver can verify the authenticity of the received message In this paper we construct one multisender authentication code frompolynomials over finite fields Some parameters and the probabilities of deceptions of this code are also computed

1 Introduction

Multisender authentication code was firstly constructed byGilbert et al [1] in 1974 Multisender authentication systemrefers towho a groupof senders cooperatively send amessageto a receiver then the receiver should be able to ascertainthat the message is authentic About this case many scholarsand researchers had made great contributions to multisenderauthentication codes such as [2ndash6]

In the actual computer network communications mul-tisender authentication codes include sequential model andsimultaneous model Sequential model is that each senderuses his own encoding rules to encode a source state orderlythe last sender sends the encoded message to the receiverand the receiver receives themessage and verifies whether themessage is legal or not Simultaneousmodel is that all sendersuse their own encoding rules to encode a source state andeach sender sends the encoded message to the synthesizerrespectively then the synthesizer forms an authenticatedmessage and verifies whether the message is legal or not Inthis paper we will adopt the second model

In a simultaneous model there are four participants agroup of senders 119880 = 119880

1 1198802 119880

119899 the key distribution

center he is responsible for the key distribution to sendersand receiver including solving the disputes between them areceiver 119877 and a synthesizer where he only runs the trustedsynthesis algorithm The code works as follows each senderand receiver has their own Cartesian authentication code

respectively Let (119878 119864119894 119879119894 119891119894) (119894 = 1 2 119899) be the sendersrsquo

Cartesian authentication code (119878 119864119877 119879 119892) be the receiverrsquos

Cartesian authentication code ℎ 1198791times 1198792times sdot sdot sdot times 119879

119899rarr

119879 be the synthesis algorithm and 120587119894

119864 rarr 119864119894be a

subkey generation algorithm where 119864 is the key set of thekey distribution center When authenticating a message thesenders and the receiver should comply with the protocolThe key distribution center randomly selects an encodingrule 119890 isin 119864 and sends 119890

119894= 120587119894(119890) to the 119894th sender 119880

119894(119894 =

1 2 119899) secretly then he calculates 119890119877by 119890 according to

an effective algorithm and secretly sends 119890119877to the receiver

119877 If the senders would like to send a source state 119904 to thereceiver 119877 119880

119894computes 119905

119894= 119891119894(119904 119890119894) (119894 = 1 2 119899) and

sends 119898119894= (119904 119905

119894) (119894 = 1 2 119899) to the synthesizer through

an open channel The synthesizer receives the message 119898119894=

(119904 119905119894) (119894 = 1 2 119899) and calculates 119905 = ℎ(119905

1 1199052 119905

119899) by the

synthesis algorithm ℎ and then sends message 119898 = (119904 119905) tothe receiver he checks the authenticity by verifying whether119905 = 119892(119904 119890

119877) or not If the equality holds the message is

authentic and is accepted Otherwise the message is rejectedWe assume that the key distribution center is credible and

though he know the sendersrsquo and receiverrsquos encoding rules hewill not participate in any communication activities Whentransmitters and receiver are disputing the key distributioncenter settles it At the same time we assume that the systemfollows the Kerckhoff principle in which except the actualused keys the other information of the whole system ispublic

2 Journal of Applied Mathematics

In a multisender authentication system we assume thatthe whole senders are cooperative to form a valid messagethat is all senders as a whole and receiver are reliable Butthere are some malicious senders who together cheat thereceiver the part of senders and receiver are not credible andthey can take impersonation attack and substitution attackIn the whole system we assume that 119880

1 1198802 119880

119899 are

senders 119877 is a receiver 119864119894is the encoding rules set of the

sender 119880119894 and 119864

119877is the decoding rules set of the receiver

119877 If the source state space 119878 and the key space 119864119877of receiver

119877 are according to a uniform distribution then the messagespace119872 and the tag space119879 are determined by the probabilitydistribution of 119878 and 119864

119877 119871 = 119894

1 1198942 119894

119897 sub 1 2 119899

119897 lt 119899 119880119871

= 1198801198941 1198801198942 119880

119894119897 119864119871

= 1198641198801198941

1198641198801198942

119864119880119894119897

Now consider that let us consider the attacks from maliciousgroups of senders Here there are two kinds of attack

The opponentrsquos impersonation attack to receiver119880119871 after

receiving their secret keys encode a message and send it tothe receiver 119880

119871are successful if the receiver accepts it as

legitimate message Denote by 119875119868the largest probability of

some opponentrsquos successful impersonation attack to receiverit can be expressed as

119875119868= max119898isin119872

1003816100381610038161003816119890119877 isin 119864119877| 119890119877sub 119898

10038161003816100381610038161003816100381610038161003816119864119877

1003816100381610038161003816

(1)

The opponentrsquos substitution attack to the receiver 119880119871

replace 119898 with another message 1198981015840 after they observe a

legitimatemessage119898119880119871are successful if the receiver accepts

it as legitimate message it can be expressed as

119875119878= max119898isin119872

max1198981015840=119898isin119872

10038161003816100381610038161003816119890119877isin 119864119877| 119890119877sub 119898119898

1015840

10038161003816100381610038161003816

1003816100381610038161003816119890119877 isin 119864119877| 119890119877sub 119898

1003816100381610038161003816

(2)

There might be 119897 malicious senders who together cheatthe receiver that is the part of senders and the receiverare not credible and they can take impersonation attackLet 119871 = 119894

1 1198942 119894

119897 sub 1 2 119899 119897 lt 119899 and 119864

119871=

1198641198801198941

1198641198801198942

119864119880119894119897

Assume that 119880119871

= 1198801198941 1198801198942 119880

119894119897

after receiving their secret keys send a message 119898 to thereceiver 119877119880

119871are successful if the receiver accepts it as legiti-

mate message Denote by 119875119880(119871) the maximum probability of

success of the impersonation attack to the receiver It can beexpressed as

119875119880(119871)

=max119890119871isin119864119871

max119890119871isin119890119880

max119898isin119872

1003816100381610038161003816119890119877isin119864119877 | 119890119877sub119898 119901 (119890119877 119890119875) =0

10038161003816100381610038161003816100381610038161003816119890119877 isin 119864

119877| 119901 (119890119877 119890119875) =0

1003816100381610038161003816

(3)

Notes119901(119890119877 119890119875) = 0 implies that any information 119904 encoded by

119890119879can be authenticated by 119890

119877

In [2] Desmedt et al gave two constructions for MRA-codes based on polynomials and finite geometries respec-tively To construct multisender or multireceiver authenti-cation by polynomials over finite fields many researchershave done much work for example [7ndash9] There are other

constructions of multisender authentication codes that aregiven in [3ndash6] The construction of authentication codesis combinational design in its nature We know that thepolynomial over finite fields can provide a better algebrastructure and is easy to count In this paper we constructone multisender authentication code from the polynomialover finite fields Some parameters and the probabilities ofdeceptions of this code are also computed We realize thegeneralization and the application of the similar idea andmethod of the paper [7ndash9]

2 Some Results about Finite Field

Let 119865119902be the finite field with 119902 elements where 119902 is a power

of a prime 119901 and 119865 is a field containing 119865119902 denote by 119865

lowast

119902be

the nonzero elements set of 119865119902 In this paper we will use the

following conclusions over finite fields

Conclusion 1 A generator 120572 of 119865lowast

119902is called a primitive

element of 119865119902

Conclusion 2 Let 120572 isin 119865119902 if some polynomials contain 120572 as

their root and their leading coefficient are 1 over 119865119902 then the

polynomial having least degree among all such polynomialsis called a minimal polynomial over 119865

119902

Conclusion 3 Let |119865| = 119902119899 then 119865 is an 119899-dimensional

vector space over 119865119902 Let 120572 be a primitive element of 119865

119902

and 119892(119909) the minimal polynomial about 120572 over 119865119902 then

dim119892(119909) = 119899 and 1 120572 1205722

120572119899minus1 is a basis of 119865 Further-

more 1 120572 1205722 120572119899minus1 is linear independent and it is equalto 120572 120572

2

120572119899minus1

120572119899 (120572 is a primitive element 120572 = 0) is also

linear independent moreover 120572119901 1205721199012

120572119901119899minus1

120572119901119899

is alsolinear independent

Conclusion 4 Consider (1199091+ 1199092+ sdot sdot sdot + 119909

119899)119898

= (1199091)119898

+

(1199092)119898

+ sdot sdot sdot + (119909119899)119898 where 119909

119894isin 119865119902 (1 le 119894 le 119899) and 119898 is a

nonnegative power of character 119901 of 119865119902

Conclusion 5 Let119898 le 119899 Then the number of119898times119899matricesof rank119898 over 119865

119902is 119902119898(119898minus1)2prod119899

119894=119899minus119898+1(119902119894

minus 1)

More results about finite fields can be found in [10ndash12]

3 Construction

Let the polynomial 119901119895(119909) = 119886

1198951119909119901119899

+ 1198861198952119909119901(119899minus1)

+ sdot sdot sdot +

119886119895119899119909119901

(1 le 119895 le 119896) where the coefficient 119886119894119897

isin 119865119902

(1 le 119897 le 119899) and these vectors by the composition of theircoefficient are linearly independent The set of source states119878 = 119865

119902 the set of 119894th transmitterrsquos encoding rules 119864

119880119894=

1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) 119909119894

isin 119865lowast

119902 (1 le 119894 le 119899) the set

of receiverrsquos encoding rules 119864119877

= 1199011(120572) 1199012(120572) 119901

119896(120572)

where 120572 is a primitive element of 119865119902 the set of 119894th transmit-

terrsquos tags 119879119894= 119905119894| 119905119894isin 119865119902 (1 le 119894 le 119899) the set of receiverrsquos

tags 119879 = 119905 | 119905 isin 119865119902

Journal of Applied Mathematics 3

Define the encoding map 119891119894 119878 times 119864

119880119894rarr 119879119894 119891119894(119904 119890119880119894) =

1199041199011(119909119894) + 1199042

1199012(119909119894) + sdot sdot sdot + 119904

119896

119901119896(119909119894) 1 le 119894 le 119899

The decoding map 119891 119878 times 119864119877

rarr 119879 119891(119904 119890119877) = 119904119901

1(120572) +

1199042

1199012(120572) + sdot sdot sdot + 119904

119896

119901119896(120572)

The synthesizing map ℎ 1198791times 1198792times sdot sdot sdot times 119879

119899rarr 119879

ℎ(1199051 1199052 119905

119899) = 1199051+ 1199052+ sdot sdot sdot + 119905

119899

The code works as followsAssume that 119902 is larger than or equal to the number of

the possible message and 119899 le 119902

31 Key Distribution The key distribution center randomlygenerates 119896 (119896 le 119899) polynomials 119901

1(119909) 1199012(119909) 119901

119896(119909)

where 119901119895(119909) = 119886

1198951119909119901119899

+ 1198861198952119909119901(119899minus1)

+ sdot sdot sdot + 119886119895119899119909119901

(1 le 119895 le 119896)and make these vectors by composed of their coefficient islinearly independent it is equivalent to the column vectors

of the matrix (

11988611 11988621 sdotsdotsdot 1198861198961

11988612 11988622 sdotsdotsdot 1198861198962

1198861119899 1198862119899 sdotsdotsdot 119886119896119899

) is linearly independent He

selects 119899 distinct nonzero elements 1199091 1199092 119909

119899isin 119865119902again

and makes 119909119894(1 le 119894 le 119899) secret then he sends privately

1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) to the sender 119880

119894(1 le 119894 le 119899) The

key distribution center also randomly chooses a primitiveelement 120572 of 119865

119902satisfying 119909

1+ 1199092+ sdot sdot sdot + 119909

119899= 120572 and sends

1199011(120572) 1199012(120572) 119901

119896(120572) to the receiver 119877

32 Broadcast If the senderswant to send a source state 119904 isin 119878

to the receiver 119877 the sender 119880119894calculates 119905

119894= 119891119894(119904 119890119880119894) =

119860119904(119909119894) = 119904119901

1(119909119894)+1199042

1199012(119909119894)+ sdot sdot sdot+119904

119896

119901119896(119909119894) 1 le 119894 le 119899 and then

sends 119860119904(119909119894) = 119905119894to the synthesizer

33 Synthesis After the synthesizer receives 1199051 1199052 119905

119899 he

calculates ℎ(1199051 1199052 119905

119899) = 1199051+ 1199052+ sdot sdot sdot + 119905

119899and then sends

119898 = (119904 119905) to the receiver 119877

34 Verification When the receiver 119877 receives 119898 = (119904 119905) hecalculates 1199051015840 = 119892(119904 119890

119877) = 119860

119904(120572) = 119904119901

1(120572) + 119904

2

1199012(120572) + sdot sdot sdot +

119904119896

119901119896(120572) If 119905 = 119905

1015840 he accepts 119905 otherwise he rejects itNext we will show that the above construction is a well

defined multisender authentication code with arbitration

Lemma 1 Let 119862119894= (119878 119864

119875119894 119879119894 119891119894) then the code is an A-code

1 le 119894 le 119899

Proof (1) For any 119890119880119894

isin 119864119880119894 119904 isin 119878 because 119864

119880119894= 1199011(119909119894)

1199012(119909119894) 119901

119896(119909119894) 119909119894isin 119865lowast

119902 so 119905

119894= 1199041199011(119909119894) + 1199042

1199012(119909119894) + sdot sdot sdot +

119904119896

119901119896(119909119894) isin 119879119894= 119865119902 Conversely for any 119905

119894isin 119879119894 choose 119890

119880119894=

1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) 119909119894isin 119865lowast

119902 where 119901

119895(119909) = 119886

1198951119909119901119899

+

1198861198952119909119901(119899minus1)

+ sdot sdot sdot + 119886119895119899119909119901

(1 le 119895 le 119896) and let 119905119894= 119891119894(119904 119890119880119894) =

1199041199011(119909119894) + 1199042

1199012(119909119894) + sdot sdot sdot + 119904

119896

119901119896(119909119894) it is equivalent to

(119909119901119899

119894 119909119901119899minus1

119894 119909

119901

119894)(

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)(

119904

1199042

119904119896

) = 119905119894

(4)

It follows that

(

(

119909119901119899

1119909119901119899minus1

1sdot sdot sdot 119909119901

1

119909119901n

2119909119901119899minus1

2sdot sdot sdot 119909119901

2

119909119901119899

119899119909119901119899minus1

119899sdot sdot sdot 119909119901

119899

)

)

times (

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)(

119904

1199042

119904119896

) = (

1199051

1199052

119905119899

)

(5)

Denote

119860 = (

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)

119883 = (

119909119901119899

1119909119901119899minus1

1sdot sdot sdot 119909119901

1

119909119901119899

2119909119901119899minus1

2sdot sdot sdot 119909119901

2

119909119901119899

119899119909119901119899minus1

119899sdot sdot sdot 119909119901

119899

)

119878 = (

119904

1199042

119904119896

) 119905 = (

1199051

1199052

119905119899

)

(6)

The above linear equation is equivalent to 119883119860119878 = 119905because the column vectors of 119860 are linearly independent119883 is equivalent to a Vandermonde matrix and 119883 is inversetherefore the above linear equation has a unique solution so119904 is only defined that is 119891

119894(1 le 119894 le 119899) is a surjection

(2) If 1199041015840 isin 119878 is another source state satisfying 1199041199011(119909119894) +

1199042

1199012(119909119894)+sdot sdot sdot+119904

119896

119901119896(119909119894) = 1199041015840

1199011(119909119894)+11990410158402

1199012(119909119894)+sdot sdot sdot+119904

1015840119896

119901119896(119909119894) =

119905119894 and it is equivalent to (119904 minus 119904

1015840

)1199011(119909119894) + (1199042

minus 11990410158402

)1199012(119909119894) + sdot sdot sdot +

(119904119896

minus 1199041015840119896

)119901119896(119909119894) = 0 then

(119909119901119899

119894 119909119901119899minus1

119894 119909

119901

119894)

times (

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)(

119904 minus 1199041015840

1199042

minus 11990410158402

119904119896

minus 1199041015840119896

) = 0

(7)

4 Journal of Applied Mathematics

Thus

(

(

119909119901119899

1119909119901119899minus1

1sdot sdot sdot 119909119901

1

119909119901n

2119909119901119899minus1

2sdot sdot sdot 119909119901

2

119909119901119899

119899119909119901119899minus1

119899sdot sdot sdot 119909119901

119899

)

)

times (

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)(

119904 minus 1199041015840

1199042

minus 11990410158402

119904119896

minus 1199041015840119896

) = (

0

0

0

)

(8)

Similar to (1) we know that the homogeneous linear equation119883119860119878 = 0 has a unique solution that is there is only zerosolution so 119904 = 119904

1015840 So 119904 is the unique source state determinedby 119890119880119894and 119905119894 thus 119862

119894(1 le 119894 le 119899) is an A-code

Lemma 2 Let 119862 = (119878 119864119877 119879 119892) then the code is an A-code

Proof (1) For any 119904 isin 119878 119890119877isin 119864119877 from the definition of 119890

119877

we assume that 119864119877

= 1199011(120572) 1199012(120572) 119901

119896(120572) where 120572 is a

primitive element of 119865119902 119892(119904 119890

119877) = 119904119901

1(120572) + 119904

2

1199012(120572) + sdot sdot sdot +

119904119896

119901119896(120572) isin 119879 = 119865

119902 on the other hand for any 119905 isin 119879 choose

119890119877= 1199011(120572) 1199012(120572) 119901

119896(120572) where 120572 is a primitive element

of 119865119902 119892(119904 119890

119877) = 119904119901

1(120572) + 119904

2

1199012(120572) + sdot sdot sdot + 119904

119896

119901119896(120572) = 119905 it is

equivalent to

(120572119901119899

120572119901119899minus1

sdot sdot sdot 120572119901

)

times (

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)(

119904

1199042

119904119896

) = 119905

119860 = (

11988611

11988621

sdot sdot sdot 1198861198961

11988612

11988622

sdot sdot sdot 1198861198962

1198861119899

1198862119899

sdot sdot sdot 119886119896119899

)

(9)

that is (120572119901119899

120572119901119899minus1

120572119901

)119860(

119904

1199042

119904119896

) = 119905 From Conclusion

3 we know that (120572119901119899

120572119901119899minus1

120572119901

) is linearly independentand the column vectors of 119860 are also linearly independenttherefore the above linear equation has unique solution so 119904

is only defined that is 119892 is a surjection

(2) If 1199041015840 is another source state satisfying 119905 = 119892(1199041015840

119890119877) then

(120572119901119899

120572119901119899minus1

120572119901

)119860(

1199041015840

11990410158402

1199041015840119896

)

= (120572119901119899

120572119901119899minus1

120572119901

)119860(

119904

1199042

119904119896

)

(10)

that is (120572119901119899

120572119901119899minus1

120572119901

)119860([[

[

1199041015840

11990410158402

1199041015840119896

]]

]

minus [

[

119904

1199042

119904119896

]

]

) = 0 Sim-

ilar to (1) we get that the homogeneous linear equation(120572119901119899

120572119901119899minus1

120572119901

)119860(1198781015840

minus 119878) = 0 has a unique solution thatis there is only zero solution so 119878 = 119878

1015840 that is 119904 = 1199041015840 So

119904 is the unique source state determined by 119890119877and 119905 thus

119862 = (119878 119864119877 119879 119892) is an A-code

At the same time for any valid119898 = (119904 119905) we have knownthat 120572 = 119909

1+ 1199092+ sdot sdot sdot + 119909

119899 and it follows that 1199051015840 = 119904119901

1(120572) +

1199042

1199012(120572)+sdot sdot sdot+119904

119896

119901119896(120572) = 119904119901

1(1199091+1199092+sdot sdot sdot+119909

119899)+1199042

1199012(1199091+1199092+

sdot sdot sdot + 119909119899) + sdot sdot sdot + 119904

119896

119901119896(1199091+ 1199092+ sdot sdot sdot + 119909

119899) We also have known

that 119901119895(119909) = 119886

1198951119909119901119899

+1198861198952119909119901(119899minus1)

+ sdot sdot sdot + 119886119895119899119909119901

(1 le 119895 le 119896) fromConclusion 4 (119909

1+ 1199092+ sdot sdot sdot + 119909

119899)119901119898

= (1199091)119901119898

+ (1199092)119901119898

+ sdot sdot sdot +

(119909119899)119901119898

where119898 is a nonnegative power of character 119901 of 119865119902

andwe get119901119895(1199091+1199092+sdot sdot sdot+119909

119899) = 119901119895(1199091)+119901119895(1199092)+sdot sdot sdot+119901

119895(119909119899)

therefore 1199051015840 = 1199041199011(120572) + 119904

2

1199012(120572) + sdot sdot sdot + 119904

119896

119901119896(120572) = (119904119901

1(1199091) +

1199042

1199012(1199091)+sdot sdot sdot+119904

119896

119901119896(1199091))+(119904119901

1(1199092)+1199042

1199012(1199092)+sdot sdot sdot+119904

119896

119901119896(1199092))+

sdot sdot sdot+(1199041199011(119909119899)+1199042

1199012(119909119899)+ sdot sdot sdot+119904

119896

119901119896(119909119899)) = 1199051+1199052+sdot sdot sdot+119905

119899= 119905

and the receiver 119877 accepts119898

From Lemmas 1 and 2 we know that such construction ofmultisender authentication codes is reasonable and there are119899 senders in this system Next we compute the parametersof this code and the maximum probability of success inimpersonation attack and substitution attack by the group ofsenders

Theorem 3 Some parameters of this construction are|119878| = 119902 |119864

119880119894| = [119902

119896(119896minus1)2

prod119899

119894=119899minus119896+1(119902119894

minus 1)] (119902minus1

1) =

[119902119896(119896minus1)2

prod119899

119894=119899minus119896+1(119902119894

minus 1)](119902 minus 1) (1 le 119894 le 119899) |119879119894| = 119902 (1 le

119894 le 119899) |119864119877| = [119902

119896(119896minus1)2

prod119899

119894=119899minus119896+1(119902119894

minus 1)]120593(119902 minus 1) |119879| = 119902Where 120593(119902 minus 1) is the 119864119906119897119890119903 119891119906119899119888119905119894119900119899 of 119902 minus 1 it representsthe number of primitive element of 119865

119902here

Proof For |119878| = 119902 |119879119894| = 119902 and |119879| = 119902 the results

are straightforward For119864119880119894 because119864

119880119894=1199011(119909119894) 1199012(119909119894)

119901119896(119909119894) 119909119894isin 119865lowast

119902 where 119901

119895(119909) = 119886

1198951119909119901119899

+ 1198861198952119909119901(119899minus1)

+ sdot sdot sdot +

119886119895119899119909119901

(1 le 119895 le 119896) and these vectors by the composition oftheir coefficient are linearly independent it is equivalent to

the columns of 119860 = (

11988611 11988621 sdotsdotsdot 1198861198961

11988612 11988622 sdotsdotsdot 1198861198962

1198861119899 1198862119899 sdotsdotsdot 119886119896119899

) is linear independent

Journal of Applied Mathematics 5

From Conclusion 5 we can conclude that the number of 119860satisfying the condition is 119902

119896(119896minus1)2

prod119899

119894=119899minus119896+1(119902119894

minus 1) On theother hand the number of distinct nonzero elements 119909

119894(1 le

119894 le 119899) in 119865119902is ( 119902minus11

) = 119902 minus 1 so |119864119880119894| = [119902

119896(119896minus1)2

prod119899

119894=119899minus119896+1(119902119894

minus

1)](119902 minus 1) For 119864119877 119864119877

= 1199011(120572) 1199012(120572) 119901

119896(120572) where 120572

is a primitive element of 119865119902 For 120572 from Conclusion 1 a

generator of119865lowast119902is called a primitive element of119865

119902 |119865lowast119902| = 119902minus1

by the theory of the group we know that the number ofgenerator of 119865lowast

119902is 120593(119902minus1) that is the number of 120572 is 120593(119902minus1)

For 1199011(119909) 1199012(119909) 119901

119896(119909) From above we have confirmed

that the number of these polynomials is 119902119896(119896minus1)2prod119899119894=119899minus119896+1

(119902119894

minus

1) therefore |119864119877| = [119902

119896(119896minus1)2

prod119899

119894=119899minus119896+1(119902119894

minus 1)]120593(119902 minus 1)

Lemma 4 For any 119898 isin 119872 the number of 119890119877contained 119898 is

120593(119902 minus 1)

Proof Let 119898 = (119904 119905) isin 119872 119890119877

= 1199011(120572) 1199012(120572) 119901

119896(120572)

where 120572 is a primitive element of 119865119902 isin 119864

119877 If 119890119877

sub 119898then 119904119901

1(120572) + 119904

2

1199012(120572) + sdot sdot sdot + 119904

119896

119901119896(120572) = 119905 hArr (120572

119901119899

120572119901119899minus1

120572119901

)119860(

119904

1199042

119904119896

) = 119905 For any 120572 suppose that there is

another 1198601015840 such that (120572

119901119899

120572119901119899minus1

120572119901

)1198601015840

(

119904

1199042

119904119896

) = 119905

then (120572119901119899

120572119901119899minus1

120572119901

)(119860 minus 1198601015840

)(

119904

1199042

119904119896

) = 0 because

120572119901119899

120572119901119899minus1

120572119901 is linearly independent so (119860minus119860

1015840

)(

119904

1199042

119904119896

) =

0 but (

119904

1199042

119904119896

) is arbitrarily therefore 119860 minus 1198601015840

= 0 that is

119860 = 1198601015840 and it follows that 119860 is only determined by 120572

Therefore as 120572 isin 119864119877 for any given 119904 and 119905 the number of

119890119877contained in119898 is 120593(119902 minus 1)

Lemma 5 For any119898 = (119904 119905) isin 119872 and1198981015840

= (1199041015840

1199051015840

) isin 119872 with119904 = 1199041015840 the number of 119890

119877contained119898 and119898

1015840 is 1

Proof Assume that 119890119877

= 1199011(120572) 1199012(120572) 119901

119896(120572) where

120572 is a primitive element of 119865119902 isin 119864

119877 If 119890119877

sub 119898 and119890119877

sub 1198981015840 then 119904119901

1(120572) + 119904

2

1199012(120572) + sdot sdot sdot + 119904

119896

119901119896(120572) = 119905 hArr

(120572119901119899

120572119901119899minus1

120572119901

)119860(

119904

1199042

119904119896

) = 119905 11990410158401199011(120572) + 119904

10158402

1199012(120572) + sdot sdot sdot +

1199041015840119896

119901119896(120572) = 119905 hArr (120572

119901119899

120572119901119899minus1

120572119901

)119860(

1199041015840

11990410158402

1199041015840119896

) = 119905 It is

equivalent to (120572119901119899

120572119901119899minus1

120572119901

)119860(

119904minus1199041015840

1199042minus11990410158402

119904119896minus1199041015840119896

) = 119905 minus 1199051015840 because

119904 = 1199041015840 so 119905 = 119905

1015840 otherwise we assume that 119905 = 1199051015840 and

since 120572119901119899

120572119901119899minus1

120572119901 and the column vectors of 119860 both

are linearly independent it forces that 119904 = 1199041015840 this is a

contradiction Therefore we get

(119905 minus 1199051015840

)minus1

[[[[

[

(120572119901119899

120572119901119899minus1

120572119901

)119860 (

119904 minus 1199041015840

1199042

minus 11990410158402

119904119896

minus 1199041015840119896

)

]]]]

]

= 1

(lowast)

since 119905 1199051015840 is given (119905 minus 119905

1015840

)minus1 is unique by equation (lowast) for

any given 119904 1199041015840 and 119905 119905

1015840 we obtain that (120572119901119899

120572119901119899minus1

120572119901

)119860 isonly determined thus the number of 119890

119877contained119898 and119898

1015840

is 1

Lemma6 For any fixed 119890119880= 1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) 119909119894isin

119865lowast

119902 (1 le 119894 le 119899) containing a given 119890

119871 then the number of 119890

119877

which is incidence with 119890119880is 120593(119902 minus 1)

Proof For any fixed 119890119880

= 1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) 119909119894isin

119865lowast

119902 (1 le 119894 le 119899) containing a given 119890

119871 we assume that

119901119895(119909119894) = 1198861198951119909119901119899

119894+1198861198952119909119901(119899minus1)

119894+sdot sdot sdot+119886

119895119899119909119901

119894(1 le 119895 le 119896 1 le 119894 le 119899)

119890119877= 1199011(120572) 1199012(120572) 119901

119896(120572) where 120572 is a primitive element

of 119865119902 From the definitions of 119890

119877and 119890119880and Conclusion 4

we can conclude that 119890119877is incidence with 119890

119880if and only if

1199091+ 1199092+ sdot sdot sdot + 119909

119899= 120572 For any 120572 since Rank(1 1 1) =

Rank(1 1 1 120572) = 1 lt 119899 so the equation1199091+1199092+sdot sdot sdot+119909

119899=

120572 always has a solution From the proof of Theorem 3 weknow the number of 119890

119877which is incident with 119890

119880(ie the

number of all 119864119877) is [119902119896(119896minus1)2prod119899

119894=119899minus119896+1(119902119894

minus 1)] 120593(119902 minus 1)

Lemma 7 For any fixed 119890119880= 1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) 119909119894isin

119865lowast

119902 (1 le 119894 le 119899) containing a given 119890

119871and119898 = (119904 119905) the num-

ber of 119890119877which is incidence with 119890

119880and contained in119898 is 1

Proof For any 119904 isin 119878 119890119877

isin 119864119877 we assume that 119890

119877=

1199011(120572) 1199012(120572) 119901

119896(120572) where 120572 is a primitive element

of 119865119902 Similar to Lemma 6 for any fixed 119890

119880=

1199011(119909119894) 1199012(119909119894) 119901

119896(119909119894) 119909119894isin 119865lowast

119902 (1 le 119894 le 119899) containing

a given 119890119871 we have known that 119890

119877is incident with 119890

119880if and

only if

1199091+ 1199092+ sdot sdot sdot + 119909

119899= 120572 (11)

Again with 119890119877sub 119898 we can get

1199041199011(120572) + 119904

2

1199012(120572) + sdot sdot sdot + 119904

119896

119901119896(120572) = 119905 (12)

By (11) and (12) and the property of 119901119895(119909) (1 le 119895 le 119896) we

have the following conclusion

1199041199011(

119899

sum

119894=1

119909119894) + 1199042

1199012(

119899

sum

119894=1

119909119894) + sdot sdot sdot + 119904

119896

119901119896(

119899

sum

119894=1

119909119894)

= 119905 lArrrArr (1199011(

119899

sum

119894=1

119909119894) 1199012(

119899

sum

119894=1

119909119894) 119901

119896(

119899

sum

119894=1

119909119894))

6 Journal of Applied Mathematics

times (

119904

1199042

119904119896

)

= 119905 lArrrArr (

119899

sum

119894=1

1199011(119909119894)

119899

sum

119894=1

1199012(119909119894)

119899

sum

119894=1

119901119896(119909119894))(

119904

s2119904119896

)

=119905lArrrArr( (

119899

sum

119894=1

119909119894)

119899

(

119899

sum

119894=1

119909119894)

119899minus1

(

119899

sum

119894=1

119909119894))119860(

119904

1199042

119904119896

)

= 119905 lArrrArr [(

119899

sum

119894=1

1199011(119909119894)

119899

sum

119894=1

1199012(119909119894)

119899

sum

119894=1

119901119896(119909119894))

minus((

119899

sum

119894=1

119909119894)

119899

(

119899

sum

119894=1

119909119894)

119899minus1

(

119899

sum

119894=1

119909119894))119860]

times (

119904

1199042

119904119896

) = 0

(13)

because 119904 is any given Similar to the proof of Lemma 4we can get (sum

119899

119894=11199011(119909119894) sum119899

119894=11199012(119909119894) sum

119899

119894=1119901119896(119909119894)) minus

((sum119899

119894=1119909119894)119899(sum119899119894=1

119909119894)119899minus1

(sum119899

119894=1119909119894))119860=0 that is ((sum119899

119894=1119909119894)119899

(sum119899

119894=1119909119894)119899minus1

(sum119899

119894=1119909119894))119860 = (sum

119899

119894=11199011(119909119894) sum119899

119894=11199012(119909119894)

sum119899

119894=1119901119896(119909119894)) but 119901

1(119909119894) 1199012(119909119894) 119901

119896(119909119894) and 119909

119894(1 le 119894 le 119899)

also are fixed thus 120572 and 119860are only determined so thenumber of 119890

119877which is incident with 119890

119880and contained in 119898

is 1

Theorem 8 In the constructed multisender authenticationcodes if the sendersrsquo encoding rules and the receiverrsquos decodingrules are chosen according to a uniform probability distribu-tion then the largest probabilities of success for different typesof deceptions respectively are

119875119868=

1

119902119896(119896minus1)2prod119899

119894=119899minus119896+1(119902119894 minus 1)

119875119878=

1

120593 (119902 minus 1)

119875119880(119871) =

1

[119902119896(119896minus1)2prod119899

119894=119899minus119896+1(119902119894 minus 1)] 120593 (119902 minus 1)

(14)

Proof By Theorem 3 and Lemma 4 we get

119875119868= max119898isin119872

1003816100381610038161003816119890119877 isin 119864119877| 119890119877sub 119898

10038161003816100381610038161003816100381610038161003816119864119877

1003816100381610038161003816

=120593 (119902 minus 1)

[119902119896(119896minus1)2prod119899

119894=119899minus119896+1(119902119894 minus 1)] 120593 (119902 minus 1)

=1

119902119896(119896minus1)2prod119899

119894=119899minus119896+1(119902119894 minus 1)

(15)

By Lemmas 4 and 5 we get

119875119878= max119898isin119872

max1198981015840=119898isin119872

10038161003816100381610038161003816119890119877isin 119864119877| 119890119877sub 119898119898

1015840

10038161003816100381610038161003816

1003816100381610038161003816119890119877 isin 119864119877| 119890119877sub 119898

1003816100381610038161003816

=1

120593 (119902 minus 1)

(16)

By Lemmas 6 and 7 we get

119875119880(119871)

=max119890119871isin119864119871

max119890119871isin119890119880

max119898isin119872

1003816100381610038161003816119890119877 isin 119864119877| 119890119877sub119898 119901 (119890

119877 119890119875) =0

10038161003816100381610038161003816100381610038161003816119890119877 isin 119864

119877| 119901 (119890119877 119890119875) =0

1003816100381610038161003816

=1

[119902119896(119896minus1)2prod119899

119894=119899minus119896+1(119902119894 minus 1)] 120593 (119902 minus 1)

(17)

Acknowledgments

This paper is supported by the NSF of China (61179026)and the Fundamental Research of the Central Universi-ties of China Civil Aviation University of Science special(ZXH2012k003)

References

[1] E N Gilbert F J MacWilliams and N J A Sloane ldquoCodeswhich detect deceptionrdquoThe Bell System Technical Journal vol53 pp 405ndash424 1974

[2] Y Desmedt Y Frankel and M Yung ldquoMulti-receivermulti-sender network security efficient authenticated multicastfeedbackrdquo in Proceedings of the the 11th Annual Conference of theIEEEComputer andCommunications Societies (Infocom rsquo92) pp2045ndash2054 May 1992

[3] K Martin and R Safavi-Naini ldquoMultisender authenticationschemes with unconditional securityrdquo in Information and Com-munications Security vol 1334 of Lecture Notes in ComputerScience pp 130ndash143 Springer Berlin Germany 1997

[4] W Ma and X Wang ldquoSeveral new constructions of multitrasmitters authentication codesrdquo Acta Electronica Sinica vol28 no 4 pp 117ndash119 2000

[5] G J Simmons ldquoMessage authentication with arbitration oftransmitterreceiver disputesrdquo in Advances in CryptologymdashEUROCRYPT rsquo87 Workshop on theTheory and Application of ofCryptographic Techniques vol 304 of Lecture Notes in ComputerScience pp 151ndash165 Springer 1988

[6] S Cheng and L Chang ldquoTwo constructions of multi-senderauthentication codes with arbitration based linear codes to bepublished in rdquoWSEAS Transactions on Mathematics vol 11 no12 2012

Journal of Applied Mathematics 7

[7] R Safavi-Naini and H Wang ldquoNew results on multi-receiver authentication codesrdquo in Advances in CryptologymdashEUROCRYPT rsquo98 (Espoo) vol 1403 of Lecture Notes in ComputSci pp 527ndash541 Springer Berlin Germany 1998

[8] R Aparna and B B Amberker ldquoMulti-sender multi-receiverauthentication for dynamic secure group communicationrdquoInternational Journal of Computer Science andNetwork Securityvol 7 no 10 pp 47ndash63 2007

[9] R Safavi-Naini and H Wang ldquoBounds and constructions formultireceiver authentication codesrdquo inAdvances in cryptologymdashASIACRYPTrsquo98 (Beijing) vol 1514 of Lecture Notes in ComputSci pp 242ndash256 Springer Berlin Germany 1998

[10] S Shen and L Chen Information and Coding Theory Sciencepress in China 2002

[11] J J Rotman Advanced Modern Algebra High Education Pressin China 2004

[12] Z Wan Geometry of Classical Group over Finite Field SciencePress in Beijing New York NY USA 2002

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 435730 8 pageshttpdxdoiorg1011552013435730

Research ArticleGeometrical and Spectral Properties of the OrthogonalProjections of the Identity

Luis Gonzaacutelez Antonio Suaacuterez and Dolores Garciacutea

Department of Mathematics Research Institute SIANI University of Las Palmas de Gran Canaria Campus de Tafira35017 Las Palmas de Gran Canaria Spain

Correspondence should be addressed to Luis Gonzalez luisglezdmaulpgces

Received 28 December 2012 Accepted 10 April 2013

Academic Editor K Sivakumar

Copyright copy 2013 Luis Gonzalez et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We analyze the best approximation119860119873 (in the Frobenius sense) to the identitymatrix in an arbitrarymatrix subspace119860119878 (119860 isin R119899times119899

nonsingular 119878 being any fixed subspace of R119899times119899) Some new geometrical and spectral properties of the orthogonal projection 119860119873are derived In particular new inequalities for the trace and for the eigenvalues of matrix119860119873 are presented for the special case that119860119873 is symmetric and positive definite

1 Introduction

The set of all 119899 times 119899 real matrices is denoted by R119899times119899 and119868 denotes the identity matrix of order 119899 In the following119860119879 and tr(119860) denote as usual the transpose and the trace

of matrix 119860 isin R119899times119899 The notations ⟨sdot sdot⟩119865and sdot

119865stand

for the Frobenius inner product and matrix norm definedon the matrix space R119899times119899 Throughout this paper the termsorthogonality angle and cosine will be used in the sense ofthe Frobenius inner product

Our starting point is the linear system

119860119909 = 119887 119860 isin R119899times119899

119909 119887 isin R119899

(1)

where 119860 is a large nonsingular and sparse matrix Theresolution of this system is usually performed by iterativemethods based on Krylov subspaces (see eg [1 2]) Thecoefficient matrix 119860 of the system (1) is often extremelyill-conditioned and highly indefinite so that in this caseKrylov subspace methods are not competitive without agood preconditioner (see eg [2 3]) Then to improve theconvergence of these Krylov methods the system (1) canbe preconditioned with an adequate nonsingular precondi-tioning matrix 119873 transforming it into any of the equivalentsystems

119873119860119909 = 119873119887

119860119873119910 = 119887 119909 = 119873119910

(2)

the so-called left and right preconditioned systems respec-tively In this paper we address only the case of the right-handside preconditioned matrices 119860119873 but analogous results canbe obtained for the left-hand side preconditioned matrices119873119860

The preconditioning of the system (1) is often performedin order to get a preconditioned matrix 119860119873 as close aspossible to the identity in some sense and the preconditioner119873 is called an approximate inverse of119860 The closeness of119860119873to 119868may bemeasured by using a suitablematrix norm like forinstance the Frobenius norm [4] In this way the problemof obtaining the best preconditioner 119873 (with respect to theFrobenius norm) of the system (1) in an arbitrary subspace119878 of R119899times119899 is equivalent to the minimization problem see forexample [5]

min119872isin119878

119860119872 minus 119868119865= 119860119873 minus 119868

119865 (3)

The solution 119873 to the problem (3) will be referred to asthe ldquooptimalrdquo or the ldquobestrdquo approximate inverse of matrix 119860in the subspace 119878 Since matrix119860119873 is the best approximationto the identity in subspace 119860119878 it will be also referred toas the orthogonal projection of the identity matrix onto thesubspace 119860119878 Although many of the results presented in thispaper are also valid for the case thatmatrix119873 is singular fromnow on we assume that the optimal approximate inverse 119873(and thus also the orthogonal projection119860119873) is a nonsingular

2 Journal of Applied Mathematics

matrix The solution 119873 to the problem (3) has been studiedas a natural generalization of the classical Moore-Penroseinverse in [6] where it has been referred to as the 119878-Moore-Penrose inverse of matrix 119860

The main goal of this paper is to derive new geometricaland spectral properties of the best approximations119860119873 (in thesense of formula (3)) to the identity matrix Such propertiescould be used to analyze the quality and theoretical effective-ness of the optimal approximate inverse119873 as preconditionerof the system (1) However it is important to highlightthat the purpose of this paper is purely theoretical and weare not looking for immediate numerical or computationalapproaches (although our theoretical results could be poten-tially applied to the preconditioning problem) In particularthe term ldquooptimal (or best) approximate inverserdquo is used inthe sense of formula (3) and not in any other sense of thisexpression

Among the many different works dealing with practicalalgorithms that can be used to compute approximate inverseswe refer the reader to for example [4 7ndash9] and to thereferences therein In [4] the author presents an exhaustivesurvey of preconditioning techniques and in particulardescribes several algorithms for computing sparse approxi-mate inverses based on Frobenius norm minimization likefor instance the well-known SPAI and FSAI algorithms Adifferent approach (which is also focused on approximateinverses based on minimizing 119860119872 minus 119868

119865) can be found

in [7] where an iterative descent-type method is used toapproximate each column of the inverse and the iterationis done with ldquosparse matrix by sparse vectorrdquo operationsWhen the system matrix is expressed in block-partitionedform some preconditioning options are explored in [8] In[9] the idea of ldquotargetrdquo matrix is introduced in the contextof sparse approximate inverse preconditioners and the gen-eralized Frobenius norms 1198612

119865119867= tr(119861119867119861119879) (119867 symmetric

positive definite) are used for minimization purposes as analternative to the classical Frobenius norm

The last results of our work are devoted to the special casethat matrix 119860119873 is symmetric and positive definite In thissense let us recall that the cone of symmetric and positivedefinite matrices has a rich geometrical structure and in thiscontext the angle that any symmetric and positive definitematrix forms with the identity plays a very important role[10] In this paper the authors extend this geometrical pointof view and analyze the geometrical structure of the subspaceof symmetric matrices of order 119899 including the location of allorthogonal matrices not only the identity matrix

This paper has been organized as follows In Section 2we present some preliminary results required to make thepaper self-contained Sections 3 and 4 are devoted to obtainnew geometrical and spectral relations respectively for theorthogonal projections 119860119873 of the identity matrix FinallySection 5 closes the paper with its main conclusions

2 Some Preliminaries

Now we present some preliminary results concerning theorthogonal projection 119860119873 of the identity onto the matrix

subspace119860119878 sub R119899times119899 For more details about these results andfor their proofs we refer the reader to [5 6 11]

Taking advantage of the prehilbertian character of thematrix Frobenius norm the solution 119873 to the problem (3)can be obtained using the orthogonal projection theoremMore precisely the matrix product 119860119873 is the orthogonalprojection of the identity onto the subspace119860119878 and it satisfiesthe conditions stated by the following lemmas see [5 11]

Lemma 1 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

0 le 1198601198732

119865= tr (119860119873) le 119899 (4)

0 le 119860119873 minus 1198682

119865= 119899 minus tr (119860119873) le 119899 (5)

An explicit formula for matrix119873 can be obtained by ex-pressing the orthogonal projection119860119873 of the identity matrixonto the subspace 119860119878 by its expansion with respect to anorthonormal basis of 119860119878 [5] This is the idea of the followinglemma

Lemma 2 Let 119860 isin R119899times119899 be nonsingular Let 119878 be a linearsubspace of R119899times119899 of dimension 119889 and 119872

1 119872

119889 a basis of 119878

such that 1198601198721 119860119872

119889 is an orthogonal basis of 119860119878 Then

the solution119873 to the problem (3) is

119873 =

119889

sum

119894=1

tr (119860119872119894)

10038171003817100381710038171198601198721198941003817100381710038171003817

2

119865

119872119894 (6)

and the minimum (residual) Frobenius norm is

119860119873 minus 1198682

119865= 119899 minus

119889

sum

119894=1

[tr (119860119872119894)]2

10038171003817100381710038171198601198721198941003817100381710038171003817

2

119865

(7)

Let us mention two possible options both taken from [5]for choosing in practice the subspace 119878 and its correspondingbasis 119872

119894119889

119894=1 The first example consists of considering the

subspace 119878 of 119899times119899matrices with a prescribed sparsity patternthat is

119878 = 119872 isin R119899times119899

119898119894119895= 0 forall (119894 119895) notin 119870

119870 sub 1 2 119899 times 1 2 119899

(8)

Then denoting by119872119894119895 the 119899 times 119899matrix whose only nonzero

entry is 119898119894119895

= 1 a basis of subspace 119878 is clearly 119872119894119895

(119894 119895) isin 119870 and then 119860119872119894119895

(119894 119895) isin 119870 will be a basisof subspace 119860119878 (since we have assumed that matrix 119860 isnonsingular) In general this basis of 119860119878 is not orthogonalso that we only need to use the Gram-Schmidt procedureto obtain an orthogonal basis of 119860119878 in order to apply theorthogonal expansion (6)

For the second example consider a linearly independentset of 119899 times 119899 real symmetric matrices 119875

1 119875

119889 and the

corresponding subspace

1198781015840

= span 1198751119860119879

119875119889119860119879

(9)

Journal of Applied Mathematics 3

which clearly satisfies

1198781015840

sube 119872 = 119875119860119879

119875 isin R119899times119899 119875119879 = 119875

= 119872 isin R119899times119899 (119860119872)119879

= 119860119872

(10)

Hence we can explicitly obtain the solution 119873 to theproblem (3) for subspace 1198781015840 from its basis 119875

1119860119879

119875119889119860119879

as follows If 119860119875

1119860119879

119860119875119889119860119879

is an orthogonal basis ofsubspace 1198601198781015840 then we just use the orthogonal expansion (6)for obtaining119873 Otherwise we use again the Gram-Schmidtprocedure to obtain an orthogonal basis of subspace1198601198781015840 andthenwe apply formula (6)The interest of this second examplestands in the possibility of using the conjugate gradientmethod for solving the preconditioned linear system whenthe symmetric matrix 119860119873 is positive definite For a moredetailed exposition of the computational aspects related tothese two examples we refer the reader to [5]

Now we present some spectral properties of the orthog-onal projection 119860119873 From now on we denote by 120582

119894119899

119894=1and

120590119894119899

119894=1the sets of eigenvalues and singular values respectively

of matrix119860119873 arranged as usual in nonincreasing order thatis

100381610038161003816100381612058211003816100381610038161003816 ge

100381610038161003816100381612058221003816100381610038161003816 ge sdot sdot sdot ge

10038161003816100381610038161205821198991003816100381610038161003816 gt 0

1205901ge 1205902ge sdot sdot sdot ge 120590

119899gt 0

(11)

The following lemma [11] provides some inequalitiesinvolving the eigenvalues and singular values of the precon-ditioned matrix 119860119873

Lemma 3 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Then

119899

sum

119894=1

1205822

119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816

2

le

119899

sum

119894=1

1205902

119894= 119860119873

2

119865

= tr (119860119873) =

119899

sum

119894=1

120582119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816 le

119899

sum

119894=1

120590119894

(12)

The following fact [11] is a direct consequence ofLemma 3

Lemma 4 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Then the smallest singular value and thesmallest eigenvaluersquos modulus of the orthogonal projection 119860119873of the identity onto the subspace 119860119878 are never greater than 1That is

0 lt 120590119899le1003816100381610038161003816120582119899

1003816100381610038161003816 le 1 (13)

The following theorem [11] establishes a tight connectionbetween the closeness of matrix 119860119873 to the identity matrixand the closeness of 120590

119899(|120582119899|) to the unity

Theorem 5 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

(1 minus1003816100381610038161003816120582119899

1003816100381610038161003816)2

le (1 minus 120590119899)2

le 119860119873 minus 1198682

119865

le 119899 (1 minus1003816100381610038161003816120582119899

1003816100381610038161003816

2

) le 119899 (1 minus 1205902

119899)

(14)

Remark 6 Theorem 5 states that the closer the smallestsingular value 120590

119899of matrix 119860119873 is to the unity the closer

matrix 119860119873 will be to the identity that is the smaller119860119873 minus 119868

119865will be and conversely The same happens with

the smallest eigenvaluersquos modulus |120582119899| of matrix119860119873 In other

words we get a good approximate inverse 119873 of 119860 when 120590119899

(|120582119899|) is sufficiently close to 1

To finish this section let us mention that recently lowerand upper bounds on the normalized Frobenius conditionnumber of the orthogonal projection 119860119873 of the identityonto the subspace 119860119878 have been derived in [12] In additionthis work proposes a natural generalization (related to anarbitrary matrix subspace 119878 of R119899times119899) of the normalizedFrobenius condition number of the nonsingular matrix 119860

3 Geometrical Properties

In this section we present some new geometrical propertiesfor matrix 119860119873 119873 being the optimal approximate inverse ofmatrix 119860 defined by (3) Our first lemma states some basicproperties involving the cosine of the angle between matrix119860119873 and the identity that is

cos (119860119873 119868) =⟨119860119873 119868⟩

119865

119860119873119865119868119865

=tr (119860119873)

119860119873119865radic119899

(15)

Lemma 7 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

cos (119860119873 119868) =tr (119860119873)

119860119873119865radic119899

=119860119873

119865

radic119899=radictr (119860119873)

radic119899 (16)

0 le cos (119860119873 119868) le 1 (17)

119860119873 minus 1198682

119865= 119899 (1 minus cos2 (119860119873 119868)) (18)

Proof First using (15) and (4) we immediately obtain (16)As a direct consequence of (16) we derive that cos(119860119873 119868) isalways nonnegative Finally using (5) and (16) we get

119860119873 minus 1198682

119865= 119899 minus tr (119860119873) = 119899 (1 minus cos2 (119860119873 119868)) (19)

and the proof is concluded

Remark 8 In [13] the authors consider an arbitrary approxi-mate inverse119876 of matrix119860 and derive the following equality

119860119876 minus 1198682

119865= (119860119876

119865minus 119868119865)2

+ 2 (1 minus cos (119860119876 119868)) 119860119876119865119868119865

(20)

4 Journal of Applied Mathematics

that is the typical decomposition (valid in any inner productspace) of the strong convergence into the convergence ofthe norms (119860119876

119865minus 119868119865)2 and the weak convergence (1 minus

cos(119860119876 119868))119860119876119865119868119865 Note that for the special case that 119876

is the optimal approximate inverse119873 defined by (3) formula(18) has stated that the strong convergence is reduced justto the weak convergence and indeed just to the cosinecos(119860119873 119868)

Remark 9 More precisely formula (18) states that the closercos(119860119873 119868) is to the unity (ie the smaller the angle ang(119860119873 119868)

is) the smaller 119860119873 minus 119868119865will be and conversely This gives

us a new measure of the quality (in the Frobenius sense) ofthe approximate inverse 119873 of matrix 119860 by comparing theminimum residual norm 119860119873 minus 119868

119865with the cosine of the

angle between 119860119873 and the identity instead of with tr(119860119873)119860119873

119865(Lemma 1) or 120590

119899 |120582119899| (Theorem 5) So for a fixed

nonsingular matrix 119860 isin R119899times119899 and for different subspaces119878 sub R119899times119899 we have

tr (119860119873) 119899 lArrrArr 119860119873119865 radic119899 lArrrArr 120590

119899 1 lArrrArr

10038161003816100381610038161205821198991003816100381610038161003816 1

lArrrArr cos (119860119873 119868) 1 lArrrArr 119860119873 minus 119868119865 0

(21)

Obviously the optimal theoretical situation corresponds tothe case

tr (119860119873) = 119899 lArrrArr 119860119873119865= radic119899 lArrrArr 120590

119899= 1 lArrrArr 120582

119899= 1

lArrrArr cos (119860119873 119868) = 1 lArrrArr 119860119873 minus 119868119865= 0

lArrrArr 119873 = 119860minus1

lArrrArr 119860minus1

isin 119878

(22)

Remark 10 Note that the ratio between cos(119860119873 119868) andcos(119860 119868) is independent of the order 119899 of matrix 119860 Indeedassuming that tr(119860) = 0 andusing (16) we immediately obtain

cos (119860119873 119868)

cos (119860 119868)=

119860119873119865 radic119899

tr (119860) 119860119865radic119899

= 119860119873119865

119860119865

tr (119860) (23)

The following lemma compares the trace and the Frobe-nius norm of the orthogonal projection 119860119873

Lemma 11 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

tr (119860119873) le 119860119873119865lArrrArr 119860119873

119865le 1 lArrrArr tr (119860119873) le 1

lArrrArr cos (119860119873 119868) le1

radic119899

(24)

tr (119860119873) ge 119860119873119865lArrrArr 119860119873

119865ge 1 lArrrArr tr (119860119873) ge 1

lArrrArr cos (119860119873 119868) ge1

radic119899

(25)

Proof Using (4) we immediately obtain the four leftmostequivalences Using (16) we immediately obtain the tworightmost equivalences

The next lemma provides us with a relationship betweenthe Frobenius norms of the inverses of matrices119860 and its bestapproximate inverse119873 in subspace 119878

Lemma 12 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

10038171003817100381710038171003817119860minus110038171003817100381710038171003817119865

10038171003817100381710038171003817119873minus110038171003817100381710038171003817119865

ge 1 (26)

Proof Using (4) we get

119860119873119865le radic119899 = 119868

119865=10038171003817100381710038171003817(119860119873) (119860119873)

minus110038171003817100381710038171003817119865

le 119860119873119865

10038171003817100381710038171003817(119860119873)minus110038171003817100381710038171003817119865

997904rArr10038171003817100381710038171003817(119860119873)minus110038171003817100381710038171003817119865

ge 1

(27)

and hence10038171003817100381710038171003817119860minus110038171003817100381710038171003817119865

10038171003817100381710038171003817119873minus110038171003817100381710038171003817119865

ge10038171003817100381710038171003817119873minus1

119860minus110038171003817100381710038171003817119865

=10038171003817100381710038171003817(119860119873)minus110038171003817100381710038171003817119865

ge 1 (28)

and the proof is concluded

The following lemma compares the minimum residualnorm 119860119873 minus 119868

119865with the distance (with respect to the

Frobenius norm) 119860minus1 minus 119873119865between the inverse of 119860 and

the optimal approximate inverse 119873 of 119860 in any subspace119878 sub R119899times119899 First note that for any two matrices 119860 119861 isin R119899times119899

(119860 nonsingular) from the submultiplicative property of theFrobenius norm we immediately get

119860119861 minus 1198682

119865=10038171003817100381710038171003817119860 (119861 minus 119860

minus1

)10038171003817100381710038171003817

2

119865

le 1198602

119865

10038171003817100381710038171003817119861 minus 119860

minus110038171003817100381710038171003817

2

119865

(29)

However for the special case that 119861 = 119873 (the solution tothe problem (3)) we also get the following inequality

Lemma 13 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

119860119873 minus 1198682

119865le 119860

119865

10038171003817100381710038171003817119873 minus 119860

minus110038171003817100381710038171003817119865 (30)

Proof Using the Cauchy-Schwarz inequality and (5) we get10038161003816100381610038161003816⟨119860minus1

minus 119873119860119879

⟩119865

10038161003816100381610038161003816le10038171003817100381710038171003817119860minus1

minus 11987310038171003817100381710038171003817119865

1003817100381710038171003817100381711986011987910038171003817100381710038171003817119865

997904rArr10038161003816100381610038161003816tr ((119860minus1 minus 119873)119860)

10038161003816100381610038161003816le10038171003817100381710038171003817119860minus1

minus 11987310038171003817100381710038171003817119865

1003817100381710038171003817100381711986011987910038171003817100381710038171003817119865

997904rArr10038161003816100381610038161003816tr (119860 (119860

minus1

minus 119873))10038161003816100381610038161003816le10038171003817100381710038171003817119873 minus Aminus110038171003817100381710038171003817119865119860119865

997904rArr |tr (119868 minus 119860119873)| le10038171003817100381710038171003817119873 minus 119860

minus110038171003817100381710038171003817119865119860119865

997904rArr 119899 minus tr (119860119873) le10038171003817100381710038171003817119873 minus 119860

minus110038171003817100381710038171003817119865119860119865

997904rArr 119860119873 minus 1198682

119865le 119860

119865

10038171003817100381710038171003817119873 minus 119860

minus110038171003817100381710038171003817119865

(31)

Journal of Applied Mathematics 5

The following extension of the Cauchy-Schwarz inequal-ity in a real or complex inner product space (119867 ⟨sdot sdot⟩) wasobtained by Buzano [14] For all 119886 119909 119887 isin 119867 we have

|⟨119886 119909⟩ sdot ⟨119909 119887⟩| le1

2(119886 119887 + |⟨119886 119887⟩|) 119909

2

(32)

Thenext lemmaprovides uswith lower and upper boundson the inner product ⟨119860119873 119861⟩

119865 for any 119899 times 119899 real matrix 119861

Lemma 14 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then for every 119861 isin R119899times119899 we have

1003816100381610038161003816⟨119860119873 119861⟩119865

1003816100381610038161003816 le1

2(radic119899119861

119865+ |tr (119861)|) (33)

Proof Using (32) for 119886 = 119868 119909 = 119860119873 119887 = 119861 and (4) we get

1003816100381610038161003816⟨119868 119860119873⟩119865sdot ⟨119860119873 119861⟩

119865

1003816100381610038161003816 le1

2(119868119865119861119865+1003816100381610038161003816⟨119868 119861⟩119865

1003816100381610038161003816) 1198601198732

119865

997904rArr1003816100381610038161003816tr (119860119873) sdot ⟨119860119873 119861⟩

119865

1003816100381610038161003816

le1

2(radic119899119861

119865+ |tr (119861)|) 119860119873

2

119865

997904rArr1003816100381610038161003816⟨119860119873 119861⟩

119865

1003816100381610038161003816 le1

2(radic119899119861

119865+ |tr (119861)|)

(34)

The next lemma provides an upper bound on the arith-metic mean of the squares of the 1198992 terms in the orthogonalprojection 119860119873 By the way it also provides us with an upperbound on the arithmetic mean of the 119899 diagonal terms inthe orthogonal projection 119860119873 These upper bounds (validfor any matrix subspace 119878) are independent of the optimalapproximate inverse119873 and thus they are independent of thesubspace 119878 and only depend on matrix 119860

Lemma 15 Let 119860 isin R119899times119899 be nonsingular with tr(119860) = 0 andlet 119878 be a linear subspace of R119899times119899 Let119873 be the solution to theproblem (3) Then

1198601198732

119865

1198992le

1198602

119865

[tr (119860)]2

tr (119860119873)

119899le

1198991198602

119865

[tr (119860)]2

(35)

Proof Using (32) for 119886 = 119860119879 119909 = 119868 and 119887 = 119860119873 and the

Cauchy-Schwarz inequality for ⟨119860119879 119860119873⟩119865and (4) we get

10038161003816100381610038161003816⟨119860119879

119868⟩119865

sdot ⟨119868 119860119873⟩119865

10038161003816100381610038161003816

le1

2(1003817100381710038171003817100381711986011987910038171003817100381710038171003817119865119860119873

119865+10038161003816100381610038161003816⟨119860119879

119860119873⟩119865

10038161003816100381610038161003816) 1198682

119865

997904rArr |tr (119860) sdot tr (119860119873)|

le119899

2(119860119865119860119873

119865+10038161003816100381610038161003816⟨119860119879

119860119873⟩119865

10038161003816100381610038161003816)

le119899

2(119860119865119860119873

119865+1003817100381710038171003817100381711986011987910038171003817100381710038171003817119865119860119873

119865)

= 119899119860119865119860119873

119865

997904rArr |tr (119860)| 1198601198732

119865le 119899119860

119865119860119873

119865

997904rArr119860119873

119865

119899le

119860119865

|tr (119860)|

997904rArr119860119873

2

119865

1198992le

1198602

119865

[tr (119860)]2997904rArr

tr (119860119873)

119899le

1198991198602

119865

[tr (119860)]2

(36)

Remark 16 Lemma 15 has the following interpretation interms of the quality of the optimal approximate inverse119873 ofmatrix 119860 in subspace 119878 The closer the ratio 119899119860

119865| tr(119860)|

is to zero the smaller tr(119860119873) will be and thus due to (5)the larger 119860119873 minus 119868

119865will be and this happens for any matrix

subspace 119878

Remark 17 By the way from Lemma 15 we obtain thefollowing inequality for any nonsingular matrix 119860 isin R119899times119899Consider any matrix subspace 119878 st 119860minus1 isin 119878 Then119873 = 119860

minus1and using Lemma 15 we get

1198601198732

119865

1198992=1198682

119865

1198992=1

nle

1198602

119865

[tr (119860)]2

997904rArr |tr (119860)| le radic119899119860119865

(37)

4 Spectral Properties

In this section we present some new spectral propertiesfor matrix 119860119873 119873 being the optimal approximate inverseof matrix 119860 defined by (3) Mainly we focus on the casethat matrix 119860119873 is symmetric and positive definite This hasbeenmotivated by the following reasonWhen solving a largenonsymmetric linear system (1) by using Krylov methodsa possible strategy consists of searching for an adequateoptimal preconditioner 119873 such that the preconditionedmatrix119860119873 is symmetric positive definite [5]This enables oneto use the conjugate gradientmethod (CG-method) which isin general a computationally efficient method for solving thenew preconditioned system [2 15]

Our starting point is Lemma 3 which has established thatthe sets of eigenvalues and singular values of any orthogonalprojection 119860119873 satisfy

119899

sum

119894=1

1205822

119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816

2

le

119899

sum

119894=1

1205902

119894

=

119899

sum

119894=1

120582119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816 le

119899

sum

119894=1

120590119894

(38)

Let us particularize (38) for some special cases

6 Journal of Applied Mathematics

First note that if 119860119873 is normal (ie for all 1 le 119894 le 119899120590119894= |120582119894| [16]) then (38) becomes

119899

sum

119894=1

1205822

119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816

2

=

119899

sum

119894=1

1205902

119894

=

119899

sum

119894=1

120582119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816 =

119899

sum

119894=1

120590119894

(39)

In particular if 119860119873 is symmetric (120590119894= |120582119894| = plusmn120582

119894isin R) then

(38) becomes119899

sum

119894=1

1205822

119894=

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816

2

=

119899

sum

119894=1

1205902

119894

=

119899

sum

119894=1

120582119894le

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816 =

119899

sum

119894=1

120590119894

(40)

In particular if 119860119873 is symmetric and positive definite (120590119894=

|120582119894| = 120582119894isin R+) then the equality holds in all (38) that is

119899

sum

119894=1

1205822

119894=

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816

2

=

119899

sum

119894=1

1205902

119894

=

119899

sum

119894=1

120582119894=

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816 =

119899

sum

119894=1

120590119894

(41)

The next lemma compares the traces of matrices 119860119873 and(119860119873)2

Lemma 18 Let 119860 isin R119899times119899 be nonsingular and let 119878 be a linearsubspace of R119899times119899 Let 119873 be the solution to the problem (3)Then

(i) for any orthogonal projection 119860119873

tr ((119860119873)2

) le 1198601198732

119865= tr (119860119873) (42)

(ii) for any symmetric orthogonal projection 11986011987310038171003817100381710038171003817(119860119873)210038171003817100381710038171003817119865

le tr ((119860119873)2

) = 1198601198732

119865= tr (119860119873) (43)

(iii) for any symmetric positive definite orthogonal projec-tion 119860119873

10038171003817100381710038171003817(119860119873)210038171003817100381710038171003817119865

le tr ((119860119873)2

) = 1198601198732

119865= tr (119860119873) le [tr (119860119873)]

2

(44)

Proof (i) Using (38) we get119899

sum

119894=1

1205822

119894le

119899

sum

119894=1

1205902

119894=

119899

sum

119894=1

120582119894 (45)

(ii) It suffices to use the obvious fact that (119860119873)2

119865

le

1198601198732

119865and the following equalities taken from (40)

119899

sum

119894=1

1205822

119894=

119899

sum

119894=1

1205902

119894=

119899

sum

119894=1

120582119894 (46)

(iii) It suffices to use (43) and the fact that (see eg [1718]) if 119875 and 119876 are symmetric positive definite matrices thentr(119875119876) le tr(119875) tr(119876) for 119875 = 119876 = 119860119873

The rest of the paper is devoted to obtain new propertiesabout the eigenvalues of the orthogonal projection119860119873 for thespecial case that this matrix is symmetric positive definite

First let us recall that the smallest singular value and thesmallest eigenvaluersquos modulus of the orthogonal projection119860119873 are never greater than 1 (see Lemma 4) The followingtheorem establishes the dual result for the largest eigenvalueof matrix 119860119873 (symmetric positive definite)

Theorem 19 Let 119860 isin R119899times119899 be nonsingular and let 119878 be alinear subspace of R119899times119899 Let 119873 be the solution to the problem(3) Suppose thatmatrix119860119873 is symmetric and positive definiteThen the largest eigenvalue of the orthogonal projection119860119873 ofthe identity onto the subspace 119860119878 is never less than 1 That is

1205901= 1205821ge 1 (47)

Proof Using (41) we get

119899

sum

119894=1

1205822

119894=

119899

sum

119894=1

120582119894997904rArr 1205822

119899minus 120582119899=

119899minus1

sum

119894=1

(120582119894minus 1205822

119894) (48)

Now since 120582119899le 1 (Lemma 4) then 1205822

119899minus 120582119899le 0 This implies

that at least one summand in the rightmost sum in (48) mustbe less than or equal to zero Suppose that such summand isthe 119896th one (1 le 119896 le 119899 minus 1) Since 119860119873 is positive definitethen 120582

119896gt 0 and thus

120582119896minus 1205822

119896le 0 997904rArr 120582

119896le 1205822

119896997904rArr 120582

119896ge 1 997904rArr 120582

1ge 1 (49)

and the proof is concluded

InTheorem 19 the assumption that matrix119860119873 is positivedefinite is essential for assuring that |120582

1| ge 1 as the following

simple counterexample shows Moreover from Lemma 4 andTheorem 19 respectively we have that the smallest and largesteigenvalues of 119860119873 (symmetric positive definite) satisfy 120582

119899le

1 and 1205821ge 1 respectively Nothing can be asserted about

the remaining eigenvalues of the symmetric positive definitematrix 119860119873 which can be greater than equal to or less thanthe unity as the same counterexample also shows

Example 20 For 119899 = 3 let

119860119896= (

3 0 0

0 119896 0

0 0 1

) 119896 isin R (50)

let 1198683be identitymatrix of order 3 and let 119878 be the subspace of

all 3times3 scalarmatrices that is 119878 = span1198683Then the solution

119873119896to the problem (3) for subspace 119878 can be immediately

obtained by using formula (6) as follows

119873119896=tr (119860119896)

10038171003817100381710038171198601198961003817100381710038171003817

2

119865

1198683=

119896 + 4

1198962 + 101198683 (51)

and then we get

119860119896119873119896=

119896 + 4

1198962 + 10119860119896=

119896 + 4

1198962 + 10(

3 0 0

0 119896 0

0 0 1

) (52)

Journal of Applied Mathematics 7

Let us arrange the eigenvalues and singular values ofmatrix 119860

119896119873119896 as usual in nonincreasing order (as shown in

(11))On one hand for 119896 = minus2 we have

119860minus2119873minus2

=1

7(

3 0 0

0 minus2 0

0 0 1

) (53)

and then

1205901=10038161003816100381610038161205821

1003816100381610038161003816 =3

7

ge 1205902=10038161003816100381610038161205822

1003816100381610038161003816 =2

7

ge 1205903=10038161003816100381610038161205823

1003816100381610038161003816 =1

7

(54)

Hence 119860minus2119873minus2

is indefinite and 1205901= |1205821| = 37 lt 1

On the other hand for 1 lt 119896 lt 3 we have (see matrix(52))

1205901= 1205821= 3

119896 + 4

1198962 + 10

ge 1205902= 1205822= 119896

119896 + 4

1198962 + 10

ge 1205903= 1205823=

119896 + 4

1198962 + 10

(55)

and then

119896 = 2 1205901= 1205821=9

7gt 1

1205902= 1205822=6

7lt 1

1205903= 1205823=3

7lt 1

119896 =5

2 1205901= 1205821=6

5gt 1

1205902= 1205822= 1

1205903= 1205823=2

5lt 1

119896 =8

3 1205901= 1205821=90

77gt 1

1205902= 1205822=80

77gt 1

1205903= 1205823=30

77lt 1

(56)

Hence for 119860119896119873119896positive definite we have (depending on 119896)

1205822lt 1 120582

2= 1 or 120582

2gt 1

The following corollary improves the lower bound zeroon both tr(119860119873) given in (4) and cos(119860119873 119868) given in (17)

Corollary 21 Let 119860 isin R119899times119899 be nonsingular and let 119878

be a linear subspace of R119899times119899 Let 119873 be the solution to the

problem (3) Suppose thatmatrix119860119873 is symmetric and positivedefinite Then

1 le 119860119873119865le tr (119860119873) = 119860119873

2

119865le 119899 (57)

cos (119860119873 119868) ge1

radic119899 (58)

Proof Denote by sdot 2the spectral norm Using the well-

known inequality sdot 2le sdot

119865[19] Theorem 19 and (4) we

get

119860119873119865ge 119860119873

2= 1205901= 1205821ge 1

997904rArr 1 le 119860119873119865le tr (119860119873) = 119860119873

2

119865le 119899

(59)

Finally (58) follows immediately from (57) and (25)

Let usmention that an upper bound on all the eigenvaluesmoduli and on all singular values of any orthogonal projec-tion 119860119873 can be immediately obtained from (38) and (4) asfollows

119899

sum

119894=1

10038161003816100381610038161205821198941003816100381610038161003816

2

le

119899

sum

119894=1

1205902

119894= 119860119873

2

119865le 119899

997904rArr1003816100381610038161003816120582119894

1003816100381610038161003816 120590119894le radic119899 forall119894 = 1 2 119899

(60)

Our last theorem improves the upper bound given in(60) for the special case that the orthogonal projection 119860119873

is symmetric positive definite

Theorem 22 Let 119860 isin R119899times119899 be nonsingular and let 119878 be alinear subspace of R119899times119899 Let 119873 be the solution to the problem(3) Suppose thatmatrix119860119873 is symmetric and positive definiteThen all the eigenvalues of matrix 119860119873 satisfy

120590119894= 120582119894le1 + radic119899

2forall119894 = 1 2 119899 (61)

Proof First note that the assertion is obvious for the smallestsingular value since |120582

119899| le 1 for any orthogonal projection

119860119873 (Lemma 4) For any eigenvalue of 119860119873 we use the factthat 119909 minus 119909

2

le 14 for all 119909 gt 0 Then from (41) we get119899

sum

119894=1

1205822

119894=

119899

sum

119894=1

120582119894

997904rArr 1205822

1minus 1205821=

119899

sum

119894=2

(120582119894minus 1205822

119894) le

119899 minus 1

4

997904rArr 1205821le1 + radic119899

2997904rArr 120582

119894le1 + radic119899

2

forall119894 = 1 2 119899

(62)

5 Conclusion

In this paper we have considered the orthogonal projection119860119873 (in the Frobenius sense) of the identity matrix onto an

8 Journal of Applied Mathematics

arbitrary matrix subspace 119860119878 (119860 isin R119899times119899 nonsingular 119878 sub

R119899times119899) Among other geometrical properties of matrix 119860119873we have established a strong relation between the qualityof the approximation 119860119873 asymp 119868 and the cosine of the angleang(119860119873 119868) Also the distance between119860119873 and the identity hasbeen related to the ratio 119899119860

119865| tr(119860)| (which is independent

of the subspace 119878) The spectral analysis has provided lowerand upper bounds on the largest eigenvalue of the symmetricpositive definite orthogonal projections of the identity

Acknowledgments

Theauthors are grateful to the anonymous referee for valuablecomments and suggestions which have improved the earlierversion of this paperThisworkwas partially supported by theldquoMinisterio de Economıa y Competitividadrdquo (Spanish Gov-ernment) and FEDER through Grant Contract CGL2011-29396-C03-01

References

[1] O Axelsson Iterative Solution Methods Cambridge UniversityPress Cambridge Mass USA 1994

[2] Y Saad Iterative Methods for Sparse Linear Systems PWS Pub-lishing Boston Mass USA 1996

[3] S-L Wu F Chen and X-Q Niu ldquoA note on the eigenvalueanalysis of the SIMPLE preconditioning for incompressibleflowrdquo Journal of Applied Mathematics vol 2012 Article ID564132 7 pages 2012

[4] M Benzi ldquoPreconditioning techniques for large linear systemsa surveyrdquo Journal of Computational Physics vol 182 no 2 pp418ndash477 2002

[5] G Montero L Gonzalez E Florez M D Garcıa and ASuarez ldquoApproximate inverse computation using Frobenius in-ner productrdquoNumerical Linear Algebra with Applications vol 9no 3 pp 239ndash247 2002

[6] A Suarez and L Gonzalez ldquoA generalization of the Moore-Penrose inverse related to matrix subspaces of 119862119899times119898rdquo AppliedMathematics andComputation vol 216 no 2 pp 514ndash522 2010

[7] E Chow and Y Saad ldquoApproximate inverse preconditioners viasparse-sparse iterationsrdquo SIAM Journal on Scientific Computingvol 19 no 3 pp 995ndash1023 1998

[8] E Chow and Y Saad ldquoApproximate inverse techniques forblock-partitioned matricesrdquo SIAM Journal on Scientific Com-puting vol 18 no 6 pp 1657ndash1675 1997

[9] R M Holland A J Wathen and G J Shaw ldquoSparse approxi-mate inverses and target matricesrdquo SIAM Journal on ScientificComputing vol 26 no 3 pp 1000ndash1011 2005

[10] R Andreani M Raydan and P Tarazaga ldquoOn the geometricalstructure of symmetric matricesrdquo Linear Algebra and Its Appli-cations vol 438 no 3 pp 1201ndash1214 2013

[11] L Gonzalez ldquoOrthogonal projections of the identity spectralanalysis and applications to approximate inverse precondition-ingrdquo SIAM Review vol 48 no 1 pp 66ndash75 2006

[12] A Suarez and L Gonzalez ldquoNormalized Frobenius conditionnumber of the orthogonal projections of the identityrdquo Journalof Mathematical Analysis and Applications vol 400 no 2 pp510ndash516 2013

[13] J-P Chehab and M Raydan ldquoGeometrical properties of theFrobenius condition number for positive definite matricesrdquo

Linear Algebra and Its Applications vol 429 no 8-9 pp 2089ndash2097 2008

[14] M L Buzano ldquoGeneralizzazione della diseguaglianza di Cau-chy-Schwarzrdquo Rendiconti del Seminario Matematico Universitae Politecnico di Torino vol 31 pp 405ndash409 1974 (Italian)

[15] M R Hestenes and E Stiefel ldquoMethods of conjugate gradientsfor solving linear systemsrdquo Journal of Research of the NationalBureau of Standards vol 49 no 6 pp 409ndash436 1952

[16] R A Horn and C R Johnson Topics in Matrix Analysis Cam-bridge University Press Cambridge UK 1991

[17] E H Lieb andWThirring ldquoInequalities for themoments of theeigenvalues of the Schrodinger Hamiltonian and their relationto Sobolev inequalitiesrdquo in Studies in Mathematical PhysicsEssays in Honor of Valentine Bargmann E Lieb B Simon andA Wightman Eds pp 269ndash303 Princeton University PressPrinceton NJ USA 1976

[18] Z Ulukok and R Turkmen ldquoOn some matrix trace inequali-tiesrdquo Journal of Inequalities and Applications vol 2010 ArticleID 201486 8 pages 2010

[19] R A Horn andC R JohnsonMatrix Analysis CambridgeUni-versity Press Cambridge UK 1985

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 489295 6 pageshttpdxdoiorg1011552013489295

Research ArticleA Parameterized Splitting Preconditioner for GeneralizedSaddle Point Problems

Wei-Hua Luo12 and Ting-Zhu Huang1

1 School of Mathematical Sciences University of Electronic Science and Technology of China Chengdu Sichuan 611731 China2 Key Laboratory of Numerical Simulation of Sichuan Province University Neijiang Normal University Neijiang Sichuan 641112 China

Correspondence should be addressed to Ting-Zhu Huang tingzhuhuang126com

Received 4 January 2013 Revised 31 March 2013 Accepted 1 April 2013

Academic Editor P N Shivakumar

Copyright copy 2013 W-H Luo and T-Z Huang This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

By using Sherman-Morrison-Woodbury formula we introduce a preconditioner based on parameterized splitting idea forgeneralized saddle point problems which may be singular and nonsymmetric By analyzing the eigenvalues of the preconditionedmatrix we find that when 120572 is big enough it has an eigenvalue at 1 with multiplicity at least n and the remaining eigenvalues are alllocated in a unit circle centered at 1 Particularly when the preconditioner is used in general saddle point problems it guaranteeseigenvalue at 1 with the same multiplicity and the remaining eigenvalues will tend to 1 as the parameter 120572 rarr 0 Consequently thiscan lead to a good convergence when some GMRES iterative methods are used in Krylov subspace Numerical results of Stokesproblems and Oseen problems are presented to illustrate the behavior of the preconditioner

1 Introduction

In some scientific and engineering applications such as finiteelement methods for solving partial differential equations [12] and computational fluid dynamics [3 4] we often considersolutions of the generalized saddle point problems of the form

(119860 119861119879

minus119861 119862)(

119909

119910) = (

119891

119892) (1)

where 119860 isin R119899times119899 119861 isin R119898times119899 (119898 le 119899) and 119862 isin R119898times119898

are positive semidefinite 119909 119891 isin R119899 and 119910 119892 isin R119898 When119862 = 0 (1) is a general saddle point problem which is also aresearching object for many authors

It is well known that when the matrices 119860 119861 and 119862

are large and sparse the iterative methods are more efficientand attractive than direct methods assuming that (1) has agood preconditioner In recent years a lot of preconditioningtechniques have arisen for solving linear system for exampleSaad [5] and Chen [6] have comprehensively surveyed someclassical preconditioning techniques including ILU pre-conditioner triangular preconditioner SPAI preconditionermultilevel recursive Schur complements preconditioner and

sparse wavelet preconditioner Particularly many precondi-tioning methods for saddle problems have been presentedrecently such as dimensional splitting (DS) [7] relaxeddimensional factorization (RDF) [8] splitting preconditioner[9] and Hermitian and skew-Hermitian splitting precondi-tioner [10]

Among these results Cao et al [9] have used splitting ideato give a preconditioner for saddle point problems where thematrix 119860 is symmetric and positive definite and 119861 is of fullrow rank According to his preconditioner the eigenvaluesof the preconditioned matrix would tend to 1 when theparameter 119905 rarr infin Consequently just as we have seen fromthose examples of [9] preconditioner has guaranteed a goodconvergence when some iterative methods were used

In this paper being motivated by [9] we use the splittingidea to present a preconditioner for the system (1) where 119860may be nonsymmetric and singular (when rank(119861) lt 119898) Wefind that when the parameter is big enough the precondi-tioned matrix has the eigenvalue at 1 with multiplicity at least119899 and the remaining eigenvalues are all located in a unit circlecentered at 1 Particularly when the precondidtioner is usedin some general saddle point problems (namely 119862 = 0) we

2 Journal of Applied Mathematics

see that the multiplicity of the eigenvalue at 1 is also at least 119899but the remaining eigenvalues will tend to 1 as the parameter120572 rarr 0

The remainder of the paper is organized as followsIn Section 2 we present our preconditioner based on thesplitting idea and analyze the bound of eigenvalues of thepreconditioned matrix In Section 3 we use some numericalexamples to show the behavior of the new preconditionerFinally we draw some conclusions and outline our futurework in Section 4

2 A Parameterized Splitting Preconditioner

Now we consider using splitting idea with a variable param-eter to present a preconditioner for the system (1)

Firstly it is evident that when 120572 = 0 the system (1) isequivalent to

(

119860 119861119879

minus119861

120572

119862

120572

)(119909

119910) = (

119891

119892

120572

) (2)

Let

119872 = (

119860 119861119879

minus119861

120572119868

) 119873 = (

0 0

0 119868 minus119862

120572

)

119883 = (119909

119910) 119865 = (

119891

119892

120572

)

(3)

Then the coefficient matrix of (2) can be expressed by119872minus119873Multiplying both sides of system (2) from the left with matrix119872minus1 we have

(119868 minus119872minus1

119873)119883 = 119872minus1

119865 (4)

Hence we obtain a preconditioned linear system from (1)using the idea of splitting and the corresponding precondi-tioner is

119867 = (

119860 119861119879

minus119861

120572119868

)

minus1

(

119868 0

0119868

120572

) (5)

Nowwe analyze the eigenvalues of the preconditioned system(4)

Theorem 1 The preconditioned matrix 119868 minus 119872minus1

119873 has aneigenvalue at 1 with multiplicity at least 119899 The remainingeigenvalues 120582 satisfy

120582 =1199041+ 1199042

1199041+ 120572

(6)

where 1199041= 120596119879

119861119860minus1

119861119879

120596 1199042= 120596119879

119862120596 and 120596 isin C119898 satisfies

(119861119860minus1

119861119879

120572+119862

120572)120596 = 120582(119868 +

119861119860minus1

119861119879

120572)120596 120596 = 1 (7)

Proof Because

119872minus1

=(

(119860 +119861119879

119861

120572)

minus1

minus119860minus1

119861119879

(119868 +119861119860minus1

119861119879

120572)

minus1

119861

120572(119860 +

119861119879

119861

120572)

minus1

(119868 +119861119860minus1

119861119879

120572)

minus1)

(8)

we can easily get

119868 minus119872minus1

119873 =(

119868 119860minus1

119861119879

(119868 +119861119860minus1

119861119879

120572)

minus1

(119868 minus119862

120572)

0 (119868 +119861119860minus1

119861119879

120572)

minus1

(119861119860minus1

119861119879

120572+119862

120572)

)

(9)

which implies the preconditioned matrix 119868 minus 119872minus1119873 has aneigenvalue at 1 with multiplicity at least 119899

For the remaining eigenvalues let

(119868 +119861119860minus1

119861119879

120572)

minus1

(119861119860minus1

119861119879

120572+119862

120572)120596 = 120582120596 (10)

with 120596 = 1 then we have

(119861119860minus1

119861119879

120572+119862

120572)120596 = 120582(119868 +

119861119860minus1

119861119879

120572)120596 (11)

By multiplying both sides of this equality from the left with120596119879 we can get

120582 =1199041+ 1199042

1199041+ 120572

(12)

This completes the proof of Theorem 1

Remark 2 FromTheorem 1 we can get that when parameter120572 is big enough the modulus of nonnil eigenvalues 120582 will belocated in interval (0 1)

Remark 3 InTheorem 1 if the matrix 119862 = 0 then for nonnileigenvalues we have

lim120572rarr0

120582 = 1 (13)

Figures 1 2 and 3 are the eigenvalues plots of the pre-conditioned matrices obtained with our preconditioner Aswe can see in the following numerical experiments thisgood phenomenon is useful for accelerating convergence ofiterative methods in Krylov subspace

Additionally for the purpose of practically executing ourpreconditioner

119867 =(

(119860 +119861119879

119861

120572)

minus1

minus119860minus1

119861119879

120572(119868 +

119861119860minus1

119861119879

120572)

minus1

119861

120572(119860 +

119861119879

119861

120572)

minus1

1

120572(119868 +

119861119860minus1

119861119879

120572)

minus1)

(14)

Journal of Applied Mathematics 3

0 02 04 06 08 12 141

0

1

2

3

4

minus1

minus2

minus3

times10minus4

= 01 120572 = 00001

(a)

0

002

004

006

008

0 02 04 06 08 12 141minus008

minus006

minus004

minus002

= 01 120572 = norm(119862 fro)20

(b)

Figure 1 Spectrum of the preconditioned steady Oseen matrix with viscosity coefficient V = 01 32 times 32 grid (a) Q2-Q1 FEM 119862 = 0 (b)Q1-P0 FEM 119862 = 0

0

0 02 04 06 08 121

02

04

06

08

1

minus02

minus04

minus06

minus08

minus1

minus02

times10minus3

= 001 120572 = 00001

(a)

0 02 04 06 08 12 141minus02

minus015

minus01

minus005

0

005

01

015

02 = 001 120572 = norm(119862 fro)20

(b)

Figure 2 Spectrum of the preconditioned steady Oseen matrix with viscosity coefficient V = 001 32 times 32 grid (a) Q2-Q1 FEM 119862 = 0 (b)Q1-P0 FEM 119862 = 0

0

2

4

6

8

0 02 04 06 08 12 141

times10minus4

minus8

minus6

minus4

minus2

= 0001 120572 = 00001

(a)

0

002

004

006

008

minus008

minus006

minus004

minus002

0 02 04 06 08 121minus02

= 0001 120572 = norm(119862 fro)20

(b)

Figure 3 Spectrum of the preconditioned steady Oseen matrix with viscosity coefficient V = 0001 32 times 32 grid (a) Q2-Q1 FEM 119862 = 0 (b)Q1-P0 FEM 119862 = 0

4 Journal of Applied Mathematics

Table 1 Preconditioned GMRES(20) on Stokes problems withdifferent grid sizes (uniform grids)

Grid 120572 its LU time its time Total time16 times 16 00001 4 00457 00267 0072432 times 32 00001 6 03912 00813 0472564 times 64 00001 9 54472 05519 59991128 times 128 00001 14 914698 57732 972430

Table 2 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 01

Grid 120572 its LU time its time Total time16 times 16 00001 3 00479 00244 0072332 times 32 00001 3 06312 00696 0700964 times 64 00001 4 132959 03811 136770128 times 128 00001 6 1305463 37727 1343190

Table 3 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 001

Grid 120572 its LU time its time Total time16 times 16 00001 2 00442 00236 0067832 times 32 00001 3 04160 00557 0471764 times 64 00001 3 73645 02623 76268128 times 128 00001 4 1694009 31709 1725718

Table 4 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 0001

Grid 120572 its LU time its time Total time16 times 16 00001 2 00481 00206 0068732 times 32 00001 3 04661 00585 0524664 times 64 00001 3 66240 02728 68969128 times 128 00001 4 1773130 29814 1802944

Table 5 Preconditioned GMRES(20) on Stokes problems withdifferent grid sizes (uniform grids)

Grid 120572 its LU time its time Total time16 times 16 10000 5 00471 00272 0074332 times 32 10000 7 03914 00906 0482064 times 64 10000 9 56107 04145 60252128 times 128 10000 14 927154 48908 976062

Table 6 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 01

Grid 120572 its LU time its time Total time16 times 16 10000 4 00458 00223 0068132 times 32 10000 4 05748 00670 0641864 times 64 10000 5 122642 04179 126821128 times 128 10000 7 1282758 16275 1299033

Table 7 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 001

Grid 120572 its LU time its time Total time16 times 16 10000 3 00458 00224 0068232 times 32 10000 3 04309 00451 0476064 times 64 10000 4 76537 01712 78249128 times 128 10000 5 1751587 34554 1786141

Table 8 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 0001

Grid 120572 its LU time its time Total time16 times 16 10000 3 00507 00212 0071932 times 32 10000 3 04735 00449 0518464 times 64 10000 4 66482 01645 68127128 times 128 10000 4 1720516 21216 1741732

Table 9 Preconditioned GMRES(20) on Stokes problems withdifferent grid sizes (uniform grids)

Grid 120572 its LU time its time Total time16 times 16 norm(119862 fro) 21 00240 00473 0071332 times 32 norm(119862 fro) 20 00890 01497 0238764 times 64 norm(119862 fro) 20 08387 06191 14578128 times 128 norm(119862 fro) 20 68917 30866 99783

Table 10 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 01

Grid 120572 its LU time its time Total time16 times 16 norm(119862 fro)20 10 00250 00302 0055232 times 32 norm(119862 fro)20 10 00816 00832 0164864 times 64 norm(119862 fro)20 12 08466 03648 12114128 times 128 norm(119862 fro)20 14 69019 20398 89417

Table 11 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 001

Grid 120572 its LU time its time Total time16 times 16 norm(119862 fro)20 7 00238 00286 0052432 times 32 norm(119862 fro)20 7 00850 00552 0140264 times 64 norm(119862 fro)20 11 08400 03177 11577128 times 128 norm(119862 fro)20 16 69537 22306 91844

Table 12 Preconditioned GMRES(20) on steady Oseen problemswith different grid sizes (uniform grids) viscosity V = 0001

Grid 120572 its LU time its time Total time16 times 16 norm(119862 fro)20 5 00245 00250 0049532 times 32 norm(119862 fro)20 5 00905 00587 0149264 times 64 norm(119862 fro)20 8 12916 03200 16116128 times 128 norm(119862 fro)20 15 104399 27468 131867

Journal of Applied Mathematics 5

Table 13 Size and number of non-nil elements of the coefficient matrix generalized by using Q2-Q1 FEM in steady Stokes problems

Grid dim(119860) nnz(119860) dim(119861) nnz(119861) nnz(119860 + (1120572)119861119879119861)16 times 16 578 times 578 6178 81 times 578 2318 3690432 times 32 2178 times 2178 28418 289 times 2178 10460 19322064 times 64 8450 times 8450 122206 1089 times 8450 44314 875966128 times 128 33282 times 33282 506376 4225 times 33282 184316 3741110

we should efficiently deal with the computation of (119868 +(119861119860minus1

119861119879

120572))minus1 This can been tackled by the well-known

Sherman-Morrison-Woodbury formula

(1198601+ 11988311198602119883119879

2)minus1

= 119860minus1

1minus 119860minus1

11198831(119860minus1

2+ 119883119879

2119860minus1

11198831)minus1

119883119879

2119860minus1

1

(15)

where 1198601isin R119899times119899 and 119860

2isin R1199031times1199031 are invertible matrices

1198831isin R119899times1199031 and 119883

2isin R119899times1199031 are any matrices and 119899 119903

1are

any positive integersFrom (15) we immediately get

(119868 +119861119860minus1

119861119879

120572)

minus1

= 119868 minus 119861(120572119860 + 119861119879

119861)minus1

119861119879

(16)

In the following numerical examples we will always use (16)to compute (119868 + (119861119860minus1119861119879120572))minus1 in (14)

3 Numerical Examples

In this sectionwe give numerical experiments to illustrate thebehavior of our preconditioner The numerical experimentsare done by using MATLAB 71 The linear systems areobtained by using finite element methods in the Stokes prob-lems and steady Oseen problems and they are respectivelythe cases of

(1) 119862 = 0 which is caused by using Q2-Q1 FEM(2) 119862 = 0 which is caused by using Q1-P0 FEM

Furthermore we compare our preconditionerwith that of[9] in the case of general saddle point problems (namely 119862 =0) For the general saddle point problem [9] has presentedthe preconditioner

= (

119860 + 119905119861119879

119861 0

minus2119861119868

119905

)

minus1

(17)

with 119905 as a parameter and has proved that when 119860 issymmetric positive definite the preconditioned matrix hasan eigenvalue 1 with multiplicity at 119899 and the remainingeigenvalues satisfy

120582 =1199051205902

119894

1 + 1199051205902

119894

(18)

lim119905rarrinfin

120582 = 1 (19)

where 120590119894 119894 = 1 2 119898 are119898 positive singular values of the

matrix 119861119860minus12All these systems can be generalized by

using IFISS software package [11] (this is a freepackage that can be downloaded from the sitehttpwwwmathsmanchesteracuksimdjsifiss) We userestarted GMRES(20) as the Krylov subspace method andwe always take a zero initial guess The stopping criterion is

100381710038171003817100381711990311989610038171003817100381710038172

1003817100381710038171003817119903010038171003817100381710038172

le 10minus6

(20)

where 119903119896is the residual vector at the 119896th iteration

In the whole course of computation we always replace(119868 + (119861119860

minus1

119861119879

120572))minus1 in (14) with (16) and use the 119871119880 factor-

ization of 119860 + (119861119879

119861120572) to tackle (119860 + (119861119879

119861120572))minus1V where

V is a corresponding vector in the iteration Concretely let119860 + (119861

119879

119861120572) = 119871119880 then we complete the matrix-vectorproduct (119860 + (119861

119879

119861120572))minus1V by 119880 119871 V in MATLAB term

In the following tables the denotation norm (119862 fro) meansthe Frobenius form of the matrix 119862 The total time is the sumof LU time and iterative time and the LU time is the time tocompute LU factorization of 119860 + (119861119879119861120572)

Case 1 (for our preconditioner) 119862 = 0 (using Q2-Q1 FEM inStokes problems and steady Oseen problems with differentviscosity coefficients The results are in Tables 1 2 3 4 )

Case 1rsquo (for preconditioner of [9]) 119862 = 0 (using Q2-Q1 FEMin Stokes problems and steady Oseen problems with differentviscosity coefficients The results are in Tables 5 6 7 8)

Case 2 119862 = 0 (using Q1-P0 FEM in Stokes problems andsteady Oseen problems with different viscosity coefficientsThe results are in Tables 9 10 11 12)

From Tables 1 2 3 4 5 6 7 and 8 we can see that theseresults are in agreement with the theoretical analyses (13) and(19) respectively Additionally comparing with the results inTables 9 10 11 and 12 we find that although the iterationsused in Case 1 (either for the preconditioner of [9] or ourpreconditioner) are less than those in Case 2 the time spentby Case 1 ismuchmore than that of Case 2This is because thedensity of the coefficient matrix generalized by Q2-Q1 FEMis much larger than that generalized by Q1-P0 FEMThis canbe partly illustrated by Tables 13 and 14 and the others can beillustrated similarly

6 Journal of Applied Mathematics

Table 14 Size and number of non-nil elements of the coefficient matrix generalized by using Q1-P0 FEM in steady Stokes problems

Grid dim(119860) nnz(119860) dim(119861) nnz(119861) nnz(119860 + (1120572)119861119879119861)16 times 16 578 times 578 3826 256 times 578 1800 707632 times 32 2178 times 2178 16818 1024 times 2178 7688 3187464 times 64 8450 times 8450 70450 4096 times 8450 31752 136582128 times 128 33282 times 33282 288306 16384 times 33282 129032 567192

4 Conclusions

In this paper we have introduced a splitting preconditionerfor solving generalized saddle point systems Theoreticalanalysis showed the modulus of eigenvalues of the precon-ditioned matrix would be located in interval (0 1) when theparameter is big enough Particularly when the submatrix119862 = 0 the eigenvalues will tend to 1 as the parameter 120572 rarr 0These performances are tested by some examples and theresults are in agreement with the theoretical analysis

There are still some future works to be done how to prop-erly choose a parameter 120572 so that the preconditioned matrixhas better propertiesHow to further precondition submatrix(119868 + (119861119860

minus1

119861119879

120572))minus1

((119861119860minus1

119861119879

120572) + (119862120572)) to improve ourpreconditioner

Acknowledgments

The authors would express their great thankfulness to thereferees and the editor Professor P N Shivakumar for theirhelpful suggestions for revising this paperThe authors wouldlike to thank H C Elman A Ramage and D J Silvester fortheir free IFISS software package This research is supportedby Chinese Universities Specialized Research Fund for theDoctoral Program (20110185110020) Sichuan Province Sci ampTech Research Project (2012GZX0080) and the Fundamen-tal Research Funds for the Central Universities

References

[1] F Brezzi andM FortinMixed and Hybrid Finite ElementMeth-ods vol 15 of Springer Series in Computational MathematicsSpringer New York NY USA 1991

[2] Z Chen Q Du and J Zou ldquoFinite element methods withmatching and nonmatching meshes for Maxwell equationswith discontinuous coefficientsrdquo SIAM Journal on NumericalAnalysis vol 37 no 5 pp 1542ndash1570 2000

[3] H C Elman D J Silvester and A J Wathen Finite Elementsand Fast Iterative Solvers With Applications in IncompressibleFluid Dynamics Numerical Mathematics and Scientific Com-putation Oxford University Press New York NY USA 2005

[4] C Cuvelier A Segal and A A van Steenhoven Finite ElementMethods and Navier-Stokes Equations vol 22 of Mathematicsand its Applications D Reidel Dordrecht The Netherlands1986

[5] Y Saad Iterative Methods for Sparse Linear Systems Society forIndustrial and Applied Mathematics Philadelphia Pa USA2nd edition 2003

[6] K Chen Matrix Preconditioning Techniques and Applicationsvol 19 ofCambridgeMonographs onApplied andComputational

Mathematics Cambridge University Press Cambridge UK2005

[7] M Benzi and X-P Guo ldquoA dimensional split preconditionerfor Stokes and linearized Navier-Stokes equationsrdquo AppliedNumerical Mathematics vol 61 no 1 pp 66ndash76 2011

[8] M Benzi M Ng Q Niu and Z Wang ldquoA relaxed dimensionalfactorization preconditioner for the incompressible Navier-Stokes equationsrdquo Journal of Computational Physics vol 230no 16 pp 6185ndash6202 2011

[9] YCaoM-Q Jiang andY-L Zheng ldquoA splitting preconditionerfor saddle point problemsrdquo Numerical Linear Algebra withApplications vol 18 no 5 pp 875ndash895 2011

[10] V Simoncini and M Benzi ldquoSpectral properties of the Her-mitian and skew-Hermitian splitting preconditioner for saddlepoint problemsrdquo SIAM Journal on Matrix Analysis and Applica-tions vol 26 no 2 pp 377ndash389 2004

[11] H C Elman A Ramage and D J Silvester ldquoAlgorithm 886IFISS a Matlab toolbox for modelling incompressible flowrdquoACM Transactions on Mathematical Software vol 33 no 2article 14 2007

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 593549 11 pageshttpdxdoiorg1011552013593549

Research ArticleSolving Optimization Problems on Hermitian Matrix Functionswith Applications

Xiang Zhang and Shu-Wen Xiang

Department of Computer Science and Information Guizhou University Guiyang 550025 China

Correspondence should be addressed to Xiang Zhang zxjnsc163com

Received 17 October 2012 Accepted 20 March 2013

Academic Editor K Sivakumar

Copyright copy 2013 X Zhang and S-W Xiang This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

We consider the extremal inertias and ranks of the matrix expressions 119891(119883 119884) = 1198603minus 119861

3119883 minus (119861

3119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast where

1198603= 119860

lowast

3 119861

3 119862

3 and 119863

3are known matrices and 119884 and 119883 are the solutions to the matrix equations 119860

1119884 = 119862

1 119884119861

1= 119863

1 and

1198602119883 = 119862

2 respectively As applications we present necessary and sufficient condition for the previous matrix function 119891(119883 119884)

to be positive (negative) non-negative (positive) definite or nonsingular We also characterize the relations between the Hermitianpart of the solutions of the above-mentioned matrix equations Furthermore we establish necessary and sufficient conditions forthe solvability of the system of matrix equations 119860

1119884 = 119862

1 119884119861

1= 119863

1 119860

2119883 = 119862

2 and 119861

3119883 + (119861

3119883)

lowast

+ 1198623119884119863

3+ (119862

3119884119863

3)lowast

= 1198603

and give an expression of the general solution to the above-mentioned system when it is solvable

1 Introduction

Throughout we denote the field of complex numbers by Cthe set of all 119898 times 119899 matrices over C by C119898times119899 and the set ofall119898 times119898Hermitian matrices by C119898times119898

ℎ The symbols 119860lowast and

R(119860) stand for the conjugate transpose the column space ofa complex matrix119860 respectively 119868

119899denotes the 119899 times 119899 identity

matrix TheMoore-Penrose inverse [1] 119860dagger of 119860 is the uniquesolution119883 to the four matrix equations

(i) 119860119883119860 = 119860

(ii) 119883119860119883 = 119883

(iii) (119860119883)lowast = 119860119883

(iv) (119883119860)lowast = 119883119860

(1)

Moreover 119871119860and 119877

119860stand for the projectors 119871

119860= 119868 minus

119860dagger

119860 119877119860= 119868 minus 119860119860

dagger induced by 119860 It is well known that theeigenvalues of a Hermitian matrix 119860 isin C119899times119899 are real and theinertia of 119860 is defined to be the triplet

I119899(119860) = 119894

+(119860) 119894

minus(119860) 119894

0(119860) (2)

where 119894+(119860) 119894

minus(119860) and 119894

0(119860) stand for the numbers of

positive negative and zero eigenvalues of119860 respectivelyThe

symbols 119894+(119860) and 119894

minus(119860) are called the positive index and

the negative index of inertia respectively For two Hermitianmatrices 119860 and 119861 of the same sizes we say 119860 ge 119861 (119860 le 119861)in the Lowner partial ordering if 119860 minus 119861 is positive (negative)semidefinite The Hermitian part of 119883 is defined as 119867(119883) =119883 + 119883

lowast We will say that 119883 is Re-nnd (Re-nonnegativesemidefinite) if 119867(119883) ge 0 119883 is Re-pd (Re-positive definite)if119867(119883) gt 0 and119883 is Re-ns if119867(119883) is nonsingular

It is well known that investigation on the solvabilityconditions and the general solution to linearmatrix equationsis very active (eg [2ndash9]) In 1999 Braden [10] gave thegeneral solution to

119861119883 + (119861119883)lowast

= 119860 (3)

In 2007 Djordjevic [11] considered the explicit solution to (3)for linear bounded operators on Hilbert spaces MoreoverCao [12] investigated the general explicit solution to

119861119883119862 + (119861119883119862)lowast

= 119860 (4)

Xu et al [13] obtained the general expression of the solutionof operator equation (4) In 2012 Wang and He [14] studied

2 Journal of Applied Mathematics

some necessary and sufficient conditions for the consistenceof the matrix equation

1198601119883 + (119860

1119883)

lowast

+ 1198611119884119862

1+ (119861

1119884119862

1)lowast

= 1198641

(5)

and presented an expression of the general solution to (5)Note that (5) is a special case of the following system

1198601119884 = 119862

1 119884119861

1= 119863

1 119860

2119883 = 119862

2

1198613119883 + (119861

3119883)

lowast

+ 1198623119884119863

3+ (119862

3119884119863

3)lowast

= 1198603

(6)

To our knowledge there has been little information about (6)One goal of this paper is to give some necessary and sufficientconditions for the solvability of the system of matrix (6) andpresent an expression of the general solution to system (6)when it is solvable

In order to find necessary and sufficient conditions for thesolvability of the system of matrix equations (6) we need toconsider the extremal ranks and inertias of (10) subject to (13)and (11)

There have been many papers to discuss the extremalranks and inertias of the following Hermitian expressions

119901 (119883) = 1198603minus 119861

3119883 minus (119861

3119883)

lowast

(7)

119892 (119884) = 119860 minus 119861119884119862 minus (119861119884119862)lowast

(8)

ℎ (119883 119884) = 1198601minus 119861

1119883119861

lowast

1minus 119862

1119884119862

lowast

1 (9)

119891 (119883 119884) = 1198603minus 119861

3119883 minus (119861

3119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

(10)

Tian has contributed much in this field One of his works[15] considered the extremal ranks and inertias of (7) Heand Wang [16] derived the extremal ranks and inertias of (7)subject to119860

1119883 = 119862

1119860

2119883119861

2= 119862

2 Liu and Tian [17] studied

the extremal ranks and inertias of (8) Chu et al [18] and Liuand Tian [19] derived the extremal ranks and inertias of (9)Zhang et al [20] presented the extremal ranks and inertias of(9) where119883 and 119884 are Hermitian solutions of

1198602119883 = 119862

2 (11)

1198841198612= 119863

2 (12)

respectively He and Wang [16] derived the extremal ranksand inertias of (10)We consider the extremal ranks and iner-tias of (10) subject to (11) and

1198601119884 = 119862

1 119884119861

1= 119863

1 (13)

which is not only the generalization of the abovematrix func-tions but also can be used to investigate the solvability con-ditions for the existence of the general solution to the system(6) Moreover it can be applied to characterize the relationsbetween Hermitian part of the solutions of (11) and (13)

The remainder of this paper is organized as follows InSection 2 we consider the extremal ranks and inertias of(10) subject to (11) and (13) In Section 3 we characterize therelations between the Hermitian part of the solution to (11)and (13) In Section 4 we establish the solvability conditionsfor the existence of a solution to (6) and obtain an expressionof the general solution to (6)

2 Extremal Ranks and Inertias of HermitianMatrix Function (10) with Some Restrictions

In this section we consider formulas for the extremal ranksand inertias of (10) subject to (11) and (13) We begin with thefollowing Lemmas

Lemma 1 (see [21]) (a) Let1198601119862

1 119861

1 and119863

1be givenThen

the following statements are equivalent

(1) system (13) is consistent(2)

1198771198601119862

1= 0 119863

1119871

1198611= 0 119860

1119863

1= 119862

11198611 (14)

(3)

119903 [1198601119862

1] = 119903 (119860

1)

[119863

1

1198611

] = 119903 (1198611)

1198601119863

1= 119862

11198611

(15)

In this case the general solution can be written as

119884 = 119860dagger

1119862

1+ 119871

1198601119863

1119861dagger

1+ 119871

1198601119881119877

1198611 (16)

where 119881 is arbitrary(b) Let 119860

2and 119862

2be given Then the following statements

are equivalent

(1) equation (11) is consistent(2)

1198771198602119862

2= 0 (17)

(3)

119903 [1198602119862

2] = 119903 (119860

2) (18)

In this case the general solution can be written as

119883 = 119860dagger

119862 + 119871119860119882 (19)

where119882 is arbitrary

Lemma 2 ([22 Lemma 15 Theorem 23]) Let 119860 isin C119898times119898

ℎ

119861 isin C119898times119899 and119863 isin C119899times119899

ℎ and denote that

119872 = [119860 119861

119861lowast

0]

119873 = [0 119876

119876lowast

0]

119871 = [119860 119861

119861lowast

119863]

119866 = [119875 119872119871

119873

119871119873119872

lowast

0]

(20)

Journal of Applied Mathematics 3

Then one has the following

(a) the following equalities hold

119894plusmn(119872) = 119903 (119861) + 119894

plusmn(119877

119861119860119877

119861) (21)

119894plusmn(119873) = 119903 (119876) (22)

(b) if R(119861) sube R(119860) then 119894plusmn(119871) = 119894

plusmn(119860) + 119894

plusmn(119863 minus

119861lowast

119860dagger

119861)) Thus 119894plusmn(119871) = 119894

plusmn(119860) if and only if R(119861) sube

R(119860) and 119894plusmn(119863 minus 119861

lowast

119860dagger

119861) = 0(c)

119894plusmn(119866) = [

[

119875 119872 0

119872lowast

0 119873lowast

0 119873 0

]

]

minus 119903 (119873) (23)

Lemma 3 (see [23]) Let 119860 isin C119898times119899 119861 isin C119898times119896 and 119862 isin C119897times119899Then they satisfy the following rank equalities

(a) 119903[119860 119861] = 119903(119860) + 119903(119864119860119861) = 119903(119861) + 119903(119864

119861119860)

(b) 119903 [ 119860119862] = 119903(119860) + 119903(119862119865

119860) = 119903(119862) + 119903(119860119865

119862)

(c) 119903 [ 119860 119861

119862 0] = 119903(119861) + 119903(119862) + 119903(119864

119861119860119865

119862)

(d) 119903[119861 119860119865119862] = 119903 [

119861 119860

0 119862] minus 119903(119862)

(e) 119903 [ 119862

119864119861119860] = 119903 [

119862 0

119860 119861] minus 119903(119861)

(f) 119903 [ 119860 119861119865119863

119864119864119862 0] = 119903 [

119860 119861 0

119862 0 119864

0 119863 0

] minus 119903(119863) minus 119903(119864)

Lemma 4 (see [15]) Let119860 isin C119898times119898

ℎ 119861 isin C119898times119899119862 isin C119899times119899

ℎ119876 isin

C119898times119899 and 119875 isin C119901times119899 be given and 119879 isin C119898times119898 be nonsingularThen one has the following

(1) 119894plusmn(119879119860119879

lowast

) = 119894plusmn(119860)

(2) 119894plusmn[119860 0

0 119862] = 119894

plusmn(119860) + 119894

plusmn(119862)

(3) 119894plusmn[

0 119876

119876lowast

0] = 119903(119876)

(4) 119894plusmn[

119860 119861119871119875

119871119875119861lowast

0] + 119903(119875) = 119894

plusmn[

119860 119861 0

119861lowast

0 119875lowast

0 119875 0

]

Lemma 5 (see [22 Lemma 14]) Let 119878 be a set consisting of(square) matrices over C119898times119898 and let 119867 be a set consisting of(square) matrices overC119898times119898

ℎ ThenThen one has the following

(a) 119878 has a nonsingularmatrix if and only ifmax119883isin119878119903(119883) =

119898(b) any 119883 isin 119878 is nonsingular if and only if min

119883isin119878119903(119883) =

119898(c) 0 isin 119878 if and only ifmin

119883isin119878119903(119883) = 0

(d) 119878 = 0 if and only ifmax119883isin119878119903(119883) = 0

(e) 119867 has a matrix 119883 gt 0 (119883 lt 0) if and only ifmax

119883isin119867119894+(119883) = 119898(max

119883isin119867119894minus(119883) = 119898)

(f) any 119883 isin 119867 satisfies 119883 gt 0 (119883 lt 0) if and only ifmin

119883isin119867119894+(119883) = 119898 (min

119883isin119867119894minus(119883) = 119898)

(g) 119867 has a matrix 119883 ge 0 (119883 le 0) if and only ifmin

119883isin119867119894minus(119883) = 0 (min

119883isin119867119894+(119883) = 0)

(h) any 119883 isin 119867 satisfies 119883 ge 0 (119883 le 0) if and only ifmax

119883isin119867119894minus(119883) = 0 (max

119883isin119867119894+(119883) = 0)

Lemma 6 (see [16]) Let 119901(119883 119884) = 119860minus119861119883minus (119861119883)lowast minus119862119884119863minus(119862119884119863)

lowast where119860119861119862 and119863 are givenwith appropriate sizesand denote that

1198721= [

[

119860 119861 119862

119861lowast

0 0

119862lowast

0 0

]

]

1198722= [

[

119860 119861 119863lowast

119861lowast

0 0

119863 0 0

]

]

1198723= [

119860 119861 119862 119863lowast

119861lowast

0 0 0]

1198724= [

[

119860 119861 119862 119863lowast

119861lowast

0 0 0

119862lowast

0 0 0

]

]

1198725= [

[

119860 119861 119862 119863lowast

119861lowast

0 0 0

119863 0 0 0

]

]

(24)

Then one has the following

(1) the maximal rank of 119901(119883 119884) is

max119883isinC119899times119898 119884

119903 [119901 (119883 119884)] = min 119898 119903 (1198721) 119903 (119872

2) 119903 (119872

3)

(25)

(2) the minimal rank of 119901(119883 119884) is

min119883isinC119899times119898 119884

119903 [119901 (119883 119884)]

= 2119903 (1198723) minus 2119903 (119861)

+max 119906++ 119906

minus V

++ V

minus 119906

++ V

minus 119906

minus+ V

+

(26)

(3) the maximal inertia of 119901(119883 119884) is

max119883isinC119899times119898119884

119894plusmn[119901 (119883 119884)] = min 119894

plusmn(119872

1) 119894

plusmn(119872

2) (27)

(4) the minimal inertias of 119901(119883 119884) is

min119883isinC119899times119898119884

119894plusmn[119901 (119883 119884)] = 119903 (119872

3) minus 119903 (119861)

+max 119894plusmn(119872

1) minus 119903 (119872

4)

119894plusmn(119872

2) minus 119903 (119872

5)

(28)

where

119906plusmn= 119894

plusmn(119872

1) minus 119903 (119872

4) V

plusmn= 119894

plusmn(119872

2) minus 119903 (119872

5) (29)

Now we present the main theorem of this section

4 Journal of Applied Mathematics

Theorem7 Let1198601isin C119898times119899119862

1isin C119898times119896119861

1isin C119896times119897119863

1isin C119899times119897

1198602isin C119905times119902 119862

2isin C119905times119901119860

3isin C

119901times119901

ℎ 119861

3isin C119901times119902 119862

3isin C119901times119899

and1198633isin C119901times119899 be given and suppose that the system of matrix

equations (13) and (11) is consistent respectively Denote the setof all solutions to (13) by 119878 and (11) by 119866 Put

1198641=

[[[[[

[

1198603

1198623119863

lowast

3119862

lowast

11198613119862

lowast

2

119862lowast

30 119860

lowast

10 0

1198621119863

3119860

10 0 0

119861lowast

30 0 0 119860

lowast

2

1198622

0 0 11986020

]]]]]

]

1198642= 119903

[[[[[

[

1198603

119863lowast

3119862

3119863

11198613119862

lowast

2

1198633

0 1198611

0 0

119863lowast

1119862

lowast

3119861lowast

10 0 0

119861lowast

30 0 0 119860

lowast

2

1198622

0 0 11986020

]]]]]

]

1198643=

[[[[[

[

1198603

1198613119862

lowast

3119863

lowast

3119862

lowast

2

119861lowast

30 0 0 119860

lowast

2

119863lowast

1119862

lowast

30 0 119861

lowast

10

1198621119863

30 119860

10 0

1198622

11986020 0 0

]]]]]

]

1198644=

[[[[[[[

[

1198603

1198613119862

3119863

lowast

3119862

lowast

2119863

lowast

3119862

lowast

1

119861lowast

30 0 0 119860

lowast

20

119862lowast

30 0 0 0 119860

lowast

1

0 0 0 119861lowast

10 0

1198621119863

30 119860

10 0 0

1198622

11986020 0 0 0

]]]]]]]

]

1198645=

[[[[[[[

[

1198603

1198613119862

3119863

lowast

3119862

lowast

2119862

3119863

1

119861lowast

30 0 0 119860

lowast

20

1198633

0 0 0 0 1198611

119863lowast

1119862

lowast

30 0 119860

lowast

10 0

0 0 11986010 0 0

1198622

11986020 0 0 0

]]]]]]]

]

(30)

Then one has the following(a) the maximal rank of (10) subject to (13) and (11) is

max119883isin119866119884isin119878

119903 [119891 (119883 119884)]

= min 119901 119903 (1198641) minus 2119903 (119860

1) minus 2119903 (119860

2)

119903 (1198642) minus 2119903 (119861

1) minus 2119903 (119860

2)

119903 (1198643) minus 2119903 (119860

2) minus 119903 (119860

1) minus 119903 (119861

1)

(31)

(b) the minimal rank of (10) subject to (13) and (11) is

min119883isin119866119884isin119878

119903 [119891 (119883 119884)]

= 2119903 (1198643) minus 2119903 [

1198613

1198602

]

+max 119903 (1198641) minus 2119903 (119864

4) 119903 (119864

2) minus 2119903 (119864

5)

119894+(119864

1) + 119894

minus(119864

2) minus 119903 (119864

4) minus 119903 (119864

5)

119894minus(119864

1) + 119894

+(119864

2) minus 119903 (119864

4) minus 119903 (119864

5)

(32)

(c) the maximal inertia of (10) subject to (13) and (11) is

max119883isin119866119884isin119878119909

119894plusmn[119891 (119883 119884)] = min 119894

plusmn(119864

1) minus 119903 (119860

1) minus 119903 (119860

2)

119894plusmn(119864

2) minus 119903 (119861

1) minus 119903 (119860

2)

(33)

(d) the minimal inertia of (10) subject to (13) and (11) is

min119883isin119866119884isin119878

119894plusmn[119891 (119883 119884)] = 119903 (119864

3) minus 119903 [

1198613

1198602

]

+max 119894plusmn(119864

1) minus 119903 (119864

4)

119894plusmn(119864

2) minus 119903 (119864

5)

(34)

Proof By Lemma 1 the general solutions to (13) and (11) canbe written as

119883 = 119860dagger

2119862

2+ 119871

1198602119882

119884 = 119860dagger

1119862

1+ 119871

1198601119863

1119861dagger

1+ 119871

1198601119885119877

1198611

(35)

where119882 and 119885 are arbitrary matrices with appropriate sizesPut

119876 = 1198613119871

1198602 119879 = 119862

3119871

1198601 119869 = 119877

1198611119863

3

119875 = 1198603minus 119861

3119860

dagger

2119862

2minus (119861

3119860

dagger

2119862

2)lowast

minus 1198623(119860

dagger

1119862

1+ 119871

1198601119863

1119861dagger

1)119863

3

minus (1198623(119860

dagger

1119862

1+ 119871

1198601119863

1119861dagger

1)119863

3)lowast

(36)

Substituting (36) into (10) yields

119891 (119883 119884) = 119875 minus 119876119882 minus (119876119882)lowast

minus 119879119885119869 minus (119879119885119869)lowast

(37)

Clearly 119875 is Hermitian It follows from Lemma 6 that

max119883isin119866119884isin119878

119903 [119891 (119883 119884)]

= max119882119885

119903 (119875 minus 119876119882 minus (119876119882)lowast

minus 119879119885119869 minus (119879119885119869)lowast

)

= min 119898 119903 (1198731) 119903 (119873

2) 119903 (119873

3)

(38)

min119883isin119866119884isin119878

119903 [119891 (119883 119884)]

= max119882119885

119903 (119875 minus 119876119882 minus (119876119882)lowast

minus 119879119885119869 minus (119879119885119869)lowast

)

= 2119903 (1198733) minus 2119903 (119876)

+max 119904++ 119904

minus 119905

++ 119905

minus 119904

++ 119905

minus 119904

minus+ 119905

+

(39)

Journal of Applied Mathematics 5

max119883isin119866119884isin119878

119894plusmn[119891 (119883 119884)]

= max119882119885

119903 (119875 minus 119876119882 minus (119876119882)lowast

minus 119879119885119869 minus (119879119885119869)lowast

)

= min 119894plusmn(119873

1) 119894

plusmn(119873

2)

(40)

min119883isin119866119884isin119878

119894plusmn[119891 (119883 119884)]

= max119882119885

119903 (119875 minus 119876119882 minus (119876119882)lowast

minus 119879119885119869 minus (119879119885119869)lowast

)

= 119903 (1198733) minus 119903 (119876) +max 119904

plusmn 119905

plusmn

(41)

where

1198731= [

[

119875 119876 119879

119876lowast

0 0

119879lowast

0 0

]

]

1198732= [

[

119875 119876 119869lowast

119876lowast

0 0

119869 0 0

]

]

1198733= [

119875 119876 119879 119869lowast

119876lowast

0 0 0]

1198734= [

[

119875 119876 119879 119869lowast

119876lowast

0 0 0

119879lowast

0 0 0

]

]

1198735= [

[

119875 119876 119879 119869lowast

119876lowast

0 0 0

119869 0 0 0

]

]

119904plusmn= 119894

plusmn(119873

1) minus 119903 (119873

4) 119905

plusmn= 119894

plusmn(119873

2) minus 119903 (119873

5)

(42)

Now we simplify the ranks and inertias of block matrices in(38)ndash(41)

By Lemma 4 blockGaussian elimination and noting that

119871lowast

119878= (119868 minus 119878

dagger

119878)lowast

= 119868 minus 119878lowast

(119878lowast

)dagger

= 119877119878lowast (43)

we have the following

119903 (1198731) = 119903[

[

119875 119876 119879

119876lowast

0 0

119879lowast

0 0

]

]

= 119903

[[[[[

[

1198603

1198623119863

lowast

3119862

lowast

11198613119862

lowast

2

119862lowast

30 119860

lowast

10 0

1198621119863

3119860

10 0 0

119861lowast

30 0 0 119860

lowast

2

1198622

0 0 11986020

]]]]]

]

minus 2119903 (1198601) minus 2119903 (119860

2)

(44)

By 11986211198611= 119860

1119863

1 we obtain

119903 (1198732) = 119903

[[

[

119875 119876 119869lowast

119876lowast

0 0

119869 0 0

]]

]

= 119903

[[[[[[[[

[

1198603

119863lowast

3119862

3119863

11198613119862

lowast

2

1198633

0 1198611

0 0

119863lowast

1119862

lowast

3119861lowast

10 0 0

119861lowast

30 0 0 119860

lowast

2

1198622

0 0 11986020

]]]]]]]]

]

minus 2119903 (1198611) minus 2119903 (119860

2)

119903 (1198733) = 119903 [

119875 119876 119879 119869lowast

119876lowast

0 0 0]

= 119903

[[[[[[[[

[

1198603

1198613119862

lowast

3119863

lowast

3119862

lowast

2

119861lowast

30 0 0 119860

lowast

2

119863lowast

1119862

lowast

30 0 119861

lowast

10

1198621119863

30 119860

10 0

1198622

11986020 0 0

]]]]]]]]

]

minus 119903 (1198611) minus 2119903 (119860

2) minus 119903 (119860

1)

119903 (1198734) = 119903

[[

[

119875 119876 119879 119869lowast

119876lowast

0 0 0

119879lowast

0 0 0

]]

]

= 119903

[[[[[[[

[

1198603

1198613119862

3119863

lowast

3119862

lowast

2119863

lowast

3119862

lowast

1

119861lowast

30 0 0 119860

lowast

20

119862lowast

30 0 0 0 119860

lowast

1

0 0 0 119861lowast

10 0

1198621119863

30 119860

10 0 0

1198622

11986020 0 0 0

]]]]]]]

]

minus 119903 (1198611) minus 2119903 (119860

2) minus 2119903 (119860

1)

119903 (1198735) = 119903[

[

119875 119876 119879 119869lowast

119876lowast

0 0 0

119879lowast

0 0 0

]

]

= 119903

[[[[[[[

[

1198603

1198613119862

3119863

lowast

3119862

lowast

2119862

3119863

1

119861lowast

30 0 0 119860

lowast

20

1198633

0 0 0 0 1198611

119863lowast

1119862

lowast

30 0 119860

lowast

10 0

0 0 11986010 0 0

1198622

11986020 0 0 0

]]]]]]]

]

minus 2119903 (1198611) minus 2119903 (119860

2) minus 119903 (119860

1)

(45)

By Lemma 2 we can get the following

119894plusmn(119873

1) = 119894

plusmn

[

[

119875 119876 119879

119876lowast

0 0

119879lowast

0 0

]

]

6 Journal of Applied Mathematics

= 119894plusmn

[[[[[

[

1198603

1198623119863

lowast

3119862

lowast

11198613119862

lowast

2

119862lowast

30 119860

lowast

10 0

1198621119863

3119860

10 0 0

119861lowast

30 0 0 119860

lowast

2

1198622

0 0 11986020

]]]]]

]

minus 119903 (1198601) minus 119903 (119860

2)

(46)

119894plusmn(119873

2) = 119894

plusmn

[

[

119875 119876 119869lowast

119876lowast

0 0

119869 0 0

]

]

= 119894plusmn

[[[[[

[

1198603

119863lowast

3119862

3119863

11198613119862

lowast

2

1198633

0 1198611

0 0

119863lowast

1119862

lowast

3119861lowast

10 0 0

119861lowast

30 0 0 119860

lowast

2

1198622

0 0 11986020

]]]]]

]

minus 119903 (1198611) minus 119903 (119860

2)

(47)

Substituting (44)-(47) into (38) and (41) yields (31)ndash(34)respectively

Corollary 8 Let 1198601 119862

1 119861

1 119863

1 119860

2 119862

2 119860

3 119861

3 119862

3 119863

3

and 119864119894 (119894 = 1 2 5) be as in Theorem 7 and suppose

that the system of matrix equations (13) and (11) is consistentrespectively Denote the set of all solutions to (13) by 119878 and (11)by 119866 Then one has the following

(a) there exist 119883 isin 119866 and 119884 isin 119878 such that 1198603minus 119861

3119883 minus

(1198613119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

gt 0 if and only if

119894+(119864

1) minus 119903 (119860

1) minus 119903 (119860

2) ge 119901

119894+(119864

2) minus 119903 (119861

1) minus 119903 (119860

2) ge 119901

(48)

(b) there exist 119883 isin 119866 and 119884 isin 119878 such that 1198603minus 119861

3119883 minus

(1198613119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

lt 0 if and only if

119894minus(119864

1) minus 119903 (119860

1) minus 119903 (119860

2) ge 119901

119894minus(119864

2) minus 119903 (119861

1) minus 119903 (119860

2) ge 119901

(49)

(c) there exist 119883 isin 119866 and 119884 isin 119878 such that 1198603minus 119861

3119883 minus

(1198613119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

ge 0 if and only if

119903 (1198643) minus 119903 [

1198613

1198602

] + 119894minus(119864

1) minus 119903 (119864

4) le 0

119903 (1198643) minus 119903 [

1198613

1198602

] + 119894minus(119864

2) minus 119903 (119864

5) le 0

(50)

(d) there exist 119883 isin 119866 and 119884 isin 119878 such that 1198603minus 119861

3119883 minus

(1198613119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

le 0 if and only if

119903 (1198643) minus 119903 [

1198613

1198602

] + 119894+(119864

1) minus 119903 (119864

4) le 0

119903 (1198643) minus 119903 [

1198613

1198602

] + 119894+(119864

2) minus 119903 (119864

5) le 0

(51)

(e) 1198603minus 119861

3119883 minus (119861

3119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

gt 0 for all119883 isin 119866 and 119884 isin 119878 if and only if

119903 (1198643) minus 119903 [

1198613

1198602

] + 119894+(119864

1) minus 119903 (119864

4) = 119901

119900119903 119903 (1198643) minus 119903 [

1198613

1198602

] + 119894+(119864

2) minus 119903 (119864

5) = 119901

(52)

(f) 1198603minus 119861

3119883 minus (119861

3119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

lt 0 for all119883 isin 119866 and 119884 isin 119878 if and only if

119903 (1198643) minus 119903 [

1198613

1198602

] + 119894minus(119864

1) minus 119903 (119864

4) = 119901

119900119903 119903 (1198643) minus 119903 [

1198613

1198602

] + 119894minus(119864

2) minus 119903 (119864

5) = 119901

(53)

(g) 1198603minus 119861

3119883 minus (119861

3119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

ge 0 for all119883 isin 119866 and 119884 isin 119878 if and only if

119894minus(119864

1) minus 119903 (119860

1) minus 119903 (119860

2) le 0

119900119903 119894minus(119864

2) minus 119903 (119861

1) minus 119903 (119860

2) le 0

(54)

(h) 1198603minus 119861

3119883 minus (119861

3119883)

lowast

minus 1198623119884119863

3minus (119862

3119884119863

3)lowast

le 0 for all119883 isin 119866 and 119884 isin 119878 if and only if

119894+(119864

1) minus 119903 (119860

1) minus 119903 (119860

2) le 0

119900119903 119894+(119864

2) minus 119903 (119861

1) minus 119903 (119860

2) le 0

(55)

(i) there exist 119883 isin 119866 and 119884 isin 119878 such that 1198603minus 119861

3119883 minus

(1198613119883)

lowast

minus1198623119884119863

3minus(119862

3119884119863

3)lowast is nonsingular if and only

if

119903 (1198641) minus 2119903 (119860

1) minus 2119903 (119860

2) ge 119901

119903 (1198642) minus 2119903 (119861

1) minus 2119903 (119860

2) ge 119901

119903 (1198643) minus 2119903 (119860

2) minus 119903 (119860

1) minus 119903 (119861

1) ge 119901

(56)

3 Relations between the Hermitian Part ofthe Solutions to (13) and (11)

Now we consider the extremal ranks and inertias of thedifference between the Hermitian part of the solutions to (13)and (11)

Theorem 9 Let 1198601isin C119898times119901 119862

1isin C119898times119901 119861

1isin C119901times119897 119863

1isin

C119901times119897 1198602isin C119905times119901 and 119862

2isin C119905times119901 be given Suppose that

the system of matrix equations (13) and (11) is consistent

Journal of Applied Mathematics 7

respectively Denote the set of all solutions to (13) by 119878 and (11)by 119866 Put

1198671=

[[[[[

[

0 119868 119862lowast

1minus119868 119862

lowast

2

119868 0 119860lowast

10 0

1198621119860

10 0 0

minus119868 0 0 0 119860lowast

2

11986220 0 119860

20

]]]]]

]

1198672= 119903

[[[[[

[

0 119868 1198631minus119868 119862

lowast

2

119868 0 1198611

0 0

119863lowast

1119861lowast

10 0 0

minus119868 0 0 0 119860lowast

2

11986220 0 119860

20

]]]]]

]

1198673=

[[[[[

[

0 minus119868 119868 119868 119862lowast

2

minus119868 0 0 0 119860lowast

2

119863lowast

10 0 119861

lowast

10

1198621

0 11986010 0

1198622119860

20 0 0

]]]]]

]

1198674=

[[[[[[[

[

0 minus119868 119868 119868 119862lowast

2119862

lowast

1

minus119868 0 0 0 119860lowast

20

119868 0 0 0 0 119860lowast

1

0 0 0 119861lowast

10 0

11986210 119860

10 0 0

1198622119860

20 0 0 0

]]]]]]]

]

1198675=

[[[[[[[

[

0 minus119868 119868 119868 119862lowast

2119863

1

minus119868 0 0 0 119860lowast

20

119868 0 0 0 0 1198611

119863lowast

10 0 119860

lowast

10 0

0 0 11986010 0 0

1198622119860

20 0 0 0

]]]]]]]

]

(57)

Then one has the following

max119883isin119866119884isin119878

119903 [(119883 + 119883lowast

) minus (119884 + 119884lowast

)]

= min 119901 119903 (1198671) minus 2119903 (119860

1) minus 2119903 (119860

2) 119903 (119867

2) minus 2119903 (119861

1)

minus2119903 (1198602) 119903 (119867

3) minus 2119903 (119860

2) minus 119903 (119860

1) minus 119903 (119861

1)

min119883isin119866119884isin119878

119903 [(119883 + 119883lowast

) minus (119884 + 119884lowast

)]

= 2119903 (1198673) minus 2119901

+max 119903 (1198671) minus 2119903 (119867

4) 119903 (119867

2) minus 2119903 (119867

5)

119894+(119867

1) + 119894

minus(119867

2) minus 119903 (119867

4) minus 119903 (119867

5)

119894minus(119867

1) + 119894

+(119867

2) minus 119903 (119867

4) minus 119903 (119867

5)

max119883isin119866119884isin119878

119894plusmn[(119883 + 119883

lowast

) minus (119884 + 119884lowast

)]

= min 119894plusmn(119867

1) minus 119903 (119860

1) minus 119903 (119860

2)

119894plusmn(119867

2) minus 119903 (119861

1) minus 119903 (119860

2)

min119883isin119866119884isin119878

119894plusmn[(119883 + 119883

lowast

) minus (119884 + 119884lowast

)]

= 119903 (1198643) minus 119901 +max 119894

plusmn(119867

1) minus 119903 (119867

4) 119894

plusmn(119867

2) minus 119903 (119867

5)

(58)

Proof By letting 1198603= 0 119861

3= minus119868 119862

3= 119868 and 119863

3= 119868 in

Theorem 7 we can get the results

Corollary 10 Let 1198601 119862

1 119861

1 119863

1 119860

2 119862

2 and 119867

119894 (119894 =

1 2 5) be as in Theorem 9 and suppose that the system ofmatrix equations (13) and (11) is consistent respectively Denotethe set of all solutions to (13) by 119878 and (11) by 119866 Then one hasthe following

(a) there exist 119883 isin 119866 and 119884 isin 119878 such that (119883 + 119883lowast

) gt

(119884 + 119884lowast

) if and only if

119894+(119867

1) minus 119903 (119860

1) minus 119903 (119860

2) ge 119901

119894+(119867

2) minus 119903 (119861

1) minus 119903 (119860

2) ge 119901

(59)

(b) there exist 119883 isin 119866 and 119884 isin 119878 such that (119883 + 119883lowast

) lt

(119884 + 119884lowast

) if and only if

119894minus(119867

1) minus 119903 (119860

1) minus 119903 (119860

2) ge 119901

119894minus(119867

2) minus 119903 (119861

1) minus 119903 (119860

2) ge 119901

(60)

(c) there exist 119883 isin 119866 and 119884 isin 119878 such that (119883 + 119883lowast

) ge

(119884 + 119884lowast

) if and only if

119903 (1198673) minus 119901 + 119894

minus(119867

1) minus 119903 (119867

4) le 0

119903 (1198673) minus 119901 + 119894

minus(119867

1) minus 119903 (119867

4) le 0

(61)

(d) there exist 119883 isin 119866 and 119884 isin 119878 such that (119883 + 119883lowast

) le

(119884 + 119884lowast

) if and only if

119903 (1198673) minus 119901 + 119894

+(119867

1) minus 119903 (119867

4) le 0

119903 (1198673) minus 119901 + 119894

+(119867

2) minus 119903 (119867

5) le 0

(62)

(e) (119883 + 119883lowast

) gt (119884 + 119884lowast

) for all 119883 isin 119866 and 119884 isin 119878 if andonly if

119903 (1198673) minus 119901 + 119894

+(119867

1) minus 119903 (119867

4) = 119901

119900119903 119903 (1198673) minus 119901 + 119894

+(119867

2) minus 119903 (119867

5) = 119901

(63)

(f) (119883 + 119883lowast

) lt (119884 + 119884lowast

) for all 119883 isin 119866 and 119884 isin 119878 if andonly if

119903 (1198673) minus 119901 + 119894

minus(119867

1) minus 119903 (119867

4) = 119901

119900119903 119903 (1198673) minus 119901 + 119894

minus(119867

2) minus 119903 (119867

5) = 119901

(64)

(g) (119883 + 119883lowast

) ge (119884 + 119884lowast

) for all 119883 isin 119866 and 119884 isin 119878 if andonly if

119894minus(119867

1) minus 119903 (119860

1) minus 119903 (119860

2) le 0

119900119903 119894minus(119867

2) minus 119903 (119861

1) minus 119903 (119860

2) le 0

(65)

8 Journal of Applied Mathematics

(h) (119883 + 119883lowast

) le (119884 + 119884lowast

) for all 119883 isin 119866 and 119884 isin 119878 if andonly if

119894+(119867

1) minus 119903 (119860

1) minus 119903 (119860

2) le 0

119900119903 119894+(119867

2) minus 119903 (119861

1) minus 119903 (119860

2) le 0

(66)

(i) there exist119883 isin 119866 and 119884 isin 119878 such that (119883 +119883lowast

) minus (119884 +

119884lowast

) is nonsingular if and only if119903 (119867

1) minus 2119903 (119860

1) minus 2119903 (119860

2) ge 119901

119903 (1198672) minus 2119903 (119861

1) minus 2119903 (119860

2) ge 119901

119903 (1198673) minus 2119903 (119860

2) minus 119903 (119860

1) minus 119903 (119861

1) ge 119901

(67)

4 The Solvability Conditions and the GeneralSolution to System (6)

We now turn our attention to (6) We in this section useTheorem 9 to give some necessary and sufficient conditionsfor the existence of a solution to (6) and present an expressionof the general solution to (6) We begin with a lemma whichis used in the latter part of this section

Lemma 11 (see [14]) Let 1198601isin C119898times1198991 119861

1isin C119898times1198992 119862

1isin

C119902times119898 and 1198641isin C119898times119898

ℎbe given Let 119860 = 119877

11986011198611 119861 = 119862

1119877

1198601

119864 = 1198771198601119864

1119877

1198601119872 = 119877

119860119861lowast119873 = 119860

lowast

119871119861 and 119878 = 119861lowast

119871119872 Then

the following statements are equivalent(1) equation (5) is consistent(2)119877

119872119877

119860119864 = 0 119877

119860119864119877

119860= 0 119871

119861119864119871

119861= 0 (68)

(3)

119903 [119864

11198611119862

lowast

1119860

1

119860lowast

10 0 0

] = 119903 [1198611119862

lowast

1119860

1] + 119903 (119860

1)

119903 [

[

11986411198611119860

1

119860lowast

10 0

119861lowast

10 0

]

]

= 2119903 [1198611119860

1]

119903 [

[

1198641119862

lowast

1119860

1

119860lowast

10 0

11986210 0

]

]

= 2119903 [119862lowast

1119860

1]

(69)

In this case the general solution of (5) can be expressed as

119884 =1

2[119860

dagger

119864119861dagger

minus 119860dagger

119861lowast

119872dagger

119864119861dagger

minus 119860dagger

119878(119861dagger

)lowast

119864119873dagger

119860lowast

119861dagger

+119860dagger

119864(119872dagger

)lowast

+ (119873dagger

)lowast

119864119861dagger

119878dagger

119878] + 1198711198601198811+ 119881

2119877

119861

+ 1198801119871

119878119871

119872+ 119877

119873119880

lowast

2119871

119872minus 119860

dagger

1198781198802119877

119873119860

lowast

119861dagger

119883 = 119860dagger

1[119864

1minus 119861

1119884119862

1minus (119861

1119884119862

1)lowast

]

minus1

2119860

dagger

1[119864

1minus 119861

1119884119862

1minus (119861

1119884119862

1)lowast

] 1198601119860

dagger

1

minus 119860dagger

1119882

1119860

lowast

1+119882

lowast

1119860

1119860

dagger

1+ 119871

1198601119882

2

(70)

where 1198801 119880

2 119881

1 119881

2119882

1 and119882

2are arbitrary matrices over

C with appropriate sizes

Now we give the main theorem of this section

Theorem 12 Let 119860119894 119862

119894 (119894 = 1 2 3) 119861

119895 and 119863

119895 (119895 = 1 3) be

given Set

119860 = 1198613119871

1198602 119861 = 119862

3119871

1198601

119862 = 1198771198611119863

3 119865 = 119877

119860119861

119866 = 119862119877119860 119872 = 119877

119865119866

lowast

119873 = 119865lowast

119871119866 119878 = 119866

lowast

119871119872

(71)

119863 = 1198603minus 119861

3119860

dagger

2119862

2minus (119861

3119860

dagger

2119862

2)lowast

minus 1198623(119860

dagger

1119862

1+ 119871

1198601119863

1119861dagger

1)119863

3

minus 119863lowast

3(119860

dagger

1119862

1+ 119871

1198601119863

1119861dagger

1)lowast

119862lowast

3

(72)

119864 = 119877119860119863119877

119860 (73)

Then the following statements are equivalent

(1) system (6) is consistent

(2) the equalities in (14) and (17) hold and

119877119872119877

119865119864 = 0 119877

119865119864119877

119865= 0 119871

119866119864119871

119866= 0 (74)

(3) the equalities in (15) and (18) hold and

119903

[[[[[

[

1198603

1198623119863

lowast

31198613119862

lowast

2

119861lowast

30 0 0 119860

lowast

2

1198621119863

3119860

10 0 0

119863lowast

1119862

lowast

30 119861

lowast

10 0

1198622

0 0 11986020

]]]]]

]

= 119903

[[[

[

1198623119863

lowast

31198613

11986010 0

0 119861lowast

10

0 0 1198602

]]]

]

+ 119903 [119860

2

1198613

]

119903

[[[[[

[

1198603

11986231198613119863

lowast

3119862

lowast

1119862

lowast

2

119862lowast

30 0 119860

lowast

10

119861lowast

30 0 0 119860

lowast

2

1198621119863

3119860

10 0 0

1198622

0 1198602

0 0

]]]]]

]

= 2119903[

[

11986231198613

11986010

0 1198602

]

]

119903

[[[[[

[

1198603

119863lowast

31198613119862

3119863

1119862

lowast

2

1198633

0 0 1198611

0

119861lowast

30 0 0 119860

lowast

2

119863lowast

1119862

lowast

3119861lowast

10 0 0

1198622

0 1198602

0 0

]]]]]

]

= 2119903[

[

119863lowast

31198613

119861lowast

10

0 1198602

]

]

(75)

In this case the general solution of system (6) can be expressedas

119883 = 119860dagger

2119862

2+ 119871

1198602119880

119884 = 119860dagger

1119862

1+ 119871

1198601119863

1119861dagger

1+ 119871

1198601119881119877

1198611

(76)

Journal of Applied Mathematics 9

where

119881 =1

2[119865

dagger

119864119866dagger

minus 119865dagger

119866lowast

119872dagger

119864119866dagger

minus 119865dagger

119878(119866dagger

)lowast

119864119873dagger

119865lowast

119866dagger

+119865dagger

119864(119872dagger

)lowast

+ (119873dagger

)lowast

119864119866dagger

119878dagger

119878] + 1198711198651198811

+ 1198812119877

119866+ 119880

1119871

119878119871

119872+ 119877

119873119880

lowast

2119871

119872minus 119865

dagger

1198781198802119877

119873119865

lowast

119866dagger

119880 = 119860dagger

[119863 minus 119861119881119862 minus (119861119881119862)lowast

]

minus1

2119860

dagger

[119863 minus 119861119881119862 minus (119861119881119862)lowast

] 119860119860dagger

minus 119860dagger

1198821119860

lowast

+119882lowast

1119860119860

dagger

+ 119871119860119882

2

(77)

where 1198801 119880

2 119881

1 119881

2119882

1 and119882

2are arbitrary matrices over

C with appropriate sizes

Proof (2)hArr (3) Applying Lemma 3 and Lemma 11 gives

119877119872119877

119865119864 = 0 lArrrArr 119903 (119877

119872119877

119865119864)

= 0 lArrrArr 119903[119863 119861 119862

lowast

119860

119860lowast

0 0 0]

= 119903 [119861 119862lowast

119860] + 119903 (119860)

lArrrArr 119903

[[[[[

[

119863 1198623119863

lowast

31198613

0

119861lowast

30 0 0 119860

lowast

2

0 11986010 0 0

0 0 119861lowast

10 0

0 0 0 11986020

]]]]]

]

= 119903

[[[

[

1198623119863

lowast

31198613

11986010 0

0 119861lowast

10

0 0 1198602

]]]

]

+ 119903 [119860

2

1198613

]

lArrrArr 119903

[[[[[

[

1198603

1198623119863

lowast

31198613119862

lowast

2

119861lowast

30 0 0 119860

lowast

2

1198621119863

3119860

10 0 0

119863lowast

1119862

lowast

30 119861

lowast

10 0

1198622

0 0 11986020

]]]]]

]

= 119903

[[[

[

1198623119863

lowast

31198613

11986010 0

0 119861lowast

10

0 0 1198602

]]]

]

+ 119903 [119860

2

1198613

]

(78)

By a similar approach we can obtain that

119877119865119864119877

119865= 0 lArrrArr 119903

[[[[[

[

1198603

11986231198613119863

lowast

3119862

lowast

1119862

lowast

2

119862lowast

30 0 119860

lowast

10

119861lowast

30 0 0 119860

lowast

2

1198621119863

3119860

10 0 0

1198622

0 1198602

0 0

]]]]]

]

= 2119903[

[

11986231198613

11986010

0 1198602

]

]

119871119866119864119871

119866= 0 lArrrArr 119903

[[[[[[[[

[

1198603

119863lowast

31198613119862

3119863

1119862

lowast

2

1198633

0 0 1198611

0

119861lowast

30 0 0 119860

lowast

2

119863lowast

1119862

lowast

3119861lowast

10 0 0

1198622

0 1198602

0 0

]]]]]]]]

]

= 2119903[[

[

119863lowast

31198613

119861lowast

10

0 1198602

]]

]

(79)

(1)hArr (2) We separate the four equations in system (6) intothree groups

1198601119884 = 119862

1 119884119861

1= 119863

1 (80)

1198602119883 = 119862

2 (81)

1198613119883 + (119861

3119883)

lowast

+ 1198623119884119863

3+ (119862

3119884119863

3)lowast

= 1198603 (82)

By Lemma 1 we obtain that system (80) is solvable if andonly if (14) (81) is consistent if and only if (17) The generalsolutions to system (80) and (81) can be expressed as (16) and(19) respectively Substituting (16) and (19) into (82) yields

119860119880 + (119860119880)lowast

+ 119861119881119862 + (119861119881119862)lowast

= 119863 (83)

Hence the system (5) is consistent if and only if (80) (81) and(83) are consistent respectively It follows fromLemma 11 that(83) is solvable if and only if

119877119872119877

119865119864 = 0 119877

119865119864119877

119865= 0 119871

119866119864119871

119866= 0 (84)

We know by Lemma 11 that the general solution of (83) canbe expressed as (77)

InTheorem 12 let 1198601and119863

1vanishThen we can obtain

the general solution to the following system

1198602119883 = 119862

2 119884119861

1= 119863

1

1198613119883 + (119861

3119883)

lowast

+ 1198623119884119863

3+ (119862

3119884119863

3)lowast

= 1198603

(85)

Corollary 13 Let 1198602 119862

2 119861

1 119863

1 119861

3 119862

3 119863

3 and 119860

3= 119860

lowast

3

be given Set

119860 = 1198613119871

1198602 119862 = 119877

1198611119863

3

119865 = 119877119860119862

3 119866 = 119862119877

119860

119872 = 119877119865119866

lowast

119873 = 119865lowast

119871119866

119878 = 119866lowast

119871119872

119863 = 1198603minus 119861

3119860

dagger

2119862

2minus (119861

3119860

dagger

2119862

2)lowast

minus 1198623119863

1119861dagger

1119863

3minus (119862

3119863

1119861dagger

1119863

3)lowast

119864 = 119877119860119863119877

119860

(86)

10 Journal of Applied Mathematics

Then the following statements are equivalent

(1) system (85) is consistent(2)

1198771198602119862

2= 0 119863

1119871

1198611= 0 119877

119872119877

119865119864 = 0 (87)

119877119865119864119877

119865= 0 119871

119866119864119871

119866= 0 (88)

(3)

119903 [1198602119862

2] = 119903 (119860

2) [

1198631

1198611

] = 119903 (1198611)

119903

[[[

[

1198603

1198623119863

lowast

31198613119862

lowast

2

119861lowast

30 0 0 119860

lowast

2

119863lowast

1119862

lowast

30 119861

lowast

10 0

1198622

0 0 11986020

]]]

]

= 119903[

[

1198623119863

lowast

31198613

0 119861lowast

10

0 0 1198602

]

]

+ 119903 [119860

2

1198613

]

119903

[[[

[

1198603119862

31198613119862

lowast

2

119862lowast

30 0 0

119861lowast

30 0 119860

lowast

2

11986220 119860

20

]]]

]

= 2119903 [119862

31198613

0 1198602

]

119903

[[[[[

[

1198603

119863lowast

31198613119862

3119863

1119862

lowast

2

1198633

0 0 1198611

0

119861lowast

30 0 0 119860

lowast

2

119863lowast

1119862

lowast

3119861lowast

10 0 0

1198622

0 1198602

0 0

]]]]]

]

= 2119903[

[

119863lowast

31198613

119861lowast

10

0 1198602

]

]

(89)

In this case the general solution of system (6) can be expressedas

119883 = 119860dagger

2119862

2+ 119871

1198602119880

119884 = 1198631119861dagger

1+ 119881119877

1198611

(90)

where

119881 =1

2[119865

dagger

119864119866dagger

minus 119865dagger

119866lowast

119872dagger

119864119866dagger

minus 119865dagger

119878(119866dagger

)lowast

119864119873dagger

119865lowast

119866dagger

+119865dagger

119864(119872dagger

)lowast

+ (119873dagger

)lowast

119864119866dagger

119878dagger

119878] + 1198711198651198811

+ 1198812119877

119866+ 119880

1119871

119878119871

119872+ 119877

119873119880

lowast

2119871

119872minus 119865

dagger

1198781198802119877

119873119865

lowast

119866dagger

119880 = 119860dagger

[119863 minus 1198623119881119862 minus (119862

3119881119862)

lowast

]

minus1

2119860

dagger

[119863 minus 1198623119881119862 minus (119862

3119881119862)

lowast

] 119860119860dagger

minus 119860dagger

1198821119860

lowast

+119882lowast

1119860119860

dagger

+ 119871119860119882

2

(91)

where 1198801 119880

2 119881

1 119881

2119882

1 and119882

2are arbitrary matrices over

C with appropriate sizes

Acknowledgments

The authors would like to thank Dr Mohamed Dr Sivaku-mar and a referee very much for their valuable suggestions

and comments which resulted in a great improvement of theoriginal paper This research was supported by the Grantsfrom the National Natural Science Foundation of China(NSFC (11161008)) and Doctoral Program Fund of Ministryof Education of PRChina (20115201110002)

References

[1] H Zhang ldquoRank equalities for Moore-Penrose inverse andDrazin inverse over quaternionrdquo Annals of Functional Analysisvol 3 no 2 pp 115ndash127 2012

[2] L Bao Y Lin and YWei ldquoA new projectionmethod for solvinglarge Sylvester equationsrdquo Applied Numerical Mathematics vol57 no 5ndash7 pp 521ndash532 2007

[3] MDehghan andMHajarian ldquoOn the generalized reflexive andanti-reflexive solutions to a system of matrix equationsrdquo LinearAlgebra and Its Applications vol 437 no 11 pp 2793ndash2812 2012

[4] F O Farid M S Moslehian Q-W Wang and Z-C Wu ldquoOnthe Hermitian solutions to a system of adjointable operatorequationsrdquo Linear Algebra and Its Applications vol 437 no 7pp 1854ndash1891 2012

[5] Z H He and Q W Wang ldquoA real quaternion matrix equationwith with applicationsrdquo Linear Multilinear Algebra vol 61 pp725ndash740 2013

[6] Q-W Wang and Z-C Wu ldquoCommon Hermitian solutionsto some operator equations on Hilbert 119862119909-modulesrdquo LinearAlgebra and Its Applications vol 432 no 12 pp 3159ndash3171 2010

[7] Q-W Wang and C-Z Dong ldquoPositive solutions to a systemof adjointable operator equations over Hilbert 119862119909-modulesrdquoLinear Algebra and Its Applications vol 433 no 7 pp 1481ndash14892010

[8] Q-W Wang J W van der Woude and H-X Chang ldquoA systemof real quaternion matrix equations with applicationsrdquo LinearAlgebra and Its Applications vol 431 no 12 pp 2291ndash2303 2009

[9] Q W Wang H-S Zhang and G-J Song ldquoA new solvable con-dition for a pair of generalized Sylvester equationsrdquo ElectronicJournal of Linear Algebra vol 18 pp 289ndash301 2009

[10] H W Braden ldquoThe equations 119860119879

119883 plusmn 119883119879

119860 = 119861rdquo SIAM Journalon Matrix Analysis and Applications vol 20 no 2 pp 295ndash3021999

[11] D S Djordjevic ldquoExplicit solution of the operator equation119860

lowast

119883 + 119883plusmn

119860 = 119861rdquo Journal of Computational and Applied Math-ematics vol 200 no 2 pp 701ndash704 2007

[12] W Cao ldquoSolvability of a quaternion matrix equationrdquo AppliedMathematics vol 17 no 4 pp 490ndash498 2002

[13] Q Xu L Sheng and Y Gu ldquoThe solutions to some operatorequationsrdquo Linear Algebra and Its Applications vol 429 no 8-9pp 1997ndash2024 2008

[14] Q-W Wang and Z-H He ldquoSome matrix equations with appli-cationsrdquo Linear and Multilinear Algebra vol 60 no 11-12 pp1327ndash1353 2012

[15] Y Tian ldquoMaximization and minimization of the rank andinertia of the Hermitian matrix expression 119860 minus 119861119883 minus (119861119883)

119909rdquoLinear Algebra and Its Applications vol 434 no 10 pp 2109ndash2139 2011

[16] Z-H He and Q-WWang ldquoSolutions to optimization problemson ranks and inertias of a matrix function with applicationsrdquoAppliedMathematics andComputation vol 219 no 6 pp 2989ndash3001 2012

[17] Y Liu andY Tian ldquoMax-min problems on the ranks and inertiasof the matrix expressions 119860 minus 119861119883119862 plusmn (119861119883119862)

lowastrdquo Journal of

Journal of Applied Mathematics 11

Optimization Theory and Applications vol 148 no 3 pp 593ndash622 2011

[18] DChu Y SHung andH JWoerdeman ldquoInertia and rank cha-racterizations of some matrix expressionsrdquo SIAM Journal onMatrix Analysis and Applications vol 31 no 3 pp 1187ndash12262009

[19] Y Liu and Y Tian ldquoA simultaneous decomposition of a mat-rix triplet with applicationsrdquoNumerical Linear Algebra with Ap-plications vol 18 no 1 pp 69ndash85 2011

[20] X Zhang Q-W Wang and X Liu ldquoInertias and ranks of someHermitian matrix functions with applicationsrdquo Central Euro-pean Journal of Mathematics vol 10 no 1 pp 329ndash351 2012

[21] Q-W Wang ldquoThe general solution to a system of real quater-nion matrix equationsrdquo Computers amp Mathematics with Appli-cations vol 49 no 5-6 pp 665ndash675 2005

[22] Y Tian ldquoEqualities and inequalities for inertias of Hermitianmatrices with applicationsrdquo Linear Algebra and Its Applicationsvol 433 no 1 pp 263ndash296 2010

[23] G Marsaglia and G P H Styan ldquoEqualities and inequalities forranks of matricesrdquo Linear and Multilinear Algebra vol 2 pp269ndash292 1974

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 230408 6 pageshttpdxdoiorg1011552013230408

Research ArticleLeft and Right Inverse Eigenpairs Problemfor 120581-Hermitian Matrices

Fan-Liang Li1 Xi-Yan Hu2 and Lei Zhang2

1 Institute of Mathematics and Physics School of Sciences Central South University of Forestry and TechnologyChangsha 410004 China

2 College of Mathematics and Econometrics Hunan University Changsha 410082 China

Correspondence should be addressed to Fan-Liang Li lfl302tomcom

Received 6 December 2012 Accepted 20 March 2013

Academic Editor Panayiotis J Psarrakos

Copyright copy 2013 Fan-Liang Li et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Left and right inverse eigenpairs problem for 120581-hermitian matrices and its optimal approximate problem are considered Based onthe special properties of 120581-hermitian matrices the equivalent problem is obtained Combining a new inner product of matricesthe necessary and sufficient conditions for the solvability of the problem and its general solutions are derived Furthermore theoptimal approximate solution and a calculation procedure to obtain the optimal approximate solution are provided

1 Introduction

Throughout this paper we use some notations as follows Let119862119899times119898 be the set of all 119899times119898 complex matrices119880119862

119899times119899119867119862119899times119899

119878119867119862119899times119899 denote the set of all 119899times119899 unitary matrices hermitian

matrices skew-hermitian matrices respectively Let 119860 119860119867and119860

+ be the conjugate conjugate transpose and theMoore-Penrose generalized inverse of 119860 respectively For 119860 119861 isin

119862119899times119898 ⟨119860 119861⟩ = re(tr(119861119867119860)) where re(tr(119861119867119860)) denotes the

real part of tr(119861119867119860) the inner product of matrices 119860 and 119861The induced matrix norm is called Frobenius norm That is119860 = ⟨119860 119860⟩

12

= (tr(119860119867119860))12

Left and right inverse eigenpairs problem is a specialinverse eigenvalue problem That is giving partial left andright eigenpairs (eigenvalue and corresponding eigenvector)(120582119894 119909119894) 119894 = 1 ℎ (120583

119895 119910119895) 119895 = 1 119897 a special matrix set

119878 finding a matrix 119860 isin 119878 such that

119860119909119894= 120582119894119909119894 119894 = 1 ℎ

119910119879

119895119860 = 120583

119895119910119879

119895 119895 = 1 119897

(1)

This problem which usually arises in perturbation analysisof matrix eigenvalues and in recursive matters has profoundapplication background [1ndash6] When the matrix set 119878 isdifferent it is easy to obtain different left and right inverse

eigenpairs problem For example we studied the left andright inverse eigenpairs problem of skew-centrosymmetricmatrices and generalized centrosymmetric matrices respec-tively [5 6] Based on the special properties of left and righteigenpairs of these matrices we derived the solvability condi-tions of the problem and its general solutions In this papercombining the special properties of 120581-hermitianmatrices anda new inner product ofmatrices we first obtain the equivalentproblem then derive the necessary and sufficient conditionsfor the solvability of the problem and its general solutions

Hill and Waters [7] introduced the following matrices

Definition 1 Let 120581 be a fixed product of disjoint transposi-tions and let119870 be the associated permutation matrix that is119870 = 119870

119867

= 119870 1198702 = 119868119899 a matrix 119860 isin 119862

119899times119899 is said to be 120581-hermitian matrices (skew 120581-hermitian matrices) if and onlyif 119886119894119895= 119886119896(119895)119896(119894)

(119886119894119895= minus119886119896(119895)119896(119894)

) 119894 119895 = 1 119899 We denote theset of 120581-hermitian matrices (skew 120581-hermitian matrices) by119870119867119862119899times119899

(119878119870119867119862119899times119899

)

FromDefinition 1 it is easy to see that hermitianmatricesand perhermitian matrices are special cases of 120581-hermitianmatrices with 119896(119894) = 119894 and 119896(119894) = 119899 minus 119894 + 1 respectivelyHermitian matrices and perhermitian matrices which areone of twelve symmetry patterns of matrices [8] are appliedin engineering statistics and so on [9 10]

2 Journal of Applied Mathematics

From Definition 1 it is also easy to prove the followingconclusions

(1) 119860 isin 119870119867119862119899times119899 if and only if 119860 = 119870119860

119867

119870(2) 119860 isin 119878119870119867119862

119899times119899 if and only if 119860 = minus119870119860119867

119870(3) If 119870 is a fixed permutation matrix then 119870119867119862

119899times119899 and119878119870119867119862

119899times119899 are the closed linear subspaces of 119862119899times119899 andsatisfy

119862119899times119899

= 119870119867119862119899times119899

⨁119878119870119867119862119899times119899

(2)

The notation 1198811oplus 1198812stands for the orthogonal direct sum of

linear subspace 1198811and 119881

2

(4) 119860 isin 119870119867119862119899times119899 if and only if there is a matrix isin

119867119862119899times119899 such that = 119870119860

(5) 119860 isin 119878119870119867119862119899times119899 if and only if there is a matrix isin

119878119867119862119899times119899 such that = 119870119860

Proof (1) From Definition 1 if 119860 = (119886119894119895) isin 119870119867119862

119899times119899 then119886119894119895= 119886119896(119895)119896(119894)

this implies119860 = 119870119860119867

119870 for119870119860119867

119870 = (119886119896(119895)119896(119894)

)(2)With the samemethod we can prove (2) So the proof

is omitted

(3) (a) For any 119860 isin 119862119899times119899 there exist 119860

1isin 119870119867119862

119899times1198991198602isin 119878119870119867119862

119899times119899 such that

119860 = 1198601+ 1198602 (3)

where 1198601= (12)(A + 119870119860

119867

119870) 1198602= (12)(119860 minus

119870119860119867

119870)(b) If there exist another 119860

1isin 119870119867119862

119899times119899 1198602

isin

119878119870119867119862119899times119899 such that

119860 = 1198601+ 1198602 (4)

(3)-(4) yields

1198601minus 1198601= minus (119860

2minus 1198602) (5)

Multiplying (5) on the left and on the right by119870 respectively and according to (1) and (2) weobtain

1198601minus 1198601= 1198602minus 1198602 (6)

Combining (5) and (6) gives1198601= 11986011198602= 1198602

(c) For any1198601isin 119870119867119862

119899times1198991198602isin 119878119870119867119862

119899times119899 we have

⟨1198601 1198602⟩ = re (tr (119860119867

21198601)) = re (tr (119870119860

119867

21198701198701198601119870))

= re (tr (minus11986011986721198601)) = minus ⟨119860

1 1198602⟩

(7)

This implies ⟨1198601 1198602⟩ = 0 Combining (a) (b)

and (c) gives (3)

(4) Let = 119870119860 if 119860 isin 119870119867119862119899times119899 then

119867

= isin 119867119862119899times119899

If 119867 = isin 119867119862119899timesn then 119860 = 119870 and 119870119860

119867

119870 = 119870119867

119870119870 =

119870 = 119860 isin 119870119867119862119899times119899

(5)With the samemethod we can prove (5) So the proofis omitted

In this paper we suppose that 119870 is a fixed permutationmatrix and assume (120582

119894 119909119894) 119894 = 1 ℎ be right eigenpairs

of 119860 (120583119895 119910119895) 119895 = 1 119897 be left eigenpairs of 119860 If we let

119883 = (1199091 119909

ℎ) isin 119862

119899timesℎ Λ = diag (1205821 120582

ℎ) isin 119862

ℎtimesℎ119884 = (119910

1 119910

119897) isin 119862119899times119897 Γ = diag(120583

1 120583

119897) isin 119862119897times119897 then the

problems studied in this paper can be described as follows

Problem 2 Giving 119883 isin 119862119899timesℎ Λ = diag(120582

1 120582

ℎ) isin 119862

ℎtimesℎ119884 isin 119862

119899times119897 Γ = diag(1205831 120583

119897) isin 119862119897times119897 find 119860 isin 119870119867119862

119899times119899 suchthat

119860119883 = 119883Λ

119884119879

119860 = Γ119884119879

(8)

Problem 3 Giving 119861 isin 119862119899times119899 find isin 119878

119864such that

10038171003817100381710038171003817119861 minus

10038171003817100381710038171003817= minforall119860isin119878119864

119861 minus 119860 (9)

where 119878119864is the solution set of Problem 2

This paper is organized as follows In Section 2 we firstobtain the equivalent problemwith the properties of119870119867119862

119899times119899

and then derive the solvability conditions of Problem 2 and itsgeneral solutionrsquos expression In Section 3 we first attest theexistence and uniqueness theorem of Problem 3 then presentthe unique approximation solution Finally we provide acalculation procedure to compute the unique approximationsolution and numerical experiment to illustrate the resultsobtained in this paper correction

2 Solvability Conditions of Problem 2

We first discuss the properties of119870119867119862119899times119899

Lemma 4 Denoting 119872 = 119870119864119870119866119864 and 119864 isin 119867119862119899times119899 one has

the following conclusions(1) If 119866 isin 119870119867119862

119899times119899 then119872 isin 119870119867119862119899times119899

(2) If 119866 isin 119878119870119867119862119899times119899 then119872 isin 119878119870119867119862

119899times119899(3) If 119866 = 119866

1+ 1198662 where 119866

1isin 119870119867119862

119899times119899 1198662isin 119878119870119867119862

119899times119899then 119872 isin 119870119867119862

119899times119899 if and only if 1198701198641198701198662119864 = 0 In

addition one has119872 = 1198701198641198701198661119864

Proof (1) 119870119872119867

119870 = 119870119864119866119867

119870119864119870119870 = 119870119864(119870119866119870)119870119864 =

119870119864119870119866119864 = 119872Hence we have119872 isin 119870119867119862

119899times119899(2) 119870119872

119867

119870 = 119870119864119866119867

119870119864119870119870 = 119870119864(minus119870119866119870)119870119864 =

minus119870119864119870119866119864 = minus119872Hence we have119872 isin 119878119870119867119862

119899times119899(3) 119872 = 119870119864119870(119866

1+ 1198662)119864 = 119870119864119870119866

1119864 + 119870119864119870119866

2119864 we

have 1198701198641198701198661119864 isin 119870119867119862

119899times119899 1198701198641198701198662119864 isin 119878119870119867119862

119899times119899 from (1)and (2) If 119872 isin 119870119867119862

119899times119899 then 119872 minus 1198701198641198701198661119864 isin 119870119867119862

119899times119899while119872minus119870119864119870119866

1119864 = 119870119864119870119866

2119864 isin 119878119870119867119862

119899times119899Therefore fromthe conclusion (3) of Definition 1 we have119870119864119870119866

2119864 = 0 that

is119872 = 1198701198641198701198661119864 On the contrary if119870119864119870119866

2119864 = 0 it is clear

that119872 = 1198701198641198701198661119864 isin 119870119867119862

119899times119899 The proof is completed

Lemma 5 Let 119860 isin 119870119867119862119899times119899 if (120582 119909) is a right eigenpair of 119860

then (120582 119870119909) is a left eigenpair of 119860

Journal of Applied Mathematics 3

Proof If (120582 119909) is a right eigenpair of 119860 then we have

119860119909 = 120582119909 (10)

From the conclusion (1) of Definition 1 it follows that

119870119860119867

119870119909 = 120582119909 (11)

This implies

(119870119909)119879

119860 = 120582(119870119909)119879

(12)

So (120582 119870119909) is a left eigenpair of 119860

From Lemma 5 without loss of the generality we mayassume that Problem 2 is as follows

119883 isin 119862119899timesℎ

Λ = diag (1205821 120582

ℎ) isin 119862ℎtimesℎ

119884 = 119870119883 isin 119862119899timesℎ

Γ = Λ isin 119862ℎtimesℎ

(13)

Combining (13) and the conclusion (4) of Definition 1 itis easy to derive the following lemma

Lemma 6 If 119883 Λ 119884 Γ are given by (13) then Problem 2 isequivalent to the following problem If 119883 Λ 119884 Γ are given by(13) find 119870119860 isin 119867119862

119899times119899 such that

119870119860119883 = 119870119883Λ (14)

Lemma 7 (see [11]) If giving119883 isin 119862119899timesℎ119861 isin 119862

119899timesℎ thenmatrixequation 119883 = 119861 has solution isin 119867119862

119899times119899 if and only if

119861 = 119861119883+

119883 119861119867

119883 = 119883119867

119861 (15)

Moreover the general solution can be expressed as

= 119861119883+

+ (119861119883+

)119867

(119868119899minus 119883119883

+

)

+ (119868119899minus 119883119883

+

) (119868119899minus 119883119883

+

) forall isin 119867119862119899times119899

(16)

Theorem 8 If119883Λ119884 Γ are given by (13) then Problem 2 hasa solution in 119870119867119862

119899times119899 if and only if

119883119867

119870119883Λ = Λ119883119867

119870119883 119883Λ = 119883Λ119883+

119883 (17)

Moreover the general solution can be expressed as

119860 = 1198600+ 119870119864119870119866119864 forall119866 isin 119870119867119862

119899times119899

(18)

where

1198600= 119883Λ119883

+

+ 119870(119883Λ119883+

)119867

119870119864 119864 = 119868119899minus 119883119883

+

(19)

Proof Necessity If there is a matrix 119860 isin 119870119867119862119899times119899 such that

(119860119883 = 119883Λ 119884119879119860 = Γ119884119879

) then from Lemma 6 there exists amatrix 119870119860 isin 119867119862

119899times119899 such that 119870119860119883 = 119870119883Λ and accordingto Lemma 7 we have

119870119883Λ = 119870119883Λ119883+

119883 (119870119883Λ)119867

119883 = 119883119867

(119870119883Λ) (20)

It is easy to see that (20) is equivalent to (17)

Sufficiency If (17) holds then (20) holds Hence matrixequation119870119860119883 = 119870119883Λ has solution119870119860 isin 119867119862

119899times119899 Moreoverthe general solution can be expressed as follows

119870119860 = 119870119883Λ119883+

+ (119870119883Λ119883+

)119867

(119868119899minus 119883119883

+

)

+ (119868119899minus 119883119883

+

) (119868119899minus 119883119883

+

) forall isin 119867119862119899times119899

(21)

Let

1198600= 119883Λ119883

+

+ 119870(119883Λ119883+

)119867

119870119864 119864 = 119868119899minus 119883119883

+

(22)

This implies 119860 = 1198600+ 119870119864119864 Combining the definition of

119870 119864 and the first equation of (17) we have

119870119860119867

0119870 = 119870(119883Λ119883

+

)119867

119870 + 119883Λ119883+

minus 119870(119883119883+

)119867

119870(119883Λ119883+

)

= 119883Λ119883+

+ 119870(119883Λ119883+

)119867

119870 minus 119870(119883Λ119883+

)119879

119870119883119883+

= 1198600

(23)

Hence1198600isin 119870119867119862

119899times119899 Combining the definition of119870 119864 (13)and (17) we have

1198600119883 = 119883Λ119883

+

119883 + 119870(119883Λ119883+

)119867

119870(119868119899minus 119883119883

+

)119883 = 119883Λ

119884119879

1198600= 119883119867

119870119883Λ119883+

+ 119883119867

119870119870(119883Λ119883+

)119867

119870119864

= Λ119883119867

119870119883119883+

+ (119870119883Λ119883+

119883)119867

(119868119899minus 119883119883

+

)

= Λ119883119867

119870119883119883+

+ Λ119883119867

119870(119868119899minus 119883119883

+

)

= Λ119883119867

119870 = Γ119884119879

(24)

Therefore 1198600is a special solution of Problem 2 Combining

the conclusion (4) of Definition 1 Lemma 4 and 119864 = 119868119899minus

119883119883+

isin 119867119862119899times119899 it is easy to prove that 119860 = 119860

0+ 119870119864119870119866119864 isin

119870119867119862119899times119899 if and only if 119866 isin 119870119867119862

119899times119899 Hence the solution setof Problem 2 can be expressed as (18)

3 An Expression of the Solution of Problem 3

From (18) it is easy to prove that the solution set 119878119864of

Problem 2 is a nonempty closed convex set if Problem 2 hasa solution in 119870119867119862

119899times119899 We claim that for any given 119861 isin 119877119899times119899

there exists a unique optimal approximation for Problem 3

Theorem9 Giving119861 isin 119862119899times119899 if the conditions of119883119884Λ Γ are

the same as those in Theorem 8 then Problem 3 has a uniquesolution isin 119878

119864 Moreover can be expressed as

= 1198600+ 119870119864119870119861

1119864 (25)

where 1198600 119864 are given by (19) and 119861

1= (12)(119861 + 119870119861

119867

119870)

Proof Denoting1198641= 119868119899minus119864 it is easy to prove thatmatrices119864

and1198641are orthogonal projectionmatrices satisfying119864119864

1= 0

It is clear that matrices 119870119864119870 and 1198701198641119870 are also orthogonal

4 Journal of Applied Mathematics

projection matrices satisfying (119870119864119870)(1198701198641119870) = 0 According

to the conclusion (3) of Definition 1 for any 119861 isin 119862119899times119899 there

exists unique

1198611isin 119870119867119862

119899times119899

1198612isin 119878119870119867119862

119899times119899 (26)

such that

119861 = 1198611+ 1198612 ⟨119861

1 1198612⟩ = 0 (27)

where

1198611=

1

2(119861 + 119870119861

119867

119870) 1198612=

1

2(119861 minus 119870119861

119867

119870) (28)

CombiningTheorem 8 for any 119860 isin 119878119864 we have

119861 minus 1198602

=1003817100381710038171003817119861 minus 119860

0minus 119870119864119870119866119864

1003817100381710038171003817

2

=10038171003817100381710038171198611 + 119861

2minus 1198600minus 119870119864119870119866119864

1003817100381710038171003817

2

=10038171003817100381710038171198611 minus 119860

0minus 119870119864119870119866119864

1003817100381710038171003817

2

+10038171003817100381710038171198612

1003817100381710038171003817

2

=1003817100381710038171003817(1198611 minus 119860

0) (119864 + 119864

1) minus 119870119864119870119866119864

1003817100381710038171003817

2

+10038171003817100381710038171198612

1003817100381710038171003817

2

=1003817100381710038171003817(1198611 minus 119860

0) 119864 minus 119870119864119870119866119864

1003817100381710038171003817

2

+1003817100381710038171003817(1198611 minus 119860

0) 1198641

1003817100381710038171003817

2

+10038171003817100381710038171198612

1003817100381710038171003817

2

=1003817100381710038171003817119870 (119864 + 119864

1)119870 (119861

1minus 1198600) 119864 minus 119870119864119870119866119864

1003817100381710038171003817

2

+1003817100381710038171003817(1198611 minus 119860

0) 1198641

1003817100381710038171003817

2

+10038171003817100381710038171198612

1003817100381710038171003817

2

=1003817100381710038171003817119870119864119870 (119861

1minus 1198600) 119864 minus 119870119864119870119866119864

1003817100381710038171003817

2

+10038171003817100381710038171198701198641119870(1198611minus 1198600) 119864

1003817100381710038171003817

2

+1003817100381710038171003817(1198611 minus 119860

0)1198641

1003817100381710038171003817

2

+10038171003817100381710038171198612

1003817100381710038171003817

2

(29)

It is easy to prove that 1198701198641198701198600119864 = 0 according to the

definitions of 1198600 119864 So we have

119861 minus 1198602

=1003817100381710038171003817119870119864119870119861

1119864 minus 119870119864119870119866119864

1003817100381710038171003817

2

+10038171003817100381710038171198701198641119870(1198611minus 1198600) 119864

1003817100381710038171003817

2

+1003817100381710038171003817(1198611 minus 119860

0)1198641

1003817100381710038171003817

2

+10038171003817100381710038171198612

1003817100381710038171003817

2

(30)

Obviously min119860isin119878119864

119861 minus 119860 is equivalent to

min119866isin119870119867119862

119899times119899

10038171003817100381710038171198701198641198701198611119864 minus 119870119864119870119866119864

1003817100381710038171003817 (31)

Since 1198641198641

= 0 (119870119864119870)(1198701198641119870) = 0 it is clear that 119866 =

1198611+ 11987011986411198701198641 for any isin 119870119867119862

119899times119899 is a solution of (31)Substituting this result to (18) we can obtain (25)

Algorithm 10 (1) Input 119883 Λ 119884 Γ according to (13) (2)Compute 119883

119867

119870119883Λ Λ119883119867

119870119883 119883Λ119883+

119883 119883Λ if (17) holdsthen continue otherwise stop (3) Compute 119860

0according to

(19) and compute 1198611according to (28) (4) According to (25)

calculate

Example 11 (119899 = 8 ℎ = 119897 = 4)

119883 =

((((

(

05661 minus02014 minus 01422119894 01446 + 02138119894 0524

minus02627 + 01875119894 05336 minus02110 minus 04370119894 minus00897 + 03467119894

minus04132 + 02409119894 00226 minus 00271119894 minus01095 + 02115119894 minus03531 minus 00642119894

minus00306 + 02109119894 minus03887 minus 00425119894 02531 + 02542119894 00094 + 02991119894

00842 minus 01778119894 minus00004 minus 03733119894 03228 minus 01113119894 01669 + 01952119894

00139 minus 03757119894 minus02363 + 03856119894 02583 + 00721119894 01841 minus 02202119894

00460 + 03276119894 minus01114 + 00654119894 minus00521 minus 02556119894 minus02351 + 03002119894

00085 minus 01079119894 00974 + 03610119894 05060 minus02901 minus 00268119894

))))

)

Λ = (

minus03967 minus 04050119894 0 0 0

0 minus03967 + 04050119894 0 0

0 0 00001 0

0 0 0 minus00001119894

)

119870 =

((((

(

0 0 0 0 0 0 0 1

0 0 0 0 0 0 1 0

0 0 0 0 0 1 0 0

0 0 0 1 0 0 0 0

0 0 0 0 1 0 0 0

0 0 1 0 0 0 0 0

0 1 0 0 0 0 0 0

1 0 0 0 0 0 0 0

))))

)

119884 = 119870119883 Γ = Λ

(32)

Journal of Applied Mathematics 5

119861 =

From the first column to the fourth column

((((

(

minus05218 + 00406119894 02267 minus 00560119894 minus01202 + 00820119894 minus00072 minus 03362119894

03909 minus 03288119894 02823 minus 02064119894 minus00438 minus 00403119894 02707 + 00547119894

02162 minus 01144119894 minus04307 + 02474119894 minus00010 minus 00412119894 02164 minus 01314119894

minus01872 minus 00599119894 minus00061 + 04698119894 03605 minus 00247119894 04251 + 01869119894

minus01227 minus 00194119894 02477 minus 00606119894 03918 + 06340119894 01226 + 00636119894

minus00893 + 04335119894 00662 + 00199119894 minus00177 minus 01412119894 04047 + 02288119894

01040 minus 02015119894 01840 + 02276119894 02681 minus 03526119894 minus05252 + 01022119894

01808 + 02669119894 02264 + 03860119894 minus01791 + 01976119894 minus00961 minus 00117119894

(33)

From the fifth column to the eighth column

minus02638 minus 04952119894 minus00863 minus 01664119894 02687 + 01958119894 minus02544 minus 01099119894

minus02741 minus 01656119894 minus00227 + 02684119894 01846 + 02456119894 minus00298 + 05163119894

minus01495 minus 03205119894 01391 + 02434119894 01942 minus 05211119894 minus03052 minus 01468119894

minus02554 + 02690119894 minus04222 minus 01080119894 02232 + 00774119894 00965 minus 00421119894

03856 minus 00619119894 01217 minus 00270119894 01106 minus 03090119894 minus01122 + 02379119894

minus01130 + 00766119894 07102 minus 00901119894 01017 + 01397119894 minus00445 + 00038119894

01216 + 00076119894 02343 minus 01772119894 05242 minus 00089119894 minus00613 + 00258119894

minus00750 minus 03581119894 00125 + 00964119894 00779 minus 01074119894 06735 minus 00266119894

))))

)

(34)

It is easy to see that matrices 119883 Λ 119884 Γ satisfy (17) Hencethere exists the unique solution for Problem 3 Using the

software ldquoMATLABrdquo we obtain the unique solution ofProblem 3

From the first column to the fourth column

((((

(

minus01983 + 00491119894 01648 + 00032119894 00002 + 01065119894 01308 + 02690119894

01071 minus 02992119894 02106 + 02381119894 minus00533 minus 03856119894 minus00946 + 00488119894

01935 minus 01724119894 minus00855 minus 00370119894 00200 minus 00665119894 minus02155 minus 00636119894

00085 minus 02373119894 minus00843 minus 01920119894 00136 + 00382119894 minus00328 minus 00000119894

minus00529 + 01703119894 01948 minus 00719119894 01266 + 01752119894 00232 minus 02351119894

00855 + 01065119894 00325 minus 02068119894 02624 + 00000119894 00136 minus 00382119894

01283 minus 01463119894 minus00467 + 00000119894 00325 + 02067119894 minus00843 + 01920119894

02498 + 00000119894 01283 + 01463119894 00855 minus 01065119894 00086 + 02373119894

(35)

From the fifth column to the eighth column

02399 minus 01019119894 01928 minus 01488119894 minus03480 minus 02574119894 01017 minus 00000119894

minus01955 + 00644119894 02925 + 01872119894 03869 + 00000119894 minus03481 + 02574119894

00074 + 00339119894 minus03132 minus 00000119894 02926 minus 01872119894 01928 + 01488119894

00232 + 02351119894 minus02154 + 00636119894 minus00946 minus 00489119894 01309 minus 02691119894

minus00545 minus 00000119894 00074 minus 00339119894 minus01955 minus 00643119894 02399 + 01019119894

01266 minus 01752119894 00200 + 00665119894 minus00533 + 03857119894 00002 minus 01065119894

01949 + 00719119894 minus00855 + 00370119894 02106 minus 02381119894 01648 minus 00032119894

minus00529 minus 01703119894 01935 + 01724119894 01071 + 02992119894 minus01983 minus 00491119894

))))

)

(36)

6 Journal of Applied Mathematics

Conflict of Interests

There is no conflict of interests between the authors

Acknowledgments

This research was supported by National Natural ScienceFoundation of China (31170532)The authors are very gratefulto the referees for their valuable comments and also thank theeditor for his helpful suggestions

References

[1] J H Wilkinson The Algebraic Eigenvalue Problem OxfordUniversity Press Oxford UK 1965

[2] A Berman and R J Plemmons Nonnegative Matrices in theMathematical Sciences vol 9 ofClassics in AppliedMathematicsSIAM Philadelphia Pa USA 1994

[3] MAravDHershkowitz VMehrmann andH Schneider ldquoTherecursive inverse eigenvalue problemrdquo SIAM Journal on MatrixAnalysis and Applications vol 22 no 2 pp 392ndash412 2000

[4] R Loewy and V Mehrmann ldquoA note on the symmetricrecursive inverse eigenvalue problemrdquo SIAM Journal on MatrixAnalysis and Applications vol 25 no 1 pp 180ndash187 2003

[5] F L Li X YHu and L Zhang ldquoLeft and right inverse eigenpairsproblem of skew-centrosymmetric matricesrdquo Applied Mathe-matics and Computation vol 177 no 1 pp 105ndash110 2006

[6] F L Li X Y Hu and L Zhang ldquoLeft and right inverseeigenpairs problem of generalized centrosymmetric matricesand its optimal approximation problemrdquo Applied Mathematicsand Computation vol 212 no 2 pp 481ndash487 2009

[7] R D Hill and S R Waters ldquoOn 120581-real and 120581-Hermitianmatricesrdquo Linear Algebra and its Applications vol 169 pp 17ndash29 1992

[8] S P Irwin ldquoMatrices with multiple symmetry propertiesapplications of centro-Hermitian and per-Hermitian matricesrdquoLinear Algebra and its Applications vol 284 no 1ndash3 pp 239ndash258 1998

[9] L C Biedenharn and J D Louck Angular Momentum inQuantum Physics vol 8 of Encyclopedia of Mathematics and itsApplications Addison-Wesley Reading Mass USA 1981

[10] WM Gibson and B R Pollard Symmetry Principles in Elemen-tary Particle Physics Cambridge University Press CambridgeUK 1976

[11] W F Trench ldquoHermitian Hermitian R-symmetric and Her-mitian R-skew symmetric Procrustes problemsrdquo Linear Algebraand its Applications vol 387 pp 83ndash98 2004

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 271978 5 pageshttpdxdoiorg1011552013271978

Research ArticleCompleting a 2 times 2 Block Matrix of Real Quaternions witha Partial Specified Inverse

Yong Lin12 and Qing-Wen Wang1

1 Department of Mathematics Shanghai University Shanghai 200444 China2 School of Mathematics and Statistics Suzhou University Suzhou 234000 China

Correspondence should be addressed to Qing-Wen Wang wqw858yahoocomcn

Received 4 December 2012 Revised 23 February 2013 Accepted 20 March 2013

Academic Editor K Sivakumar

Copyright copy 2013 Y Lin and Q-W Wang This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited

This paper considers a completion problem of a nonsingular 2 times 2 block matrix over the real quaternion algebra H Let1198981 1198982 1198991 1198992be nonnegative integers119898

1+ 1198982= 1198991+ 1198992= 119899 gt 0 and 119860

12isin H1198981times1198992 119860

21isin H1198982times1198991 119860

22isin H1198982times1198992 119861

11isin H1198991times1198981

be given We determine necessary and sufficient conditions so that there exists a variant block entry matrix11986011isin H1198981times1198991 such that

119860 = (11986011 11986012

11986021 11986022) isin H119899times119899 is nonsingular and 119861

11is the upper left block of a partitioning of 119860minus1 The general expression for 119860

11is also

obtained Finally a numerical example is presented to verify the theoretical findings

1 Introduction

The problem of completing a block-partitioned matrix of aspecified type with some of its blocks given has been studiedby many authors Fiedler and Markham [1] considered thefollowing completion problem over the real number field RSuppose119898

1 1198982 1198991 1198992are nonnegative integers119898

1+1198982=

1198991+1198992= 119899 gt 0 119860

11isin R1198981times1198991 119860

12isin R1198981times1198992 119860

21isin R1198982times1198991

and 11986122isin R1198992times1198982 Determine a matrix 119860

22isin R1198982times1198992 such

that

119860 = (11986011

11986012

11986021

11986022

) (1)

is nonsingular and 11986122is the lower right block of a partition-

ing of 119860minus1 This problem has the form of

(11986011

11986012

11986021

)

minus1

= (

11986122

) (2)

and the solution and the expression for 11986022were obtained in

[1] Dai [2] considered this form of completion problemswithsymmetric and symmetric positive definite matrices over R

Some other particular forms for 2times2 block matrices overR have also been examined (see eg [3]) such as

(11986011

11986012

11986021

)

minus1

= (11986111

)

(11986011

)

minus1

= (

11986122

)

(11986011

11986022

)

minus1

= ( 11986112

11986121

)

(3)

The real quaternion matrices play a role in computerscience quantum physics and so on (eg [4ndash6]) Quaternionmatrices are receiving much attention as witnessed recently(eg [7ndash9]) Motivated by the work of [1 10] and keepingsuch applications of quaternionmatrices in view in this paperwe consider the following completion problem over the realquaternion algebra

H = 1198860+ 1198861119894 + 1198862119895 + 1198863119896 |

1198942

= 1198952

= 1198962

= 119894119895119896 = minus1 and 1198860 1198861 1198862 1198863isin R

(4)

Problem 1 Suppose 1198981 1198982 1198991 1198992are nonnegative inte-

gers 1198981+ 1198982= 1198991+ 1198992= 119899 gt 0 and 119860

12isin H1198981times1198992

2 Journal of Applied Mathematics

11986021isin H1198982times1198991 119860

22isin 1198771198982times1198992 119861

11isin H1198991times1198981 Find a matrix

11986011isin H1198981times1198991 such that

119860 = (11986011

11986012

11986021

11986022

) isin H119899times119899 (5)

is nonsingular and11986111is the upper left block of a partitioning

of 119860minus1 That is

( 119860

12

11986021

11986022

)

minus1

= (11986111

) (6)

where H119898times119899 denotes the set of all 119898 times 119899matrices over H and119860minus1 denotes the inverse matrix of 119860

Throughout over the real quaternion algebra H wedenote the identity matrix with the appropriate size by 119868 thetranspose of 119860 by 119860119879 the rank of 119860 by 119903(119860) the conjugatetranspose of119860 by119860lowast = (119860)119879 a reflexive inverse of a matrix119860over H by 119860+ which satisfies simultaneously 119860119860+119860 = 119860 and119860+

119860119860+

= 119860+Moreover119871

119860= 119868minus119860

+

119860 119877119860= 119868minus119860119860

+ where119860+ is an arbitrary but fixed reflexive inverse of 119860 Clearly 119871

119860

and119877119860are idempotent and each is a reflexive inverse of itself

R(119860) denotes the right column space of the matrix 119860The rest of this paper is organized as follows In Section 2

we establish some necessary and sufficient conditions to solveProblem 1 over H and the general expression for 119860

11is also

obtained In Section 3 we present a numerical example toillustrate the developed theory

2 Main Results

In this section we begin with the following lemmas

Lemma 1 (singular-value decomposition [9]) Let 119860 isin H119898times119899

be of rank 119903 Then there exist unitary quaternion matrices 119880 isin

H119898times119898 and 119881 isin H119899times119899 such that

119880119860119881 = (1198631199030

0 0) (7)

where 119863119903= diag(119889

1 119889

119903) and the 119889

119895rsquos are the positive

singular values of 119860

Let H119899119888denote the collection of column vectors with 119899

components of quaternions and 119860 be an 119898 times 119899 quaternionmatrix Then the solutions of 119860119909 = 0 form a subspace of H119899

119888

of dimension 119899(119860) We have the following lemma

Lemma 2 Let

(11986011

11986012

11986021

11986022

) (8)

be a partitioning of a nonsingular matrix 119860 isin H119899times119899 and let

(11986111

11986112

11986121

11986122

) (9)

be the corresponding (ie transpose) partitioning of 119860minus1 Then119899(11986011) = 119899(119861

22)

Proof It is readily seen that

(11986122

11986121

11986112

11986111

)

(11986022

11986021

11986012

11986011

)

(10)

are inverse to each other so we may suppose that 119899(11986011) lt

119899(11986122)

If 119899(11986122) = 0 necessarily 119899(119860

11) = 0 and we are finished

Let 119899(11986122) = 119888 gt 0 then there exists a matrix 119865 with 119888 right

linearly independent columns such that 11986122119865 = 0 Then

using

1198601111986112+ 1198601211986122= 0 (11)

we have

1198601111986112119865 = 0 (12)

From

1198602111986112+ 1198602211986122= 119868 (13)

we have

1198602111986112119865 = 119865 (14)

It follows that the rank 119903(11986112119865) ge 119888 In view of (12) this

implies

119899 (11986011) ge 119903 (119861

12119865) ge 119888 = 119899 (119861

22) (15)

Thus

119899 (11986011) = 119899 (119861

22) (16)

Lemma 3 (see [10]) Let 119860 isin H119898times119899 119861 isin H119901times119902 119863 isin H119898times119902 beknown and 119883 isin H119899times119901 unknown Then the matrix equation

119860119883119861 = 119863 (17)

is consistent if and only if

119860119860+

119863119861+

119861 = 119863 (18)

In that case the general solution is

119883 = 119860+

119863119861+

+ 1198711198601198841+ 1198842119877119861 (19)

where11988411198842are any matrices with compatible dimensions over

H

By Lemma 1 let the singular value decomposition of thematrix 119860

22and 119861

11in Problem 1 be

11986022= 119876(

Λ 0

0 0)119877lowast

(20)

11986111= 119880(

Σ 0

0 0)119881lowast

(21)

Journal of Applied Mathematics 3

where Λ = diag(1205821 1205822 120582

119904) is a positive diagonal matrix

120582119894= 0 (119894 = 1 119904) are the singular values of 119860

22 119904 =

119903(11986022) Σ = diag(120590

1 1205902 120590

119903) is a positive diagonal matrix

120590119894= 0 (119894 = 1 119903) are the singular values of 119861

11and 119903 =

119903(11986111)119876 = (119876

11198762) isin H1198982times1198982 119877 = (119877

11198772) isin H1198992times1198992

119880 = (11988011198802) isin H1198991times1198991 119881 = (119881

11198812) isin H1198981times1198981 are unitary

quaternion matrices where 1198761isin H1198982times119904 119877

1isin H1198992times119904 119880

1isin

H1198991times119903 and 1198811isin H1198981times119903

Theorem 4 Problem 1 has a solution if and only if thefollowing conditions are satisfied

(a) 119903 ( 1198601211986022

) = 1198992

(b) 1198992minus 119903(119860

22) = 119898

1minus 119903(11986111) that is 119899

2minus 119904 = 119898

1minus 119903

(c) R(1198602111986111) subR(119860

22)

(d) R(119860lowast

12119861lowast

11) subR(119860

lowast

22)

In that case the general solution has the form of

11986011= 119861+

11+ 11986012119877(

Λminus1

119876lowast

1119860211198801Σ 0

119867 minus(119881lowast

2119860121198772)minus1)

times 119881lowast

119861+

11+ 119884 minus 119884119861

11119861+

11

(22)

where 119867 is an arbitrary matrix in H(1198992minus119904)times119903 and 119884 is anarbitrary matrix in H1198981times1198991

Proof If there exists an 1198981times 1198991matrix 119860

11such that 119860 is

nonsingular and 11986111is the corresponding block of 119860minus1 then

(a) is satisfied From 119860119861 = 119861119860 = 119868 we have that

1198602111986111+ 1198602211986121= 0

1198611111986012+ 1198611211986022= 0

(23)

so that (c) and (d) are satisfiedBy (11) we have

119903 (11986022) + 119899 (119860

22) = 1198992 119903 (119861

11) + 119899 (119861

11) = 119898

1 (24)

From Lemma 2 Notice that ( 11986011 1198601211986021 11986022

) is the correspondingpartitioning of 119861minus1 we have

119899 (11986111) = 119899 (119860

22) (25)

implying that (b) is satisfiedConversely from (c) we know that there exists a matrix

119870 isin H1198992times1198981 such that

1198602111986111= 11986022119870 (26)

Let

11986121= minus119870 (27)

From (20) (21) and (26) we have

11986021119880(

Σ 0

0 0)119881lowast

= 119876(Λ 0

0 0)119877lowast

119870 (28)

It follows that

119876lowast

11986021119880(

Σ 0

0 0)119881lowast

119881 = 119876lowast

119876(Λ 0

0 0)119877lowast

119870119881 (29)

This implies that

(

119876lowast

1119860211198801119876lowast

1119860211198802

119876lowast

2119860211198801119876lowast

2119860211198802

)(Σ 0

0 0)

= (Λ 0

0 0)(

119877lowast

11198701198811119877lowast

11198701198812

119877lowast

21198701198811119877lowast

21198701198812

)

(30)

Comparing corresponding blocks in (30) we obtain

119876lowast

2119860211198801= 0 (31)

Let 119877lowast119870119881 = From (29) (30) we have

= (Λminus1

119876lowast

1119860211198801Σ 0

119867 11987022

)

119867 isin H(1198992minus119904)times119903 119870

22isin H(1198992minus119904)times(1198981minus119903)

(32)

In the same way from (d) we can obtain

119881lowast

1119860121198772= 0 (33)

Notice that ( 1198601211986022

) in (a) is a full column rank matrix By (20)(21) and (33) we have

(0 119876lowast

119881lowast

0)(

11986012

11986022

)119877 = (

Λ 0

0 0

119881lowast

1119860121198771119881lowast

1119860121198772

119881lowast

2119860121198771119881lowast

2119860121198772

) (34)

so that

1198992= 119903(

11986012

11986022

) = 119903((0 119876lowast

119881lowast

0)(

11986012

11986022

)119877)

= 119903(

Λ 0

0 0

119881lowast

1119860121198771119881lowast

1119860121198772

119881lowast

2119860121198771119881lowast

2119860121198772

)

= 119903 (Λ) + 119903 (119881lowast

2119860121198772)

= 119904 + 119903 (119881lowast

2119860121198772)

(35)

It follows from (b) and (35) that 1198811198792119860121198772is a full column

rank matrix so it is nonsingularFrom 119860119861 = 119868 we have the following matrix equation

1198601111986111+ 1198601211986121= 119868 (36)

that is

1198601111986111= 119868 minus 119860

1211986121 119868 isin H

1198981times1198981 (37)

4 Journal of Applied Mathematics

where 11986111 11986012

were given 11986121

= minus119870 (from (27)) ByLemma 3 the matrix equation (37) has a solution if and onlyif

(119868 minus 1198601211986121) 119861+

1111986111= 119868 minus 119860

1211986121 (38)

By (21) (27) (32) and (33) we have that (38) is equivalent to

(119868 + 11986012119870)119881(

Σminus1

0

0 0)119880lowast

119880(Σ 0

0 0)119881lowast

= 119868 + 11986012119870 (39)

We simplify the equation aboveThe left hand side reduces to(119868 + 119860

12119870)1198811119881lowast

1and so we have

119860121198701198811119881lowast

1minus 11986012119870 = 119868 minus 119881

1119881lowast

1 (40)

So

11986012119877119881lowast

1198811119881lowast

1minus 11986012119877119881lowast

= (11988111198812) (119881lowast

1

119881lowast

2

) minus 1198811119881lowast

1

(41)

This implies that

11986012119877(

119881lowast

11198811

119881lowast

21198811

)119881lowast

1minus 11986012119877(

119881lowast

1

119881lowast

2

) = 1198812119881lowast

2 (42)

so that

11986012119877(

119868

0)119881lowast

1minus 11986012119877(

119881lowast

1

119881lowast

2

) = 1198812119881lowast

2 (43)

So

minus11986012119877(

0

119881lowast

2

) = 1198812119881lowast

2 (44)

and hence

minus (119860121198771119860121198772) (Λminus1

119876lowast

1119860211198801Σ 0

119867 11987022

)(0

119881lowast

2

) = 1198812119881lowast

2

(45)

Finally we obtain

11986012119877211987022119881lowast

2= minus1198812119881lowast

2 (46)

Multiplying both sides of (46) by 119881lowast from the left consider-ing (33) and the fact that 119881lowast

2119860121198772is nonsingular we have

11987022= minus(119881

lowast

2119860121198772)minus1

(47)

From Lemma 3 (38) (47) Problem 1 has a solution and thegeneral solution is

11986011= 119861+

11+ 11986012119877(

Λminus1

119876lowast

1119860211198801Σ 0

119867 minus(119881lowast

2119860121198772)minus1)

times 119881lowast

119861+

11+ 119884 minus 119884119861

11119861+

11

(48)

where 119867 is an arbitrary matrix in H(1198992minus119904)times119903 and 119884 is anarbitrary matrix in H1198981times1198991

3 An Example

In this section we give a numerical example to illustrate thetheoretical results

Example 5 Consider Problem 1 with the parameter matricesas follows

11986012= (

2 + 1198951

2119896

minus119896 1 +1

2119895

)

11986021= (

3

2+1

2119894 minus

1

2119895 minus

1

2119896

1

2119895 +

1

2119896

3

2+1

2119894

)

11986022= (

2 119894

2119895 119896) 119861

11= (

1 119894

119895 119896)

(49)

It is easy to show that (c) (d) are satisfied and that

1198992= 119903(

11986012

11986022

) = 2

1198992minus 119903 (119860

22) = 119898

1minus 119903 (119861

11) = 0

(50)

so (a) (b) are satisfied too Therefore we have

119861+

11= (

1

2minus1

2119895

minus1

2119894 minus

1

2119896

)

11986022= 119876(

Λ 0

0 0)119877lowast

11986111= 119880(

Σ 0

0 0)119881lowast

(51)

where

119876 =1

radic2

(1 119894

119895 119896) Λ = (

2radic2 0

0 radic2)

119877 = (1 0

0 1) 119880 =

1

radic2

(1 119894

119895 119896)

Σ = (radic2 0

0 radic2) 119881 = (

1 0

0 1)

(52)

We also have

1198761=

1

radic2

(1 119894

119895 119896) 119877

1= (

1 0

0 1)

1198801=

1

radic2

(1 119894

119895 119896) 119881

1= (

1 0

0 1)

(53)

Journal of Applied Mathematics 5

By Theorem 4 for an arbitrary matrices 119884 isin H2times2 wehave

11986011= 119861+

11+ 11986012119877 (Λminus1

119876lowast

1119860211198801Σ)119881lowast

119861+

11+ 119884 minus 119884119861

11119861+

11

= (

3

2+1

4119895 +

1

4119896

3

4+1

4119894 minus

3

2119895

1

2minus 119894 +

1

4119895 minus

1

41198961

4minus3

4119894 minus

1

2119895 minus 119896

)

(54)

it follows that

119860 =((

(

3

2

+

1

4

119895 +

1

4

119896

3

4

+

1

4

119894 minus

3

2

119895 2 + 119895

1

2

119896

1

2

minus 119894 +

1

4

119895 minus

1

4

119896

1

4

minus

3

4

119894 minus

1

2

119895 minus 119896 minus119896 1 +

1

2

119895

3

2

+

1

2

119894 minus

1

2

119895 minus

1

2

119896 2 119894

1

2

119895 +

1

2

119896

3

2

+

1

2

119894 2119895 119896

))

)

119860minus1

=(

1 119894 minus1 minus1

119895 119896 0 minus1

minus1 03

4

1

2minus3

4119895

minus1 minus11

2minus 119894

1

2minus1

2119894 minus

1

2119895 minus 119896

)

(55)

The results verify the theoretical findings of Theorem 4

Acknowledgments

The authors would like to give many thanks to the refereesand Professor K C Sivakumar for their valuable suggestionsand comments which resulted in a great improvement ofthe paper This research was supported by Grants from theKey Project of Scientific Research Innovation Foundation ofShanghai Municipal Education Commission (13ZZ080) theNational Natural Science Foundation of China (11171205) theNatural Science Foundation of Shanghai (11ZR1412500) theDiscipline Project at the corresponding level of Shanghai (A13-0101-12-005) and Shanghai Leading Academic DisciplineProject (J50101)

References

[1] M Fiedler and T L Markham ldquoCompleting a matrix whencertain entries of its inverse are specifiedrdquo Linear Algebra andIts Applications vol 74 pp 225ndash237 1986

[2] H Dai ldquoCompleting a symmetric 2 times 2 block matrix and itsinverserdquo Linear Algebra and Its Applications vol 235 pp 235ndash245 1996

[3] W W Barrett M E Lundquist C R Johnson and H JWoerdeman ldquoCompleting a block diagonal matrix with apartially prescribed inverserdquoLinearAlgebra and Its Applicationsvol 223224 pp 73ndash87 1995

[4] S L Adler Quaternionic Quantum Mechanics and QuantumFields Oxford University Press New York NY USA 1994

[5] A Razon and L P Horwitz ldquoUniqueness of the scalar productin the tensor product of quaternion Hilbert modulesrdquo Journalof Mathematical Physics vol 33 no 9 pp 3098ndash3104 1992

[6] J W Shuai ldquoThe quaternion NN model the identification ofcolour imagesrdquo Acta Comput Sinica vol 18 no 5 pp 372ndash3791995 (Chinese)

[7] F Zhang ldquoOn numerical range of normal matrices of quater-nionsrdquo Journal of Mathematical and Physical Sciences vol 29no 6 pp 235ndash251 1995

[8] F Z Zhang Permanent Inequalities and Quaternion Matrices[PhD thesis] University of California Santa Barbara CalifUSA 1993

[9] F Zhang ldquoQuaternions and matrices of quaternionsrdquo LinearAlgebra and Its Applications vol 251 pp 21ndash57 1997

[10] Q-W Wang ldquoThe general solution to a system of real quater-nion matrix equationsrdquo Computers amp Mathematics with Appli-cations vol 49 no 5-6 pp 665ndash675 2005

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 178209 7 pageshttpdxdoiorg1011552013178209

Research ArticleFuzzy Approximate Solution of Positive Fully Fuzzy LinearMatrix Equations

Xiaobin Guo1 and Dequan Shang2

1 College of Mathematics and Statistics Northwest Normal University Lanzhou 730070 China2Department of Public Courses Gansu College of Traditional Chinese Medicine Lanzhou 730000 China

Correspondence should be addressed to Dequan Shang gxbglz163com

Received 22 November 2012 Accepted 19 January 2013

Academic Editor Panayiotis J Psarrakos

Copyright copy 2013 X Guo and D ShangThis is an open access article distributed under theCreativeCommonsAttribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

The fuzzy matrix equations otimesotimes = in which and are119898times119898 119899times119899 and119898times119899 nonnegative LR fuzzy numbers matricesrespectively are investigated The fuzzy matrix systems is extended into three crisp systems of linear matrix equations accordingto arithmetic operations of LR fuzzy numbers Based on pseudoinverse of matrix the fuzzy approximate solution of original fuzzysystems is obtained by solving the crisp linear matrix systems In addition the existence condition of nonnegative fuzzy solution isdiscussed Two examples are calculated to illustrate the proposed method

1 Introduction

Since many real-world engineering systems are too com-plex to be defined in precise terms imprecision is ofteninvolved in any engineering design process Fuzzy systemshave an essential role in this fuzzy modeling which canformulate uncertainty in actual environment Inmanymatrixequations some or all of the system parameters are vagueor imprecise and fuzzy mathematics is a better tool thancrisp mathematics for modeling these problems and hencesolving a fuzzy matrix equation is becomingmore importantThe concept of fuzzy numbers and arithmetic operationswith these numbers were first introduced and investigatedby Zadeh [1 2] Dubois and Prade [3] and Nahmias [4]A different approach to fuzzy numbers and the structureof fuzzy number spaces was given by Puri and Ralescu [5]Goetschel Jr and Voxman [6] and Wu and Ma [7 8]

In the past decades many researchers have studied thefuzzy linear equations such as fuzzy linear systems (FLS) dualfuzzy linear systems (DFLS) general fuzzy linear systems(GFLS) fully fuzzy linear systems (FFLS) dual fully fuzzylinear systems (DFFLS) and general dual fuzzy linear systems(GDFLS) These works were performed mainly by Friedmanet al [9 10] Allahviranloo et al [11ndash17] Abbasbandy et al

[18ndash21]Wang andZheng [22 23] andDehghan et al [24 25]The general method they applied is the fuzzy linear equationswere converted to a crisp function system of linear equationswith high order according to the embedding principlesand algebraic operations of fuzzy numbers Then the fuzzysolution of the original fuzzy linear systems was derived fromsolving the crisp function linear systemsHowever for a fuzzymatrix equationwhich always has awide use in control theoryand control engineering few works have been done in thepast In 2009 Allahviranloo et al [26] discussed the fuzzy lin-ear matrix equations (FLME) of the form 119860119861 = in whichthematrices119860 and119861 are known119898times119898 and 119899times119899 real matricesrespectively is a given 119898 times 119899 fuzzy matrix By using theparametric form of fuzzy number they derived necessaryand sufficient conditions for the existence condition of fuzzysolutions and designed a numerical procedure for calculatingthe solutions of the fuzzy matrix equations In 2011 Guoet al [27ndash29] investigated a class of fuzzy matrix equations119860 = by means of the block Gaussian elimination methodand studied the least squares solutions of the inconsistentfuzzy matrix equation 119860 = by using generalized inversesof the matrix and discussed fuzzy symmetric solutions offuzzy matrix equations 119860 = What is more there are two

2 Journal of Applied Mathematics

shortcomings in the above fuzzy systems The first is that inthese fuzzy linear systems and fuzzy matrix systems the fuzzyelements were denoted by triangular fuzzy numbers so theextended model equations always contain parameter 119903 0 le

119903 le 1 which makes their computation especially numericalimplementation inconvenient in some sense The other oneis that the weak fuzzy solution of fuzzy linear systems119860 =

does not exist sometimes see [30]To make the multiplication of fuzzy numbers easy and

handle the fully fuzzy systems Dubois and Prade [3] intro-duced the LR fuzzy number in 1978 We know that triangularfuzzy numbers and trapezoidal fuzzy numbers [31] are justspecious cases of LR fuzzy numbers In 2006 Dehghan etal [25] discussed firstly computational methods for fullyfuzzy linear systems = whose coefficient matrix andthe right-hand side vector are LR fuzzy numbers In 2012Allahviranloo et al studied LR fuzzy linear systems [32]by the linear programming with equality constraints Otadiand Mosleh considered the nonnegative fuzzy solution [33]of fully fuzzy matrix equations = by employinglinear programming with equality constraints at the sameyear Recently Guo and Shang [34] investigated the fuzzyapproximate solution of LR fuzzy Sylvester matrix equations119860 + 119861 = In this paper we propose a general model forsolving the fuzzy linearmatrix equation otimesotimes = where and are119898times119898 119899times119899 and119898times119899 nonnegative LR fuzzynumbersmatrices respectivelyThemodel is proposed in thisway that is we extend the fuzzy linear matrix system intoa system of linear matrix equations according to arithmeticoperations of LR fuzzy numbers The LR fuzzy solution ofthe original matrix equation is derived from solving crispsystems of linearmatrix equationsThe structure of this paperis organized as follows

In Section 2 we recall the LR fuzzy numbers andpresent the concept of fully fuzzy linear matrix equationThecomputing model to the positive fully fuzzy linear matrixequation is proposed in detail and the fuzzy approximatesolution of the fuzzy linear matrix equation is obtained byusing pseudo-inverse in Section 3 Some examples are givento illustrate our method in Section 4 and the conclusion isdrawn in Section 5

2 Preliminaries

21 The LR Fuzzy Number

Definition 1 (see [1]) A fuzzy number is a fuzzy set like 119906

119877 rarr 119868 = [0 1] which satisfies the following

(1) 119906 is upper semicontinuous(2) 119906 is fuzzy convex that is 119906(120582119909 + (1 minus 120582)119910) ge

min119906(119909) 119906(119910) for all 119909 119910 isin 119877 120582 isin [0 1](3) 119906 is normal that is there exists 119909

0isin 119877 such that

119906(1199090) = 1

(4) supp 119906 = 119909 isin 119877 | 119906(119909) gt 0 is the support of the 119906and its closure cl(supp 119906) is compact

Let 1198641 be the set of all fuzzy numbers on 119877

Definition 2 (see [3]) A fuzzy number is said to be an LRfuzzy number if

120583(119909) =

119871(119898 minus 119909

120572) 119909 le 119898 120572 gt 0

119877(119909 minus 119898

120573) 119909 ge 119898 120573 gt 0

(1)

where 119898 120572 and 120573 are called the mean value left and rightspreads of respectively The function 119871(sdot) which is calledleft shape function satisfies

(1) 119871(119909) = 119871(minus119909)(2) 119871(0) = 1 and 119871(1) = 0(3) 119871(119909) is nonincreasing on [0infin)

The definition of a right shape function 119877(sdot) is similar tothat of 119871(sdot)

Clearly = (119898 120572 120573)LR is positive (negative) if and onlyif119898 minus 120572 gt 0 (119898 + 120573 lt 0)

Also two LR fuzzy numbers = (119898 120572 120573)LR and =

(119899 120574 120575)LR are said to be equal if and only if 119898 = 119899 120572 = 120574and 120573 = 120575

Definition 3 (see [5]) For arbitrary LR fuzzy numbers =

(119898 120572 120573)LR and = (119899 120574 120575)LR we have the following

(1) Addition

oplus = (119898 120572 120573)LR oplus (119899 120574 120575)LR = (119898 + 119899 120572 + 120574 120573 + 120575)LR

(2)

(2) Multiplication(i) If gt 0 and gt 0 then

otimes = (119898 120572 120573)LR otimes (119899 120574 120575)LR

cong (119898119899119898120574 + 119899120572119898120575 + 119899120573)LR(3)

(ii) if lt 0 and gt 0 then

otimes = (119898 120572 120573)RL otimes (119899 120574 120575)LR

cong (119898119899 119899120572 minus 119898120575 119899120573 minus 119898120574)RL(4)

(iii) if lt 0 and lt 0 then

otimes = (119898 120572 120573)RL otimes (119899 120574 120575)LR

cong (119898119899 minus119898120575 minus 119899120573 minus119898120574 minus 119899120572)RL(5)

(3) Scalar multiplication

120582 times = 120582 times (119898 120572 120573)LR

= (120582119898 120582120572 120582120573)LR 120582 ge 0

(120582119898 minus120582120573 minus120582120572)RL 120582 lt 0

(6)

Journal of Applied Mathematics 3

22 The LR Fuzzy Matrix

Definition 4 Amatrix = (119894119895) is called an LR fuzzy matrix

if each element 119894119895of is an LR fuzzy number

will be positive (negative) and denoted by gt 0 ( lt

0) if each element 119894119895of is positive (negative) where

119894119895=

(119886119894119895 119886119897

119894119895 119886119903

119894119895)LR Up to the rest of this paper we use positive

LR fuzzy numbers and formulas given in Definition 3 Forexample we represent 119898 times 119899 LR fuzzy matrix = (

119894119895) that

119894119895= (119886119894119895 120572119894119895 120573119894119895)LR with new notation = (119860119872119873) where

119860 = (119886119894119895) 119872 = (120572

119894119895) and 119873 = (120573

119894119895) are three 119898 times 119899 crisp

matrices In particular an 119899 dimensions LR fuzzy numbersvector can be denoted by (119909 119909119897 119909119903) where 119909 = (119909

119894) 119909119897 =

(119909119897

119894) and 119909119903 = (119909

119903

119894) are three 119899 dimensions crisp vectors

Definition 5 Let = (119894119895) and = (

119894119895) be two 119898 times 119899 and

119899 times 119901 fuzzy matrices we define otimes = = (119894119895) which is an

119898 times 119901 fuzzy matrix where

119894119895=

oplus

sum

119896=12119899

119894119896otimes 119887119896119895 (7)

23 The Fully Fuzzy Linear Matrix Equation

Definition 6 Thematrix system

(

11

12

sdot sdot sdot 1119898

21

22

sdot sdot sdot 2119898

1198981

1198982

sdot sdot sdot 119898119898

)otimes(

11

12

sdot sdot sdot 1119899

21

22

sdot sdot sdot 2119899

1198981

1198982

sdot sdot sdot 119898119899

)

otimes(

11

11988712

sdot sdot sdot 1198871119899

21

11988722

sdot sdot sdot 1198872119899

1198991

1198992

sdot sdot sdot 119899119899

)

=(

11

12

sdot sdot sdot 1119899

21

22

sdot sdot sdot 2119899

1198981

1198982

sdot sdot sdot 119898119899

)

(8)

where 119894119895 1 le 119894 119895 le 119898

119894119895 1 le 119894 119895 le 119899 and

119894119895 1 le 119894 le 119898

1 le 119895 le 119899 are LR fuzzy numbers is called an LR fully fuzzylinear matrix equation (FFLME)

Using matrix notation we have

otimes otimes = (9)

A fuzzy numbers matrix

= (119909119894119895 119910119894119895 119911119894119895)LR 1 le 119894 le 119898 1 le 119895 le 119899 (10)

is called an LR fuzzy approximate solution of fully fuzzy linearmatrix equation (8) if satisfies (9)

Up to the rest of this paper we will discuss the nonnega-tive solution = (119883 119884 119885) ge 0 of FFLME otimes otimes =

where = (119860119872119873) ge 0 = (119861 119864 119865) ge 0 and =

(119862 119866119867) ge 0

3 Method for Solving FFLME

First we extend the fuzzy linear matrix system (9) into threesystems of linear matrix equations according to the LR fuzzynumber and its arithmetic operations

Theorem 7 The fuzzy linear matrix system (9) can beextended into the following model

119860119883119861 = 119862

119860119883119864 + 119860119884119861 +119872119883119861 = 119866

119860119883119865 + 119860119885119861 + 119873119883119861 = 119867

(11)

Proof We denote = (119860119872119873) = (119861 119864 119865) and =

(119862 119866119867) and assume = (119883 119884 119885) ge 0 then

= (119860119872119873) otimes (119883 119884 119885) otimes (119861 119864 119865)

= (119860119883119860119884 +119872119883119860119885 + 119873119883) otimes (119861 119864 119865)

= (119860119883119861119860119883119864 + 119860119884119861 +119872119883119861119860119883119865 + 119860119885119861 + 119873119883119861)

= (119862 119866119867)

(12)

according tomultiplication of nonnegative LR fuzzy numbersof Definition 2 Thus we obtain a model for solving FFLME(9) as follows

119860119883119861 = 119862

119860119883119864 + 119860119884119861 +119872119883119861 = 119866

119860119883119865 + 119860119885119861 + 119873119883119861 = 119867

(13)

Secondly in order to solve the fuzzy linear matrix equa-tion (9) we need to consider the crisp systems of linearmatrix equation (11) Supposing119860 and119861 are nonsingular crispmatrices we have

119883 = 119860minus1

119862119861minus1

119884 = 119860minus1

119866 minus 119883119864119861minus1

minus 119860minus1

119872119883

119885 = 119860minus1

119867 minus 119883119865119861minus1

minus 119860minus1

119873119883

(14)

that is

119883 = 119860minus1

119862119861minus1

119884 = 119860minus1

119866 minus 119860minus1

119862119861minus1

119864119861minus1

minus 119860minus1

119872119860minus1

119862119861minus1

119885 = 119860minus1

119867 minus 119860minus1

119862119861minus1

119865119861minus1

minus 119860minus1

119873119860minus1

119862119861minus1

(15)

Definition 8 Let = (119883 119884 119885) be an LR fuzzy matrix If(119883 119884 119885) is an exact solution of (11) such that 119883 ge 0 119884 ge 0119885 ge 0 and 119883 minus 119884 ge 0 we call = (119883 119884 119885) a nonnegativeLR fuzzy approximate solution of (9)

4 Journal of Applied Mathematics

Theorem 9 Let = (119860119872119873) = (119861 119864 119865) and =

(119862 119866119867) be three nonnegative fuzzymatrices respectively andlet 119860 and 119861 be the product of a permutation matrix by adiagonal matrix with positive diagonal entries Moreover let119866119861 ge 119862119861

minus1

119864 + 119872119860minus1

119862 119867119861 ge 119862119861minus1

119865 + 119873119860minus1

119862 and119862 + 119862119861

minus1

119864 +119872119860minus1

119862 ge 119866119861 Then the systems otimes otimes =

has nonnegative fuzzy solutions

Proof Our hypotheses on 119860 and 119861 imply that 119860minus1 and119861minus1 exist and they are nonnegative matrices Thus 119883 =

119860minus1

119862119861minus1

ge 0On the other hand because 119866119861 ge 119862119861

minus1

119864 + 119872119860minus1

119862 and119867119861 ge 119862119861

minus1

119865 + 119873119860minus1

119862 so with 119884 = 119860minus1

(119866119861 minus 119862119861minus1

119864 minus

119872119860minus1

119862)119861minus1 and 119885 = 119860

minus1

(119867119861 minus 119862119861minus1

119865 minus 119873119860minus1

119862)119861minus1 we

have 119884 ge 0 and 119885 ge 0 Thus = (119883 119884 119885) is a fuzzy matrixwhich satisfies otimes otimes = Since119883 minus 119884 = 119860

minus1

(119862 minus 119866119861 +

119862119861minus1

119864 + 119872119860minus1

119862)119861minus1 the positivity property of can be

obtained from the condition 119862+119862119861minus1119864+119872119860minus1

119862 ge 119866119861

When 119860 or 119861 is a singular crisp matrix the followingresult is obvious

Theorem 10 (see [35]) For linearmatrix equations119860119883119861 = 119862where 119860 isin 119877

119898times119898 119861 isin 119877119899times119899 and 119862 isin 119877

119898times119899 Then

119883 = 119860dagger

119862119861dagger (16)

is its minimal norm least squares solutionBy the pseudoinverse of matrices we solve model (11) and

obtain its minimal norm least squares solution as follows

119883 = 119860dagger

119862119861dagger

119884 = 119860dagger

119866 minus 119883119864119861dagger

minus 119860dagger

119872119883

119885 = 119860dagger

119867 minus 119883119865119861dagger

minus 119860dagger

119873119883

(17)

that is

119883 = 119860dagger

119862119861dagger

119884 = 119860dagger

119866 minus 119860dagger

119862119861dagger

119864119861dagger

minus 119860dagger

119872119860dagger

119862119861dagger

119885 = 119860dagger

119867 minus 119860dagger

119862119861dagger

119865119861dagger

minus 119860dagger

119873119860dagger

119862119861dagger

(18)

Definition 11 Let = (119883 119884 119885) be an LR fuzzy matrix If(119883 119884 119885) is a minimal norm least squares solution of (11) suchthat119883 ge 0119884 ge 0119885 ge 0 and119883minus119884 ge 0 we call = (119883 119884 119885)

a nonnegative LR fuzzy minimal norm least squares solutionof (9)

At last we give a sufficient condition for nonnegativefuzzy minimal norm least squares solution of FFLME (9) inthe same way

Theorem 12 Let 119860dagger and 119861dagger be nonnegative matrices More-over let 119866119861 ge 119862119861

dagger

119864 + 119872119860dagger

119862 119867119861 ge 119862119861dagger

119865 + 119873119860dagger

119862 and119862+119862119861

dagger

119864+119872119860dagger

119862 ge 119866119861 Then the systems otimes otimes = hasnonnegative fuzzy minimal norm least squares solutions

Proof Since 119860dagger and 119861dagger are nonnegative matrices we have

119883 = 119860dagger

119862119861dagger

ge 0Now that 119866119861 ge 119862119861

dagger

119864+119872119860dagger

119862 and119867119861 ge 119862119861dagger

119865+119873119860dagger

119862therefore with 119884 = 119860

dagger

(119866119861 minus 119862119861dagger

119864 minus 119872119860dagger

119862)119861dagger and 119885 =

119860dagger

(119867119861 minus 119862119861dagger

119865 minus 119873119860dagger

119862)119861dagger we have 119884 ge 0 and 119885 ge 0 Thus

= (119883 119884 119885) is a fuzzy matrix which satisfies otimes otimes =

Since 119883 minus 119884 = 119860dagger

(119862 minus 119866119861 + 119862119861dagger

119864 + 119872119860dagger

119862)119861dagger

the nonnegativity property of can be obtained from thecondition 119862 + 119862119861

dagger

119864 +119872119860dagger

119862 ge 119866119861

The following Theorems give some results for such 119878minus1

and 119878dagger to be nonnegative As usual (sdot)⊤ denotes the transposeof a matrix (sdot)

Theorem 13 (see [36]) The inverse of a nonnegative matrix119860 is nonnegative if and only if 119860 is a generalized permutationmatrix

Theorem 14 (see [37]) Let119860 be an119898 times119898 nonnegativematrixwith rank 119903 Then the following assertions are equivalent

(a) 119860dagger ge 0

(b) There exists a permutation matrix 119875 such that 119875119860 hasthe form

119875119860 =(

1198781

1198782

119878119903

119874

) (19)

where each 119878119894has rank 1 and the rows of 119878

119894are

orthogonal to the rows of 119878119895 whenever 119894 = 119895 the zero

matrix may be absent

(c) 119860dagger = (119870119875⊤119870119876⊤

119870119875⊤119870119875⊤ ) for some positive diagonal matrix 119870

In this case

(119875 + 119876)dagger

= 119870(119875 + 119876)⊤

(119875 minus 119876)dagger

= 119870(119875 minus 119876)⊤

(20)

4 Numerical Examples

Example 15 Consider the fully fuzzy linear matrix system

((2 1 1)LR (1 0 1)LR(1 0 0)LR (2 1 0)LR

) otimes (11

12

13

21

22

23

)

otimes (

(1 0 1)LR (2 1 1)LR (1 1 0)LR(2 1 0)LR (1 0 0)LR (2 1 1)LR(1 0 0)LR (2 1 1)LR (1 0 1)LR

)

= ((17 12 23)LR (22 22 29)LR (17 16 28)LR(19 15 13)LR (23 23 16)LR (19 16 17)LR

)

(21)

Journal of Applied Mathematics 5

ByTheorem 7 the model of the above fuzzy linear matrixsystem is made of the following three crisp systems of linearmatrix equations

(2 1

1 2)(

11990911

11990912

11990913

11990921

11990922

11990923

)(

1 2 1

2 1 2

1 2 1

) = (17 22 17

19 23 19)

(2 1

1 2)(

11990911

11990912

11990913

11990921

11990922

11990923

)(

0 1 1

1 0 1

0 1 0

)

+ (2 1

1 2)(

11991011

11991012

11991013

11991021

11991022

11991023

)(

1 2 1

2 1 2

1 2 1

)

+ (1 0

0 1)(

11990911

11990912

11990913

11990921

11990922

11990923

)(

1 2 1

2 1 2

1 2 1

) = (12 22 16

15 23 20)

(2 1

1 2)(

11990911

11990912

11990913

11990921

11990922

11990923

)(

1 1 0

0 0 1

0 1 1

)

+ (2 1

1 2)(

11991111

11991112

11991113

11991121

11991122

11991123

)(

1 2 1

2 1 2

1 2 1

)

+ (1 1

0 0)(

11990911

11990912

11990913

11990921

11990922

11990923

)(

1 2 1

2 1 2

1 2 1

) = (23 29 28

13 16 17)

(22)

Now that the matrix 119861 is singular according to formula(18) the solutions of the above three systems of linear matrixequations are as follows

119883 = 119860dagger

119862119861dagger

= (2 1

1 2)

dagger

(17 22 17

19 23 19)(

1 2 1

2 1 2

1 2 1

)

dagger

= (15000 10000 15000

15000 20000 15000)

119884 = 119860dagger

119866 minus 119883119864119861dagger

minus 119860dagger

119872119883 = (2 1

1 2)

dagger

(12 22 16

19 23 19)

minus 119883(

0 1 1

1 0 1

0 1 0

)(

1 2 1

2 1 2

1 2 1

)

dagger

minus (2 1

1 2)

dagger

(1 0

0 1)119883

= (10333 06667 08725

11250 16553 12578)

119885 = 119860dagger

119867 minus 119883119865119861dagger

minus 119860dagger

119873119883 = (2 1

1 2)

dagger

(23 29 28

13 16 17)

minus 119883(

1 1 0

0 0 1

0 1 1

)(

1 2 1

2 1 2

1 2 1

)

dagger

minus (2 1

1 2)

dagger

(1 1

0 0)119883

= (83333 116667 93333

14167 13333 24167)

(23)

By Definition 11 we know that the original fuzzy linearmatrix equations have a nonnegative LR fuzzy solution

= (11

12

13

21

22

23

) = ((15000 10333 83333)LR (10000 06667 116667)LR (15000 08725 93333)LR(15000 11250 14167)LR (20000 16553 13333)LR (15000 12578 24167)LR

) (24)

since119883 ge 0 119884 ge 0 119885 ge 0 and119883 minus 119884 ge 0

Example 16 Consider the following fuzzy matrix system

((1 1 0)LR (2 0 1)LR(2 0 1)LR (1 0 0)LR

)(11

12

21

22

)((1 0 1)LR (2 1 0)LR(2 1 1)LR (3 1 2)LR

)

= ((15 14 18)LR (25 24 24)LR(12 10 12)LR (20 17 15)LR

)

(25)

ByTheorem 7 the model of the above fuzzy linear matrixsystem is made of following three crisp systems of linear

matrix equations

(2 1

1 2)(

11990911

11990912

11990921

11990922

)(1 2

2 3) = (

15 25

10 20)

(2 1

1 2)(

11990911

11990912

11990921

11990922

)(0 1

1 1) + (

2 1

1 2)(

11991011

11991012

11991021

11991022

)(1 2

2 3)

+ (1 0

0 0)(

11990911

11990912

11990921

11990922

)(1 2

2 3) = (

14 24

10 17)

(2 1

1 2)(

11990911

11990912

11990921

11990922

)(1 0

1 2) + (

2 1

1 2)(

11991111

11991112

11991121

11991122

)(1 2

2 3)

+ (0 1

1 0)(

11990911

11990912

11990921

11990922

)(1 2

2 3) = (

18 24

12 15)

(26)

6 Journal of Applied Mathematics

By the same way we obtain that the solutions of the abovethree systems of linear matrix equations are as follows

119883 = 119860minus1

119862119861minus1

= (1 1

2 2)

119884 = 119860minus1

119866 minus 119883119864119861minus1

minus 119860minus1

119872119883 = (0 1

0 1)

119885 = 119860minus1

119867 minus 119883119865119861minus1

minus 119860minus1

119873119883 = (0 0

1 0)

(27)

Since 119883 ge 0 119884 ge 0 119885 ge 0 and 119883 minus 119884 ge 0 we know thatthe original fuzzy linear matrix equations have a nonnegativeLR fuzzy solution given by

= (11

12

21

22

)

= ((1000 1000 0000)LR (1000 1000 0000)LR(2000 0000 1000)LR (2000 1000 0000)LR

)

(28)

5 Conclusion

In this work we presented a model for solving fuzzy linearmatrix equations otimes otimes = in which and are119898 times 119898 and 119899 times 119899 fuzzy matrices respectively and isan 119898 times 119899 arbitrary LR fuzzy numbers matrix The modelwas made of three crisp systems of linear equations whichdetermined the mean value and the left and right spreads ofthe solution The LR fuzzy approximate solution of the fuzzylinear matrix equation was derived from solving the crispsystems of linear matrix equations In addition the existencecondition of strong LR fuzzy solution was studied Numericalexamples showed that ourmethod is feasible to solve this typeof fuzzy matrix equations Based on LR fuzzy numbers andtheir operations we can investigate all kinds of fully fuzzymatrix equations in future

Acknowledgments

The work is supported by the Natural Scientific Funds ofChina (no 71061013) and the Youth Research Ability Projectof Northwest Normal University (NWNU-LKQN-1120)

References

[1] L A Zadeh ldquoFuzzy setsrdquo Information and Computation vol 8pp 338ndash353 1965

[2] L A Zadeh ldquoThe concept of a linguistic variable and itsapplication to approximate reasoning Irdquo Information Sciencevol 8 pp 199ndash249 1975

[3] D Dubois and H Prade ldquoOperations on fuzzy numbersrdquoInternational Journal of Systems Science vol 9 no 6 pp 613ndash626 1978

[4] S Nahmias ldquoFuzzy variablesrdquo Fuzzy Sets and Systems AnInternational Journal in Information Science and Engineeringvol 1 no 2 pp 97ndash110 1978

[5] M L Puri and D A Ralescu ldquoDifferentials of fuzzy functionsrdquoJournal of Mathematical Analysis and Applications vol 91 no 2pp 552ndash558 1983

[6] R Goetschel Jr and W Voxman ldquoElementary fuzzy calculusrdquoFuzzy Sets and Systems vol 18 no 1 pp 31ndash43 1986

[7] C X Wu and M Ma ldquoEmbedding problem of fuzzy numberspace Irdquo Fuzzy Sets and Systems vol 44 no 1 pp 33ndash38 1991

[8] C X Wu and M Ma ldquoEmbedding problem of fuzzy numberspace IIIrdquo Fuzzy Sets and Systems vol 46 no 2 pp 281ndash2861992

[9] M Friedman M Ming and A Kandel ldquoFuzzy linear systemsrdquoFuzzy Sets and Systems vol 96 no 2 pp 201ndash209 1998

[10] M Ma M Friedman and A Kandel ldquoDuality in fuzzy linearsystemsrdquo Fuzzy Sets and Systems vol 109 no 1 pp 55ndash58 2000

[11] T Allahviranloo and M Ghanbari ldquoOn the algebraic solutionof fuzzy linear systems based on interval theoryrdquo AppliedMathematical Modelling vol 36 no 11 pp 5360ndash5379 2012

[12] T Allahviranloo and S Salahshour ldquoFuzzy symmetric solutionsof fuzzy linear systemsrdquo Journal of Computational and AppliedMathematics vol 235 no 16 pp 4545ndash4553 2011

[13] T Allahviranloo and M Ghanbari ldquoA new approach to obtainalgebraic solution of interval linear systemsrdquo Soft Computingvol 16 no 1 pp 121ndash133 2012

[14] T Allahviranloo ldquoNumericalmethods for fuzzy systemof linearequationsrdquo Applied Mathematics and Computation vol 155 no2 pp 493ndash502 2004

[15] T Allahviranloo ldquoSuccessive over relaxation iterative methodfor fuzzy system of linear equationsrdquo Applied Mathematics andComputation vol 162 no 1 pp 189ndash196 2005

[16] T Allahviranloo ldquoThe Adomian decomposition method forfuzzy system of linear equationsrdquo Applied Mathematics andComputation vol 163 no 2 pp 553ndash563 2005

[17] RNuraei T Allahviranloo andMGhanbari ldquoFinding an innerestimation of the so- lution set of a fuzzy linear systemrdquoAppliedMathematical Modelling vol 37 no 7 pp 5148ndash5161 2013

[18] S Abbasbandy R Ezzati and A Jafarian ldquoLU decompositionmethod for solving fuzzy system of linear equationsrdquo AppliedMathematics and Computation vol 172 no 1 pp 633ndash6432006

[19] S Abbasbandy A Jafarian and R Ezzati ldquoConjugate gradientmethod for fuzzy symmetric positive definite system of linearequationsrdquo Applied Mathematics and Computation vol 171 no2 pp 1184ndash1191 2005

[20] S Abbasbandy M Otadi andM Mosleh ldquoMinimal solution ofgeneral dual fuzzy linear systemsrdquo Chaos Solitons amp Fractalsvol 37 no 4 pp 1113ndash1124 2008

[21] B Asady S Abbasbandy and M Alavi ldquoFuzzy general linearsystemsrdquo Applied Mathematics and Computation vol 169 no 1pp 34ndash40 2005

[22] K Wang and B Zheng ldquoInconsistent fuzzy linear systemsrdquoApplied Mathematics and Computation vol 181 no 2 pp 973ndash981 2006

[23] B Zheng and K Wang ldquoGeneral fuzzy linear systemsrdquo AppliedMathematics and Computation vol 181 no 2 pp 1276ndash12862006

[24] M Dehghan and B Hashemi ldquoIterative solution of fuzzy linearsystemsrdquo Applied Mathematics and Computation vol 175 no 1pp 645ndash674 2006

[25] M Dehghan B Hashemi and M Ghatee ldquoSolution of the fullyfuzzy linear systems using iterative techniquesrdquo Chaos Solitonsamp Fractals vol 34 no 2 pp 316ndash336 2007

Journal of Applied Mathematics 7

[26] T Allahviranloo N Mikaeilvand and M Barkhordary ldquoFuzzylinear matrix equationrdquo Fuzzy Optimization and Decision Mak-ing vol 8 no 2 pp 165ndash177 2009

[27] X Guo and Z Gong ldquoBlock Gaussian elimination methodsfor fuzzy matrix equationsrdquo International Journal of Pure andApplied Mathematics vol 58 no 2 pp 157ndash168 2010

[28] Z Gong and X Guo ldquoInconsistent fuzzy matrix equationsand its fuzzy least squares solutionsrdquo Applied MathematicalModelling vol 35 no 3 pp 1456ndash1469 2011

[29] X Guo and D Shang ldquoFuzzy symmetric solutions of fuzzymatrix equationsrdquo Advances in Fuzzy Systems vol 2012 ArticleID 318069 9 pages 2012

[30] T Allahviranloo M Ghanbari A A Hosseinzadeh E Haghiand R Nuraei ldquoA note on lsquoFuzzy linear systemsrsquordquo Fuzzy Sets andSystems vol 177 pp 87ndash92 2011

[31] C-T Yeh ldquoWeighted semi-trapezoidal approximations of fuzzynumbersrdquo Fuzzy Sets and Systems vol 165 pp 61ndash80 2011

[32] T Allahviranloo F H Lotfi M K Kiasari and M KhezerlooldquoOn the fuzzy solution of LR fuzzy linear systemsrdquo AppliedMathematical Modelling vol 37 no 3 pp 1170ndash1176 2013

[33] MOtadi andMMosleh ldquoSolving fully fuzzymatrix equationsrdquoApplied Mathematical Modelling vol 36 no 12 pp 6114ndash61212012

[34] X B Guo and D Q Shang ldquoApproximate solutions of LR fuzzySylvester matrix equationsrdquo Journal of Applied Mathematics Inpress

[35] A Ben-Israel and T N E Greville Generalized Inverses Theoryand Applications Springer-Verlag New York NY USA 2ndedition 2003

[36] A Berman and R J Plemmons Nonnegative Matrices in theMathematical Sciences Academic Press New York NY USA1979

[37] R J Plemmons ldquoRegular nonnegative matricesrdquo Proceedings ofthe American Mathematical Society vol 39 pp 26ndash32 1973

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 514984 9 pageshttpdxdoiorg1011552013514984

Research ArticleRanks of a Constrained Hermitian MatrixExpression with Applications

Shao-Wen Yu

Department of Mathematics East China University of Science and Technology Shanghai 200237 China

Correspondence should be addressed to Shao-Wen Yu yushaowenecusteducn

Received 12 November 2012 Accepted 4 January 2013

Academic Editor Yang Zhang

Copyright copy 2013 Shao-Wen YuThis is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We establish the formulas of the maximal and minimal ranks of the quaternion Hermitian matrix expression 1198624minus 1198604119883119860lowast

4where

119883 is a Hermitian solution to quaternion matrix equations 1198601119883 = 119862

1 1198831198611= 1198622 and 119860

3119883119860lowast

3= 1198623 As applications we give a new

necessary and sufficient condition for the existence of Hermitian solution to the system of matrix equations 1198601119883 = 119862

11198831198611= 1198622

1198603119883119860lowast

3= 1198623 and 119860

4119883119860lowast

4= 1198624 which was investigated by Wang and Wu 2010 by rank equalities In addition extremal ranks of

the generalized Hermitian Schur complement 1198624minus 1198604119860sim

3119860lowast

4with respect to a Hermitian g-inverse 119860sim

3of 1198603 which is a common

solution to quaternion matrix equations 1198601119883 = 119862

1and119883119861

1= 1198622 are also considered

1 Introduction

Throughout this paper we denote the real number field byRthe complex number field by C the set of all 119898 times 119899 matricesover the quaternion algebra

H = 1198860+ 1198861119894 + 1198862119895 + 1198863119896 | 1198942

= 1198952

= 1198962

= 119894119895119896 = minus1 1198860 1198861 1198862 1198863isin R

(1)

by H119898times119899 the identity matrix with the appropriate size by 119868the column right space the row left space of a matrix 119860 overH by R(119860) N(119860) respectively the dimension of R(119860) bydimR(119860) a Hermitian g-inverse of a matrix 119860 by 119883 = 119860

∽

which satisfies 119860119860∽119860 = 119860 and 119883 = 119883lowast and the Moore-

Penrose inverse of matrix119860 overH by119860dagger which satisfies fourPenrose equations 119860119860dagger119860 = 119860 119860

dagger

119860119860dagger

= 119860dagger

(119860119860dagger

)lowast

=

119860119860dagger

and (119860dagger

119860)lowast

= 119860dagger

119860 In this case 119860dagger is unique and(119860dagger

)lowast

= (119860lowast

)dagger Moreover 119877

119860and 119871

119860stand for the two

projectors 119871119860

= 119868 minus 119860dagger

119860 119877119860

= 119868 minus 119860119860dagger induced by 119860

Clearly119877119860and 119871

119860are idempotent Hermitian and119877

119860= 119871119860lowast

By [1] for a quaternion matrix 119860 dimR(119860) = dimN(119860)dimR(119860) is called the rank of a quaternion matrix 119860 anddenoted by 119903(119860)

Mitra [2] investigated the system of matrix equations

1198601119883 = 119862

1 119883119861

1= 1198622 (2)

Khatri and Mitra [3] gave necessary and sufficient con-ditions for the existence of the common Hermitian solutionto (2) and presented an explicit expression for the generalHermitian solution to (2) by generalized inverses Using thesingular value decomposition (SVD) Yuan [4] investigatedthe general symmetric solution of (2) over the real numberfield R By the SVD Dai and Lancaster [5] considered thesymmetric solution of equation

119860119883119860lowast

= 119862 (3)

over R which was motivated and illustrated with an inverseproblem of vibration theory Groszlig [6] Tian and Liu [7]gave the solvability conditions for Hermitian solution and itsexpressions of (3) over C in terms of generalized inversesrespectively Liu Tian and Takane [8] investigated ranksof Hermitian and skew-Hermitian solutions to the matrixequation (3) By using the generalized SVD Chang andWang[9] examined the symmetric solution to the matrix equations

1198603119883119860lowast

3= 1198623 119860

4119883119860lowast

4= 1198624

(4)

2 Journal of Applied Mathematics

over R Note that all the matrix equations mentioned aboveare special cases of

1198601119883 = 119862

1 119883119861

1= 1198622

1198603119883119860lowast

3= 1198623 119860

4119883119860lowast

4= 1198624

(5)

Wang and Wu [10] gave some necessary and sufficientconditions for the existence of the common Hermitiansolution to (5) for operators between Hilbert Clowast-modulesby generalized inverses and range inclusion of matrices Inview of the complicated computations of the generalizedinverses of matrices we naturally hope to establish a morepractical necessary and sufficient condition for system (5)over quaternion algebra to have Hermitian solution by rankequalities

As is known to us solutions tomatrix equations and ranksof solutions to matrix equations have been considered previ-ously by many authors [10ndash34] and extremal ranks of matrixexpressions can be used to characterize their rank invariancenonsingularity range inclusion and solvability conditionsof matrix equations Tian and Cheng [35] investigated themaximal and minimal ranks of 119860 minus 119861119883119862 with respect to 119883with applications Tian [36] gave the maximal and minimalranks of 119860

1minus 11986111198831198621subject to a consistent matrix equation

11986121198831198622= 1198602 Tian and Liu [7] established the solvability

conditions for (4) to have a Hermitian solution over C bythe ranks of coefficientmatricesWang and Jiang [20] derivedextreme ranks of (skew)Hermitian solutions to a quaternionmatrix equation 119860119883119860

lowast

+ 119861119884119861lowast

= 119862 Wang Yu and Lin[31] derived the extremal ranks of 119862

4minus 11986041198831198614subject to a

consistent system of matrix equations

1198601119883 = 119862

1 119883119861

1= 1198622 119860

31198831198613= 1198623

(6)

over H and gave a new solvability condition to system

1198601119883 = 119862

1 119883119861

1= 1198622

11986031198831198613= 1198623 119860

41198831198614= 1198624

(7)

In matrix theory and its applications there are manymatrix expressions that have symmetric patterns or involveHermitian (skew-Hermitian) matrices For example

119860 minus 119861119883119861lowast

119860 minus 119861119883 plusmn 119883lowast

119861lowast

119860 minus 119861119883119861lowast

minus 119862119884119862lowast

119860 minus 119861119883119862 plusmn (119861119883119862)lowast

(8)

where 119860 = plusmn119860lowast

119861 and 119862 are given and 119883 and 119884 arevariable matrices In recent papers [7 8 37 38] Liu and Tianconsidered some maximization and minimization problemson the ranks of Hermitian matrix expressions (8)

Define a Hermitian matrix expression

119891 (119883) = 1198624minus 1198604119883119860lowast

4 (9)

where 1198624= 119862lowast

4 we have an observation that by investigating

extremal ranks of (9) where 119883 is a Hermitian solution to asystem of matrix equations

1198601119883 = 119862

1 119883119861

1= 1198622 119860

3119883119860lowast

3= 1198623 (10)

A new necessary and sufficient condition for system (5) tohave Hermitian solution can be given by rank equalitieswhich is more practical than one given by generalizedinverses and range inclusion of matrices

It is well known that Schur complement is one of themostimportant matrix expressions in matrix theory there havebeen many results in the literature on Schur complementsand their applications [39ndash41] Tian [36 42] has investigatedthe maximal and minimal ranks of Schur complements withapplications

Motivated by the workmentioned above we in this paperinvestigate the extremal ranks of the quaternion Hermitianmatrix expression (9) subject to the consistent system ofquaternion matrix equations (10) and its applications InSection 2 we derive the formulas of extremal ranks of (9)with respect to Hermitian solution of (10) As applications inSection 3 we give a new necessary and sufficient conditionfor the existence of Hermitian solution to system (5) byrank equalities In Section 4 we derive extremal ranks ofgeneralized Hermitian Schur complement subject to (2) Wealso consider the rank invariance problem in Section 5

2 Extremal Ranks of (9) Subject to System (10)Corollary 8 in [10] over Hilbert Clowast-modules can be changedinto the following lemma over H

Lemma 1 Let 1198601 1198621

isin H119898times119899 1198611 1198622

isin H119899times119904 1198603

isin

H119903times119899 1198623isin H119903times119903 be given and 119865 = 119861

lowast

11198711198601 119872 = 119878119871

119865 119878 =

11986031198711198601 119863 = 119862

lowast

2minus119861lowast

1119860dagger

11198621 119869 = 119860

dagger

11198621+119865dagger

119863 119866 = 1198623minus1198603(119869+

1198711198601119871lowast

119865119869lowast

)119860lowast

3 then the following statements are equivalent

(1) the system (10) have a Hermitian solution(2) 1198623= 119862lowast

3

11986011198622= 11986211198611 119860

1119862lowast

1= 1198621119860lowast

1 119861

lowast

11198622= 119862lowast

21198611 (11)

11987711986011198621= 0 119877

119865119863 = 0 119877

119872119866 = 0 (12)

(3) 1198623= 119862lowast

3 the equalities in (11) hold and

119903 [11986011198621] = 119903 (119860

1) 119903 [

11986011198621

119861lowast

1119862lowast

2

] = 119903 [1198601

119861lowast

1

]

119903[[

[

11986011198621119860lowast

3

119861lowast

1119862lowast

2119860lowast

3

1198603

1198623

]]

]

= 119903[

[

1198601

119861lowast

1

1198603

]

]

(13)

In that case the general Hermitian solution of (10) can beexpressed as

119883 = 119869 + 1198711198601119871119865119869lowast

+ 1198711198601119871119865119872dagger

119866(119872dagger

)lowast

1198711198651198711198601

+ 11987111986011198711198651198711198721198811198711198651198711198601+ 1198711198601119871119865119881lowast

1198711198721198711198651198711198601

(14)

where 119881 is Hermitian matrix over H with compatible size

Journal of Applied Mathematics 3

Lemma 2 (see Lemma 24 in [24]) Let 119860 isin H119898times119899 119861 isin

H119898times119896 119862 isin H119897times119899 119863 isin H119895times119896 and 119864 isin H119897times119894 Then the followingrank equalities hold

(a) 119903(119862119871119860) = 119903 [

119860

119862] minus 119903(119860)

(b) 119903 [ 119861 119860119871119862 ] = 119903 [119861 119860

0 119862] minus 119903(119862)

(c) 119903 [ 119862119877119861119860

] = 119903 [119862 0

119860 119861] minus 119903(119861)

(d) 119903 [ 119860 119861119871119863119877119864119862 0

] = 119903 [119860 119861 0

119862 0 119864

0 119863 0

] minus 119903(119863) minus 119903(119864)

Lemma 2 plays an important role in simplifying ranks ofvarious block matrices

Liu and Tian [38] has given the following lemma over afield The result can be generalized to H

Lemma 3 Let 119860 = plusmn119860lowast

isin H119898times119898 119861 isin H119898times119899 and 119862 isin H119901times119898

be given then

max119883isinH119899times119901

119903 [119860 minus 119861119883119862 ∓ (119861119883119862)lowast

]

= min119903 [119860 119861 119862lowast

] 119903 [119860 119861

119861lowast

0] 119903 [

119860 119862lowast

119862 0]

min119883isinH119899times119901

119903 [119860 minus 119861119883119862 ∓ (119861119883119862)lowast

]

= 2119903 [119860 119861 119862lowast

] +max 1199041 1199042

(15)

where

1199041= 119903 [

119860 119861

119861lowast

0] minus 2119903 [

119860 119861 119862lowast

119861lowast

0 0]

1199042= 119903 [

119860 119862lowast

119862 0] minus 2119903 [

119860 119861 119862lowast

119862 0 0]

(16)

IfR(119861) sube R(119862lowast

)

max119883

119903 [119860 minus 119861119883119862 minus (119861119883119862)lowast

] = min119903 [119860 119862lowast

] 119903 [119860 119861

119861lowast

0]

max119883

119903 [119860 minus 119861119883119862 minus (119861119883119862)lowast

] = min119903 [119860 119862lowast

] 119903 [119860 119861

119861lowast

0]

(17)

Now we consider the extremal ranks of the matrixexpression (9) subject to the consistent system (10)

Theorem 4 Let 1198601 1198621 1198611 1198622 1198603 and 119862

3be defined as

Lemma 11198624isin H119905times119905 and 119860

4isin H119905times119899Then the extremal ranks

of the quaternionmatrix expression119891(119883) defined as (9) subjectto system (10) are the following

max 119903 [119891 (119883)] = min 119886 119887 (18)

where

119886 = 119903

[[[[

[

1198624

1198604

119862lowast

2119860lowast

4119861lowast

1

1198621119860lowast

41198601

]]]]

]

minus 119903 [119861lowast

1

1198601

]

119887 = 119903

[[[[[[[[

[

0 119860lowast

4119860lowast

31198611

119860lowast

1

11986041198624

0 0 0

1198603

0 minus1198623

minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]]

]

minus 2119903[

[

1198603

119861lowast

1

1198601

]

]

(19)

min 119903 [119891 (119883)] = 2119903

[[[[

[

1198624

1198604

119862lowast

2119860lowast

4119861lowast

1

1198621119860lowast

41198601

]]]]

]

+ 119903

[[[[[[[[[

[

0 119860lowast

4119860lowast

31198611

119860lowast

1

11986041198624

0 0 0

1198603

0 minus1198623

minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]]]

]

minus 2119903

[[[[[[[[[

[

0 119860lowast

41198611

119860lowast

1

11986041198624

0 0

1198603

0 minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus11986211198611minus1198621119860lowast

1

]]]]]]]]]

]

(20)

Proof By Lemma 1 the general Hermitian solution of thesystem (10) can be expressed as

119883 = 119869 + 1198711198601119871119865119869lowast

+ 1198711198601119871119865119872dagger

119866(119872dagger

)lowast

1198711198651198711198601

+ 11987111986011198711198651198711198721198811198711198651198711198601+ 1198711198601119871119865119881lowast

1198711198721198711198651198711198601

(21)

where 119881 is Hermitian matrix over H with appropriate sizeSubstituting (21) into (9) yields

119891 (119883) = 1198624minus 1198604(119869 + 119871

1198601119871119865119869lowast

+ 1198711198601119871119865119872dagger

119866(119872dagger

)lowast

1198711198651198711198601)119860

lowast

4

minus 119860411987111986011198711198651198711198721198811198711198651198711198601119860lowast

4

minus 11986041198711198601119871119865119881lowast

1198711198721198711198651198711198601119860lowast

4

(22)

4 Journal of Applied Mathematics

Put

1198624minus 1198604(119869 + 119871

1198601119871119865119869lowast

+ 1198711198601119871119865119872dagger

119866(119872dagger

)lowast

1198711198651198711198601)119860lowast

4= 119860

119869 + 1198711198601119871119865119869lowast

+ 1198711198601119871119865119872dagger

119866(119872dagger

)lowast

1198711198651198711198601

= 1198691015840

11986041198711198601119871119865119871119872= 119873

1198711198651198711198601119860lowast

4= 119875

(23)

then

119891 (119883) = 119860 minus 119873119881119875 minus (119873119881119875)lowast

(24)

Note that119860 = 119860lowast andR(119873) sube R(119875

lowast

) Thus applying (17) to(24) we get the following

max 119903 [119891 (119883)] = max119881

119903 (119860 minus 119873119881119875 minus (119873119881119875)lowast

)

= min119903 [119860 119875lowast

] 119903 [119860 119873

119873lowast

0]

min 119903 [119891 (119883)] = min119881

119903 (119860 minus 119873119881119875 minus (119873119881119875)lowast

)

= 2119903 [119860 119875lowast

] + 119903 [119860 119873

119873lowast

0] minus 2119903 [

119860 119873

119875 0]

(25)

Now we simplify the ranks of block matrices in (25)In view of Lemma 2 block Gaussian elimination (11)

(12) and (23) we have the following

119903 (119865) = 119903 (119861lowast

11198711198601) = 119903 [

119861lowast

1

1198601

] minus 119903 (1198601)

119903 (119872) = 119903 (119878119871119865) = 119903 [

119878

119865] minus 119903 (119865)

= 119903 [

11986031198711198601

119861lowast

11198711198601

] minus 119903 (119865)

= 119903[

[

1198603

119861lowast

1

1198601

]

]

minus 119903 (1198601) minus 119903 (119865)

119903 [119860 119875lowast

] = 119903 [1198624minus 1198604119869119860lowast

4119875lowast

]

= 119903 [1198624minus 1198604119869119860lowast

411986041198711198601

0 119865] minus 119903 (119865)

= 119903[

[

1198624minus 1198604119869119860lowast

41198604

0 119861lowast

1

0 1198601

]

]

minus 119903 (119865) minus 119903 (1198601)

= 119903

[[[

[

1198624

1198604

119862lowast

2119860lowast

4119861lowast

1

1198621119860lowast

41198601

]]]

]

minus 119903 [119861lowast

1

1198601

]

119903 [119860 119873

119873lowast

0] = 119903 [

1198624minus 11986041198691015840

119860lowast

411986041198711198601119871119865119871119872

119877119872lowast119877119865lowast119877119860lowast

1

119860lowast

40

]

= 119903

[[[[[[

[

1198624minus 11986041198691015840

119860lowast

41198604

0 0 0

119860lowast

40 119860lowast

31198611119860lowast

1

0 1198603

0 0 0

0 119861lowast

10 0 0

0 1198601

0 0 0

]]]]]]

]

minus 2119903 (119872) minus 2119903 (119865) minus 2119903 (1198601)

= 119903

[[[[[[[[[[[[

[

1198624

1198604

0 0 0

119860lowast

40 119860lowast

31198611119860lowast

1

11986031198691015840

119860lowast

41198603

0 0 0

119861lowast

11198691015840

119860lowast

4119861lowast

10 0 0

11986011198691015840

119860lowast

41198601

0 0 0

]]]]]]]]]]]]

]

minus 2119903[

[

1198603

119861lowast

1

1198601

]

]

= 119903

[[[[[[[[[

[

11986241198604

0 0 0

119860lowast

40 119860

lowast

31198611

119860lowast

1

0 1198603

minus1198623

minus11986031198622minus1198603119862lowast

1

0 119861lowast

1minus119862lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

0 1198601minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]]]

]

minus 2119903[

[

1198603

119861lowast

1

1198601

]

]

= 119903

[[[[[[[[[

[

0 119860lowast

4119860lowast

31198611

119860lowast

1

11986041198624

0 0 0

1198603

0 minus1198623

minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]]]

]

minus 2119903[

[

1198603

119861lowast

1

1198601

]

]

119903 [119860 119873

119875 0] = 119903 [

1198624minus 11986041198691015840

119860lowast

411986041198711198601119871119865119871119872

119877119865lowast119877119860lowast

1

119860lowast

40

]

= 119903

[[[[[[[[[[

[

11986241198604

0 0

119860lowast

40 119861

1119860lowast

1

0 1198603minus11986031198622minus1198603119862lowast

1

0 119861lowast

1minus119862lowast

21198611minus119862lowast

2119860lowast

1

0 1198601minus11986211198611minus1198621119860lowast

1

]]]]]]]]]]

]

minus 119903[

[

1198603

119861lowast

1

1198601

]

]

minus 119903 [119861lowast

1

1198601

]

Journal of Applied Mathematics 5

= 119903

[[[[[[[[[[

[

0 119860lowast

41198611

119860lowast

1

11986041198624

0 0

1198603

0 minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus11986211198611minus1198621119860lowast

1

]]]]]]]]]]

]

minus 119903[[

[

1198603

119861lowast

1

1198601

]]

]

minus 119903 [

119861lowast

1

1198601

]

(26)

Substituting (26) into (25) yields (18) and (20)

In Theorem 4 letting 1198624vanish and 119860

4be 119868 with

appropriate size respectively we have the following

Corollary 5 Assume that 1198601 1198621

isin H119898times119899 11986111198622

isin

H119899times119904 1198603isin H119903times119899 and 119862

3isin H119903times119903 are given then the maximal

and minimal ranks of the Hermitian solution 119883 to the system(10) can be expressed as

max 119903 (119883) = min 119886 119887 (27)

where

119886 = 119899 + 119903 [119862lowast

2

1198621

] minus 119903 [119861lowast

1

1198601

]

119887 = 2119899 + 119903

[[[[

[

1198623

119860311986221198603119862lowast

1

119862lowast

2119860lowast

3119862lowast

21198611119862lowast

2119860lowast

1

1198621119860lowast

3119862111986111198621119860lowast

1

]]]]

]

minus 2119903[

[

1198603

119861lowast

1

1198601

]

]

min 119903 (119883) = 2119903 [

119862lowast

2

1198621

]

+ 119903

[[[

[

1198623

119860311986221198603119862lowast

1

119862lowast

2119860lowast

3119862lowast

21198611119862lowast

2119860lowast

1

1198621119860lowast

3119862111986111198621119860lowast

1

]]]

]

minus 2119903

[[[[

[

119860311986221198603119862lowast

1

119862lowast

21198611119862lowast

2119860lowast

1

119862111986111198621119860lowast

1

]]]]

]

(28)

InTheorem 4 assuming that 1198601 1198611 1198621 and 119862

2vanish

we have the following

Corollary 6 Suppose that the matrix equation 1198603119883119860lowast

3= 1198623

is consistent then the extremal ranks of the quaternion matrixexpression 119891(119883) defined as (9) subject to 119860

3119883119860lowast

3= 1198623are the

following

max 119903 [119891 (119883)]

= min

119903 [11986241198604] 119903 [

[

0 119860lowast

4119860lowast

3

11986041198624

0

1198603

0 minus1198623

]

]

minus 2119903 (1198603)

min 119903 [119891 (119883)] = 2119903 [11986241198604]

+ 119903[

[

0 119860lowast

4119860lowast

3

11986041198624

0

1198603

0 minus1198623

]

]

minus 2119903[

[

0 119860lowast

4

11986041198624

1198603

0

]

]

(29)

3 A Practical Solvability Condition forHermitian Solution to System (5)

In this section we use Theorem 4 to give a necessary andsufficient condition for the existence of Hermitian solutionto system (5) by rank equalities

Theorem 7 Let 1198601 1198621

isin H119898times119899 11986111198622

isin H119899times119904 1198603

isin

H119903times119899 1198623isin H119903times119903 119860

4isin H119905times119899 and 119862

4isin H119905times119905be given then

the system (5) have Hermitian solution if and only if 1198623= 119862lowast

3

(11) (13) hold and the following equalities are all satisfied

119903 [11986041198624] = 119903 (119860

4) (30)

119903

[[[

[

1198624

1198604

119862lowast

2119860lowast

4119861lowast

1

1198621119860lowast

41198601

]]]

]

= 119903[

[

1198604

119861lowast

1

1198601

]

]

(31)

119903

[[[[[[[[

[

0 119860lowast

4119860lowast

31198611

119860lowast

1

11986041198624

0 0 0

1198603

0 minus1198623

minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]]

]

= 2119903

[[[[

[

1198604

1198603

119861lowast

1

1198601

]]]]

]

(32)

Proof It is obvious that the system (5) have Hermitiansolution if and only if the system (10) haveHermitian solutionand

min 119903 [119891 (119883)] = 0 (33)

where 119891(119883) is defined as (9) subject to system (10) Let1198830be

aHermitian solution to the system (5) then1198830is aHermitian

solution to system (10) and1198830satisfies119860

41198830119860lowast

4= 1198624 Hence

Lemma 1 yields 1198623

= 119862lowast

3 (11) (13) and (30) It follows

6 Journal of Applied Mathematics

from

[[[[[[

[

119868 0 0 0 0

0 119868 0 0 0

119860311988300 119868 0 0

119861lowast

111988300 0 119868 0

119860111988300 0 0 119868

]]]]]]

]

times

[[[[[[[

[

0 119860lowast

4119860lowast

31198611

119860lowast

1

11986041198624

0 0 0

1198603

0 minus1198623

minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]

]

times

[[[[[

[

119868 minus1198830119860lowast

40 0 0

0 119868 0 0 0

0 0 119868 0 0

0 0 0 119868 0

0 0 0 0 119868

]]]]]

]

=

[[[[[

[

0 119860lowast

4119860lowast

31198611119860lowast

1

1198604

0 0 0 0

1198603

0 0 0 0

119861lowast

10 0 0 0

1198601

0 0 0 0

]]]]]

]

(34)

that (32) holds Similarly we can obtain (31)Conversely assume that 119862

3= 119862lowast

3 (11) (13) hold then by

Lemma 1 system (10) have Hermitian solution By (20) (31)-(32) and

119903

[[[[[[[

[

0 119860lowast

41198611

119860lowast

1

11986041198624

0 0

1198603

0 minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus11986211198611minus1198621119860lowast

1

]]]]]]]

]

ge 119903

[[[

[

1198604

1198603

119861lowast

1

1198601

]]]

]

+ 119903[

[

1198604

119861lowast

1

1198601

]

]

(35)

we can get

min 119903 [119891 (119883)] le 0 (36)

However

min 119903 [119891 (119883)] ge 0 (37)

Hence (33) holds implying that the system (5) have Hermi-tian solution

ByTheorem 7 we can also get the following

Corollary 8 Suppose that 1198603 1198623 1198604 and 119862

4are those in

Theorem 7 then the quaternion matrix equations 1198603119883119860lowast

3=

1198623and 119860

4119883119860lowast

4= 1198624have common Hermitian solution if and

only if (30) hold and the following equalities are satisfied

119903 [11986031198623] = 119903 (119860

3)

119903 [

[

0 119860lowast

4119860lowast

3

11986041198624

0

1198603

0 minus1198623

]

]

= 2119903 [1198603

1198604

]

(38)

Corollary 9 Suppose that11986011198621isin H119898times119899119861

11198622isin H119899times119904 and

119860 119861 isin H119899times119899 are Hermitian Then 119860 and 119861 have a common

Hermitian g-inverse which is a solution to the system (2) if andonly if (11) holds and the following equalities are all satisfied

119903[[

[

11986011198621119860

119861lowast

1119862lowast

2119860

119860 119860

]]

]

= 119903[

[

1198601

119861lowast

1

119860

]

]

119903[[

[

11986011198621119861

119861lowast

1119862lowast

2119861

119861 119861

]]

]

= 119903[

[

1198601

119861lowast

1

119861

]

]

(39)

119903

[[[[[[[

[

0 119861 119860 1198611

119860lowast

1

119861 119861 0 0 0

119860 0 minus119860 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

2119860 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

1119860 minus119862

11198611minus1198621119860lowast

1

]]]]]]]

]

= 2119903

[[[[

[

119861

119860

119861lowast

1

1198601

]]]]

]

(40)

4 Extremal Ranks of Schur ComplementSubject to (2)

As is well known for a given block matrix

119872 = [119860 119861

119861lowast

119863] (41)

where 119860 and 119863 are Hermitian quaternion matrices withappropriate sizes then the Hermitian Schur complement of119860 in119872 is defined as

119878119860= 119863 minus 119861

lowast

119860sim

119861 (42)

where 119860sim is a Hermitian g-inverse of 119860 that is 119860sim isin 119883 |

119860119883119860 = 119860119883 = 119883lowast

Now we use Theorem 4 to establish the extremal ranks

of 119878119860given by (42) with respect to 119860sim which is a solution to

system (2)

Theorem 10 Suppose 1198601 1198621isin H119898times119899 119861

1 1198622isin H119899times119904 119863 isin

H119905times119905 119861 isin H119899times119905 and 119860 isin H119899times119899 are given and system (2)is consistent then the extreme ranks of 119878

119860given by (42) with

respect to 119860sim which is a solution of (2) are the following

max1198601119860sim=1198621

119860sim1198611=1198622

119903 (119878119860) = min 119886 119887

(43)

Journal of Applied Mathematics 7

where

119886 = 119903[

[

119863 119861lowast

119862lowast

2119861 119861lowast

1

1198621119861 1198601

]

]

minus 119903 [119861lowast

1

1198601

]

119887 = 119903

[[[[[[

[

0 119861 119860 1198611

119860lowast

1

119861lowast

119863 0 0 0

119860 0 minus119860 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

2119860 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

1119860 minus119862

11198611minus1198621119860lowast

1

]]]]]]

]

minus 2119903[

[

119860

119861lowast

1

1198601

]

]

min1198601119860sim=1198621

119860sim1198611=1198622

119903 (119878119860) = 2119903[

[

119863 119861lowast

119862lowast

2119861 119861lowast

1

1198621119861 1198601

]

]

+ 119903

[[[[[[[[

[

0 119861 119860 1198611

119860lowast

1

119861lowast

119863 0 0 0

119860 0 minus119860 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

2119860 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

1119860 minus119862

11198611minus1198621119860lowast

1

]]]]]]]]

]

minus 2119903

[[[[[[[[

[

0 119861 1198611

119860lowast

1

119861lowast

119863 0 0

119860 0 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

11198611minus1198621119860lowast

1

]]]]]]]]

]

(44)

Proof It is obvious that

max1198601119860sim=1198621 119860

sim1198611=1198622

119903 (119863 minus 119861lowast

119860sim

119861)

= max1198601119883=1198621 1198831198611=1198622 119860119883119860=119860

119903 (119863 minus 119861lowast

119883119861)

min1198601119860sim=1198621 119860

sim1198611=1198622

119903 (119863 minus 119861lowast

119860sim

119861)

= min1198601119883=1198621 1198831198611=1198622 119860119883119860=119860

119903 (119863 minus 119861lowast

119883119861)

(45)

Thus in Theorem 4 and its proof letting 1198603= 119860lowast

3= 1198623= 119860

1198604= 119861lowast

and 1198624= 119863 we can easily get the proof

In Theorem 10 let 1198601 1198621 1198611 and 119862

2vanish Then we

can easily get the following

Corollary 11 The extreme ranks of 119878119860given by (42) with

respect to 119860sim are the following

max119860sim119903 (119878119860) = min

119903 [119863 119861lowast

] 119903 [

[

0 119861 119860

119861lowast

119863 0

119860 0 minus119860

]

]

minus 2119903 (119860)

min119860sim119903 (119878119860) = 2119903 [119863 119861

lowast

] + 119903[

[

0 119861 119860

119861lowast

119863 0

119860 0 minus119860

]

]

minus 2119903[

[

0 119861

119861lowast

119863

119860 0

]

]

(46)

5 The Rank Invariance of (9)As another application of Theorem 4 we in this sectionconsider the rank invariance of thematrix expression (9) withrespect to the Hermitian solution of system (10)

Theorem 12 Suppose that (10) have Hermitian solution thenthe rank of 119891(119883) defined by (9) with respect to the Hermitiansolution of (10) is invariant if and only if

119903

[[[[[[[[

[

0 119860lowast

41198611

119860lowast

1

11986041198624

0 0

1198603

0 minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus11986211198611minus1198621119860lowast

1

]]]]]]]]

]

= 119903[[

[

1198624

1198604

119862lowast

2119860lowast

4119861lowast

1

1198621119860lowast

41198601

]]

]

+ 119903[

[

1198603

119861lowast

1

1198601

]

]

119903

[[[[[[[[

[

0 119860lowast

4119860lowast

31198611

119860lowast

1

11986041198624

0 0 0

1198603

0 minus1198623

minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

2119860lowast

3minus119862lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus1198621119860lowast

3minus11986211198611minus1198621119860lowast

1

]]]]]]]]

]

+ 119903 [119861lowast

1

1198601

]

= 119903

[[[[[[

[

0 119860lowast

41198611

119860lowast

1

11986041198624

0 0

1198603

0 minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus11986211198611minus1198621119860lowast

1

]]]]]]

]

+ 119903[

[

1198603

119861lowast

1

1198601

]

]

(47)

or

119903

[[[[[[[

[

0 119860lowast

41198611

119860lowast

1

11986041198624

0 0

1198603

0 minus11986031198622minus1198603119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

1198601

0 minus11986211198611minus1198621119860lowast

1

]]]]]]]

]

= 119903[[

[

1198624

1198604

119862lowast

2119860lowast

4119861lowast

1

1198621119860lowast

41198601

]]

]

+ 119903[[

[

1198603

119861lowast

1

1198601

]]

]

(48)

Proof It is obvious that the rank of 119891(119883) with respect toHermitian solution of system (10) is invariant if and only if

max 119903 [119891 (119883)] minusmin 119903 [119891 (119883)] = 0 (49)

By (49) Theorem 4 and simplifications we can get (47)and (48)

8 Journal of Applied Mathematics

Corollary 13 The rank of 119878119860defined by (42) with respect to

119860sim which is a solution to system (2) is invariant if and only if

119903

[[[[[[

[

0 119861 1198611

119860lowast

1

119861lowast

119863 0 0

119860 0 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

11198611minus1198621119860lowast

1

]]]]]]

]

= 119903[

[

119863 119861lowast

119862lowast

2119861 119861lowast

1

1198621119861 1198601

]

]

+ 119903[

[

119860

119861lowast

1

1198601

]

]

119903

[[[[[[[[

[

0 119861 119860 1198611

119860lowast

1

119861lowast

119863 0 0 0

119860 0 minus119860 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

2119860 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

1119860 minus119862

11198611minus1198621119860lowast

1

]]]]]]]]

]

+ 119903 [119861lowast

1

1198601

]

= 119903

[[[[[[

[

0 119861 1198611

119860lowast

1

119861lowast

119863 0 0

119860 0 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

11198611minus1198621119860lowast

1

]]]]]]

]

+ 119903[

[

119860

119861lowast

1

1198601

]

]

(50)

or

119903

[[[[[[

[

0 119861 1198611

119860lowast

1

119861lowast

119863 0 0

119860 0 minus1198601198622

minus119860119862lowast

1

119861lowast

10 minus119862

lowast

21198611minus119862lowast

2119860lowast

1

11986010 minus119862

11198611minus1198621119860lowast

1

]]]]]]

]

= 119903[

[

119863 119861lowast

119862lowast

2119861 119861lowast

1

1198621119861 1198601

]

]

+ 119903[

[

119860

119861lowast

1

1198601

]

]

(51)

Acknowledgments

This research was supported by the National Natural ScienceFoundation of China Tian Yuan Foundation (11226067) theFundamental Research Funds for the Central Universities(WM1214063) and China Postdoctoral Science Foundation(2012M511014)

References

[1] T W Hungerford Algebra Springer New York NY USA 1980[2] S K Mitra ldquoThe matrix equations 119860119883 = 119862 119883119861 = 119863rdquo Linear

Algebra and its Applications vol 59 pp 171ndash181 1984[3] C G Khatri and S K Mitra ldquoHermitian and nonnegative

definite solutions of linear matrix equationsrdquo SIAM Journal onApplied Mathematics vol 31 no 4 pp 579ndash585 1976

[4] Y X Yuan ldquoOn the symmetric solutions of matrix equation(119860119883119883119862) = (119861119863)rdquo Journal of East China Shipbuilding Institutevol 15 no 4 pp 82ndash85 2001 (Chinese)

[5] H Dai and P Lancaster ldquoLinear matrix equations from aninverse problem of vibration theoryrdquo Linear Algebra and itsApplications vol 246 pp 31ndash47 1996

[6] J Groszlig ldquoA note on the general Hermitian solution to 119860119883119860lowast =119861rdquo Malaysian Mathematical Society vol 21 no 2 pp 57ndash621998

[7] Y G Tian and Y H Liu ldquoExtremal ranks of some symmetricmatrix expressions with applicationsrdquo SIAM Journal on MatrixAnalysis and Applications vol 28 no 3 pp 890ndash905 2006

[8] Y H Liu Y G Tian and Y Takane ldquoRanks of Hermitian andskew-Hermitian solutions to the matrix equation 119860119883119860lowast = 119861rdquoLinear Algebra and its Applications vol 431 no 12 pp 2359ndash2372 2009

[9] X W Chang and J S Wang ldquoThe symmetric solution of thematrix equations 119860119883 + 119884119860 = 119862 119860119883119860119879 + 119861119884119861

119879

= 119862 and(119860119879

119883119860 119861119879

119883119861) = (119862119863)rdquo Linear Algebra and its Applicationsvol 179 pp 171ndash189 1993

[10] Q-W Wang and Z-C Wu ldquoCommon Hermitian solutionsto some operator equations on Hilbert 119862lowast-modulesrdquo LinearAlgebra and its Applications vol 432 no 12 pp 3159ndash3171 2010

[11] F O Farid M S Moslehian Q-W Wang and Z-C Wu ldquoOnthe Hermitian solutions to a system of adjointable operatorequationsrdquo Linear Algebra and its Applications vol 437 no 7pp 1854ndash1891 2012

[12] Z-H He and Q-WWang ldquoSolutions to optimization problemson ranks and inertias of a matrix function with applicationsrdquoAppliedMathematics andComputation vol 219 no 6 pp 2989ndash3001 2012

[13] Q-W Wang ldquoThe general solution to a system of real quater-nion matrix equationsrdquo Computers amp Mathematics with Appli-cations vol 49 no 5-6 pp 665ndash675 2005

[14] Q-W Wang ldquoBisymmetric and centrosymmetric solutions tosystems of real quaternion matrix equationsrdquo Computers ampMathematics with Applications vol 49 no 5-6 pp 641ndash6502005

[15] Q W Wang ldquoA system of four matrix equations over vonNeumann regular rings and its applicationsrdquo Acta MathematicaSinica vol 21 no 2 pp 323ndash334 2005

[16] Q-W Wang ldquoA system of matrix equations and a linear matrixequation over arbitrary regular rings with identityrdquo LinearAlgebra and its Applications vol 384 pp 43ndash54 2004

[17] Q-WWang H-X Chang andQNing ldquoThe common solutionto six quaternion matrix equations with applicationsrdquo AppliedMathematics and Computation vol 198 no 1 pp 209ndash2262008

[18] Q-W Wang H-X Chang and C-Y Lin ldquoP-(skew)symmetriccommon solutions to a pair of quaternion matrix equationsrdquoApplied Mathematics and Computation vol 195 no 2 pp 721ndash732 2008

[19] Q W Wang and Z H He ldquoSome matrix equations withapplicationsrdquo Linear and Multilinear Algebra vol 60 no 11-12pp 1327ndash1353 2012

[20] Q W Wang and J Jiang ldquoExtreme ranks of (skew-)Hermitiansolutions to a quaternion matrix equationrdquo Electronic Journal ofLinear Algebra vol 20 pp 552ndash573 2010

[21] Q-W Wang and C-K Li ldquoRanks and the least-norm of thegeneral solution to a system of quaternion matrix equationsrdquoLinear Algebra and its Applications vol 430 no 5-6 pp 1626ndash1640 2009

[22] Q-W Wang X Liu and S-W Yu ldquoThe common bisymmetricnonnegative definite solutions with extreme ranks and inertiasto a pair of matrix equationsrdquo Applied Mathematics and Com-putation vol 218 no 6 pp 2761ndash2771 2011

Journal of Applied Mathematics 9

[23] Q-W Wang F Qin and C-Y Lin ldquoThe common solution tomatrix equations over a regular ring with applicationsrdquo IndianJournal of Pure andAppliedMathematics vol 36 no 12 pp 655ndash672 2005

[24] Q-W Wang G-J Song and C-Y Lin ldquoExtreme ranks of thesolution to a consistent system of linear quaternion matrixequations with an applicationrdquo Applied Mathematics and Com-putation vol 189 no 2 pp 1517ndash1532 2007

[25] Q-W Wang G-J Song and C-Y Lin ldquoRank equalities relatedto the generalized inverse 119860

(2)

119879119878with applicationsrdquo Applied

Mathematics and Computation vol 205 no 1 pp 370ndash3822008

[26] Q W Wang G J Song and X Liu ldquoMaximal and minimalranks of the common solution of some linear matrix equationsover an arbitrary division ring with applicationsrdquo AlgebraColloquium vol 16 no 2 pp 293ndash308 2009

[27] Q-W Wang Z-C Wu and C-Y Lin ldquoExtremal ranks ofa quaternion matrix expression subject to consistent systemsof quaternion matrix equations with applicationsrdquo AppliedMathematics and Computation vol 182 no 2 pp 1755ndash17642006

[28] Q W Wang and S W Yu ldquoRanks of the common solution tosome quaternion matrix equations with applicationsrdquo Bulletinof Iranian Mathematical Society vol 38 no 1 pp 131ndash157 2012

[29] Q W Wang S W Yu and W Xie ldquoExtreme ranks of realmatrices in solution of the quaternion matrix equation 119860119883119861 =

119862with applicationsrdquoAlgebra Colloquium vol 17 no 2 pp 345ndash360 2010

[30] Q-W Wang S-W Yu and Q Zhang ldquoThe real solutions toa system of quaternion matrix equations with applicationsrdquoCommunications in Algebra vol 37 no 6 pp 2060ndash2079 2009

[31] Q-W Wang S-W Yu and C-Y Lin ldquoExtreme ranks of alinear quaternionmatrix expression subject to triple quaternionmatrix equations with applicationsrdquo Applied Mathematics andComputation vol 195 no 2 pp 733ndash744 2008

[32] Q-W Wang and F Zhang ldquoThe reflexive re-nonnegativedefinite solution to a quaternion matrix equationrdquo ElectronicJournal of Linear Algebra vol 17 pp 88ndash101 2008

[33] Q W Wang X Zhang and Z H He ldquoOn the Hermitianstructures of the solution to a pair of matrix equationsrdquo Linearand Multilinear Algebra vol 61 no 1 pp 73ndash90 2013

[34] X Zhang Q-W Wang and X Liu ldquoInertias and ranks ofsome Hermitian matrix functions with applicationsrdquo CentralEuropean Journal of Mathematics vol 10 no 1 pp 329ndash3512012

[35] Y G Tian and S Z Cheng ldquoThemaximal andminimal ranks of119860 minus 119861119883119862 with applicationsrdquo New York Journal of Mathematicsvol 9 pp 345ndash362 2003

[36] Y G Tian ldquoUpper and lower bounds for ranks of matrixexpressions using generalized inversesrdquo Linear Algebra and itsApplications vol 355 pp 187ndash214 2002

[37] Y H Liu and Y G Tian ldquoMore on extremal ranks of thematrix expressions119860minus119861119883plusmn119883

lowast

119861lowast with statistical applicationsrdquo

Numerical Linear Algebra with Applications vol 15 no 4 pp307ndash325 2008

[38] Y H Liu and Y G Tian ldquoMax-min problems on the ranksand inertias of the matrix expressions 119860 minus 119861119883119862 plusmn (119861119883119862)

lowast withapplicationsrdquo Journal of Optimization Theory and Applicationsvol 148 no 3 pp 593ndash622 2011

[39] T Ando ldquoGeneralized Schur complementsrdquo Linear Algebra andits Applications vol 27 pp 173ndash186 1979

[40] D Carlson E Haynsworth and T Markham ldquoA generalizationof the Schur complement by means of the Moore-Penroseinverserdquo SIAM Journal onAppliedMathematics vol 26 pp 169ndash175 1974

[41] M Fiedler ldquoRemarks on the Schur complementrdquo Linear Algebraand its Applications vol 39 pp 189ndash195 1981

[42] Y G Tian ldquoMore on maximal and minimal ranks of Schurcomplements with applicationsrdquo Applied Mathematics andComputation vol 152 no 3 pp 675ndash692 2004

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 217540 14 pageshttpdxdoiorg1011552013217540

Research ArticleAn Efficient Algorithm for the Reflexive Solution of theQuaternion Matrix Equation 119860119883119861 + 119862119883119867119863 = 119865

Ning Li12 Qing-Wen Wang1 and Jing Jiang3

1 Department of Mathematics Shanghai University Shanghai 200444 China2 School of Mathematics and Quantitative Economics Shandong University of Finance and Economics Jinan 250002 China3Department of Mathematics Qilu Normal University Jinan 250013 China

Correspondence should be addressed to Qing-Wen Wang wqw858yahoocomcn

Received 3 October 2012 Accepted 4 December 2012

Academic Editor P N Shivakumar

Copyright copy 2013 Ning Li et al This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We propose an iterative algorithm for solving the reflexive solution of the quaternion matrix equation 119860119883119861 + 119862119883119867

119863 = 119865 Whenthe matrix equation is consistent over reflexive matrix X a reflexive solution can be obtained within finite iteration steps in theabsence of roundoff errors By the proposed iterative algorithm the least Frobenius norm reflexive solution of the matrix equationcan be derived when an appropriate initial iterative matrix is chosen Furthermore the optimal approximate reflexive solution to agiven reflexive matrix119883

0can be derived by finding the least Frobenius norm reflexive solution of a new corresponding quaternion

matrix equation Finally two numerical examples are given to illustrate the efficiency of the proposed methods

1 Introduction

Throughout the paper the notations R119898times119899 and H119898times119899 repre-sent the set of all 119898 times 119899 real matrices and the set of all 119898 times 119899

matrices over the quaternion algebraH = 1198861+ 119886

2119894+ 119886

3119895+ 119886

4119896 |

1198942

= 1198952

= 1198962

= 119894119895119896 = minus1 1198861 119886

2 119886

3 119886

4isin R We denote the

identity matrix with the appropriate size by 119868 We denote theconjugate transpose the transpose the conjugate the tracethe column space the real part the 119898119899 times 1 vector formedby the vertical concatenation of the respective columnsof a matrix 119860 by 119860

119867

119860119879

119860 tr(119860) 119877(119860)Re(119860) and vec(119860)respectivelyThe Frobenius norm of119860 is denoted by 119860 thatis 119860 = radictr(119860119867119860) Moreover119860otimes119861 and119860⊙119861 stand for theKronecker matrix product and Hadmard matrix product ofthe matrices 119860 and 119861

Let 119876 isin H119899times119899 be a generalized reflection matrix that is119876

2

= 119868 and 119876119867

= 119876 A matrix 119860 is called reflexive withrespect to the generalized reflection matrix 119876 if 119860 = 119876119860119876It is obvious that any matrix is reflexive with respect to 119868Let RH119899times119899

(119876) denote the set of order 119899 reflexive matriceswith respect to 119876 The reflexive matrices with respect to ageneralized reflection matrix 119876 have been widely used inengineering and scientific computations [1 2]

In the field of matrix algebra quaternion matrix equa-tions have received much attention Wang et al [3] gavenecessary and sufficient conditions for the existence andthe representations of P-symmetric and P-skew-symmetricsolutions of quaternion matrix equations 119860

119886119883 = 119862

119886and

119860119887119883119861

119887= 119862

119887 Yuan and Wang [4] derived the expressions of

the least squares 120578-Hermitian solution with the least normand the expressions of the least squares anti-120578-Hermitiansolution with the least norm for the quaternion matrixequation 119860119883119861 + 119862119883119863 = 119864 Jiang and Wei [5] derivedthe explicit solution of the quaternion matrix equation 119883 minus

119860119883119861 = 119862 Li and Wu [6] gave the expressions of symmetricand skew-antisymmetric solutions of the quaternion matrixequations 119860

1119883 = 119862

1and 119883119861

3= 119862

3 Feng and Cheng [7]

gave a clear description of the solution set of the quaternionmatrix equation 119860119883 minus 119883119861 = 0

The iterative method is a very important method tosolve matrix equations Peng [8] constructed a finite iterationmethod to solve the least squares symmetric solutions oflinear matrix equation 119860119883119861 = 119862 Also Peng [9ndash11] presentedseveral efficient iteration methods to solve the constrainedleast squares solutions of linear matrix equations 119860119883119861 =

119862 and 119860119883119861 + 119862119884119863 = 119864 by using Paigersquos algorithm [12]

2 Journal of Applied Mathematics

as the frame method Duan et al [13ndash17] proposed iterativealgorithms for the (Hermitian) positive definite solutionsof some nonlinear matrix equations Ding et al proposedthe hierarchical gradient-based iterative algorithms [18] andhierarchical least squares iterative algorithms [19] for solvinggeneral (coupled) matrix equations based on the hierarchi-cal identification principle [20] Wang et al [21] proposedan iterative method for the least squares minimum-normsymmetric solution of 119860X119861 = 119864 Dehghan and Hajarianconstructed finite iterative algorithms to solve several linearmatrix equations over (anti)reflexive [22ndash24] generalizedcentrosymmetric [25 26] and generalized bisymmetric [2728] matrices Recently Wu et al [29ndash31] proposed iterativealgorithms for solving various complex matrix equations

However to the best of our knowledge there has beenlittle information on iterative methods for finding a solutionof a quaternionmatrix equation Due to the noncommutativemultiplication of quaternions the study of quaternionmatrixequations is more complex than that of real and complexequations Motivated by the work mentioned above andkeeping the interests and wide applications of quaternionmatrices in view (eg [32ndash45]) we in this paper consideran iterative algorithm for the following two problems

Problem 1 For given matrices 119860119862 isin H119898times119899

119861 119863 isin

H119899times119901

119865 isin H119898times119901 and the generalized reflectionmatrix119876 find119883 isin RH119899times119899

(119876) such that

119860119883119861 + 119862119883119867

119863 = 119865 (1)

Problem 2 When Problem 1 is consistent let its solutionset be denoted by 119878

119867 For a given reflexive matrix 119883

0isin

RH119899times119899

(119876) find119883 isin RH119899times119899

(119876) such that10038171003817100381710038171003817119883 minus 119883

0

10038171003817100381710038171003817= min

119883isin119878119867

1003817100381710038171003817119883 minus 1198830

1003817100381710038171003817 (2)

The remainder of this paper is organized as followsIn Section 2 we give some preliminaries In Section 3 weintroduce an iterative algorithm for solving Problem 1 Thenwe prove that the given algorithm can be used to obtain areflexive solution for any initial matrix within finite stepsin the absence of roundoff errors Also we prove that theleast Frobenius norm reflexive solution can be obtained bychoosing a special kind of initial matrix In addition theoptimal reflexive solution of Problem 2 by finding the leastFrobenius norm reflexive solution of a new matrix equationis given In Section 4 we give two numerical examples toillustrate our results In Section 5 we give some conclusionsto end this paper

2 Preliminary

In this section we provide some results which will playimportant roles in this paper First we give a real innerproduct for the space H119898times119899 over the real field R

Theorem 3 In the space H119898times119899 over the field R a real innerproduct can be defined as

⟨119860 119861⟩ = Re [tr (119861119867

119860)] (3)

for 119860 119861 isin H119898times119899 This real inner product space is denoted as(H119898times119899

R ⟨sdot sdot⟩)

Proof (1) For 119860 isin H119898times119899 let 119860 = 1198601+ 119860

2119894 + 119860

3119895 + 119860

4119896 then

⟨119860 119860⟩ = Re [tr (119860119867

119860)]

= tr (119860119879

1119860

1+ 119860

119879

2119860

2+ 119860

119879

3119860

3+ 119860

119879

4119860

4)

(4)

It is obvious that ⟨119860 119860⟩ gt 0 and ⟨119860 119860⟩ = 0 hArr 119860 = 0(2) For 119860 119861 isin H119898times119899 let 119860 = 119860

1+ 119860

2119894 + 119860

3119895 + 119860

4119896 and

119861 = 1198611+ 119861

2119894 + 119861

3119895 + 119861

4119896 then we have

⟨119860 119861⟩ = Re [tr (119861119867

119860)]

= tr (119861119879

1119860

1+ 119861

119879

2119860

2+ 119861

119879

3119860

3+ 119861

119879

4119860

4)

= tr (119860119879

1119861

1+ 119860

119879

2119861

2+ 119860

119879

3119861

3+ 119860

119879

4119861

4)

= Re [tr (119860119867

119861)] = ⟨119861 119860⟩

(5)

(3) For 119860 119861 119862 isin H119898times119899

⟨119860 + 119861 119862⟩ = Re tr [119862119867

(119860 + 119861)]

= Re [tr (119862119867

119860 + 119862119867

119861)]

= Re [tr (119862119867

119860)] + Re [tr (119862119867

119861)]

= ⟨119860 119862⟩ + ⟨119861 119862⟩

(6)

(4) For 119860 119861 isin H119898times119899 and 119886 isin R

⟨119886119860 119861⟩ = Re tr [119861119867

(119886119860)] = Re [tr (119886119861119867

119860)]

= 119886Re [tr (119861119867

119860)] = 119886 ⟨119860 119861⟩

(7)

All the above arguments reveal that the space H119898times119899 overfield R with the inner product defined in (3) is an innerproduct space

Let sdot Δrepresent the matrix norm induced by the inner

product ⟨sdot sdot⟩ For an arbitrary quaternion matrix 119860 isin H119898times119899it is obvious that the following equalities hold

119860Δ= radic⟨119860119860⟩ = radicRe [tr (119860119867119860)] = radictr (119860119867119860) = 119860

(8)

which reveals that the induced matrix norm is exactly theFrobenius norm For convenience we still use sdot to denotethe induced matrix norm

Let 119864119894119895

denote the 119898 times 119899 quaternion matrix whose(119894 119895) entry is 1 and the other elements are zeros In innerproduct space (H119898times119899

R ⟨sdot sdot⟩) it is easy to verify that 119864119894119895 119864

119894119895119894

119864119894119895119895 119864

119894119895119896 119894 = 1 2 119898 119895 = 1 2 119899 is an orthonormal

basis which reveals that the dimension of the inner productspace (H119898times119899

R ⟨sdot sdot⟩) is 4119898119899Next we introduce a real representation of a quaternion

matrix

Journal of Applied Mathematics 3

For an arbitrary quaternion matrix 119872 = 1198721+ 119872

2119894 +

1198723119895+ 119872

4119896 a map 120601(sdot) fromH119898times119899 toR4119898times4119899 can be defined

as

120601 (119872) =

[[[

[

1198721

minus1198722

minus1198723

minus1198724

1198722

1198721

minus1198724

1198723

1198723

1198724

1198721

minus1198722

1198724

minus1198723

1198722

1198721

]]]

]

(9)

Lemma4 (see [41]) Let119872 and119873 be two arbitrary quaternionmatrices with appropriate size The map 120601(sdot) defined by (9)satisfies the following properties

(a) 119872 = 119873 hArr 120601(119872) = 120601(119873)

(b) 120601(119872 + 119873) = 120601(119872) + 120601(119873) 120601(119872119873) = 120601(119872)120601(119873)

120601(119896119872) = 119896120601(119872) 119896 isin R

(c) 120601(119872119867

) = 120601119879

(119872)

(d) 120601(119872) = 119879minus1

119898120601(119872)119879

119899= 119877

minus1

119898120601(119872)119877

119899= 119878

minus1

119898120601(119872)119878

119899

where

119879119905=

[[[

[

0 minus119868119905

0 0

119868119905

0 0 0

0 0 0 119868119905

0 0 minus119868119905

0

]]]

]

119877119905= [

0 minus1198682119905

1198682119905

0]

119878119905=

[[[

[

0 0 0 minus119868119905

0 0 119868119905

0

0 minus119868119905

0 0

119868119905

0 0 0

]]]

]

119905 = 119898 119899

(10)

By (9) it is easy to verify that

1003817100381710038171003817120601 (119872)1003817100381710038171003817 = 2 119872 (11)

Finally we introduce the commutation matrixA commutation matrix 119875(119898 119899) is a 119898119899 times 119898119899 matrix

which has the following explicit form

119875 (119898 119899) =

119898

sum

119894=1

119899

sum

119895=1

119864119894119895otimes 119864

119879

119894119895= [119864

119879

119894119895] 119894=1119898

119895=1119899

(12)

Moreover 119875(119898 119899) is a permutation matrix and 119875(119898 119899) =

119875119879

(119899119898) = 119875minus1

(119899119898) We have the following lemmas on thecommutation matrix

Lemma 5 (see [46]) Let 119860 be a 119898 times 119899 matrix There is acommutation matrix 119875(119898 119899) such that

vec (119860119879

) = 119875 (119898 119899) vec (119860) (13)

Lemma 6 (see [46]) Let 119860 be a 119898 times 119899 matrix and 119861 a 119901 times

119902 matrix There exist two commutation matrices 119875(119898 119901) and119875(119899 119902) such that

119861 otimes 119860 = 119875119879

(119898 119901) (119860 otimes 119861) 119875 (119899 119902) (14)

3 Main Results

31 The Solution of Problem 1 In this subsection we willconstruct an algorithm for solving Problem 1 Then somelemmaswill be given to analyse the properties of the proposedalgorithm Using these lemmas we prove that the proposedalgorithm is convergent

Algorithm 7 (Iterative algorithm for Problem 1)

(1) Choose an initial matrix119883(1) isin RH119899times119899

(119876)(2) Calculate

119877 (1) = 119865 minus 119860119883 (1) 119861 minus 119862119883119867

(1)119863

119875 (1) =1

2(119860

119867

119877 (1) 119861119867

+ 119863119877119867

(1) 119862

+ 119876119860119867

119877 (1) 119861119867

119876 + 119876119863119877119867

(1) 119862119876)

119896 = 1

(15)

(3) If 119877(119896) = 0 then stop and 119883(119896) is the solution ofProblem 1 else if 119877(119896) =0 and 119875(119896) = 0 then stopand Problem 1 is not consistent else 119896 = 119896 + 1

(4) Calculate

119883 (119896) = 119883 (119896 minus 1) +119877 (119896 minus 1)

2

119875 (119896 minus 1)2119875 (119896 minus 1)

119877 (119896) = 119877 (119896 minus 1) minus119877 (119896 minus 1)

2

119875 (119896 minus 1)2

times (119860119875 (119896 minus 1) 119861 + 119862119875119867

(119896 minus 1)119863)

119875 (119896) =1

2(119860

119867

119877 (119896) 119861119867

+ 119863119877119867

(119896) 119862

+ 119876119860119867

119877 (119896) 119861119867

119876 +119876119863119877119867

(119896) 119862119876)

+119877 (119896)

2

119877 (119896 minus 1)2119875 (119896 minus 1)

(16)

(5) Go to Step (3)

Lemma 8 Assume that the sequences 119877(119894) and 119875(119894)

are generated by Algorithm 7 then ⟨119877(119894) 119877(119895)⟩ =

0 and ⟨119875(119894) 119875(119895)⟩ = 0 for 119894 119895 = 1 2

119894 =119895

Proof Since ⟨119877(119894) 119877(119895)⟩ = ⟨119877(119895) 119877(119894)⟩ and⟨119875(119894) 119875(119895)⟩ = ⟨119875(119895) 119875(119894)⟩ for 119894 119895 = 1

2 we only need to prove that ⟨119877(119894)

119877(119895)⟩ = 0 and ⟨119875(119894) 119875(119895)⟩ = 0 for 1 le 119894 lt 119895Now we prove this conclusion by induction

Step 1 We show that

⟨119877 (119894) 119877 (119894 + 1)⟩ = 0

⟨119875 (119894) 119875 (119894 + 1)⟩ = 0 for 119894 = 1 2

(17)

4 Journal of Applied Mathematics

We also prove (17) by induction When 119894 = 1 we have

⟨119877 (1) 119877 (2)⟩

= Re tr [119877119867

(2) 119877 (1)]

= Retr[(119877 (1) minus119877 (1)

2

119875 (1)2(119860119875 (1) 119861 + 119862119875

119867

(1)119863))

119867

times 119877 (1) ]

= 119877 (1)2

minus119877 (1)

2

119875 (1)2

times Re tr [119875119867

(1) (119860119867

119877 (1) 119861119867

+ 119863119877119867

(1) 119862)]

= 119877(1)2

minus119877(1)

2

119875(1)2

timesRetr [119875119867

(1) ((119860119867

119877 (1) 119861119867

+119863119877119867

(1) 119862+119876119860119867

119877 (1)

times 119861119867

119876 + 119876119863119877119867

(1) 119862119876) times (2)minus1

)]

= 119877 (1)2

minus119877 (1)

2

119875 (1)2119875 (1)

2

= 0

(18)

Also we can write

⟨119875 (1) 119875 (2)⟩

= Re tr [119875119867

(2) 119875 (1)]

= Retr[(1

2(119860

119867

119877 (2) 119861119867

+ 119863119877119867

(2) 119862

+ 119876119860119867

119877 (2) 119861119867

119876 + 119876119863119877119867

(2) 119862119876)

+119877 (2)

2

119877 (1)2119875 (1))

119867

119875 (1)]

=119877 (2)

2

119877 (1)2119875 (1)

2

+ Re tr [119875119867

(1) times ((119860119867

119877 (2) 119861119867

+ 119863119877119867

(2) 119862

+ 119876119860119867

119877 (2) 119861119867

119876

+119876119863119877119867

(2) 119862119876) times (2)minus1

)]

=119877 (2)

2

119877 (1)2119875 (1)

2

+ Re tr [119875119867

(1) (119860119867

119877 (2) 119861119867

+ 119863119877119867

(2) 119862)]

=119877 (2)

2

119877 (1)2119875 (1)

2

+ Re tr [119877119867

(2) (119860119875 (1) 119861 + 119862119875119867

(1)119863)]

=119877 (2)

2

119877 (1)2119875 (1)

2

+119875 (1)

2

119877 (1)2Re tr [119877119867

(2) (119877 (1) minus 119877 (2))]

=119877 (2)

2

119877 (1)2119875 (1)

2

minus119875 (1)

2

119877 (1)2119877 (2)

2

= 0

(19)

Now assume that conclusion (17) holds for 1 le 119894 le 119904 minus 1 then

⟨119877 (119904) 119877 (119904 + 1)⟩

= Re tr [119877119867

(119904 + 1) 119877 (119904)]

= Retr[(119877 (119904) minus119877 (119904)

2

119875 (119904)2(119860119875 (119904) 119861 + 119862119875

119867

(119904)119863))

119867

times 119877 (119904) ]

= 119877 (s)2 minus119877 (119904)

2

119875 (119904)2

times Re tr [119875119867

(119904) (119860119867

119877 (119904) 119861119867

+ 119863119877119867

(119904) 119862)]

= 119877 (119904)2

minus119877 (119904)

2

119875 (119904)2

times Re tr [119875119867

(119904) times ((119860119867

119877 (119904) 119861119867

+ 119863119877119867

(119904) 119862

+ 119876119860119867

119877 (119904) 119861119867

119876

+ 119876119863119877119867

(119904) 119862119876) times (2)minus1

)]

= 119877 (119904)2

minus119877 (119904)

2

119875 (119904)2

times Retr[119875119867

(119904) (119875 (119904) minus119877 (119904)

2

119877 (119904 minus 1)2119875 (119904 minus 1))]

= 0

(20)

Journal of Applied Mathematics 5

And it can also be obtained that

⟨119875 (119904) 119875 (119904 + 1)⟩

= Re tr [119875119867

(119904 + 1) 119875 (119904)]

= Retr[(((119860119867

119877 (119904 + 1) 119861119867

+ 119863119877119867

(119904 + 1) 119862

+ 119876119860119867

119877 (119904 + 1) 119861119867

119876

+ 119876119863119877119867

(119904 + 1) 119862119876) times (2)minus1

)

+119877 (119904 + 1)

2

119877 (119904)2

119875(119904))

119867

119875 (119904)]

=119877 (119904 + 1)

2

119877 (119904)2

119875 (119904)2

+ Re tr [119875119867

(119904) (119860119867

119877 (119904 + 1) 119861119867

+ 119863119877119867

(119904 + 1) 119862)]

=119877 (119904 + 1)

2

119877 (119904)2

119875 (119904)2

+ Re tr [119877119867

(119904 + 1) (119860119875 (119904) 119861 + 119862119875119867

(119904)119863)]

=119877 (119904 + 1)

2

119877 (119904)2

119875 (119904)2

+119875 (119904)

2

119877 (119904)2Re tr [119877119867

(119904 + 1) (119877 (119904) minus 119877 (119904 + 1))]

= 0

(21)

Therefore the conclusion (17) holds for 119894 = 1 2 Step 2 Assume that ⟨119877(119894) 119877(119894+119903)⟩ = 0 and ⟨119875(119894) 119875(119894+119903)⟩ = 0

for 119894 ge 1 and 119903 ge 1 We will show that

⟨119877 (119894) 119877 (119894 + 119903 + 1)⟩ = 0 ⟨119875 (119894) 119875 (119894 + 119903 + 1)⟩ = 0

(22)

We prove the conclusion (22) in two substepsSubstep 21 In this substep we show that

⟨119877 (1) 119877 (119903 + 2)⟩ = 0 ⟨119875 (1) 119875 (119903 + 2)⟩ = 0 (23)

It follows from Algorithm 7 that

⟨119877 (1) 119877 (119903 + 2)⟩

= Re tr [119877119867

(119903 + 2) 119877 (1)]

= Retr[(119877 (119903 + 1) minus119877 (119903 + 1)

2

119875 (119903 + 1)2

times (119860119875 (119903 + 1) 119861+119862119875119867

(119903 + 1)119863))

119867

119877 (1)]

= Re tr [119877119867

(119903 + 1) 119877 (1)] minus119877 (119903 + 1)

2

119875 (119903 + 1)2

times Re tr [119875119867

(119903 + 1) (119860119867

119877 (1) 119861119867

+ 119863119877119867

(1) 119862)]

= Re tr [119877119867

(119903 + 1) 119877 (1)] minus119877 (119903 + 1)

2

119875 (119903 + 1)2

times Re tr [119875119867

(119903 + 1) ((119860119867

119877 (1) 119861119867

+ 119863119877119867

(1) 119862

+ 119876119860119867

119877 (1) 119861119867

119876

+ 119876119863119877119867

(1) 119862119876) times (2)minus1

)]

= Re tr [119877119867

(119903 + 1) 119877 (1)]

minus119877 (119903 + 1)

2

119875 (119903 + 1)2Re tr [119875119867

(119903 + 1) 119875 (1)]

= 0

(24)

Also we can write

⟨119875 (1) 119875 (119903 + 2)⟩

= Re tr [119875119867

(119903 + 2) 119875 (1)]

= Retr[( ((119860119867

119877 (119903 + 2) 119861119867

+ 119863119877119867

(119903 + 2) 119862

+119876119860119867

119877 (119903 + 2) 119861119867

119876 + 119876119863119877119867

(119903 + 2) 119862119876)

times (2)minus1

) +119877(119903 + 2)

2

119877(119903 + 1)2119875(119903 + 1))

119867

119875 (1)]

= Re tr [(119860119867

119877 (119903 + 2) 119861119867

+ 119863119877119867

(119903 + 2) 119862)119867

119875 (1)]

+119877 (119903 + 2)

2

119877 (119903 + 1)2Re tr [119875119867

(119903 + 1) 119875 (1)]

= Re tr [119877119867

(119903 + 2) (119860119875 (1) 119861 + 119862119875119867

(1)119863)]

+119877 (119903 + 2)

2

119877 (119903 + 1)2Re tr [119875119867

(119903 + 1) 119875 (1)]

6 Journal of Applied Mathematics

=119875 (1)

2

119877 (1)2Re tr [119877119867

(119903 + 2) (119877 (1) minus 119877 (2))]

+119877 (119903 + 2)

2

119877 (119903 + 1)2Re tr [119875119867

(119903 + 1) 119875 (1)]

= 0

(25)

Substep 22 In this substep we prove the conclusion (22) inStep 2 It follows from Algorithm 7 that

⟨119877 (119894) 119877 (119894 + 119903 + 1)⟩

= Re tr [119877119867

(119894 + 119903 + 1) 119877 (119894)]

= Retr[(119877 (119894 + 119903) minus119877 (119894 + 119903)

2

119875 (119894 + 119903)2

times (119860119875 (119894 + 119903) 119861 + 119862119875119867

(119894 + 119903)119863))

119867

119877 (119894) ]

= Re tr [119877119867

(119894 + 119903) 119877 (119894)] minus119877 (119894 + 119903)

2

119875 (119894 + 119903)2

times Re tr [119875119867

(119894 + 119903) (119860119867

119877 (119894) 119861119867

+ 119863119877119867

(119894) 119862)]

= minus119877 (119894 + 119903)

2

119875 (119894 + 119903)2

times Re tr [119875119867

(119894 + 119903) ((119860119867

119877 (119894) 119861119867

+ 119863119877119867

(119894) 119862

+ 119876119860119867

119877 (119894) 119861119867

119876

+ 119876119863119877119867

(119894) 119862119876) times (2)minus1

)]

= minus119877 (119894 + 119903)

2

119875 (119894 + 119903)2

times Retr[119875119867

(119894 + 119903) (119875 (119894) minus119877 (119894)

2

119877 (119894 minus 1)2119875 (119894 minus 1))]

=119877 (119894 + 119903)

2

119877 (119894)2

119875 (119894 + 119903)2

119877 (119894 minus 1)2Re tr [119875119867

(119894 + 119903) 119875 (119894 minus 1)]

⟨119875 (119894) 119875 (119894 + 119903 + 1)⟩

= Re tr [119875119867

(119894 + 119903 + 1) 119875 (119894)]

= Retr[(1

2(119860

119867

119877 (119894 + 119903 + 1) 119861119867

+ 119863119877119867

(119894 + 119903 + 1) 119862

+ 119876119860119867

119877 (119894 + 119903 + 1) 119861119867

119876

+ 119876119863119877119867

(119894 + 119903 + 1) 119862119876)

+119877(119894 + 119903 + 1)

2

119877(119894 + 119903)2

119875(119894 + 119903))

119867

119875 (119894)]

= Re tr [(119860119867

119877 (119894 + 119903 + 1) 119861119867

+119863119877119867

(119894+119903+1) 119862)119867

119875 (119894)]

= Re tr [119877119867

(119894 + 119903 + 1) (119860119875 (119894)B + 119862119875119867

(119894) 119863)]

=119875 (119894)

2

119877 (119894)2Re tr [119877119867

(119894 + 119903 + 1) (119877 (119894) minus 119877 (119894 + 1))]

=119875 (119894)

2

119877 (119894)2Re tr [119877119867

(119894 + 119903 + 1) 119877 (119894)]

minus119875 (119894)

2

119877 (119894)2Re tr [119877119867

(119894 + 119903 + 1) 119877 (119894 + 1)]

=119875 (119894)

2

119877 (119894 + 119903)2

119877 (119894)2

119877 (119894)2

119875 (119894 + 119903)2

119877 (119894 minus 1)2

times Re tr [119875119867

(119894 + 119903) 119875 (119894 minus 1)]

(26)

Repeating the above process (26) we can obtain

⟨119877 (119894) 119877 (119894 + 119903 + 1)⟩ = sdot sdot sdot = 120572Re tr [119875119867

(119903 + 2) 119875 (1)]

⟨119875 (119894) 119875 (119894 + 119903 + 1)⟩ = sdot sdot sdot = 120573Re tr [119875119867

(119903 + 2) 119875 (1)]

(27)

Combining these two relations with (24) and (25) it impliesthat (22) holds So by the principle of induction we knowthat Lemma 8 holds

Lemma 9 Assume that Problem 1 is consistent and let 119883 isin

RH119899times119899

(119876) be its solution Then for any initial matrix 119883(1) isin

RH119899times119899

(119876) the sequences 119877(119894) 119875(119894) and 119883(119894) generatedby Algorithm 7 satisfy

⟨119875 (119894) 119883 minus 119883 (119894)⟩ = 119877 (119894)2

119894 = 1 2 (28)

Proof We also prove this conclusion by inductionWhen 119894 = 1 it follows from Algorithm 7 that

⟨119875 (1) 119883 minus 119883 (1)⟩

= Re tr [(119883 minus 119883 (1))119867

119875 (1)]

= Re tr [(119883 minus 119883 (1))119867

times ((119860119867

119877 (1) 119861119867

+ 119863119877119867

(1) 119862+119876119860119867

119877 (1) 119861119867

119876

+ 119876119863119877119867

(1) 119862119876) times (2)minus1

) ]

Journal of Applied Mathematics 7

= Re tr [(119883 minus 119883 (1))119867

(119860119867

119877 (1) 119861119867

+ 119863119877119867

(1) 119862)]

= Retr[119877119867

(1) (119860 (119883 minus 119883 (1)) 119861+119862 (119883 minus 119883 (1))119867

119863)]

= Re tr [119877119867

(1) 119877 (1)]

= 119877 (1)2

(29)

This implies that (28) holds for 119894 = 1Now it is assumed that (28) holds for 119894 = 119904 that is

⟨119875 (119904) 119883 minus 119883 (119904)⟩ = 119877 (119904)2

(30)

Then when 119894 = 119904 + 1

⟨119875 (119904 + 1) 119883 minus 119883 (119904 + 1)⟩

= Re tr [(119883 minus 119883 (119904 + 1))119867

119875 (119904 + 1)]

= Retr[(119883 minus 119883 (119904 + 1))119867

times (1

2(119860

119867

119877 (119904 + 1) 119861119867

+ 119863119877119867

(119904 + 1) 119862

+ 119876119860119867

119877 (119904 + 1) 119861119867

119876

+ 119876119863119877119867

(119904 + 1) 119862119876)

+119877(119904 + 1)

2

119877(119904)2

119875 (119904))]

= Re tr [(119883 minus 119883 (119904 + 1))119867

times (119860119867

119877 (119904 + 1) 119861119867

+ 119863119877119867

(119904 + 1) 119862) ]

+119877 (119904 + 1)

2

119877 (119904)2

Re tr [(119883 minus 119883 (119904 + 1))119867

119875 (119904)]

= Re tr [119877119867

(119904 + 1)

times (119860 (119883 minus 119883 (119904 + 1)) 119861

+ 119862 (119883 minus 119883 (119904 + 1))119867

119863)]

+119877 (119904 + 1)

2

119877 (119904)2

Re tr [(119883 minus 119883 (119904))119867

119875 (119904)]

minus Re tr [(119883 (119904 + 1) minus 119883 (119904))119867

119875 (119904)]

= 119877 (119904 + 1)2

+119877 (119904 + 1)

2

119877 (119904)2

times 119877 (119904)2

minus119877 (119904)

2

119875 (119904)2Re tr [119875119867

(119904) 119875 (119904)]

= 119877 (119904 + 1)2

(31)

Therefore Lemma 9 holds by the principle of induction

From the above two lemmas we have the followingconclusions

Remark 10 If there exists a positive number 119894 such that119877(119894) =0 and 119875(119894) = 0 then we can get from Lemma 9 thatProblem 1 is not consistent Hence the solvability of Problem1 can be determined by Algorithm 7 automatically in theabsence of roundoff errors

Theorem 11 Suppose that Problem 1 is consistentThen for anyinitial matrix119883(1) isin RH119899times119899

(119876) a solution of Problem 1 can beobtained within finite iteration steps in the absence of roundofferrors

Proof In Section 2 it is known that the inner productspace (H119898times119901

119877 ⟨sdot sdot⟩) is 4119898119901-dimensional According toLemma 9 if 119877(119894) =0 119894 = 1 2 4119898119901 then we have119875(119894) =0 119894 = 1 2 4119898119901 Hence 119877(4119898119901+1) and 119875(4119898119901+1)

can be computed From Lemma 8 it is not difficult to get

⟨119877 (119894) 119877 (119895)⟩ = 0 for 119894 119895 = 1 2 4119898119901 119894 =119895 (32)

Then 119877(1) 119877(2) 119877(4119898119901) is an orthogonal basis of theinner product space (H119898times119901

119877 ⟨sdot sdot⟩) In addition we can getfrom Lemma 8 that

⟨119877 (119894) 119877 (4119898119901 + 1)⟩ = 0 for 119894 = 1 2 4119898119901 (33)

It follows that 119877(4119898119901+1) = 0 which implies that119883(4119898119901+1)

is a solution of Problem 1

32 The Solution of Problem 2 In this subsection firstly weintroduce some lemmas Then we will prove that the leastFrobenius norm reflexive solution of (1) can be derived bychoosing a suitable initial iterative matrix Finally we solveProblem 2 by finding the least Frobenius norm reflexivesolution of a new-constructed quaternion matrix equation

Lemma 12 (see [47]) Assume that the consistent system oflinear equations119872119910 = 119887 has a solution 119910

0isin 119877(119872

119879

) then 1199100is

the unique least Frobenius norm solution of the system of linearequations

Lemma 13 Problem 1 is consistent if and only if the system ofquaternion matrix equations

119860119883119861 + 119862119883119867

119863 = 119865

119860119876119883119876119861 + 119862119876119883119867

119876119863 = 119865

(34)

8 Journal of Applied Mathematics

is consistent Furthermore if the solution sets of Problem 1 and(34) are denoted by 119878

119867and 119878

1

119867 respectively then we have 119878

119867sube

1198781

119867

Proof First we assume that Problem 1 has a solution 119883 By119860119883119861 + 119862119883

119867

119863 = 119865 and 119876119883119876 = 119883 we can obtain 119860119883119861 +

119862119883119867

119863 = 119865 and 119860119876119883119876119861 + 119862119876119883119867

119876119863 = 119865 which impliesthat 119883 is a solution of quaternion matrix equations (34) and119878119867

sube 1198781

119867

Conversely suppose (34) is consistent Let119883 be a solutionof (34) Set 119883

119886= (119883 + 119876119883119876)2 It is obvious that 119883

119886isin

RH119899times119899

(119876) Now we can write

119860119883119886119861 + 119862119883

119867

119886119863

=1

2[119860 (119883 + 119876119883119876)119861 + 119862(119883 + 119876119883119876)

119867

119863]

=1

2[119860119883119861 + 119862119883

119867

119863 + 119860119876119883119876119861 + 119862119876119883119867

119876119863]

=1

2[119865 + 119865]

= 119865

(35)

Hence 119883119886is a solution of Problem 1 The proof is completed

Lemma 14 The system of quaternion matrix equations (34) isconsistent if and only if the system of real matrix equations

120601 (119860) [119883119894119895]4 times 4

120601 (119861) + 120601 (119862) [119883119894119895]119879

4 times 4

120601 (119863) = 120601 (119865)

120601 (119860) 120601 (119876) [119883119894119895]4 times 4

120601 (119876) 120601 (119861)

+ 120601 (119862) 120601 (119876) [119883119894119895]119879

4 times 4

120601 (119876) 120601 (119863) = 120601 (119865)

(36)

is consistent where 119883119894119895

isin H119899times119899

119894 119895 = 1 2 3 4 are submatri-ces of the unknown matrix Furthermore if the solution sets of(34) and (36) are denoted by 119878

1

119867and 119878

2

119877 respectively then we

have 120601(1198781

119867) sube 119878

2

119877

Proof Suppose that (34) has a solution

119883 = 1198831+ 119883

2119894 + 119883

3119895 + 119883

4119896 (37)

Applying (119886) (119887) and (119888) in Lemma 4 to (34) yields

120601 (119860) 120601 (119883) 120601 (119861) + 120601 (119862) 120601119879

(119883) 120601 (119863) = 120601 (119865)

120601 (119860) 120601 (119876) 120601 (119883) 120601 (119876) 120601 (119861)

+ 120601 (119862) 120601 (119876) 120601119879

(119883) 120601 (119876) 120601 (119863) = 120601 (119865)

(38)

which implies that 120601(119883) is a solution of (36) and 120601(1198781

119867) sube 119878

2

119877

Conversely suppose that (36) has a solution

119883 = [119883119894119895]4 times 4

(39)

By (119889) in Lemma 4 we have that

119879minus1

119898120601 (119860) 119879

119899119883119879

minus1

119899120601 (119861) 119879

119901

+ 119879minus1

119898120601 (119862) 119879

119899119883

119879

119879minus1

119899120601 (119863)119879

119901= 119879

minus1

119898120601 (119865) 119879

119901

119877minus1

119898120601 (119860) 119877

119899119883119877

minus1

119899120601 (119861) 119877

119901

+ 119877minus1

119898120601 (119862) 119877

119899119883

119879

119877minus1

119899120601 (119863) 119877

119901= 119877

minus1

119898120601 (119865) 119877

119901

119878minus1

119898120601 (119860) 119878

119899119883119878

minus1

119899120601 (119861) 119878

119901

+ Sminus1

119898120601 (119862) 119878

119899119883

119879

119878minus1

119899120601 (119863) 119878

119901= 119878

minus1

119898120601 (119865) 119878

119901

119879minus1

119898120601 (119860) 120601 (119876) 119879

119899119883119879

minus1

119899120601 (119876) 120601 (119861) 119879

119901

+ 119879minus1

119898120601 (119862) 120601 (119876) 119879

119899119883

119879

119879minus1

119899120601 (119876) 120601 (119863) 119879

119901= 119879

minus1

119898120601 (119865) 119879

119901

119877minus1

119898120601 (119860) 120601 (119876) 119877

119899119883119877

minus1

119899120601 (119876) 120601 (119861) 119877

119901

+ 119877minus1

119898120601 (119862) 120601 (119876) 119877

119899119883

119879

119877minus1

119899120601 (119876) 120601 (119863) 119877

119901= 119877

minus1

119898120601 (119865) 119877

119901

119878minus1

119898120601 (119860) 120601 (119876) 119878

119899119883119878

minus1

119899120601 (119876) 120601 (119861) 119878

119901

+ 119878minus1

119898120601 (119862) 120601 (119876) 119878

119899119883

119879

119878minus1

119899120601 (119876) 120601 (119863) 119878

119901= 119878

minus1

119898120601 (119865) 119878

119901

(40)

Hence

120601 (119860) 119879119899119883119879

minus1

119899120601 (119861) + 120601 (119862) (119879

119899119883119879

minus1

119899)119879

120601 (119863) = 120601 (119865)

120601 (119860) 119877119899119883119877

minus1

119899120601 (119861) + 120601 (119862) (R

119899119883119877

minus1

119899)119879

120601 (119863) = 120601 (119865)

120601 (119860) 119878119899119883119878

minus1

119899120601 (119861) + 120601 (119862) (119878

119899119883119878

minus1

119899)119879

120601 (119863) = 120601 (119865)

120601 (119860) 120601 (119876) 119879119899119883119879

minus1

119899120601 (119876) 120601 (119861)

+ 120601 (119862) 120601 (119876) (119879119899119883119879

minus1

119899)119879

120601 (119876) 120601 (119863) = 120601 (119865)

120601 (119860) 120601 (119876) 119877119899119883119877

minus1

119899120601 (119876) 120601 (119861)

+ 120601 (119862) 120601 (119876) (119877119899119883119877

minus1

119899)119879

120601 (119876) 120601 (119863) = 120601 (119865)

120601 (119860) 120601 (119876) 119878119899119883119878

minus1

119899120601 (119876) 120601 (119861)

+ 120601 (119862) 120601 (119876) (119878119899119883119878

minus1

119899)119879

120601 (119876) 120601 (119863) = 120601 (119865)

(41)

which implies that 119879119899119883119879

minus1

119899 119877

119899119883119877

minus1

119899 and 119878

119899119883119878

minus1

119899are also

solutions of (36) Thus

1

4(119883 + 119879

119899119883119879

minus1

119899+ 119877

119899119883119877

minus1

119899+ 119878

119899119883119878

minus1

119899) (42)

Journal of Applied Mathematics 9

is also a solution of (36) where

119883 + 119879119899119883119879

minus1

119899+ 119877

119899119883119877

minus1

119899+ 119878

119899119883119878

minus1

119899

= [119883119894119895]4 times 4

119894 119895 = 1 2 3 4

11988311

= 11988311

+ 11988322

+ 11988333

+ 11988344

11988312

= 11988312

minus 11988321

+ 11988334

minus 11988343

11988313

= 11988313

minus 11988324

minus 11988331

+ 11988342

11988314

= 11988314

+ 11988323

minus 11988332

minus 11988341

11988321

= minus11988312

+ 11988321

minus 11988334

+ 11988343

11988322

= 11988311

+ 11988322

+ 11988333

+ 11988344

11988323

= 11988314

+ 11988323

minus 11988332

minus 11988341

11988324

= minus11988313

+ 11988324

+ 11988331

minus 11988342

11988331

= minus11988313

+ 11988324

+ 11988331

minus 11988342

11988332

= minus11988314

minus 11988323

+ 11988332

+ 11988341

11988333

= 11988311

+ 11988322

+ 11988333

+ 11988344

11988334

= 11988312

minus 11988321

+ 11988334

minus 11988343

11988341

= minus11988314

minus 11988323

+ 11988332

+ 11988341

11988342

= 11988313

minus 11988324

minus 11988331

+ 11988342

11988343

= minus11988312

+ 11988321

minus 11988334

+ 11988343

11988344

= 11988311

+ 11988322

+ 11988333

+ 11988344

(43)

Let

119883 =1

4(119883

11+ 119883

22+ 119883

33+ 119883

44)

+1

4(minus119883

12+ 119883

21minus 119883

34+ 119883

43) 119894

+1

4(minus119883

13+ 119883

24+ 119883

31minus 119883

42) 119895

+1

4(minus119883

14minus 119883

23+ 119883

32+ 119883

41) 119896

(44)

Then it is not difficult to verify that

120601 (119883) =1

4(119883 + 119879

119899119883119879

minus1

119899+ 119877

119899119883119877

minus1

119899+ 119878

119899119883119878

minus1

119899) (45)

We have that 119883 is a solution of (34) by (119886) in Lemma 4 Theproof is completed

0 5 10 15 20

0

2

4

Iteration number k

minus12

minus10

minus8

minus6

minus4

minus2

log10r (k)

Figure 1 The convergence curve for the Frobenius norm of theresiduals from Example 18

0 5 10 15 20

0

2

4

Iteration number k

minus12

minus10

minus8

minus6

minus4

minus2

log10r (k)

Figure 2 The convergence curve for the Frobenius norm of theresiduals from Example 19

Lemma 15 There exists a permutation matrix P(4n4n) suchthat (36) is equivalent to

[120601

119879

(119861) otimes 120601 (119860) + (120601119879

(119863) otimes 120601 (119862)) 119875 (4119899 4119899)

120601119879

(119861) 120601 (119876) otimes 120601 (119860) 120601 (119876) + (120601119879

(119863) 120601 (119876) otimes 120601 (119862) 120601 (119876)) 119875 (4119899 4119899)

]

times vec ([119883119894119895]4 times 4

) = [vec (120601 (119865))

vec (120601 (119865))]

(46)

Lemma 15 is easily proven through Lemma 5 So we omitit here

10 Journal of Applied Mathematics

Theorem 16 When Problem 1 is consistent let its solution setbe denoted by 119878

119867 If

∘

119883 isin 119878119867 and

∘

119883 can be expressed as

∘

119883 = 119860119867

119866119861119867

+ 119863119866119867

119862 + 119876119860119867

119866119861119867

119876

+ 119876119863119866119867

119862119876 119866 isin H119898times119901

(47)

then∘

119883 is the least Frobenius norm solution of Problem 1

Proof By (119886) (119887) and (119888) in Lemmas 4 5 and 6 we have that

vec(120601(

∘

119883))

= vec (120601119879

(119860) 120601 (119866) 120601119879

(119861) + 120601 (119863) 120601119879

(119866) 120601 (119862)

+ 120601 (119876) 120601119879

(119860) 120601 (119866) 120601119879

(119861) 120601 (119876)

+ 120601 (119876) 120601 (119863) 120601119879

(119866) 120601 (119862) 120601 (119876))

= [120601 (119861) otimes 120601119879

(119860) + (120601119879

(119862) otimes 120601 (119863)) 119875 (4119898 4119901)

120601 (119876) 120601 (119861) otimes 120601 (119876) 120601119879

(119860)

+ (120601 (119876) 120601119879

(119862) otimes 120601 (119876) 120601 (119863)) 119875 (4119898 4119901)]

times [vec (120601 (119866))

vec (120601 (119866))]

= [120601

119879

(119861) otimes 120601(119860) + (120601119879

(119863) otimes 120601(119862))119875(4119899 4119899)

120601119879

(119861)120601(119876) otimes 120601(119860)120601(119876) + (120601119879

(119863)120601(119876) otimes 120601(119862)120601(119876))119875(4119899 4119899)

]

119879

times [vec (120601 (119866))

vec (120601 (119866))]

isin 119877 [120601

119879

(119861) otimes 120601 (119860)

120601119879

(119861) 120601 (119876) otimes 120601 (119860) 120601 (119876)

+ (120601119879

(119863) otimes 120601 (119862)) 119875 (4119899 4119899)

+ (120601119879

(119863) 120601 (119876) otimes 120601 (119862) 120601 (119876)) 119875 (4119899 4119899)]

119879

(48)

By Lemma 12 120601(∘

119883) is the least Frobenius norm solution ofmatrix equations (46)

Noting (11) we derive from Lemmas 13 14 and 15 that∘

119883

is the least Frobenius norm solution of Problem 1

From Algorithm 7 it is obvious that if we consider

119883(1) = 119860119867

119866119861119867

+ 119863119866119867

119862 + 119876119860119867

119866119861119867

119876

+ 119876119863119866119867

119862119876 119866 isin H119898times119901

(49)

then all119883(119896) generated by Algorithm 7 can be expressed as

119883 (119896) = 119860119867

119866119896119861

119867

+ 119863119866119867

119896119862 + 119876119860

119867

119866119896119861

119867

119876

+ 119876119863119866119867

119896119862119876 119866

119896isin H

119898times119901

(50)

Using the above conclusion and consideringTheorem 16we propose the following theorem

Theorem 17 Suppose that Problem 1 is consistent Let theinitial iteration matrix be

119883 (1) = 119860119867

119866119861H+ 119863119866

119867

119862 + 119876119860119867

119866119861119867

119876 + 119876119863119866119867

119862119876 (51)

where 119866 is an arbitrary quaternion matrix or especially119883(1) = 0 then the solution 119883

lowast generated by Algorithm 7 isthe least Frobenius norm solution of Problem 1

Now we study Problem 2 When Problem 1 is consistentthe solution set of Problem 1 denoted by 119878

119867is not empty

Then For a given reflexive matrix1198830isin RH119899times119899

(119876)

119860119883119861 + 119862119883119867

119863 = 119865 lArrrArr 119860(119883 minus 1198830) 119861 + 119862(119883 minus 119883

0)119867

119863

= 119865 minus 1198601198830119861 minus 119862119883

119867

0119863

(52)

Let 119883 = 119883 minus 1198830and 119865 = 119865 minus 119860119883

0119861 minus 119862119883

119867

0119863 then Problem

2 is equivalent to finding the least Frobenius norm reflexivesolution of the quaternion matrix equation

119860119883119861 + 119862119883119867

119863 = 119865 (53)

By using Algorithm 7 let the initial iteration matrix 119883(1) =

119860119867

119866119861119867

+ 119863119866119867

119862 + 119876119860119867

119866119861119867

119876 + 119876119863119866119867

119862Q where 119866 is anarbitrary quaternion matrix in H119898times119901 or especially 119883(1) = 0we can obtain the least Frobenius norm reflexive solution119883

lowast

of (53) Then we can obtain the solution of Problem 2 whichis

119883 = 119883lowast

+ 1198830 (54)

Journal of Applied Mathematics 11

4 Examples

In this section we give two examples to illustrate the effi-ciency of the theoretical results

Example 18 Consider the quaternion matrix equation

119860119883119861 + 119862119883119867

119863 = 119865 (55)with

119860 = [1 + 119894 + 119895 + 119896 2 + 2119895 + 2119896 119894 + 3119895 + 3119896 2 + 119894 + 4119895 + 4119896

3 + 2 119894 + 2 119895 + 119896 3 + 3119894 + 119895 1 + 7119894 + 3119895 + 119896 6 minus 7119894 + 2119895 minus 4119896]

119861 =

[[[

[

minus2 + 119894 + 3119895 + 119896 3 + 3119894 + 4119895 + 119896

5 minus 4119894 minus 6119895 minus 5119896 minus1 minus 5119894 + 4119895 + 3119896

minus1 + 2119894 + 119895 + 119896 2 + 2119895 + 6119896

3 minus 2119894 + 119895 + 119896 4 + 119894 minus 2119895 minus 4119896

]]]

]

119862 = [1 + 119894 + 119895 + 119896 2 + 2119894 + 2119895 119896 2 + 2119894 + 2119895 + 119896

minus1 + 119894 + 3119895 + 3119896 minus2 + 3119894 + 4119895 + 2119896 6 minus 2119894 + 6119895 + 4119896 3 + 119894 + 119895 + 5119896]

119863 =

[[[

[

minus1 + 4119895 + 2119896 1 + 2119894 + 3119895 + 3119896

7 + 6119894 + 5119895 + 6119896 3 + 7119894 minus 119895 + 9119896

4 + 119894 + 6119895 + 119896 7 + 2119894 + 9119895 + 119896

1 + 119894 + 3119895 minus 3119896 1 + 3119894 + 2119895 + 2119896

]]]

]

119865 = [1 + 3119894 + 2119895 + 119896 2 minus 2119894 + 3119895 + 3119896

minus1 + 4119894 + 2119895 + 119896 minus2 + 2119894 minus 1119895]

(56)

We apply Algorithm 7 to find the reflexive solution with res-pect to the generalized reflection matrix

119876 =

[[[

[

028 0 096119896 0

0 minus1 00 0

minus096119896 0 minus028 0

0 0 0 minus1

]]]

]

(57)

For the initial matrix119883(1) = 119876 we obtain a solution that is

119883lowast

= 119883 (20)

=[[

[

027257 minus 023500119894 + 0034789119895 + 0054677119896 0085841 minus 0064349119894 + 010387119895 minus 026871119896

0028151 minus 0013246119894 minus 0091097119895 + 0073137119896 minus046775 + 0048363119894 minus 0015746119895 + 021642119896

0043349 + 0079178119894 + 0085167119895 minus 084221119896 035828 minus 013849119894 minus 0085799119895 + 011446119896

013454 minus 0028417119894 minus 0063010119895 minus 010574119896 010491 minus 0039417119894 + 018066119895 minus 0066963119896

minus0043349 + 0079178119894 + 0085167119895 + 084221119896 minus0014337 minus 014868119894 minus 0011146119895 minus 000036397119896

0097516 + 012146119894 minus 0017662119895 minus 0037534119896 minus0064483 + 0070885119894 minus 0089331119895 + 0074969119896

minus021872 + 018532119894 + 0011399119895 + 0029390119896 000048529 + 0014861119894 minus 019824119895 minus 0019117119896

minus014099 + 0084013119894 minus 0037890119895 minus 017938119896 minus063928 minus 0067488119894 + 0042030119895 + 010106119896

]]

]

(58)

with corresponding residual 119877(20) = 22775 times 10minus11 The

convergence curve for the Frobenius norm of the residuals119877(119896) is given in Figure 1 where 119903(119896) = 119877(119896)

Example 19 In this example we choose the matrices119860 119861 119862 119863 119865 and 119876 as same as in Example 18 Let

1198830= [

[

1 minus036119894 minus 075119896 0 minus05625119894 minus 048119896

036119894 128 048119895 minus096119895

0 1 minus 048119895 1 064 minus 075119895

048119896 096119895 064 072

]

]

isin RH119899times119899

(119876)

(59)

In order to find the optimal approximation reflexive solutionto the given matrix 119883

0 let 119883 = 119883 minus 119883

0and 119865 = 119865 minus

1198601198830119861 minus 119862119883

119867

0119863 Now we can obtain the least Frobenius

norm reflexive solution119883lowast of the quaternionmatrix equation

119860119883119861 + 119862 119883119867

119863 = 119865 by choosing the initial matrix 119883(1) = 0that is

12 Journal of Applied Mathematics

119883lowast

= 119883 (20)

=

[[[

[

minus047933 + 0021111119894 + 011703119895 minus 0066934119896 minus052020 minus 019377119894 + 017420119895 minus 059403119896

minus031132 minus 011705119894 minus 033980119895 + 011991119896 minus086390 minus 021715119894 minus 0037089119895 + 040609119896

minus024134 minus 0025941119894 minus 031930119895 minus 026961119896 minus044552 + 013065119894 + 014533119895 + 039015119896

minus011693 minus 027043119894 + 022718119895 minus 00000012004119896 minus044790 minus 031260119894 minus 10271119895 + 027275119896

024134 minus 0025941119894 minus 031930119895 + 026961119896 0089010 minus 067960119894 + 043044119895 minus 0045772119896

minus0089933 minus 025485119894 + 0087788119895 minus 023349119896 minus017004 minus 037655119894 + 080481119895 + 037636119896

minus063660 + 016515119894 minus 013216119895 + 0073845119896 minus0034329 + 032283119894 + 050970119895 minus 0066757119896

000000090029 + 017038119894 + 020282119895 minus 0087697119896 minus044735 + 021846119894 minus 021469119895 minus 015347119896

]]]

]

(60)

with corresponding residual 119877(20) = 43455 times 10minus11 The

convergence curve for the Frobenius norm of the residuals119877(119896) is given in Figure 2 where 119903(119896) = 119877(119896)

Therefore the optimal approximation reflexive solutionto the given matrix119883

0is

119883 = 119883lowast

+ 1198830

=

[[[

[

052067 + 0021111119894 + 011703119895 minus 0066934119896 minus052020 minus 055377119894 + 017420119895 minus 13440119896

minus031132 + 024295119894 minus 033980119895 + 011991119896 041610 minus 021715119894 minus 0037089119895 + 040609119896

minus024134 minus 0025941119894 minus 031930119895 minus 026961119896 055448 + 013065119894 minus 033467119895 + 039015119896

minus011693 minus 027043119894 + 022718119895 + 048000119896 minus044790 minus 031260119894 minus 0067117119895 + 027275119896

024134 minus 0025941119894 minus 031930119895 + 026961119896 0089010 minus 12421119894 + 043044119895 minus 052577119896

minus0089933 minus 025485119894 + 056779119895 minus 023349119896 minus017004 minus 037655119894 minus 015519119895 + 037636119896

036340 + 016515119894 minus 013216119895 + 0073845119896 060567 + 032283119894 minus 024030119895 minus 0066757119896

064000 + 017038119894 + 020282119895 minus 0087697119896 027265 + 021846119894 minus 021469119895 minus 015347119896

]]]

]

(61)

The results show that Algorithm 7 is quite efficient

5 Conclusions

In this paper an algorithm has been presented for solving thereflexive solution of the quaternion matrix equation 119860119883119861 +

119862119883119867

119863 = 119865 By this algorithm the solvability of the problemcan be determined automatically Also when the quaternionmatrix equation 119860119883119861 + 119862119883

119867

119863 = 119865 is consistent overreflexive matrix 119883 for any reflexive initial iterative matrixa reflexive solution can be obtained within finite iterationsteps in the absence of roundoff errors It has been proventhat by choosing a suitable initial iterative matrix we canderive the least Frobenius norm reflexive solution of thequaternion matrix equation 119860119883119861 + 119862119883

119867

119863 = 119865 throughAlgorithm 7 Furthermore by using Algorithm 7 we solvedProblem 2 Finally two numerical examples were given toshow the efficiency of the presented algorithm

Acknowledgments

This research was supported by Grants from InnovationProgram of Shanghai Municipal Education Commission(13ZZ080) the National Natural Science Foundation ofChina (11171205) theNatural Science Foundation of Shanghai

(11ZR1412500) the Discipline Project at the correspondinglevel of Shanghai (A 13-0101-12-005) and Shanghai LeadingAcademic Discipline Project (J50101)

References

[1] H C Chen and A H Sameh ldquoAmatrix decompositionmethodfor orthotropic elasticity problemsrdquo SIAM Journal on MatrixAnalysis and Applications vol 10 no 1 pp 39ndash64 1989

[2] H C Chen ldquoGeneralized reflexive matrices special propertiesand applicationsrdquo SIAM Journal on Matrix Analysis and Appli-cations vol 19 no 1 pp 140ndash153 1998

[3] Q W Wang H X Chang and C Y Lin ldquo119875-(skew)symmetriccommon solutions to a pair of quaternion matrix equationsrdquoApplied Mathematics and Computation vol 195 no 2 pp 721ndash732 2008

[4] S F Yuan and Q W Wang ldquoTwo special kinds of least squaressolutions for the quaternionmatrix equation119860119883119861+119862119883119863 = 119864rdquoElectronic Journal of Linear Algebra vol 23 pp 257ndash274 2012

[5] T S Jiang and M S Wei ldquoOn a solution of the quaternionmatrix equation 119883 minus 119860119883119861 = 119862 and its applicationrdquo ActaMathematica Sinica (English Series) vol 21 no 3 pp 483ndash4902005

[6] Y T Li and W J Wu ldquoSymmetric and skew-antisymmetricsolutions to systems of real quaternion matrix equationsrdquoComputers amp Mathematics with Applications vol 55 no 6 pp1142ndash1147 2008

Journal of Applied Mathematics 13

[7] L G Feng and W Cheng ldquoThe solution set to the quaternionmatrix equation119860119883minus119883119861 = 0rdquo Algebra Colloquium vol 19 no1 pp 175ndash180 2012

[8] Z Y Peng ldquoAn iterative method for the least squares symmetricsolution of the linear matrix equation 119860119883119861 = 119862rdquo AppliedMathematics and Computation vol 170 no 1 pp 711ndash723 2005

[9] Z Y Peng ldquoA matrix LSQR iterative method to solve matrixequation 119860119883119861 = 119862rdquo International Journal of ComputerMathematics vol 87 no 8 pp 1820ndash1830 2010

[10] Z Y Peng ldquoNew matrix iterative methods for constraintsolutions of the matrix equation 119860119883119861 = 119862rdquo Journal ofComputational and Applied Mathematics vol 235 no 3 pp726ndash735 2010

[11] Z Y Peng ldquoSolutions of symmetry-constrained least-squaresproblemsrdquo Numerical Linear Algebra with Applications vol 15no 4 pp 373ndash389 2008

[12] C C Paige ldquoBidiagonalization of matrices and solutions of thelinear equationsrdquo SIAM Journal on Numerical Analysis vol 11pp 197ndash209 1974

[13] X F Duan A P Liao and B Tang ldquoOn the nonlinear matrixequation 119883 minus sum

119898

119894=1119860

119909

2119886119894119883

120575119894119860119894= 119876rdquo Linear Algebra and Its

Applications vol 429 no 1 pp 110ndash121 2008[14] X F Duan and A P Liao ldquoOn the existence of Hermitian

positive definite solutions of thematrix equation119883119904

+119860lowast

119883minus119905

119860 =

119876rdquo Linear Algebra and Its Applications vol 429 no 4 pp 673ndash687 2008

[15] X FDuan andA P Liao ldquoOn the nonlinearmatrix equation119883+

119860lowast

119883minus119902

119860 = 119876(119902 ge 1)rdquo Mathematical and Computer Modellingvol 49 no 5-6 pp 936ndash945 2009

[16] X F Duan and A P Liao ldquoOn Hermitian positive definitesolution of the matrix equation119883minussum

119898

119894=1119860

lowast

119894119883

119903

119860119894= 119876rdquo Journal

of Computational and Applied Mathematics vol 229 no 1 pp27ndash36 2009

[17] X F Duan CM Li and A P Liao ldquoSolutions and perturbationanalysis for the nonlinear matrix equation119883+sum

119898

119894=1119860

lowast

119894119883

minus1

119860119894=

119868rdquo Applied Mathematics and Computation vol 218 no 8 pp4458ndash4466 2011

[18] F Ding P X Liu and J Ding ldquoIterative solutions of thegeneralized Sylvester matrix equations by using the hierarchicalidentification principlerdquo Applied Mathematics and Computa-tion vol 197 no 1 pp 41ndash50 2008

[19] F Ding and T Chen ldquoIterative least-squares solutions ofcoupled Sylvester matrix equationsrdquo Systems amp Control Lettersvol 54 no 2 pp 95ndash107 2005

[20] FDing andTChen ldquoHierarchical gradient-based identificationof multivariable discrete-time systemsrdquo Automatica vol 41 no2 pp 315ndash325 2005

[21] M H Wang M S Wei and S H Hu ldquoAn iterative method forthe least-squares minimum-norm symmetric solutionrdquo CMESComputer Modeling in Engineering amp Sciences vol 77 no 3-4pp 173ndash182 2011

[22] M Dehghan and M Hajarian ldquoAn iterative algorithm for thereflexive solutions of the generalized coupled Sylvester matrixequations and its optimal approximationrdquoAppliedMathematicsand Computation vol 202 no 2 pp 571ndash588 2008

[23] M Dehghan and M Hajarian ldquoFinite iterative algorithmsfor the reflexive and anti-reflexive solutions of the matrixequation119860

1119883

1119861

1+119860

2119883

2119861

2= 119862rdquoMathematical and Computer

Modelling vol 49 no 9-10 pp 1937ndash1959 2009[24] M Hajarian and M Dehghan ldquoSolving the generalized

Sylvester matrix equation119901

sum

119894=1

119860119894119883119861

119894+

119902

sum

119895=1

119862119895119884119863

119895= 119864 over

reflexive and anti-reflexive matricesrdquo International Journal ofControl and Automation vol 9 no 1 pp 118ndash124 2011

[25] M Dehghan and M Hajarian ldquoAn iterative algorithm forsolving a pair of matrix equations 119860119884119861 = 119864 119862119884119863 = 119865

over generalized centro-symmetric matricesrdquo Computers ampMathematics with Applications vol 56 no 12 pp 3246ndash32602008

[26] M Dehghan and M Hajarian ldquoAnalysis of an iterative algo-rithm to solve the generalized coupled Sylvester matrix equa-tionsrdquo Applied Mathematical Modelling vol 35 no 7 pp 3285ndash3300 2011

[27] M Dehghan and M Hajarian ldquoThe general coupled matrixequations over generalized bisymmetric matricesrdquo Linear Alge-bra and Its Applications vol 432 no 6 pp 1531ndash1552 2010

[28] M Dehghan and M Hajarian ldquoAn iterative method for solvingthe generalized coupled Sylvester matrix equations over gener-alized bisymmetric matricesrdquo Applied Mathematical Modellingvol 34 no 3 pp 639ndash654 2010

[29] A G Wu B Li Y Zhang and G R Duan ldquoFinite iterativesolutions to coupled Sylvester-conjugate matrix equationsrdquoApplied Mathematical Modelling vol 35 no 3 pp 1065ndash10802011

[30] A G Wu L L Lv and G R Duan ldquoIterative algorithmsfor solving a class of complex conjugate and transpose matrixequationsrdquo Applied Mathematics and Computation vol 217 no21 pp 8343ndash8353 2011

[31] AGWu L L Lv andMZHou ldquoFinite iterative algorithms forextended Sylvester-conjugate matrix equationsrdquo Mathematicaland Computer Modelling vol 54 no 9-10 pp 2363ndash2384 2011

[32] N L Bihan and J Mars ldquoSingular value decomposition ofmatrices of Quaternions matrices a new tool for vector-sensorsignal processingrdquo Signal Process vol 84 no 7 pp 1177ndash11992004

[33] N L Bihan and S J Sangwine ldquoJacobi method for quaternionmatrix singular value decompositionrdquoAppliedMathematics andComputation vol 187 no 2 pp 1265ndash1271 2007

[34] F O Farid Q W Wang and F Zhang ldquoOn the eigenvalues ofquaternion matricesrdquo Linear and Multilinear Algebra vol 59no 4 pp 451ndash473 2011

[35] S de Leo and G Scolarici ldquoRight eigenvalue equation inquaternionic quantummechanicsrdquo Journal of Physics A vol 33no 15 pp 2971ndash2995 2000

[36] S J Sangwine and N L Bihan ldquoQuaternion singular valuedecomposition based on bidiagonalization to a real or com-plex matrix using quaternion Householder transformationsrdquoApplied Mathematics and Computation vol 182 no 1 pp 727ndash738 2006

[37] C C Took D P Mandic and F Zhang ldquoOn the unitarydiagonalisation of a special class of quaternion matricesrdquoApplied Mathematics Letters vol 24 no 11 pp 1806ndash1809 2011

[38] QWWang H X Chang and Q Ning ldquoThe common solutionto six quaternion matrix equations with applicationsrdquo AppliedMathematics and Computation vol 198 no 1 pp 209ndash2262008

[39] Q W Wang J W van der Woude and H X Chang ldquoA systemof real quaternion matrix equations with applicationsrdquo LinearAlgebra and Its Applications vol 431 no 12 pp 2291ndash2303 2009

[40] Q W Wang S W Yu and W Xie ldquoExtreme ranks of realmatrices in solution of the quaternion matrix equation 119860119883119861 =

119862with applicationsrdquoAlgebra Colloquium vol 17 no 2 pp 345ndash360 2010

14 Journal of Applied Mathematics

[41] Q W Wang and J Jiang ldquoExtreme ranks of (skew-)Hermitiansolutions to a quaternion matrix equationrdquo Electronic Journal ofLinear Algebra vol 20 pp 552ndash573 2010

[42] Q W Wang X Liu and S W Yu ldquoThe common bisymmetricnonnegative definite solutions with extreme ranks and inertiasto a pair of matrix equationsrdquo Applied Mathematics and Com-putation vol 218 no 6 pp 2761ndash2771 2011

[43] Q W Wang Y Zhou and Q Zhang ldquoRanks of the commonsolution to six quaternion matrix equationsrdquo Acta Mathemati-cae Applicatae Sinica (English Series) vol 27 no 3 pp 443ndash4622011

[44] F Zhang ldquoGersgorin type theorems for quaternionic matricesrdquoLinear Algebra and Its Applications vol 424 no 1 pp 139ndash1532007

[45] F Zhang ldquoQuaternions and matrices of quaternionsrdquo LinearAlgebra and Its Applications vol 251 pp 21ndash57 1997

[46] R A Horn and C R Johnson Topics in Matrix AnalysisCambridge University Press Cambridge UK 1991

[47] Y X Peng X Y Hu and L Zhang ldquoAn iterationmethod for thesymmetric solutions and the optimal approximation solutionof the matrix equation 119860119883119861 = 119862rdquo Applied Mathematics andComputation vol 160 no 3 pp 763ndash777 2005

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2013 Article ID 961568 6 pageshttpdxdoiorg1011552013961568

Research ArticleThe Optimization on Ranks and Inertias of a QuadraticHermitian Matrix Function and Its Applications

Yirong Yao

Department of Mathematics Shanghai University Shanghai 200444 China

Correspondence should be addressed to Yirong Yao yryaostaffshueducn

Received 3 December 2012 Accepted 9 January 2013

Academic Editor Yang Zhang

Copyright copy 2013 Yirong Yao This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We solve optimization problems on the ranks and inertias of the quadratic Hermitian matrix function 119876 minus 119883119875119883lowast subject to a

consistent system of matrix equations 119860119883 = 119862 and 119883119861 = 119863 As applications we derive necessary and sufficient conditions for thesolvability to the systems of matrix equations and matrix inequalities 119860119883 = 119862119883119861 = 119863 and 119883119875119883lowast = (gt lt ge le)119876 in the Lownerpartial ordering to be feasible respectively The findings of this paper widely extend the known results in the literature

1 Introduction

Throughout this paper we denote the complex number fieldby C The notations C119898times119899 and C119898times119898

ℎstand for the sets

of all 119898 times 119899 complex matrices and all 119898 times 119898 complexHermitian matrices respectivelyThe identity matrix with anappropriate size is denoted by 119868 For a complex matrix 119860the symbols 119860lowast and 119903(119860) stand for the conjugate transposeand the rank of119860 respectivelyTheMoore-Penrose inverse of119860 isin C119898times119899 denoted by119860dagger is defined to be the unique solution119883 to the following four matrix equations

(1) 119860119883119860 = 119860 (2) 119883119860119883 = 119883

(3) (119860119883)lowast

= 119860119883 (4) (119883119860)lowast

= 119883119860

(1)

Furthermore 119871119860and 119877

119860stand for the two projectors 119871

119860=

119868 minus 119860dagger

119860 and 119877119860= 119868 minus 119860119860

dagger induced by 119860 respectively It isknown that 119871

119860= 119871lowast

119860and 119877

119860= 119877lowast

119860 For 119860 isin C119898times119898

ℎ its inertia

I119899(119860) = (119894

+(119860) 119894minus(119860) 1198940(119860)) (2)

is the triple consisting of the numbers of the positive nega-tive and zero eigenvalues of 119860 counted with multiplicitiesrespectively It is easy to see that 119894

+(119860) + 119894

minus(119860) = 119903(119860) For

two Hermitian matrices 119860 and 119861 of the same sizes we say119860 gt 119861 (119860 ge 119861) in the Lowner partial ordering if 119860 minus 119861 ispositive (nonnegative) definite

The investigation on maximal and minimal ranks andinertias of linear and quadratic matrix function is activein recent years (see eg [1ndash24]) Tian [21] considered themaximal and minimal ranks and inertias of the Hermitianquadratic matrix function

ℎ (119883) = 119860119883119861119883lowast

119860lowast

+ 119860119883119862 + 119862lowast

119883lowast

119860lowast

+ 119863 (3)

where 119861 and 119863 are Hermitian matrices Moreover Tian [22]investigated the maximal and minimal ranks and inertias ofthe quadratic Hermitian matrix function

119891 (119883) = 119876 minus 119883119875119883lowast (4)

such that 119860119883 = 119862The goal of this paper is to give the maximal andminimal

ranks and inertias of the matrix function (4) subject to theconsistent system of matrix equations

119860119883 = 119862 119883119861 = 119863 (5)

where 119876 isin C119899times119899ℎ

119875 isin C119901times119901

ℎare given complex matrices

As applications we consider the necessary and sufficient

2 Journal of Applied Mathematics

conditions for the solvability to the systems of matrix equa-tions and inequality

119860119883 = 119862 119883119861 = 119863 119883119875119883lowast

= 119876

119860119883 = 119862 119883119861 = 119863 119883119875119883lowast

gt 119876

119860119883 = 119862 119883119861 = 119863 119883119875119883lowast

lt 119876

119860119883 = 119862 119883119861 = 119863 119883119875119883lowast

ge 119876

119860119883 = 119862 119883119861 = 119863 119883119875119883lowast

le 119876

(6)

in the Lowner partial ordering to be feasible respectively

2 The Optimization on Ranks and Inertiasof (4) Subject to (5)

In this section we consider the maximal and minimal ranksand inertias of the quadratic Hermitian matrix function (4)subject to (5) We begin with the following lemmas

Lemma 1 (see [3]) Let 119860 isin C119898times119898ℎ

119861 isin C119898times119901 and 119862 isin C119902times119898

be given and denote

1198751= [

119860 119861

119861lowast

0] 119875

2= [

119860 119862lowast

119862 0]

1198753= [

119860 119861 119862lowast

119861lowast

0 0] 119875

4= [

119860 119861 119862lowast

119862 0 0]

(7)

Then

max119884isinC119901times119902

119903 [119860 minus 119861119884119862 minus (119861119884119862)lowast

]

= min 119903 [119860 119861 119862lowast

] 119903 (1198751) 119903 (119875

2)

min119884isinC119901times119902

119903 [119860 minus 119861119884119862 minus (119861119884119862)lowast

]

= 2119903 [119860 119861 119862lowast

]

+max 119908++ 119908minus 119892++ 119892minus 119908++ 119892minus 119908minus+ 119892+

max119884isinC119901times119902

119894plusmn[119860 minus 119861119884119862 minus (119861119884119862)

lowast

] = min 119894plusmn(1198751) 119894plusmn(1198752)

min119884isinC119901times119902

119894plusmn[119860 minus 119861119884119862 minus (119861119884119862)

lowast

]

= 119903 [119860 119861 119862lowast

] +max 119894plusmn(1198751) minus 119903 (119875

3) 119894plusmn(1198752) minus 119903 (119875

4)

(8)

where

119908plusmn= 119894plusmn(1198751) minus 119903 (119875

3) 119892

plusmn= 119894plusmn(1198752) minus 119903 (119875

4) (9)

Lemma 2 (see [4]) Let 119860 isin C119898times119899 119861 isin C119898times119896 119862 isin C119897times119899 119863 isin

C119898times119901 119864 isin C119902times119899 119876 isin C1198981times119896 and 119875 isin C119897times1198991 be given Then

(1) 119903 (119860) + 119903 (119877119860119861) = 119903 (119861) + 119903 (119877

119861119860) = 119903 [119860 119861]

(2) 119903 (119860) + 119903 (119862119871119860) = 119903 (119862) + 119903 (119860119871

119862) = 119903 [

119860

119862]

(3) 119903 (119861) + 119903 (119862) + 119903 (119877119861119860119871119862) = 119903 [

119860 119861

119862 0]

(4) 119903 (119875) + 119903 (119876) + 119903 [119860 119861119871

119876

119877119875119862 0

] = 119903[

[

119860 119861 0

119862 0 119875

0 119876 0

]

]

(5) 119903 [119877119861119860119871119862119877119861119863

119864119871119862

0] + 119903 (119861) + 119903 (119862) = 119903[

[

119860 119863 119861

119864 0 0

119862 0 0

]

]

(10)

Lemma 3 (see [23]) Let 119860 isin C119898times119898ℎ

119861 isin C119898times119899 119862 isin C119899times119899ℎ

119876 isin C119898times119899 and 119875 isin C119901times119899 be given and 119879 isin C119898times119898 benonsingular Then

(1) 119894plusmn(119879119860119879

lowast

) = 119894plusmn(119860)

(2) 119894plusmn[119860 0

0 119862] = 119894plusmn(119860) + 119894

plusmn(119862)

(3) 119894plusmn[0 119876

119876lowast

0] = 119903 (119876)

(4) 119894plusmn[

119860 119861119871119875

119871119875119861lowast

0] + 119903 (119875) = 119894

plusmn

[

[

119860 119861 0

119861lowast

0 119875lowast

0 119875 0

]

]

(11)

Lemma 4 Let 119860 119862 119861 and 119863 be given Then the followingstatements are equivalent

(1) System (5) is consistent

(2) Let

119903 [119860 119862] = 119903 (119860) [119863

119861] = 119903 (119861) 119860119863 = 119862119861 (12)

In this case the general solution can be written as

119883 = 119860dagger

119862 + 119871119860119863119861dagger

+ 119871119860119881119877119861 (13)

where 119881 is an arbitrary matrix over C with appropriate size

Now we give the fundamental theorem of this paper

Journal of Applied Mathematics 3

Theorem5 Let119891(119883) be as given in (4) and assume that119860119883 =

119862 and 119883119861 = 119863 in (5) is consistent Then

max119860119883=119862119883119861=119863

119903 (119876 minus 119883119875119883lowast

)

= min

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861)

minus 119903 (119875) 2119899 + 119903 (119860119876119860lowast

minus 119862119875119862lowast

)

minus2119903 (119860) 119903 [

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 2119903 (119861)

min119860119883=119862119883119861=119863

119903 (119876 minus 119883119875119883lowast

)

= 2119899 + 2119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 2119903 (119860) minus 2119903 (119861) minus 119903 (119875)

+max 119904++ 119904minus 119905++ 119905minus 119904++ 119905minus 119904minus+ 119905+

max119860119883=119862119883119861=119863

119894plusmn(119876 minus 119883119875119883

lowast

)

= min

119899 + 119894plusmn(119860119876119860

lowast

minus 119862119875119862lowast

)

minus119903 (119860) 119894plusmn

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 119903 (119861)

(14)

min119860119883=119862119883119861=119863

119894plusmn(119876 minus 119883119875119883

lowast

)

= 119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861)

minus 119894plusmn(119875) +max 119904

plusmn 119905plusmn

(15)

where

119904plusmn=minus119899+119903 (119860)minus119894

∓(119875)+119894

plusmn(119860119876119860

lowast

minus119862119875119862lowast

)minus119903 [119862119875 119860119876119860

lowast

119861lowast

119863lowast

119860lowast]

119905plusmn=minus119899+119903 (119860)minus119894

∓(119875)+119894

plusmn

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus[

[

0 119875 119861

119860119876 119862119875 0

119863lowast

119861lowast

0

]

]

(16)

Proof It follows from Lemma 4 that the general solution of(4) can be expressed as

119883 = 1198830+ 119871119860119881119877119861 (17)

where 119881 is an arbitrary matrix over C and 1198830is a special

solution of (5) Then

119876 minus 119883119875119883lowast

= 119876 minus (1198830+ 119871119860119881119877119861) 119875(119883

0+ 119871119860119881119877119861)lowast

(18)

Note that

119903 [119876 minus (1198830+ 119871119860119881119877119861) 119875(119883

0+ 119871119860119881119877119861)lowast

]

= 119903 [119876 (119883

0+ 119871119860119881119877119861) 119875

119875(1198830+ 119871119860119881119877119861)lowast

119875] minus 119903 (119875)

= 119903 [[119876 119883

0119875

119875119883lowast

0119875] + [

119871119860

0]119881 [0 119877

119861119875]

+([119871119860

0]119881 [0 119877

119861119875])

lowast

] minus 119903 (119875)

(19)

119894plusmn[119876 minus (119883

0+ 119871119860119881119877119861) 119875(119883

0+ 119871119860119881119877119861)lowast

]

= 119894plusmn[

119876 (1198830+ 119871119860119881119877119861) 119875

119875(1198830+ 119871119860119881119877119861)lowast

119875] minus 119894plusmn(119875)

= 119894plusmn[[

119876 1198830119875

119875119883lowast

0119875] + [

119871119860

0]119881 [0 119877

119861119875]

+([119871119860

0]119881 [0 119877

119861119875])

lowast

] minus 119894plusmn(119875)

(20)

Let

119902 (119881) = [119876 119883

0119875

119875119883lowast

0119875] + [

119871119860

0]119881 [0 119877

119861119875]

+ ([119871119860

0]119881 [0 119877

119861119875])

lowast

(21)

Applying Lemma 1 to (19) and (20) yields

max119881

119903 [119902 (119881)] = min 119903 (119872) 119903 (1198721) 119903 (119872

2)

min119881

119903 [119902 (119881)]

= 2119903 (119872) +max 119904++ 119904minus 119905++ 119905minus 119904++ 119905minus 119904minus+ 119905+

max119881

119894plusmn[119902 (119881)] = min 119894

plusmn(1198721) 119894plusmn(1198722)

min119881

119894plusmn[119902 (119881)] = 119903 (119872) +max 119904

plusmn 119905plusmn

(22)

where

119872=[119876 119883

0119875 119871119860

0

119875119883lowast

0119875 0 119875119877

119861

] 1198721=[

[

119876 1198830119875 119871119860

119875119883lowast

0119875 0

119871119860

0 0

]

]

1198722=[

[

119876 1198830119875 0

119875119883lowast

0119875 119875119877

119861

0 119877119861119875 0

]

]

1198723=[

[

119876 1198830119875 119871119860

0

119875119883lowast

0119875 0 119875119877

119861

119871119860

0 0 0

]

]

1198724= [

[

119876 1198830119875 119871119860

0

119875119883lowast

0119875 0 119875119877

119861

0 119877119861119875 0 0

]

]

119904plusmn= 119894plusmn(1198721) minus 119903 (119872

3) 119905

plusmn= 119894plusmn(1198722) minus 119903 (119872

4)

(23)

4 Journal of Applied Mathematics

Applying Lemmas 2 and 3 elementary matrix operations andcongruence matrix operations we obtain

119903 (119872) = 119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861)

119903 (1198721) = 2119899 + 119903 (119860119876119860

lowast

minus 119862119875119862lowast

) minus 2119903 (119860) + 119903 (119875)

119894plusmn(1198721) = 119899 + 119894

plusmn(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 (119860) + 119894plusmn(119875)

119903 (1198722) = 119903[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 2119903 (119861) + 119903 (119875)

119894plusmn(1198722) = 119894plusmn

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 119903 (119861) + 119894plusmn(119875)

119903 (1198723) = 2119899 + 119903 (119875) minus 2119903 (119860) minus 119903 (119861) + 119903 [

119862119875 119860119876119860lowast

119861lowast

119863lowast

119860lowast]

119903 (1198724) = 119899 + 119903 (119875) + 119903[

[

0 119875 119861

119860119876 119862119875 0

119863lowast

119861lowast

0

]

]

minus 2119903 (119861) minus 119903 (119860)

(24)

Substituting (24) into (22) we obtain the results

Using immediately Theorem 5 we can easily get thefollowing

Theorem 6 Let 119891(119883) be as given in (4) 119904plusmnand let 119905

plusmnbe as

given in Theorem 5 and assume that 119860119883 = 119862 and 119883119861 = 119863 in(5) are consistent Then we have the following

(a) 119860119883 = 119862 and 119883119861 = 119863 have a common solution suchthat 119876 minus 119883119875119883

lowast

ge 0 if and only if

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894minus(119875) + 119904

minusle 0

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894minus(119875) + 119905

minusle 0

(25)

(b) 119860119883 = 119862 and 119883119861 = 119863 have a common solution suchthat 119876 minus 119883119875119883

lowast

le 0 if and only if

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894+(119875) + 119904

+le 0

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894+(119875) + 119905

+le 0

(26)

(c) 119860119883 = 119862 and 119883119861 = 119863 have a common solution suchthat 119876 minus 119883119875119883

lowast

gt 0 if and only if

119894+(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 (119860) ge 0

119894+

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 119903 (119861) ge 119899

(27)

(d) 119860119883 = 119862 and 119883119861 = 119863 have a common solution suchthat 119876 minus 119883119875119883

lowast

lt 0 if and only if

119894minus(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 (119860) ge 0

119894minus

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 119903 (119861) ge 119899

(28)

(e) All common solutions of 119860119883 = 119862 and 119883119861 = 119863 satisfy119876 minus 119883119875119883

lowast

ge 0 if and only if

119899 + 119894minus(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 (119860) = 0

or 119894minus

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 119903 (119861) = 0

(29)

(f) All common solutions of 119860119883 = 119862 and 119883119861 = 119863 satisfy119876 minus 119883119875119883

lowast

le 0 if and only if

119899 + 119894+(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 (119860) = 0

or 119894+

[

[

119876 0 119863

0 minus119875 119861

119863lowast

119861lowast

0

]

]

minus 119903 (119861) = 0

(30)

(g) All common solutions of 119860119883 = 119862 and 119883119861 = 119863 satisfy119876 minus 119883119875119883

lowast

gt 0 if and only if

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894+(119875) + 119904

+= 119899

(31)or

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894+(119875) + 119905

+= 119899

(32)

(h) All common solutions of 119860119883 = 119862 and 119883119861 = 119863 satisfy119876 minus 119883119875119883

lowast

lt 0 if and only if

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894minus(119875) + 119904

minus= 119899

(33)or

119899 + 119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus 119903 (119860) minus 119903 (119861) minus 119894minus(119875) + 119905

minus= 119899

(34)

Journal of Applied Mathematics 5

(i) 119860119883 = 119862 119883119861 = 119863 and 119876 = 119883119875119883lowast have a common

solution if and only if

2119899+2119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus2119903 (119860)minus2119903 (119861)minus119903 (119875)+119904++119904minusle0

2119899+2119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus2119903 (119860)minus2119903 (119861)minus119903 (119875)+119905++119905minusle0

2119899+2119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus2119903 (119860)minus2119903 (119861)minus119903 (119875) + 119904++119905minusle0

2119899+2119903[

[

0 119875 119875

119860119876 119862119875 0

minus119863lowast

0 119861lowast

]

]

minus2119903 (119860)minus2119903 (119861)minus119903 (119875) + 119904minus+119905+le0

(35)

Let 119875 = 119868 in Theorem 5 we get the following corollary

Corollary 7 Let 119876 isin C119899times119899 119860 119861 119862 and119863 be given Assumethat (5) is consistent Denote

1198791= [

119862 119860119876

119861lowast

119863lowast] 119879

2= 119860119876119860

lowast

minus 119862119862lowast

1198793= [

119876 119863

119863lowast

119861lowast

119861] 119879

4= [

119862 119860119876119860lowast

119861lowast

119863lowast

119860lowast]

1198795= [

119862119861 119860119876

119861lowast

119861 119863lowast]

(36)

Then

max119860119883=119862119883119861=119863

119903 (119876 minus 119883119883lowast

)

= min 119899 + 119903 (1198791) minus 119903 (119860) minus 119903 (119861) 2119899 + 119903 (119879

2)

minus2119903 (119860) 119899 + 119903 (1198793) minus 2119903 (119861)

min119860119883=119862119883119861=119863

119903 (119876 minus 119883119883lowast

)

= 2119903 (1198791) +max 119903 (119879

2) minus 2119903 (119879

4) minus119899 + 119903 (119879

3)

minus 2119903 (1198795) 119894+(1198792) + 119894minus(1198793)

minus 119903 (1198794) minus 119903 (119879

5) minus119899 + 119894

minus(1198792)

+119894+(1198793) minus 119903 (119879

4) minus 119903 (119879

5)

max119860119883=119862119883119861=119863

119894+(119876 minus 119883119883

lowast

)

= min 119899 + 119894+(1198792) minus 119903 (119860) 119894

+(1198793) minus 119903 (119861)

max119860119883=119862119883119861=119863

119894minus(119876 minus 119883119883

lowast

)

= min 119899 + 119894minus(1198792) minus 119903 (119860) 119899 + 119894

minus(1198793) minus 119903 (119861)

min119860119883=119862119883119861=119863

119894+(119876 minus 119883119883

lowast

)

= 119903 (1198791) +max 119894

+(1198792) minus 119903 (119879

4) 119894+(1198793) minus 119899 minus 119903 (119879

5)

min119860119883=119862119883119861=119863

119894minus(119876 minus 119883119883

lowast

)

= 119903 (1198791) +max 119894

minus(1198792) minus 119903 (119879

4) 119894minus(1198793) minus 119903 (119879

5)

(37)

Remark 8 Corollary 7 is one of the results in [24]

Let 119861 and 119863 vanish in Theorem 5 then we can obtainthe maximal and minimal ranks and inertias of (4) subjectto 119860119883 = 119862

Corollary 9 Let 119891(119883) be as given in (4) and assume that119860119883 = 119862 is consistent Then

max119860119883=119862

119903 (119876 minus 119883119875119883lowast

)

= min 119899 + 119903 [119860119876 119862119875] minus 119903 (119860) minus 119903 (119861)

2119899 + 119903 (119860119876119860lowast

minus 119862119875119862lowast

) minus 2119903 (119860) 119903 (119876) + 119903 (119875)

min119860119883=119862

119903 (119876 minus 119883119875119883lowast

)

= 2119899 + 2119903 [119860119876 119862119875] minus 2119903 (119860)

+max 119904++ 119904minus 119905++ 119905minus 119904++ 119905minus 119904minus+ 119905+

max119860119883=119862

119894plusmn(119876 minus 119883119875119883

lowast

)

= min 119899 + 119894plusmn(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 (119860) 119894plusmn(119876) + 119894

∓(119875)

min119860119883=119862

119894plusmn(119876 minus 119883119875119883

lowast

)

= 119899 + 119903 [119860119876 119862119875] minus 119903 (119860) + 119894∓(119875) +max 119904

plusmn 119905plusmn

(38)

where

119904plusmn= minus 119899 + 119903 (119860) minus 119894

∓(119875)

+ 119894plusmn(119860119876119860

lowast

minus 119862119875119862lowast

) minus 119903 [119862119875 119860119876119860lowast

]

119905plusmn= minus119899 + 119903 (119860) + 119894

plusmn(119876) minus 119903 (119875) minus [119860119876 119862119875]

(39)

Remark 10 Corollary 9 is one of the results in [22]

References

[1] D L Chu Y S Hung and H J Woerdeman ldquoInertia and rankcharacterizations of somematrix expressionsrdquo SIAM Journal onMatrix Analysis and Applications vol 31 no 3 pp 1187ndash12262009

[2] Z H He and Q W Wang ldquoSolutions to optimization problemson ranks and inertias of a matrix function with applicationsrdquo

6 Journal of Applied Mathematics

AppliedMathematics andComputation vol 219 no 6 pp 2989ndash3001 2012

[3] Y Liu andY Tian ldquoMax-min problems on the ranks and inertiasof the matrix expressions119860minus119861119883119862plusmn(119861119883119862)lowast with applicationsrdquoJournal of OptimizationTheory and Applications vol 148 no 3pp 593ndash622 2011

[4] G Marsaglia and G P H Styan ldquoEqualities and inequalities forranks of matricesrdquo Linear and Multilinear Algebra vol 2 pp269ndash292 197475

[5] X Zhang Q W Wang and X Liu ldquoInertias and ranks ofsome Hermitian matrix functions with applicationsrdquo CentralEuropean Journal of Mathematics vol 10 no 1 pp 329ndash3512012

[6] Q W Wang Y Zhou and Q Zhang ldquoRanks of the commonsolution to six quaternion matrix equationsrdquo Acta Mathemati-cae Applicatae Sinica vol 27 no 3 pp 443ndash462 2011

[7] Q Zhang and Q W Wang ldquoThe (119875 119876)-(skew)symmetricextremal rank solutions to a system of quaternion matrixequationsrdquo Applied Mathematics and Computation vol 217 no22 pp 9286ndash9296 2011

[8] D Z Lian QWWang and Y Tang ldquoExtreme ranks of a partialbanded block quaternion matrix expression subject to somematrix equationswith applicationsrdquoAlgebraColloquium vol 18no 2 pp 333ndash346 2011

[9] H S Zhang andQWWang ldquoRanks of submatrices in a generalsolution to a quaternion system with applicationsrdquo Bulletin ofthe Korean Mathematical Society vol 48 no 5 pp 969ndash9902011

[10] Q W Wang and J Jiang ldquoExtreme ranks of (skew-)Hermitiansolutions to a quaternion matrix equationrdquo Electronic Journal ofLinear Algebra vol 20 pp 552ndash573 2010

[11] I A Khan Q W Wang and G J Song ldquoMinimal ranks ofsome quaternionmatrix expressions with applicationsrdquoAppliedMathematics and Computation vol 217 no 5 pp 2031ndash20402010

[12] Q Wang S Yu and W Xie ldquoExtreme ranks of real matricesin solution of the quaternion matrix equation 119860119883119861 = 119862 withapplicationsrdquo Algebra Colloquium vol 17 no 2 pp 345ndash3602010

[13] Q W Wang H S Zhang and G J Song ldquoA new solvable con-dition for a pair of generalized Sylvester equationsrdquo ElectronicJournal of Linear Algebra vol 18 pp 289ndash301 2009

[14] Q W Wang S W Yu and Q Zhang ldquoThe real solutions toa system of quaternion matrix equations with applicationsrdquoCommunications in Algebra vol 37 no 6 pp 2060ndash2079 2009

[15] Q W Wang and C K Li ldquoRanks and the least-norm of thegeneral solution to a system of quaternion matrix equationsrdquoLinear Algebra and its Applications vol 430 no 5-6 pp 1626ndash1640 2009

[16] Q W Wang H S Zhang and S W Yu ldquoOn solutions to thequaternionmatrix equation119860119883119861+119862119884119863 = 119864rdquoElectronic Journalof Linear Algebra vol 17 pp 343ndash358 2008

[17] Q W Wang G J Song and X Liu ldquoMaximal and minimalranks of the common solution of some linear matrix equationsover an arbitrary division ring with applicationsrdquo AlgebraColloquium vol 16 no 2 pp 293ndash308 2009

[18] Q W Wang G J Song and C Y Lin ldquoRank equalities relatedto the generalized inverse 119860

(2)

119879119878with applicationsrdquo Applied

Mathematics and Computation vol 205 no 1 pp 370ndash3822008

[19] Q W Wang S W Yu and C Y Lin ldquoExtreme ranks of alinear quaternionmatrix expression subject to triple quaternionmatrix equations with applicationsrdquo Applied Mathematics andComputation vol 195 no 2 pp 733ndash744 2008

[20] Q W Wang G J Song and C Y Lin ldquoExtreme ranks ofthe solution to a consistent system of linear quaternion matrixequations with an applicationrdquo Applied Mathematics and Com-putation vol 189 no 2 pp 1517ndash1532 2007

[21] Y Tian ldquoFormulas for calculating the extremum ranks andinertias of a four-term quadratic matrix-valued function andtheir applicationsrdquo Linear Algebra and its Applications vol 437no 3 pp 835ndash859 2012

[22] Y Tian ldquoSolving optimization problems on ranks and inertiasof some constrained nonlinearmatrix functions via an algebraiclinearization methodrdquo Nonlinear Analysis Theory Methods ampApplications vol 75 no 2 pp 717ndash734 2012

[23] Y Tian ldquoMaximization and minimization of the rank andinertia of the Hermitianmatrix expression119860minus119861119883minus(119861119883)lowast withapplicationsrdquo Linear Algebra and its Applications vol 434 no10 pp 2109ndash2139 2011

[24] Q W Wang X Zhang and Z H He ldquoOn the Hermitianstructures of the solution to a pair of matrix equationsrdquo Linearand Multilinear Algebra vol 61 no 1 pp 73ndash90 2013

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2012 Article ID 471573 19 pagesdoi1011552012471573

Research ArticleConstrained Solutions of a System ofMatrix Equations

Qing-Wen Wang1 and Juan Yu1 2

1 Department of Mathematics Shanghai University 99 Shangda Road Shanghai 200444 China2 Department of Basic Mathematics China University of Petroleum Qingdao 266580 China

Correspondence should be addressed to Qing-Wen Wang wqw858yahoocomcn

Received 26 September 2012 Accepted 7 December 2012

Academic Editor Panayiotis J Psarrakos

Copyright q 2012 Q-W Wang and J Yu This is an open access article distributed under theCreative Commons Attribution License which permits unrestricted use distribution andreproduction in any medium provided the original work is properly cited

We derive the necessary and sufficient conditions of and the expressions for the orthogonalsolutions the symmetric orthogonal solutions and the skew-symmetric orthogonal solutions ofthe system of matrix equations AX = B and XC = D respectively When the matrix equationsare not consistent the least squares symmetric orthogonal solutions and the least squares skew-symmetric orthogonal solutions are respectively given As an auxiliary an algorithm is providedto compute the least squares symmetric orthogonal solutions and meanwhile an example ispresented to show that it is reasonable

1 Introduction

Throughout this paper the following notations will be used Rmtimesn OR

ntimesn SRntimesn and ASR

ntimesn

denote the set of all m times n real matrices the set of all n times n orthogonal matrices the set of alln times n symmetric matrices and the set of all n times n skew-symmetric matrices respectively In isthe identity matrix of order n (middot)T and tr(middot) represent the transpose and the trace of the realmatrix respectively middot stands for the Frobenius norm induced by the inner product Thefollowing two definitions will also be used

Definition 11 (see [1]) A real matrix X isin Rntimesn is said to be a symmetric orthogonal matrix if

XT = X and XTX = In

Definition 12 (see [2]) A real matrix X isin R2mtimes2m is called a skew-symmetric orthogonal

matrix if XT = minusX and XTX = In

The set of all n times n symmetric orthogonal matrices and the set of all 2m times 2m skew-symmetric orthogonal matrices are respectively denoted by SOR

ntimesn and SSOR2mtimes2m Since

2 Journal of Applied Mathematics

the linear matrix equation(s) and its optimal approximation problem have great applicationsin structural design biology control theory and linear optimal control and so forth seefor example [3ndash5] there has been much attention paid to the linear matrix equation(s) Thewell-known system of matrix equations

AX = B XC = D (11)

as one kind of linear matrix equations has been investigated by many authors and a seriesof important and useful results have been obtained For instance the system (11) withunknown matrix X being bisymmetric centrosymmetric bisymmetric nonnegative definiteHermitian and nonnegative definite and (PQ)-(skew) symmetric has been respectivelyinvestigated by Wang et al [6 7] Khatri and Mitra [8] and Zhang and Wang [9] Of courseif the solvability conditions of system (11) are not satisfied we may consider its least squaressolution For example Li et al [10] presented the least squares mirrorsymmetric solutionYuan [11] got the least-squares solution Some results concerning the system (11) can also befound in [12ndash18]

Symmetric orthogonal matrices and skew-symmetric orthogonal matrices play impor-tant roles in numerical analysis and numerical solutions of partial differential equationsPapers [1 2] respectively derived the symmetric orthogonal solution X of the matrixequation XC = D and the skew-symmetric orthogonal solution X of the matrix equationAX = B Motivated by the work mentioned above we in this paper will respectively studythe orthogonal solutions symmetric orthogonal solutions and skew-symmetric orthogonalsolutions of the system (11) Furthermore if the solvability conditions are not satisfiedthe least squares skew-symmetric orthogonal solutions and the least squares symmetricorthogonal solutions of the system (11) will be also given

The remainder of this paper is arranged as follows In Section 2 some lemmas areprovided to give the main results of this paper In Sections 3 4 and 5 the necessary andsufficient conditions of and the expression for the orthogonal the symmetric orthogonaland the skew-symmetric orthogonal solutions of the system (11) are respectively obtainedIn Section 6 the least squares skew-symmetric orthogonal solutions and the least squaressymmetric orthogonal solutions of the system (11) are presented respectively In additionan algorithm is provided to compute the least squares symmetric orthogonal solutions andmeanwhile an example is presented to show that it is reasonable Finally in Section 7 someconcluding remarks are given

2 Preliminaries

In this section we will recall some lemmas and the special C-S decomposition which will beused to get the main results of this paper

Lemma 21 (see [1 Lemmas 1 and 2]) Given C isin R2mtimesn D isin R

2mtimesn The matrix equation YC =D has a solution Y isin OR

2mtimes2m if and only if DTD = CTC Let the singular value decompositions ofC and D be respectively

C = U

(Π 00 0

)V T D = W

(Π 00 0

)V T (21)

Journal of Applied Mathematics 3

where

Π = diag(σ1 σk) gt 0 k = rank(C) = rank(D)

U =(U1 U2

) isin OR2mtimes2m U1 isin R

2mtimesk

W =(W1 W2

) isin OR2mtimes2m W1 isin R

2mtimesk V =(V1 V2

) isin ORntimesn V1 isin R

ntimesk

(22)

Then the orthogonal solutions of YC = D can be described as

Y = W

(Ik 00 P

)UT (23)

where P isin OR(2mminusk)times(2mminusk) is arbitrary

Lemma 22 (see [2 Lemmas 1 and 2]) GivenA isin Rntimesm B isin R

ntimesm The matrix equationAX = Bhas a solution X isin OR

mtimesm if and only if AAT = BBT Let the singular value decompositions of Aand B be respectively

A = U

(Σ 00 0

)V T B = U

(Σ 00 0

)QT (24)

where

Σ = diag(δ1 δl) gt 0 l = rank(A) = rank(B) U =(U1 U2

) isin ORntimesn U1 isin R

ntimesl

V =(V1 V2

) isin ORmtimesm V1 isin R

mtimesl Q =(Q1 Q2

) isin ORmtimesm Q1 isin R

mtimesl(25)

Then the orthogonal solutions of AX = B can be described as

X = V

(Il 00 W

)QT (26)

whereW isin OR(mminusl)times(mminusl) is arbitrary

Lemma 23 (see [2 Theorem 1]) If

X =(X11 X12

X21 X22

)isin OR

2mtimes2m X11 isin ASRktimesk (27)

then the C-S decomposition of X can be expressed as

(D1 00 D2

)T(X11 X12

X21 X22

)(D1 00 R2

)=(Σ11 Σ12

Σ21 Σ22

) (28)

4 Journal of Applied Mathematics

where D1 isin ORktimesk D2 R2 isin OR

(2mminusk)times(2mminusk)

Σ11 =

⎛⎜⎝

I 0 00 C 00 0 0

⎞⎟⎠ Σ12 =

⎛⎝0 0 0

0 S 00 0 I

⎞⎠

Σ21 =

⎛⎝0 0 0

0 S 00 0 I

⎞⎠ Σ22 =

⎛⎜⎝

I 0 00 minusCT 00 0 0

⎞⎟⎠

I = diag(I1 Ir1

) I1 = middot middot middot = Ir1 =

(0 1minus1 0

) C = diag(C1 Cl1)

Ci =(

0 ciminusci 0

) i = 1 l1 S = diag(S1 Sl1)

Si =(si 00 si

) ST

i Si + CTi Ci = I2 i = 1 l1

2r1 + 2l1 = rank(X11) Σ21 = ΣT12 k minus 2r1 = rank(X21)

(29)

Lemma 24 (see [1 Theorem 1]) If

K =(K11 K12

K21 K22

)isin OR

mtimesm K11 isin SRltimesl (210)

then the C-S decomposition of K can be described as

(D1 00 D2

)T(K11 K12

K21 K22

)(D1 00 R2

)=(Π11 Π12

Π21 Π22

) (211)

where D1 isin ORltimesl D2 R2 isin OR

(mminusl)times(mminusl)

Π11 =

⎛⎜⎝

I 0 00 C 00 0 0

⎞⎟⎠ Π12 =

⎛⎝0 0 0

0 S 00 0 I

⎞⎠

Π21 =

⎛⎝0 0 0

0 S 00 0 I

⎞⎠ Π22 =

⎛⎜⎝

I 0 00 minusC 00 0 0

⎞⎟⎠

I = diag(i1 ir) ij = plusmn1 j = 1 r

S = diag(s1prime slprime) C = diag(c1prime clprime)

s2iprime =radic

1 minus c2iprime i = 1prime lprime r + lprime = rank(K11) Π21 = ΠT

12

(212)

Journal of Applied Mathematics 5

Remarks 25 In order to know the C-S decomposition of an orthogonal matrix with ak times k leading (skew-) symmetric submatrix for details one can deeply study the proof ofTheorem 1 in [1] and [2]

Lemma 26 Given A isin Rntimesm B isin R

ntimesm Then the matrix equation AX = B has a solution X isinSOR

mtimesm if and only ifAAT = BBT andABT = BAT When these conditions are satisfied the generalsymmetric orthogonal solutions can be expressed as

X = V

(I2lminusr 0

0 G

)QT (213)

where

Q = JQdiag(ID2) isin ORmtimesm V = JV diag(I R2) isin OR

mtimesm

J =

⎛⎝I 0 0

0 0 I0 I 0

⎞⎠

(214)

and G isin SOR(mminus2l+r)times(mminus2l+r) is arbitrary

Proof The Necessity Assume X isin SORmtimesm is a solution of the matrix equation AX = B then

we have

BBT = AXXTAT = AAT

BAT = AXAT = AXTAT = ABT(215)

The Sufficiency Since the equality AAT = BBT holds then by Lemma 22 the singularvalue decompositions of A and B can be respectively expressed as (24) Moreover thecondition ABT = BAT means

U

(Σ 00 0

)V TQ

(Σ 00 0

)UT = U

(Σ 00 0

)QTV

(Σ 00 0

)UT (216)

which can be written as

V T1 Q1 = QT

1V1 (217)

From Lemma 22 the orthogonal solutions of the matrix equation AX = B can be describedas (26) Now we aim to find that X in (26) is also symmetric Suppose that X is symmetricthen we have

(Il 00 WT

)V TQ = QTV

(Il 00 W

) (218)

6 Journal of Applied Mathematics

together with the partitions of the matrices Q and V in Lemma 22 we get

V T1 Q1 = QT

1V1 (219)

QT1V2W = V T

1 Q2 (220)

QT2 V2W = WTV T

2 Q2 (221)

By (217) we can get (219) Now we aim to find the orthogonal solutions of the systemof matrix equations (220) and (221) Firstly we obtain from (220) that QT

1V2(QT1V2)

T =V T

1 Q2(V T1 Q2)

T then by Lemma 22 (220) has an orthogonal solution W By (217) the l times lleading principal submatrix of the orthogonal matrix V TQ is symmetric Then we have fromLemma 24

V T1 Q2 = D1Π12R

T2 (222)

QT1V2 = D1Π12D

T2 (223)

V T2 Q2 = D2Π22R

T2 (224)

From (220) (222) and (223) the orthogonal solution W of (220) is

W = D2

⎛⎝G 0 0

0 I 00 0 I

⎞⎠RT

2 (225)

where G isin OR(mminus2l+r)times(mminus2l+r) is arbitrary Combining (221) (224) and (225) yields GT =

G that is G is a symmetric orthogonal matrix Denote

V = V diag(ID2) Q = Qdiag(I R2) (226)

then the symmetric orthogonal solutions of the matrix equation AX = B can be expressed as

X = V

⎛⎝I 0 0

0 G 00 0 I

⎞⎠QT (227)

Let the partition matrix J be

J =

⎛⎝I 0 0

0 0 I0 I 0

⎞⎠ (228)

Journal of Applied Mathematics 7

compatible with the block matrix

⎛⎝I 0 0

0 G 00 0 I

⎞⎠ (229)

Put

V = JV Q = JQ (230)

then the symmetric orthogonal solutions of the matrix equation AX = B can be described as(213)

Setting A = CT B = DT and X = YT in [2 Theorem 2] and then by Lemmas 21 and23 we can have the following result

Lemma 27 Given C isin R2mtimesn D isin R

2mtimesn Then the equation has a solution Y isin SSOR2mtimes2m if

and only ifDTD = CTC andDTC = minusCTD When these conditions are satisfied the skew-symmetricorthogonal solutions of the matrix equation YC = D can be described as

Y = W

(I 00 H

)UT (231)

where

W = J primeW diag(IminusD2) isin OR2mtimes2m U = J primeUdiag(I R2) isin OR

2mtimes2m

J prime =

⎛⎝I 0 0

0 0 I0 I 0

⎞⎠

(232)

and H isin SSOR2rtimes2r is arbitrary

3 The Orthogonal Solutions of the System (11)

The following theorems give the orthogonal solutions of the system (11)

Theorem 31 Given A B isin Rntimesm and C D isin R

mtimesn suppose the singular value decompositions ofA and B are respectively as (24) Denote

QTC =(C1

C2

) V TD =

(D1

D2

) (31)

8 Journal of Applied Mathematics

where C1 D1 isin Rltimesn and C2 D2 isin R

(mminusl)timesn Let the singular value decompositions of C2 and D2 berespectively

C2 = U

(Π 00 0

)V T D2 = W

(Π 00 0

)V T (32)

where U W isin OR(mminusl)times(mminusl) V isin OR

ntimesn Π isin Rktimesk is diagonal whose diagonal elements are

nonzero singular values of C2 or D2 Then the system (11) has orthogonal solutions if and only if

AAT = BBT C1 = D1 DT2 D2 = CT

2 C2 (33)

In which case the orthogonal solutions can be expressed as

X = V

(Ik+l 00 Gprime

)QT (34)

where

V = V

(Il 00 W

)isin OR

mtimesm Q = Q

(Il 00 U

)isin OR

mtimesm (35)

and Gprime isin OR(mminuskminusl)times(mminuskminusl) is arbitrary

Proof Let the singular value decompositions of A and B be respectively as (24) Since thematrix equation AX = B has orthogonal solutions if and only if

AAT = BBT (36)

then by Lemma 22 its orthogonal solutions can be expressed as (26) Substituting (26) and(31) into the matrix equation XC = D we have C1 = D1 and WC2 = D2 By Lemma 21 thematrix equation WC2 = D2 has orthogonal solution W if and only if

DT2 D2 = CT

2C2 (37)

Let the singular value decompositions of C2 and D2 be respectively

C2 = U

(Π 00 0

)V T D2 = W

(Π 00 0

)V T (38)

where U W isin OR(mminusl)times(mminusl) V isin OR

ntimesn Π isin Rktimesk is diagonal whose diagonal elements are

nonzero singular values of C2 or D2 Then the orthogonal solutions can be described as

W = W

(Ik 00 Gprime

)UT (39)

Journal of Applied Mathematics 9

where Gprime isin OR(mminuskminusl)times(mminuskminusl) is arbitrary Therefore the common orthogonal solutions of the

system (11) can be expressed as

X = V

(Il 00 W

)QT = V

(Il 00 W

)⎛⎝Il 0 0

0 Ik 00 0 Gprime

⎞⎠(Il 00 UT

)QT = V

(Ik+l 00 Gprime

)QT (310)

where

V = V

(Il 00 W

)isin OR

mtimesm Q = Q

(Il 00 U

)isin OR

mtimesm (311)

and Gprime isin OR(mminuskminusl)times(mminuskminusl) is arbitrary

The following theorem can be shown similarly

Theorem 32 GivenA B isin Rntimesm and C D isin R

mtimesn let the singular value decompositions of C andD be respectively as (21) Partition

AW =(A1 A2

) BU =

(B1 B2

) (312)

where A1 B1 isin Rntimesk A2 B2 isin R

ntimes(mminusk) Assume the singular value decompositions of A2 and B2

are respectively

A2 = U

(Σ 00 0

)V T B2 = U

(Σ 00 0

)QT (313)

where V Q isin OR(mminusk)times(mminusk) U isin OR

ntimesn Σ isin Rlprimetimeslprime is diagonal whose diagonal elements are

nonzero singular values of A2 or B2 Then the system (11) has orthogonal solutions if and onlyif

DTD = CTC A1 = B1 A2AT2 = B2B

T2 (314)

In which case the orthogonal solutions can be expressed as

X = W

(Ik+lprime 0

0 H prime

)UT (315)

where

W = W

(Ik 00 W

)isin OR

mtimesm U = U

(Ik 00 U

)isin OR

mtimesm (316)

and H prime isin OR(mminuskminuslprime)times(mminuskminuslprime) is arbitrary

10 Journal of Applied Mathematics

4 The Symmetric Orthogonal Solutions of the System (11)

We now present the symmetric orthogonal solutions of the system (11)

Theorem 41 Given AB isin Rntimesm CD isin R

mtimesn Let the symmetric orthogonal solutions of thematrix equation AX = B be described as in Lemma 26 Partition

QTC =(Cprime

1Cprime

2

) V TD =

(Dprime

1Dprime

2

) (41)

where Cprime1 D

prime1 isin R

(2lminusr)timesn Cprime2 D

prime2 isin R

(mminus2l+r)timesn Then the system (11) has symmetric orthogonalsolutions if and only if

AAT = BBT ABT = BAT Cprime1 = Dprime

1 DprimeT2 Dprime

2 = CprimeT2 Cprime

2 DprimeT2 Cprime

2 = CprimeT2 Dprime

2 (42)

In which case the solutions can be expressed as

X = V

(I 00 Gprimeprime

)QT (43)

where

V = V

(I2lminusr 0

0 W

)isin OR

mtimesm Q = Q

(I2lminusr 0

0 U

)isin OR

mtimesm (44)

and Gprimeprime isin SOR(mminus2l+rminus2lprime+r prime)times(mminus2l+rminus2lprime+r prime) is arbitrary

Proof From Lemma 26 we obtain that the matrix equation AX = B has symmetricorthogonal solutions if and only if AAT = BBT and ABT = BAT When these conditionsare satisfied the general symmetric orthogonal solutions can be expressed as

X = V

(I2lminusr 0

0 G

)QT (45)

where G isin SOR(mminus2l+r)times(mminus2l+r) is arbitrary Q isin OR

mtimesm V isin ORmtimesm Inserting (41) and (45)

into the matrix equation XC = D we get Cprime1 = Dprime

1 and GCprime2 = Dprime

2 By [1 Theorem 2] thematrix equation GCprime

2 = Dprime2 has a symmetric orthogonal solution if and only if

DprimeT2 Dprime

2 = CprimeT2 Cprime

2 DprimeT2 Cprime

2 = CprimeT2 Dprime

2 (46)

In which case the solutions can be described as

G = W

(I2lprimeminusr prime 0

0 Gprimeprime

)UT (47)

Journal of Applied Mathematics 11

where Gprimeprime isin SOR(mminus2l+rminus2lprime+r prime)times(mminus2l+rminus2lprime+r prime) is arbitrary W isin OR

(mminus2l+r)times(mminus2l+r) and U isinOR

(mminus2l+r)times(mminus2l+r) Hence the system (11) has symmetric orthogonal solutions if and onlyif all equalities in (42) hold In which case the solutions can be expressed as

X = V

(I2lminusr 0

0 G

)QT = V

(I2lminusr 0

0 W

)⎛⎝I2lminusr 0

0(I 00 Gprimeprime

)⎞⎠(I2lminusr 0

0 UT

)QT (48)

that is the expression in (43)

The following theorem can also be obtained by the method used in the proof ofTheorem 41

Theorem 42 Given AB isin Rntimesm CD isin R

mtimesn Let the symmetric orthogonal solutions of thematrix equation XC = D be described as

X = M

(I2kminusr 0

0 G

)NT (49)

where M N isin ORmtimesm G isin SOR

(mminus2k+r)times(mminus2k+r) Partition

AM =(M1 M2

) BN =

(N1 N2

) M1N1 isin R

ntimes(2kminusr) M2N2 isin Rntimes(mminus2k+r) (410)

Then the system (11) has symmetric orthogonal solutions if and only if

DTD = CTC DTC = CTD M1 = N1 M2MT2 = N2N

T2 M2N

T2 = N2M

T2

(411)

In which case the solutions can be expressed as

X = M

(I 00 H primeprime

)NT (412)

where

M = M

(I2kminusr 0

0 W1

)isin OR

mtimesm N = N

(I2kminusr 0

0 U1

)isin OR

mtimesm (413)

and H primeprime isin SOR(mminus2k+rminus2kprime+r prime)times(mminus2k+rminus2kprime+r prime) is arbitrary

12 Journal of Applied Mathematics

5 The Skew-Symmetric Orthogonal Solutions of the System (11)

In this section we show the skew-symmetric orthogonal solutions of the system (11)

Theorem 51 Given AB isin Rntimes2m CD isin R

2mtimesn Suppose the matrix equation AX = B has skew-symmetric orthogonal solutions with the form

X = V1

(I 00 H

)QT

1 (51)

whereH isin SSOR2rtimes2r is arbitrary V1 Q1 isin OR

2mtimes2m Partition

QT1C =

(Q1

Q2

) V T

1 D =(V1

V2

) (52)

where Q1 V1 isin R(2mminus2r)timesn Q2 V2 isin R

2rtimesn Then the system (11) has skew-symmetric orthogonalsolutions if and only if

AAT = BBT ABT = minusBAT Q1 = V1 QT2 Q2 = V T

2 V2 QT2V2 = minusV T

2 Q2 (53)

In which case the solutions can be expressed as

X = V1

(I 00 J prime

)QT

1 (54)

where

V = V1

(I2mminus2r 0

0 W

)isin OR

2mtimes2m Q = Q1

(I2mminus2r 0

0 U

)isin OR

2mtimes2m (55)

and J prime isin SSOR2kprimetimes2kprime

is arbitrary

Proof By [2 Theorem 2] the matrix equation AX = B has the skew-symmetric orthogonalsolutions if and only if AAT = BBT and ABT = minusBAT When these conditions are satisfiedthe general skew-symmetric orthogonal solutions can be expressed as (51) Substituting (51)and (52) into the matrix equation XC = D we get Q1 = V1 and HQ2 = V2 From Lemma 27equation HQ2 = V2 has a skew-symmetric orthogonal solution H if and only if

QT2 Q2 = V T

2 V2 QT2V2 = minusV T

2 Q2 (56)

When these conditions are satisfied the solution can be described as

H = W

(I 00 J prime

)UT (57)

Journal of Applied Mathematics 13

where J prime isin SSOR2kprimetimes2kprime

is arbitrary W U isin OR2rtimes2r Inserting (57) into (51) yields that the

system (11) has skew-symmetric orthogonal solutions if and only if all equalities in (53)hold In which case the solutions can be expressed as (54)

Similarly the following theorem holds

Theorem 52 Given AB isin Rntimes2m CD isin R

2mtimesn Suppose the matrix equation XC = D hasskew-symmetric orthogonal solutions with the form

X = W

(I 00 K

)UT (58)

where K isin SSOR2ptimes2p is arbitrary W U isin OR

2mtimes2m Partition

AW =(W1 W2

) BU =

(U1 U2

) (59)

where W1 U1 isin Rntimes(2mminus2p) W2 U2 isin R

ntimes2p Then the system (11) has skew-symmetric orthogonalsolutions if and only if

DTD = CTC DTC = minusCTD W1 = U1 W2WT2 = U2U

T2 U2W

T2 = minusW2U

T2

(510)

In which case the solutions can be expressed as

X = W1

(I 00 J primeprime

)UT

1 (511)

where

W1 = W

(I2mminus2p 0

0 W1

)isin OR

2mtimes2m U1 =

(I2mminus2p 0

0 UT1

)U isin OR

2mtimes2m (512)

and J primeprime isin SSOR2qtimes2q is arbitrary

6 The Least Squares (Skew-) Symmetric Orthogonal Solutions ofthe System (11)

If the solvability conditions of a system of matrix equations are not satisfied it is natural toconsider its least squares solution In this section we get the least squares (skew-) symmetricorthogonal solutions of the system (11) that is seek X isin SSOR

2mtimes2m(SORntimesn) such that

minXisinSSOR2mtimes2m(SORntimesn)

AX minus B2 + XC minusD2 (61)

14 Journal of Applied Mathematics

With the help of the definition of the Frobenius norm and the properties of the skew-symmetric orthogonal matrix we get that

AX minus B2 + XC minusD2 = A2 + B2 + C2 + D2 minus 2 tr(XT(ATB +DCT

)) (62)

Let

ATB +DCT = K

KT minusK

2= T

(63)

Then it follows from the skew-symmetric matrix X that

tr(XT(ATB +DCT

))= tr(XTK

)= tr(minusXK) = tr

(X

(KT minusK

2

))= tr(XT) (64)

Therefore (61) holds if and only if (64) reaches its maximum Now we pay our attention tofind the maximum value of (64) Assume the eigenvalue decomposition of T is

T = E

(Λ 00 0

)ET (65)

with

Λ = diag(Λ1 Λl) Λi =(

0 αi

minusαi 0

) αi gt 0 i = 1 l 2l = rank(T) (66)

Denote

ETXE =(X11 X12

X21 X22

) (67)

partitioned according to

(Λ 00 0

) (68)

then (64) has the following form

tr(XT) = tr(ETXE

(Λ 00 0

))= tr(X11Λ) (69)

Journal of Applied Mathematics 15

Thus by

X11 = I = diag(I1 Il

) (610)

where

Ii =(

0 minus11 0

) i = 1 l (611)

Equation (69) gets its maximum Since ETXE is skew-symmetric it follows from

X = E

(I 00 G

)ET (612)

where G isin SSOR(2mminus2l)times(2mminus2l) is arbitrary that (61) obtains its minimum Hence we have the

following theorem

Theorem 61 Given AB isin Rntimes2m and CD isin R

2mtimesn denote

ATB +DCT = KKT minusK

2= T (613)

and let the spectral decomposition of T be (65) Then the least squares skew-symmetric orthogonalsolutions of the system (11) can be expressed as (612)

If X in (61) is a symmetric orthogonal matrix then by the definition of the Frobeniusnorm and the properties of the symmetric orthogonal matrix (62) holds Let

ATB +DCT = HHT +H

2= N (614)

Then we get that

minXisinSORntimesn

AX minus B2 + XC minusD2 = minXisinSORntimesn

[A2 + B2 + C2 + D2 minus 2 tr(XN)

] (615)

Thus (615) reaches its minimum if and only if tr(XN) obtains its maximum Now we focuson finding the maximum value of tr(XN) Let the spectral decomposition of the symmetricmatrix N be

N = M

(Σ 00 0

)MT (616)

16 Journal of Applied Mathematics

where

Σ =(Σ+ 00 Σminus

) Σ+ = diag(λ1 λs)

λ1 ge λ2 ge middot middot middot ge λs gt 0 Σminus = diag(λs+1 λt)

λt le λtminus1 le middot middot middot le λs+1 lt 0 t = rank(N)

(617)

Denote

MTXM =

(X11 X12

X21 X22

)(618)

being compatible with

(Σ 00 0

) (619)

Then

tr(XN) = tr(MTXM

(Σ 00 0

))= tr

((X11 X12

X21 X22

)(Σ 00 0

))= tr(X11Σ

) (620)

Therefore it follows from

X11 = I =(Is 00 minusItminuss

)(621)

that (620) reaches its maximum Since MTXM is a symmetric orthogonal matrix then whenX has the form

X = M

(I 00 L

)MT (622)

where L isin SOR(nminust)times(nminust) is arbitrary (615) gets its minimum Thus we obtain the following

theorem

Theorem 62 Given AB isin Rntimesm CD isin R

mtimesn denote

ATB +DCT = HHT +H

2= N (623)

and let the eigenvalue decomposition of N be (616) Then the least squares symmetric orthogonalsolutions of the system (11) can be described as (622)

Journal of Applied Mathematics 17

Algorithm 63 Consider the following

Step 1 Input AB isin Rntimesm and CD isin R

mtimesn

Step 2 Compute

ATB +DCT = H

N =HT +H

2

(624)

Step 3 Compute the spectral decomposition of N with the form (616)

Step 4 Compute the least squares symmetric orthogonal solutions of (11) according to(622)

Example 64 Assume

A =

⎛⎜⎜⎜⎜⎜⎝

122 84 minus56 63 94 107118 29 85 69 96 minus78106 23 115 78 67 8936 78 49 119 94 5945 67 78 31 56 116

⎞⎟⎟⎟⎟⎟⎠

B =

⎛⎜⎜⎜⎜⎜⎝

85 94 36 78 63 4727 36 79 94 56 7837 67 86 98 34 29minus43 62 57 74 54 9529 39 minus52 63 78 46

⎞⎟⎟⎟⎟⎟⎠

C =

⎛⎜⎜⎜⎜⎜⎜⎜⎝

54 64 37 56 9736 42 78 minus63 7867 35 minus46 29 28minus27 72 108 37 3819 39 82 56 11289 78 94 79 56

⎞⎟⎟⎟⎟⎟⎟⎟⎠

D =

⎛⎜⎜⎜⎜⎜⎜⎜⎝

79 95 54 28 8687 26 67 84 8157 39 minus29 52 1948 58 18 minus72 5828 79 45 67 9695 41 34 98 39

⎞⎟⎟⎟⎟⎟⎟⎟⎠

(625)

It can be verified that the given matrices ABC and D do not satisfy the solvabilityconditions in Theorem 41 or Theorem 42 So we intend to derive the least squares symmetricorthogonal solutions of the system (11) By Algorithm 63 we have the following results

(1) the least squares symmetric orthogonal solution

X =

⎛⎜⎜⎜⎜⎜⎜⎜⎝

001200 006621 minus024978 027047 091302 minus016218006621 065601 022702 minus055769 024587 037715minus024978 022702 080661 040254 004066 minus026784027047 minus055769 040254 006853 023799 062645091302 024587 004066 023799 minus012142 minus018135minus016218 037715 minus026784 062645 minus0181354 057825

⎞⎟⎟⎟⎟⎟⎟⎟⎠

(626)

(2)

minXisinSORmtimesm

AX minus B2 + XC minusD2 = 198366

∥∥∥XTX minus I6

∥∥∥ = 284882 times 10minus15∥∥∥XT minusX

∥∥∥ = 000000(627)

18 Journal of Applied Mathematics

Remark 65 (1) There exists a unique symmetric orthogonal solution such that (61) holds ifand only if the matrix

N =HT +H

2 (628)

where

H = ATB +DCT (629)

is invertible Example 64 just illustrates it(2) The algorithm about computing the least squares skew-symmetric orthogonal

solutions of the system (11) can be shown similarly we omit it here

7 Conclusions

This paper is devoted to giving the solvability conditions of and the expressions ofthe orthogonal solutions the symmetric orthogonal solutions and the skew-symmetricorthogonal solutions to the system (11) respectively and meanwhile obtaining the leastsquares symmetric orthogonal and skew-symmetric orthogonal solutions of the system (11)In addition an algorithm and an example have been provided to compute its least squaressymmetric orthogonal solutions

Acknowledgments

This research was supported by the grants from Innovation Program of Shanghai MunicipalEducation Commission (13ZZ080) the National Natural Science Foundation of China(11171205) the Natural Science Foundation of Shanghai (11ZR1412500) the DisciplineProject at the corresponding level of Shanghai (A 13-0101-12-005) and Shanghai LeadingAcademic Discipline Project (J50101)

References

[1] C J Meng and X Y Hu ldquoThe inverse problem of symmetric orthogonal matrices and its optimalapproximationrdquo Mathematica Numerica Sinica vol 28 pp 269ndash280 2006 (Chinese)

[2] C J Meng X Y Hu and L Zhang ldquoThe skew-symmetric orthogonal solutions of the matrix equationAX = Brdquo Linear Algebra and Its Applications vol 402 no 1ndash3 pp 303ndash318 2005

[3] D L Chu H C Chan and D W C Ho ldquoRegularization of singular systems by derivative andproportional output feedbackrdquo SIAM Journal on Matrix Analysis and Applications vol 19 no 1 pp21ndash38 1998

[4] K T Joseph ldquoInverse eigenvalue problem in structural designrdquo AIAA Journal vol 30 no 12 pp2890ndash2896 1992

[5] A Jameson and E Kreinder ldquoInverse problem of linear optimal controlrdquo SIAM Journal on Controlvol 11 no 1 pp 1ndash19 1973

[6] Q W Wang ldquoBisymmetric and centrosymmetric solutions to systems of real quaternion matrixequationsrdquo Computers and Mathematics with Applications vol 49 no 5-6 pp 641ndash650 2005

[7] Q W Wang X Liu and S W Yu ldquoThe common bisymmetric nonnegative definite solutions withextreme ranks and inertias to a pair of matrix equationsrdquo Applied Mathematics and Computation vol218 pp 2761ndash2771 2011

Journal of Applied Mathematics 19

[8] C G Khatri and S K Mitra ldquoHermitian and nonnegative definite solutions of linear matrixequationsrdquo SIAM Journal on Applied Mathematics vol 31 pp 578ndash585 1976

[9] Q Zhang and Q W Wang ldquoThe (PQ)-(skew)symmetric extremal rank solutions to a system ofquaternion matrix equationsrdquo Applied Mathematics and Computation vol 217 no 22 pp 9286ndash92962011

[10] F L Li X Y Hu and L Zhang ldquoLeast-squares mirrorsymmetric solution for matrix equations (AX =BXC = D)rdquo Numerical Mathematics vol 15 pp 217ndash226 2006

[11] Y X Yuan ldquoLeast-squares solutions to the matrix equations AX = B and XC = Drdquo AppliedMathematics and Computation vol 216 no 10 pp 3120ndash3125 2010

[12] A Dajic and J J Koliha ldquoPositive solutions to the equations AX = C and XB = D for Hilbert spaceoperatorsrdquo Journal of Mathematical Analysis and Applications vol 333 no 2 pp 567ndash576 2007

[13] F L Li X Y Hu and L Zhang ldquoThe generalized reflexive solution for a class of matrix equations(AX = BXC = D)rdquo Acta Mathematica Scientia vol 28 no 1 pp 185ndash193 2008

[14] S Kumar Mitra ldquoThe matrix equations AX = CXB = Drdquo Linear Algebra and Its Applications vol 59pp 171ndash181 1984

[15] Y Qiu Z Zhang and J Lu ldquoThe matrix equations AX = BXC = D with PX = sXP constraintrdquoApplied Mathematics and Computation vol 189 no 2 pp 1428ndash1434 2007

[16] Y Y Qiu and A D Wang ldquoLeast squares solutions to the equations AX = BXC = B with someconstraintsrdquo Applied Mathematics and Computation vol 204 no 2 pp 872ndash880 2008

[17] Q W Wang X Zhang and Z H He ldquoOn the Hermitian structures of the solution to a pair of matrixequationsrdquo Linear Multilinear Algebra vol 61 no 1 pp 73ndash90 2013

[18] Q X Xu ldquoCommon Hermitian and positive solutions to the adjointable operator equations AX =CXB = Drdquo Linear Algebra and Its Applications vol 429 no 1 pp 1ndash11 2008

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2012 Article ID 398085 14 pagesdoi1011552012398085

Research ArticleOn the Hermitian R-Conjugate Solution ofa System of Matrix Equations

Chang-Zhou Dong1 Qing-Wen Wang2 and Yu-Ping Zhang3

1 School of Mathematics and Science Shijiazhuang University of Economics ShijiazhuangHebei 050031 China

2 Department of Mathematics Shanghai University Shanghai Shanghai 200444 China3 Department of Mathematics Ordnance Engineering College Shijiazhuang Hebei 050003 China

Correspondence should be addressed to Qing-Wen Wang wqw858yahoocomcn

Received 3 October 2012 Accepted 26 November 2012

Academic Editor Yang Zhang

Copyright q 2012 Chang-Zhou Dong et al This is an open access article distributed under theCreative Commons Attribution License which permits unrestricted use distribution andreproduction in any medium provided the original work is properly cited

Let R be an n by n nontrivial real symmetric involution matrix that is R = Rminus1 = RT = In An n times n

complex matrix A is termed R-conjugate if A = RAR where A denotes the conjugate of A Wegive necessary and sufficient conditions for the existence of the Hermitian R-conjugate solutionto the system of complex matrix equations AX = C andXB = D and present an expression ofthe Hermitian R-conjugate solution to this system when the solvability conditions are satisfiedIn addition the solution to an optimal approximation problem is obtained Furthermore the leastsquares Hermitian R-conjugate solution with the least norm to this system mentioned above isconsidered The representation of such solution is also derived Finally an algorithm and numericalexamples are given

1 Introduction

Throughout we denote the complex m times n matrix space by Cmtimesn the real m times n matrix space

by Rmtimesn and the set of all matrices in R

mtimesn with rank r by Rmtimesnr The symbols IAAT Alowast Adagger

and A stand for the identity matrix with the appropriate size the conjugate the transposethe conjugate transpose the Moore-Penrose generalized inverse and the Frobenius norm ofA isin C

mtimesn respectively We use Vn to denote the ntimes n backward matrix having the elements 1along the southwest diagonal and with the remaining elements being zeros

Recall that an n times n complex matrix A is centrohermitian if A = VnAVn Centro-hermitian matrices and related matrices such as k-Hermitian matrices Hermitian Toeplitzmatrices and generalized centrohermitian matrices appear in digital signal processing andothers areas (see [1ndash4]) As a generalization of a centrohermitian matrix and related matrices

2 Journal of Applied Mathematics

Trench [5] gave the definition of R-conjugate matrix A matrix A isin Cntimesn is R-conjugate if

A = RAR where R is a nontrivial real symmetric involution matrix that is R = Rminus1 = RT andR= In At the same time Trench studied the linear equation Az = w for R-conjugate matricesin [5] where zw are known column vectors

Investigating the matrix equation

AX = B (11)

with the unknown matrix X being symmetric reflexive Hermitian-generalized Hamiltonianand repositive definite is a very active research topic [6ndash14] As a generalization of (11) theclassical system of matrix equations

AX = C XB = D (12)

has attracted many authorrsquos attention For instance [15] gave the necessary and sufficientconditions for the consistency of (12) [16 17] derived an expression for the general solutionby using singular value decomposition of a matrix and generalized inverses of matricesrespectively Moreover many results have been obtained about the system (12) with variousconstraints such as bisymmetric Hermitian positive semidefinite reflexive and generalizedreflexive solutions (see [18ndash28]) To our knowledge so far there has been little investigationof the Hermitian R-conjugate solution to (12)

Motivated by the work mentioned above we investigate Hermitian R-conjugatesolutions to (12) We also consider the optimal approximation problem

∥∥∥X minus E∥∥∥ = min

XisinSX

X minus E (13)

where E is a given matrix in Cntimesn and SX the set of all Hermitian R-conjugate solutions to

(12) In many cases the system (12) has not Hermitian R-conjugate solution Hence we needto further study its least squares solution which can be described as follows Let RHC

ntimesn

denote the set of all Hermitian R-conjugate matrices in Cntimesn

SL =X | min

XisinRHCntimesn

(AX minus C2 + XB minusD2

) (14)

Find X isin Cntimesn such that

∥∥∥X∥∥∥ = minXisinSL

X (15)

In Section 2 we present necessary and sufficient conditions for the existence of theHermitian R-conjugate solution to (12) and give an expression of this solution when thesolvability conditions are met In Section 3 we derive an optimal approximation solution to(13) In Section 4 we provide the least squares Hermitian R-conjugate solution to (15) InSection 5 we give an algorithm and a numerical example to illustrate our results

Journal of Applied Mathematics 3

2 R-Conjugate Hermitian Solution to (12)

In this section we establish the solvability conditions and the general expression for theHermitian R-conjugate solution to (12)

We denote RCntimesn and RHC

ntimesn the set of all R-conjugate matrices and Hermitian R-conjugate matrices respectively that is

RCntimesn =

A | A = RAR

HRCntimesn =

A | A = RARA = Alowast

(21)

where R is n times n nontrivial real symmetric involution matrixChang et al in [29] mentioned that for nontrivial symmetric involution matrix R isin

Rntimesn there exist positive integer r and n times n real orthogonal matrix [P Q] such that

R =[P Q

]⎡⎣Ir 0

0 minusInminusr

⎤⎦⎡⎣P

T

QT

⎤⎦ (22)

where P isin Rntimesr Q isin R

ntimes(nminusr) By (22)

RP = P RQ = minusQ PTP = Ir QTQ = Inminusr PTQ = 0 QTP = 0 (23)

Throughout this paper we always assume that the nontrivial symmetric involutionmatrix R is fixed which is given by (22) and (23) Now we give a criterion of judging amatrix is R-conjugate Hermitian matrix

Theorem 21 A matrix K isin HRCntimesn if and only if there exists a symmetric matrix H isin R

ntimesn suchthat K = ΓHΓlowast where

Γ =[P iQ

] (24)

with PQ being the same as (22)

Proof If K isin HRCntimesn then K = RKR By (22)

K = RKR =[P Q

]⎡⎣Ir 0

0 minusInminusr

⎤⎦⎡⎣P

T

QT

⎤⎦K[P Q

]⎡⎣Ir 0

0 minusInminusr

⎤⎦⎡⎣P

T

QT

⎤⎦ (25)

4 Journal of Applied Mathematics

which is equivalent to

⎡⎣P

T

QT

⎤⎦K[P Q

]

=

⎡⎣Ir 0

0 minusInminusr

⎤⎦⎡⎣P

T

QT

⎤⎦K[P Q

]⎡⎣Ir 0

0 minusInminusr

⎤⎦

(26)

Suppose that

⎡⎣P

T

QT

⎤⎦K[P Q

]=

⎡⎣K11 K12

K21 K22

⎤⎦ (27)

Substituting (27) into (26) we obtain

⎡⎣K11 K12

K21 K22

⎤⎦ =

⎡⎣Ir 0

0 minusInminusr

⎤⎦⎡⎣K11 K12

K21 K22

⎤⎦⎡⎣Ir 0

0 minusInminusr

⎤⎦ =

⎡⎣ K11 minusK12

minusK21 K22

⎤⎦ (28)

Hence K11 = K11 K12 = minusK12 K21 = minusK21 K22 = K22 that is K11 iK12 iK21 K22 are realmatrices If we denote M = iK12 N = minusiK21 then by (27)

K =[P Q

]⎡⎣K11 K12

K21 K22

⎤⎦⎡⎣P

T

QT

⎤⎦ =

[P iQ

]⎡⎣K11 M

N K22

⎤⎦⎡⎣ PT

minusiQT

⎤⎦ (29)

Let Γ = [P iQ] and

H =

⎡⎣K11 M

N K22

⎤⎦ (210)

Then K can be expressed as ΓHΓlowast where Γ is unitary matrix and H is a real matrix ByK = Klowast

ΓHTΓlowast = Klowast = K = ΓHΓlowast (211)

we obtain H = HT

Journal of Applied Mathematics 5

Conversely if there exists a symmetric matrix H isin Rntimesn such that K = ΓHΓlowast then it

follows from (23) that

RKR = RΓHΓlowastR = R[P iQ

]H

⎡⎣ PT

minusiQT

⎤⎦R =

[P minusiQ]H

⎡⎣ PT

iQT

⎤⎦ = ΓHΓlowast = K

Klowast = ΓHTΓlowast = ΓHΓlowast = K

(212)

that is K isin HRCntimesn

Theorem 21 implies that an arbitrary complex Hermitian R-conjugate matrix is equiv-alent to a real symmetric matrix

Lemma 22 For any matrix A isin Cmtimesn A = A1 + iA2 where

A1 =A +A

2 A2 =

A minusA

2i (213)

Proof For any matrix A isin Cmtimesn it is obvious that A = A1 + iA2 where A1 A2 are defined

as (213) Now we prove that the decomposition A = A1 + iA2 is unique If there exist B1 B2

such that A = B1 + iB2 then

A1 minus B1 + i(B2 minusA2) = 0 (214)

It follows from A1 A2 B1 andB2 are real matrix that

A1 = B1 A2 = B2 (215)

Hence A = A1 + iA2 holds where A1 A2 are defined as (213)

By Theorem 21 for X isin HRCntimesn we may assume that

X = ΓYΓlowast (216)

where Γ is defined as (24) and Y isin Rntimesn is a symmetric matrix

Suppose that AΓ = A1 + iA2 isin Cmtimesn CΓ = C1 + iC2 isin C

mtimesn ΓlowastB = B1 + iB2 isin Cntimesl and

ΓlowastD = D1 + iD2 isin Cntimesl where

A1 =AΓ +AΓ

2 A2 =

AΓ minusAΓ2i

C1 =CΓ + CΓ

2 C2 =

CΓ minus CΓ2i

B1 =ΓlowastB + ΓlowastB

2 B2 =

ΓlowastB minus ΓlowastB2i

D1 =ΓlowastD + ΓlowastD

2 D2 =

ΓlowastD minus ΓlowastD2i

(217)

6 Journal of Applied Mathematics

Then system (12) can be reduced into

(A1 + iA2)Y = C1 + iC2 Y (B1 + iB2) = D1 + iD2 (218)

which implies that

⎡⎣A1

A2

⎤⎦Y =

⎡⎣C1

C2

⎤⎦ Y

[B1 B2

]=[D1 D2

] (219)

Let

F =

⎡⎣A1

A2

⎤⎦ G =

⎡⎣C1

C2

⎤⎦ K =

[B1 B2

]

L =[D1 D2

] M =

⎡⎣ F

KT

⎤⎦ N =

⎡⎣G

LT

⎤⎦

(220)

Then system (12) has a solution X in HRCntimesn if and only if the real system

MY = N (221)

has a symmetric solution Y in Rntimesn

Lemma 23 (Theorem 1 in [7]) Let A isin Rmtimesn The SVD of matrix A is as follows

A = U

⎡⎣Σ 0

0 0

⎤⎦V T (222)

where U = [U1 U2] isin Rmtimesm and V = [V1 V2] isin R

ntimesn are orthogonal matrices Σ = diag(σ1 σr) σi gt 0 (i = 1 r) r = rank(A) U1 isin R

mtimesr V1 isin Rntimesr Then (11) has a symmetric

solution if and only if

ABT = BAT UT2 B = 0 (223)

In that case it has the general solution

X = V1Σminus1UT1 B + V2V

T2 B

TU1Σminus1V T1 + V2GVT

2 (224)

where G is an arbitrary (n minus r) times (n minus r) symmetric matrix

By Lemma 23 we have the following theorem

Journal of Applied Mathematics 7

Theorem 24 Given A isin Cmtimesn C isin C

mtimesn B isin Cntimesl and D isin C

ntimesl Let A1 A2 C1C2B1 B2 D1 D2 F G K L M andN be defined in (217) (220) respectively Assume thatthe SVD of M isin R

(2m+2l)timesn is as follows

M = U

⎡⎣M1 0

0 0

⎤⎦V T (225)

where U = [U1 U2] isin R(2m+2l)times(2m+2l) and V = [V1 V2] isin R

ntimesn are orthogonal matrices M1 =diag(σ1 σr) σi gt 0 (i = 1 r) r = rank(M) U1 isin R

(2m+2l)timesr V1 isin Rntimesr Then system

(12) has a solution in HRCntimesn if and only if

MNT = NMT UT2 N = 0 (226)

In that case it has the general solution

X = Γ(V1M

minus11 UT

1 N + V2VT2 N

TU1Mminus11 V T

1 + V2GVT2

)Γlowast (227)

where G is an arbitrary (n minus r) times (n minus r) symmetric matrix

3 The Solution of Optimal Approximation Problem (13)

When the set SX of all Hermitian R-conjugate solution to (12) is nonempty it is easy to verifySX is a closed set Therefore the optimal approximation problem (13) has a unique solutionby [30]

Theorem 31 GivenA isin Cmtimesn C isin C

mtimesn B isin Cntimesl D isin C

ntimesl E isin Cntimesn and E1 = (12)(ΓlowastEΓ+

ΓlowastEΓ) Assume SX is nonempty then the optimal approximation problem (13) has a unique solutionX and

X = Γ(V1M

minus11 UT

1 N + V2VT2 N

TU1Mminus11 V T

1 + V2VT2 E1V2V

T2

)Γlowast (31)

Proof Since SX is nonempty X isin SX has the form of (227) By Lemma 22 ΓlowastEΓ can be writtenas

ΓlowastEΓ = E1 + iE2 (32)

where

E1 =12

(ΓlowastEΓ + ΓlowastEΓ

) E2 =

12i

(ΓlowastEΓ minus ΓlowastEΓ

) (33)

8 Journal of Applied Mathematics

According to (32) and the unitary invariance of Frobenius norm

X minus E =∥∥∥Γ(V1M

minus1UT1N + V2V

T2 N

TU1Mminus1V T

1 + V2GVT2

)Γlowast minus E

∥∥∥=∥∥∥(E1 minus V1M

minus1UT1N minus V2V

T2 N

TU1Mminus1V T

1 minus V2GVT2

)+ iE2

∥∥∥(34)

We get

X minus E2 =∥∥∥E1 minus V1M

minus1UT1 N minus V2V

T2 N

TU1Mminus1V T

1 minus V2GVT2

∥∥∥2+ E22 (35)

Then minXisinSXX minus E is consistent if and only if there exists G isin R(nminusr)times(nminusr) such that

min∥∥∥E1 minus V1M

minus1UT1 N minus V2V

T2 N

TU1Mminus1V T

1 minus V2GVT2

∥∥∥ (36)

For the orthogonal matrix V

∥∥∥E1 minus V1Mminus1UT

1N minus V2VT2 N

TU1Mminus1V T

1 minus V2GVT2

∥∥∥2

=∥∥∥V T (E1 minus V1M

minus1UT1 N minus V2V

T2 N

TU1Mminus1V T

1 minus V2GVT2 )V

∥∥∥2

=∥∥∥V T

1

(E1 minus V1M

minus1UT1 N)V1

∥∥∥2+∥∥∥V T

1 (E1 minus V1Mminus1UT

1 N)V2

∥∥∥2

+∥∥∥V T

2

(E1 minus V2V

T2 N

TU1Mminus1V T

1

)V1

∥∥∥2+∥∥∥V T

2

(E1 minus V2GVT

2

)V2

∥∥∥

(37)

Therefore

min∥∥∥E1 minus V1M

minus1UT1 N minus V2V

T2 N

TU1Mminus1V T

1 minus V2GVT2

∥∥∥ (38)

is equivalent to

G = V T2 E1V2 (39)

Substituting (39) into (227) we obtain (31)

4 The Solution of Problem (15)

In this section we give the explicit expression of the solution to (15)

Theorem 41 Given A isin Cmtimesn C isin C

mtimesn B isin Cntimesl and D isin C

ntimesl Let A1 A2 C1 C2B1 B2 D1 D2 F G K L M and N be defined in (217) (220) respectively Assume that the

Journal of Applied Mathematics 9

SVD of M isin R(2m+2l)timesn is as (225) and system (12) has not a solution in HRC

ntimesn Then X isin SL

can be expressed as

X = ΓV

⎡⎣Mminus1

1 UT1NV1 Mminus1

1 UT1NV2

V T2 N

TU1Mminus11 Y22

⎤⎦V TΓlowast (41)

where Y22 isin R(nminusr)times(nminusr) is an arbitrary symmetric matrix

Proof It yields from (217)ndash(221) and (225) that

AX minus C2 + XB minusD2 = MY minusN2

=

∥∥∥∥∥∥U⎡⎣M1 0

0 0

⎤⎦V TY minusN

∥∥∥∥∥∥2

=

∥∥∥∥∥∥⎡⎣M1 0

0 0

⎤⎦V TYV minusUTNV

∥∥∥∥∥∥2

(42)

Assume that

V TYV =

⎡⎣Y11 Y12

Y21 Y22

⎤⎦ Y11 isin R

rtimesr Y22 isin R(nminusr)times(nminusr) (43)

Then we have

AX minus C2 + XB minusD2

=∥∥∥M1Y11 minusUT

1 NV1

∥∥∥2+∥∥∥M1Y12 minusUT

1NV2

∥∥∥2

+∥∥∥UT

2NV1

∥∥∥2+∥∥∥UT

2 NV2

∥∥∥2

(44)

Hence

min(AX minus C2 + XB minusD2

)(45)

is solvable if and only if there exist Y11 Y12 such that

∥∥∥M1Y11 minusUT1 NV1

∥∥∥2= min

∥∥∥M1Y12 minusUT1NV2

∥∥∥2= min

(46)

10 Journal of Applied Mathematics

It follows from (46) that

Y11 = Mminus11 UT

1 NV1

Y12 = Mminus11 UT

1 NV2(47)

Substituting (47) into (43) and then into (216) we can get that the form of elements in SL is(41)

Theorem 42 Assume that the notations and conditions are the same as Theorem 41 Then

∥∥∥X∥∥∥ = minXisinSL

X (48)

if and only if

X = ΓV

⎡⎣Mminus1

1 UT1NV1 Mminus1

1 UT1NV2

V T2 N

TU1Mminus11 0

⎤⎦V TΓlowast (49)

Proof In Theorem 41 it implies from (41) that minXisinSLX is equivalent to X has theexpression (41) with Y22 = 0 Hence (49) holds

5 An Algorithm and Numerical Example

Base on the main results of this paper we in this section propose an algorithm for finding thesolution of the approximation problem (13) and the least squares problem with least norm(15) All the tests are performed by MATLAB 65 which has a machine precision of around10minus16

Algorithm 51 (1) Input A isin Cmtimesn C isin C

mtimesn B isin Cntimesl D isin C

ntimesl(2) Compute A1 A2 C1 C2 B1 B2 D1 D2 F G K L M andN by (217) and

(220)(3) Compute the singular value decomposition of M with the form of (225)(4) If (226) holds then input E isin C

ntimesn and compute the solution X of problem (13)according (31) else compute the solution X to problem (15) by (49)

To show our algorithm is feasible we give two numerical example Let an nontrivialsymmetric involution be

R =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 0 0 0

0 minus1 0 0

0 0 1 0

0 0 0 minus1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦ (51)

Journal of Applied Mathematics 11

We obtain [P Q] in (22) by using the spectral decomposition of R then by (24)

Γ =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

1 0 0 0

0 0 i 0

0 1 0 0

0 0 0 i

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦ (52)

Example 52 Suppose A isin C2times4 C isin C

2times4 B isin C4times3 D isin C

4times3 and

A =

⎡⎣333 minus 5987i 45i 721 minusi

0 minus066i 7694 1123i

⎤⎦

C =

⎡⎣02679 minus 00934i 00012 + 40762i minus00777 minus 01718i minus12801i

02207 minus01197i 00877 07058i

⎤⎦

B =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

4 + 12i 2369i 4256 minus 5111i

4i 466i 821 minus 5i

0 483i 56 + i

222i minus4666 7i

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

D =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

00616 + 01872i minus00009 + 01756i 16746 minus 00494i

00024 + 02704i 01775 + 04194i 07359 minus 06189i

minus00548 + 03444i 00093 minus 03075i minus04731 minus 01636i

00337i 01209 minus 01864i minus02484 minus 38817i

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦

(53)

We can verify that (226) holds Hence system (12) has an Hermitian R-conjugate solutionGiven

E =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

735i 8389i 99256 minus 651i minus46i

155 456i 771 minus 75i i

5i 0 minus4556i minus799

422i 0 51i 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦ (54)

12 Journal of Applied Mathematics

Applying Algorithm 51 we obtain the following

X =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

15597 00194i 28705 00002i

minus00194i 90001 02005i minus39997

28705 minus02005i minus00452 79993i

minus00002i minus39997 minus79993i 56846

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦ (55)

Example 52 illustrates that we can solve the optimal approximation problem withAlgorithm 51 when system (12) have Hermitian R-conjugate solutions

Example 53 Let A B andC be the same as Example 52 and let D in Example 52 be changedinto

D =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

00616 + 01872i minus00009 + 01756i 16746 + 00494i

00024 + 02704i 01775 + 04194i 07359 minus 06189i

minus00548 + 03444i 00093 minus 03075i minus04731 minus 01636i

00337i 01209 minus 01864i minus02484 minus 38817i

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦ (56)

We can verify that (226) does not hold By Algorithm 51 we get

X =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

052 22417i 04914 03991i

minus22417i 86634 01921i minus28232

04914 minus01921i 01406 13154i

minus03991i minus28232 minus13154i 63974

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦ (57)

Example 53 demonstrates that we can get the least squares solution with Algo-rithm 51 when system (12) has not Hermitian R-conjugate solutions

Acknowledgments

This research was supported by the Grants from the Key Project of Scientific ResearchInnovation Foundation of Shanghai Municipal Education Commission (13ZZ080) theNational Natural Science Foundation of China (11171205) the Natural Science Foundationof Shanghai (11ZR1412500) the PhD Programs Foundation of Ministry of Education ofChina (20093108110001) the Discipline Project at the corresponding level of Shanghai (A13-0101-12-005) Shanghai Leading Academic Discipline Project (J50101) the Natural ScienceFoundation of Hebei province (A2012403013) and the Natural Science Foundation of Hebeiprovince (A2012205028) The authors are grateful to the anonymous referees for their helpfulcomments and constructive suggestions

Journal of Applied Mathematics 13

References

[1] R D Hill R G Bates and S R Waters ldquoOn centro-Hermitian matricesrdquo SIAM Journal on MatrixAnalysis and Applications vol 11 no 1 pp 128ndash133 1990

[2] R Kouassi P Gouton and M Paindavoine ldquoApproximation of the Karhunen-Loeve tranformationand Its application to colour imagesrdquo Signal Processing Image Communication vol 16 pp 541ndash5512001

[3] A Lee ldquoCentro-Hermitian and skew-centro-Hermitian matricesrdquo Linear Algebra and Its Applicationsvol 29 pp 205ndash210 1980

[4] Z-Y Liu H-D Cao and H-J Chen ldquoA note on computing matrix-vector products with generalizedcentrosymmetric (centrohermitian) matricesrdquo Applied Mathematics and Computation vol 169 no 2pp 1332ndash1345 2005

[5] W F Trench ldquoCharacterization and properties of matrices with generalized symmetry or skewsymmetryrdquo Linear Algebra and Its Applications vol 377 pp 207ndash218 2004

[6] K-W E Chu ldquoSymmetric solutions of linear matrix equations by matrix decompositionsrdquo LinearAlgebra and Its Applications vol 119 pp 35ndash50 1989

[7] H Dai ldquoOn the symmetric solutions of linear matrix equationsrdquo Linear Algebra and Its Applicationsvol 131 pp 1ndash7 1990

[8] F J H Don ldquoOn the symmetric solutions of a linear matrix equationrdquo Linear Algebra and ItsApplications vol 93 pp 1ndash7 1987

[9] I Kyrchei ldquoExplicit representation formulas for the minimum norm least squares solutions of somequaternion matrix equationsrdquo Linear Algebra and Its Applications vol 438 no 1 pp 136ndash152 2013

[10] Y Li Y Gao and W Guo ldquoA Hermitian least squares solution of the matrix equation AXB = Csubject to inequality restrictionsrdquo Computers amp Mathematics with Applications vol 64 no 6 pp 1752ndash1760 2012

[11] Z-Y Peng and X-Y Hu ldquoThe reflexive and anti-reflexive solutions of the matrix equation AX = BrdquoLinear Algebra and Its Applications vol 375 pp 147ndash155 2003

[12] L Wu ldquoThe re-positive definite solutions to the matrix inverse problem AX = Brdquo Linear Algebra andIts Applications vol 174 pp 145ndash151 1992

[13] S F Yuan Q W Wang and X Zhang ldquoLeast-squares problem for the quaternion matrix equationAXB + CYD = E over different constrained matricesrdquo Article ID 722626 International Journal ofComputer Mathematics In press

[14] Z-Z Zhang X-Y Hu and L Zhang ldquoOn the Hermitian-generalized Hamiltonian solutions of linearmatrix equationsrdquo SIAM Journal on Matrix Analysis and Applications vol 27 no 1 pp 294ndash303 2005

[15] F Cecioni ldquoSopra operazioni algebricherdquo Annali della Scuola Normale Superiore di Pisa vol 11 pp17ndash20 1910

[16] K-W E Chu ldquoSingular value and generalized singular value decompositions and the solution oflinear matrix equationsrdquo Linear Algebra and Its Applications vol 88-89 pp 83ndash98 1987

[17] S K Mitra ldquoA pair of simultaneous linear matrix equations A1XB1 = C1 A2XB2 = C2 and a matrixprogramming problemrdquo Linear Algebra and Its Applications vol 131 pp 107ndash123 1990

[18] H-X Chang and Q-W Wang ldquoReflexive solution to a system of matrix equationsrdquo Journal of ShanghaiUniversity vol 11 no 4 pp 355ndash358 2007

[19] A Dajic and J J Koliha ldquoEquations ax = c and xb = d in rings and rings with involution withapplications to Hilbert space operatorsrdquo Linear Algebra and Its Applications vol 429 no 7 pp 1779ndash1809 2008

[20] A Dajic and J J Koliha ldquoPositive solutions to the equations AX = C and XB = D for Hilbert spaceoperatorsrdquo Journal of Mathematical Analysis and Applications vol 333 no 2 pp 567ndash576 2007

[21] C-Z Dong Q-W Wang and Y-P Zhang ldquoThe common positive solution to adjointable operatorequations with an applicationrdquo Journal of Mathematical Analysis and Applications vol 396 no 2 pp670ndash679 2012

[22] X Fang J Yu and H Yao ldquoSolutions to operator equations on Hilbert Clowast-modulesrdquo Linear Algebraand Its Applications vol 431 no 11 pp 2142ndash2153 2009

[23] C G Khatri and S K Mitra ldquoHermitian and nonnegative definite solutions of linear matrixequationsrdquo SIAM Journal on Applied Mathematics vol 31 no 4 pp 579ndash585 1976

[24] I Kyrchei ldquoAnalogs of Cramerrsquos rule for the minimum norm least squares solutions of some matrixequationsrdquo Applied Mathematics and Computation vol 218 no 11 pp 6375ndash6384 2012

[25] F L Li X Y Hu and L Zhang ldquoThe generalized reflexive solution for a class of matrix equations(AX = CXB = D)rdquo Acta Mathematica Scientia vol 28B no 1 pp 185ndash193 2008

14 Journal of Applied Mathematics

[26] Q-W Wang and C-Z Dong ldquoPositive solutions to a system of adjointable operator equations overHilbert Clowast-modulesrdquo Linear Algebra and Its Applications vol 433 no 7 pp 1481ndash1489 2010

[27] Q Xu ldquoCommon Hermitian and positive solutions to the adjointable operator equations AX = CXB = Drdquo Linear Algebra and Its Applications vol 429 no 1 pp 1ndash11 2008

[28] F Zhang Y Li and J Zhao ldquoCommon Hermitian least squares solutions of matrix equations A1XAlowast1 =

B1 and A2XAlowast2 = B2 subject to inequality restrictionsrdquo Computers ampMathematics with Applications vol

62 no 6 pp 2424ndash2433 2011[29] H-X Chang Q-W Wang and G-J Song ldquo(R S)-conjugate solution to a pair of linear matrix

equationsrdquo Applied Mathematics and Computation vol 217 no 1 pp 73ndash82 2010[30] E W Cheney Introduction to Approximation Theory McGraw-Hill New York NY USA 1966

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2012 Article ID 784620 13 pagesdoi1011552012784620

Research ArticlePerturbation Analysis for the Matrix EquationX minussumm

i=1 Alowasti XAi +

sumnj=1 B

lowastj XBj = I

Xue-Feng Duan1 2 and Qing-Wen Wang3

1 School of Mathematics and Computational Science Guilin University of Electronic TechnologyGuilin 541004 China

2 Research Center on Data Science and Social Computing Guilin University of Electronic TechnologyGuilin 541004 China

3 Department of Mathematics Shanghai University Shanghai 200444 China

Correspondence should be addressed to Qing-Wen Wang wqw858yahoocomcn

Received 18 September 2012 Accepted 19 November 2012

Academic Editor Yang Zhang

Copyright q 2012 X-F Duan and Q-W Wang This is an open access article distributed underthe Creative Commons Attribution License which permits unrestricted use distribution andreproduction in any medium provided the original work is properly cited

We consider the perturbation analysis of the matrix equation X minussummi=1 A

lowasti XAi +

sumnj=1 B

lowastj XBj = I

Based on the matrix differentiation we first give a precise perturbation bound for the positivedefinite solution A numerical example is presented to illustrate the sharpness of the perturbationbound

1 Introduction

In this paper we consider the matrix equation

X minusmsumi=1

Alowasti XAi +

nsumj=1

Blowastj XBj = I (11)

where A1 A2 Am B1 B2 Bn are n times n complex matrices I is an n times n identity matrixm n are nonnegative integers and the positive definite solution X is practical interestHere Alowast

i and Blowasti denote the conjugate transpose of the matrices Ai and Bi respectively

Equation (11) arises in solving some nonlinear matrix equations with Newton method Seefor example the nonlinear matrix equation which appears in Sakhnovich [1] Solving these

2 Journal of Applied Mathematics

nonlinear matrix equations gives rise to (11) On the other hand (11) is the general case ofthe generalized Lyapunov equation

MYSlowast + SYMlowast +tsum

k=1

NkYNlowastk + CClowast = 0 (12)

whose positive definite solution is the controllability Gramian of the bilinear control system(see [2 3] for more details)

Mx(t) = Sx(t) +tsum

k=1

Nkx(t)uk(t) + Cu(t) (13)

Set

E =1radic2(M minus S + I) F =

1radic2(M + S + I) G = S minus I Q = CClowast

X = Qminus12YQminus12 Ai = Qminus12NiQ12 i = 1 2 t

At+1 = Qminus12FQ12 At+2 = Qminus12GQ12

B1 = Qminus12EQ12 B2 = Qminus12CQ12

(14)

then (12) can be equivalently written as (11) with m = t + 2 and n = 2Some special cases of (11) have been studied Based on the kronecker product and

fixed point theorem in partially ordered sets Reurings [4] and Ran and Reurings [5 6] gavesome sufficient conditions for the existence of a unique positive definite solution of the linearmatrix equations X minussumm

i=1 Alowasti XAi = I and X +

sumnj=1 B

lowasti XBi = I And the expressions for these

unique positive definite solutions were also derived under some constraint conditions Forthe general linear matrix equation (11) Reurings [[4] Page 61] pointed out that it is hard tofind sufficient conditions for the existence of a positive definite solution because the map

G(X) = I +msumi=1

Alowasti XAi minus

nsumj=1

Blowastj XBj (15)

is not monotone and does not map the set of n times n positive definite matrices into itselfRecently Berzig [7] overcame these difficulties by making use of Bhaskar-Lakshmikanthamcoupled fixed point theorem and gave a sufficient condition for (11) existing a uniquepositive definite solution An iterative method was constructed to compute the uniquepositive definite solution and the error estimation was given too

Recently the matrix equations of the form (11) have been studied by many authors(see [8ndash14]) Some numerical methods for solving the well-known Lyapunov equationX + ATXA = Q such as Bartels-Stewart method and Hessenberg-Schur method have beenproposed in [12] Based on the fixed point theorem the sufficient and necessary conditions forthe existence of a positive definite solution of the matrix equation Xs plusmnAlowastXminustA = Q s t isin Nhave been given in [8 9] The fixed point iterative method and inversion-free iterative method

Journal of Applied Mathematics 3

were developed for solving the matrix equations X plusmn AlowastXminusαA = Q α gt 0 in [13 14] Bymaking use of the fixed point theorem of mixed monotone operator the matrix equationX minussumm

i=1 Alowasti X

δiAi = Q 0 lt |δi| lt 1 was studied in [10] and derived a sufficient condition forthe existence of a positive definite solution Assume that F maps positive definite matriceseither into positive definite matrices or into negative definite matrices the general nonlinearmatrix equation X +AlowastF(X)A = Q was studied in [11] and the fixed point iterative methodwas constructed to compute the positive definite solution under some additional conditions

Motivated by the works and applications in [2ndash6] we continue to study the matrixequation (11) Based on a new mathematical tool (ie the matrix differentiation) we firstlygive a differential bound for the unique positive definite solution of (11) and then use it toderive a precise perturbation bound for the unique positive definite solution A numericalexample is used to show that the perturbation bound is very sharp

Throughout this paper we write B gt 0 (B ge 0) if the matrix B is positive definite(semidefinite) If B minus C is positive definite (semidefinite) then we write B gt C (B ge C) Ifa positive definite matrix X satisfies B le X le C we denote that X isin [BC] The symbolsλ1(B) and λn(B) denote the maximal and minimal eigenvalues of an n times n Hermitian matrixB respectively The symbol Hntimesn stands for the set of n times n Hermitian matrices The symbolB denotes the spectral norm of the matrix B

2 Perturbation Analysis for the Matrix Equation (11)

Based on the matrix differentiation we firstly give a differential bound for the unique positivedefinite solution XU of (11) and then use it to derive a precise perturbation bound for XU inthis section

Definition 21 ([15] Definition 36) Let F = (fij)mtimesn then the matrix differentiation of F isdF = (dfij)mtimesn For example let

F =(

s + t s2 minus 2t2s + t3 t2

) (21)

Then

dF =(

ds + dt 2sds minus 2dt2ds + 3t2dt 2tdt

) (22)

Lemma 22 ([15] Theorem 32) The matrix differentiation has the following properties

(1) d(F1 plusmn F2) = dF1 plusmn dF2

(2) d(kF) = k(dF) where k is a complex number

(3) d(Flowast) = (dF)lowast

(4) d(F1F2F3) = (dF1)F2F3 + F1(dF2)F3 + F1F2(dF3)

(5) dFminus1 = minusFminus1(dF)Fminus1

(6) dF = 0 where F is a constant matrix

4 Journal of Applied Mathematics

Lemma 23 ([7] Theorem 31) If

msumi=1

Alowasti Ai lt

12I

nsumj=1

Blowastj Bj lt

12I (23)

then (11) has a unique positive definite solution XU and

XU isin⎡⎣I minus 2

nsumj=1

Blowastj Bj I + 2

msumi=1

Alowasti Ai

⎤⎦ (24)

Theorem 24 If

msumi=1

Ai2 lt12

nsumj=1

∥∥Bj

∥∥2lt

12 (25)

then (11) has a unique positive definite solution XU and it satisfies

dXU le2(

1 + 2summ

i=1 Ai2)[summ

i=1(AidAi) +sumn

j=1(∥∥Bj

∥∥∥∥dBj

∥∥)]

1 minussummi=1 Ai2 minussumn

j=1

∥∥Bj

∥∥2 (26)

Proof Since

λ1(Alowast

i Ai

) le ∥∥Alowasti Ai

∥∥ le Ai2 i = 1 2 m

λ1

(Blowastj Bj

)le∥∥∥Blowast

j Bj

∥∥∥ le ∥∥Bj

∥∥2 j = 1 2 n

(27)

then

Alowasti Ai le λ1

(Alowast

i Ai

)I le ∥∥Alowast

i Ai

∥∥I le Ai2I i = 1 2 m

Blowastj Bj le λ1

(Blowastj Bj

)I le∥∥∥Blowast

j Bj

∥∥∥ le ∥∥Bj

∥∥2I j = 1 2 n

(28)

consequently

msumi=1

Alowasti Ai le

msumi=1

Ai2I

nsumj=1

Blowastj Bj le

nsumj=1

∥∥Bj

∥∥2I

(29)

Journal of Applied Mathematics 5

Combining (25)ndash(29) we have

msumi=1

Alowasti Ai le

msumi=1

Ai2I lt12I

nsumj=1

Blowastj Bj le

nsumj=1

∥∥Bj

∥∥2I lt

12I

(210)

Then by Lemma 23 we obtain that (11) has a unique positive definite solution XU whichsatisfies

XU isin⎡⎣I minus 2

nsumj=1

Blowastj Bj I + 2

msumi=1

Alowasti Ai

⎤⎦ (211)

Noting that XU is the unique positive definite solution of (11) then

XU minusmsumi=1

Alowasti XUAi +

nsumj=1

Blowastj XUBj = I (212)

It is known that the elements of XU are differentiable functions of the elements of Ai and BiDifferentiating (212) and by Lemma 22 we have

dXU minusmsumi=1

[(dAlowast

i

)XUAi +Alowast

i (dXU)Ai +Alowasti XU(dAi)

]

+nsumj=1

[(dBlowast

j

)XUBj + Blowast

j (dXU)Bj + Blowastj XU

(dBj

)]= 0

(213)

which implies that

dXU minusmsumi=1

Alowasti (dXU)Ai +

nsumj=1

Blowastj (dXU)Bj

=msumi=1

(dAlowast

i

)XUAi +

msumi=1

Alowasti XU(dAi) minus

nsumj=1

(dBlowast

j

)XUBj minus

nsumj=1

Blowastj XU

(dBj

)

(214)

6 Journal of Applied Mathematics

By taking spectral norm for both sides of (214) we obtain that

∥∥∥∥∥∥dXU minusmsumi=1

Alowasti (dXU)Ai +

nsumj=1

Blowastj (dXU)Bj

∥∥∥∥∥∥

=

∥∥∥∥∥∥msumi=1

(dAlowast

i

)XUAi +

msumi=1

Alowasti XU(dAi) minus

nsumj=1

(dBlowast

j

)XUBj minus

nsumj=1

Blowastj XU

(dBj

)∥∥∥∥∥∥

le∥∥∥∥∥

msumi=1

(dAlowast

i

)XUAi

∥∥∥∥∥ +∥∥∥∥∥

msumi=1

Alowasti XU(dAi)

∥∥∥∥∥ +∥∥∥∥∥∥

nsumj=1

(dBlowast

j

)XUBj

∥∥∥∥∥∥ +∥∥∥∥∥∥

nsumj=1

Blowastj XU

(dBj

)∥∥∥∥∥∥

lemsumi=1

∥∥(dAlowasti

)XUAi

∥∥ + msumi=1

∥∥Alowasti XU(dAi)

∥∥ + nsumj=1

∥∥∥(dBlowastj

)XUBj

∥∥∥ +nsumj=1

∥∥∥Blowastj XU

(dBj

)∥∥∥

lemsumi=1

∥∥dAlowasti

∥∥XUAi +msumi=1

∥∥Alowasti

∥∥XUdAi +nsumj=1

∥∥∥dBlowastj

∥∥∥XU∥∥Bj

∥∥ + nsumj=1

∥∥∥Blowastj

∥∥∥XU∥∥dBj

∥∥

= 2msumi=1

(AiXUdAi) + 2nsumj=1

(∥∥Bj

∥∥XU∥∥dBj

∥∥)

= 2

⎡⎣ msum

i=1

(AidAi) +nsumj=1

(∥∥Bj

∥∥∥∥dBj

∥∥)⎤⎦XU

(215)

and noting (211) we obtain that

XU le∥∥∥∥∥I + 2

msumi=1

Alowasti Ai

∥∥∥∥∥ le 1 + 2msumi=1

Ai2 (216)

Then

∥∥∥∥∥∥dXU minusmsumi=1

Alowasti (dXU)Ai +

nsumj=1

Blowastj (dXU)Bj

∥∥∥∥∥∥

le 2

⎡⎣ msum

i=1

(AidAi) +nsumj=1

(∥∥Bj

∥∥∥∥dBj

∥∥)⎤⎦XU

le 2

(1 + 2

msumi=1

Ai2

)⎡⎣ msum

i=1

(AidAi) +nsumj=1

(∥∥Bj

∥∥∥∥dBj

∥∥)⎤⎦

(217)

Journal of Applied Mathematics 7∥∥∥∥∥∥dXU minus

msumi=1

Alowasti (dXU)Ai +

nsumj=1

Blowastj (dXU)Bj

∥∥∥∥∥∥

ge dXU minus∥∥∥∥∥

msumi=1

Alowasti (dXU)Ai

∥∥∥∥∥ minus∥∥∥∥∥∥

nsumj=1

Blowastj (dXU)Bj

∥∥∥∥∥∥

ge dXU minusmsumi=1

∥∥Alowasti (dXU)Ai

∥∥ minus nsumj=1

∥∥∥Blowastj (dXU)Bj

∥∥∥

ge dXU minusmsumi=1

∥∥Alowasti

∥∥dXUAi minusnsumj=1

∥∥∥Blowastj

∥∥∥dXU∥∥Bj

∥∥

=

⎛⎝1 minus

msumi=1

Ai2 minusnsumj=1

∥∥Bj

∥∥2

⎞⎠dXU

(218)

Due to (25) we have

1 minusmsumi=1

Ai2 minusnsumj=1

∥∥Bj

∥∥2gt 0 (219)

Combining (217) (218) and noting (219) we have

⎛⎝1 minus

msumi=1

Ai2 minusnsumj=1

∥∥Bj

∥∥2

⎞⎠dXU

le∥∥∥∥∥∥dXU minus

msumi=1

Alowasti (dXU)Ai +

nsumj=1

Blowastj (dXU)Bj

∥∥∥∥∥∥

le 2

(1 + 2

msumi=1

Ai2

)⎡⎣ msum

i=1

(AidAi) +nsumj=1

(∥∥Bj

∥∥∥∥dBj

∥∥)⎤⎦

(220)

which implies that

dXU le2(

1 + 2summ

i=1 Ai2)[summ

i=1(AidAi) +sumn

j=1(∥∥Bj

∥∥∥∥dBj

∥∥)]

1 minussummi=1 Ai2 minussumn

j=1

∥∥Bj

∥∥2 (221)

8 Journal of Applied Mathematics

Theorem 25 Let A1 A2 Am B1 B2 Bn be perturbed matrices of A1 A2 AmB1 B2 Bn in (11) and Δi = Ai minusAi i = 1 2 m Δj = Bj minus Bj j = 1 2 n If

msumi=1

Ai2 lt12

nsumj=1

∥∥Bj

∥∥2lt

12 (222)

2msumi=1

(AiΔi) +msumi=1

Ai2 lt12minus

msumi=1

Ai2 (223)

2nsumj=1

(∥∥Bj

∥∥∥∥Δj

∥∥) + nsumj=1

∥∥Δj

∥∥2lt

12minus

nsumj=1

∥∥Bj

∥∥2 (224)

then (11) and its perturbed equation

X minusmsumi=1

Alowasti XAi +

nsumj=1

Blowastj XBj = I (225)

have unique positive definite solutions XU and XU respectively which satisfy

∥∥∥XU minusXU

∥∥∥ le Serr (226)

where

Serr =2[1 + 2

summi=1 (Ai + Δi)2

][summi=1 (Ai + Δi)2Δi +

sumnj=1(∥∥Bj

∥∥ + ∥∥Δj

∥∥)2∥∥Δj

∥∥]

1 minussummi=1 (Ai + Δi)2 minussumn

j=1(∥∥Bj

∥∥ + ∥∥Δj

∥∥)2

(227)

Proof By (222) and Theorem 24 we know that (11) has a unique positive definite solutionXU And by (223) we have

msumi=1

∥∥∥Ai

∥∥∥2=

msumi=1

Ai + Δi2 lemsumi=1

(Ai + Δi)2

=msumi=1

(Ai2 + 2AiΔi + Δi2

)

=msumi=1

Ai2 + 2msumi=1

(AiΔi) +msumi=1

Δi2

ltmsumi=1

Ai2 +12minus

msumi=1

Ai2 =12

(228)

Journal of Applied Mathematics 9

similarly by (224) we have

nsumj=1

∥∥∥Bj

∥∥∥2lt

12 (229)

By (228) (229) and Theorem 24 we obtain that the perturbed equation (225) has a uniquepositive definite solution XU

Set

Ai(t) = Ai + tΔi Bj(t) = Bj + tΔj t isin [0 1] (230)

then by (223) we have

msumi=1

Ai(t)2 =msumi=1

Ai + tΔi2 lemsumi=1

(Ai + tΔi)2

=msumi=1

(Ai2 + 2tAiΔi + t2Δi2

)

lemsumi=1

(Ai2 + 2AiΔi + Δi2

)

=msumi=1

Ai2 + 2msumi=1

(AiΔi) +msumi=1

Δi2

ltmsumi=1

Ai2 +12minus

msumi=1

Ai2 =12

(231)

similarly by (224) we have

nsumj=1

∥∥Bj(t)∥∥2 =

msumi=1

∥∥Bj + tΔj

∥∥2lt

12 (232)

Therefore by (231) (232) and Theorem 24 we derive that for arbitrary t isin [0 1] thematrix equation

X minusmsumi=1

Alowasti (t)XAi(t) +

nsumj=1

Blowastj (t)XBj(t) = I (233)

has a unique positive definite solution XU(t) especially

XU(0) = XU XU(1) = XU (234)

10 Journal of Applied Mathematics

From Theorem 24 it follows that

∥∥∥XU minusXU

∥∥∥ = XU(1) minusXU(0) =

∥∥∥∥∥int1

0dXU(t)

∥∥∥∥∥ leint1

0dXU(t)

leint1

0

2(

1 + 2summ

i=1 Ai(t)2)[summ

i=1(Ai(t)dAi(t)) +sumn

j=1(∥∥Bj(t)

∥∥∥∥dBj(t)∥∥)]

1 minussummi=1 Ai(t)2 minussumn

j=1

∥∥Bj(t)∥∥2

leint1

0

2(

1 + 2summ

i=1 Ai(t)2)[summ

i=1(Ai(t)Δidt) +sumn

j=1(∥∥Bj(t)

∥∥∥∥Δj

∥∥dt)]

1 minussummi=1 Ai(t)2 minussumn

j=1

∥∥Bj(t)∥∥2

=int1

0

2(

1 + 2summ

i=1 Ai(t)2)[summ

i=1(Ai(t)Δi) +sumn

j=1(∥∥Bj(t)

∥∥∥∥Δj

∥∥)]

1 minussummi=1 Ai(t)2 minussumn

j=1

∥∥Bj(t)∥∥2

dt

(235)

Noting that

Ai(t) = Ai + tΔi le Ai + tΔi i = 1 2 m∥∥Bj(t)

∥∥ =∥∥Bj + tΔi

∥∥ le ∥∥Bj

∥∥ + tΔi j = 1 2 n(236)

and combining Mean Value Theorem of Integration we have

∥∥∥XU minusXU

∥∥∥

leint1

0

2(

1 + 2summ

i=1 Ai(t)2)[summ

i=1(Ai(t)Δi) +sumn

j=1(∥∥Bj(t)

∥∥∥∥Δj

∥∥)]

1 minussummi=1 Ai(t)2 minussumn

j=1

∥∥Bj(t)∥∥2

dt

leint1

0

2[1 + 2

summi=1 (Ai + tΔi)2

][summi=1 (Ai + tΔi)2Δi +

sumnj=1(∥∥Bj

∥∥ + t∥∥Δj

∥∥)2∥∥Δj

∥∥]

1 minussummi=1 (Ai + tΔi)2 minussumn

j=1(∥∥Bj

∥∥ + t∥∥Δj

∥∥)2dt

=2[1 + 2

summi=1 (Ai + ξΔi)2

][summi=1 (Ai + ξΔi)2Δi +

sumnj=1(∥∥Bj

∥∥ + ξ∥∥Δj

∥∥)2∥∥Δj

∥∥]

1 minussummi=1 (Ai + ξΔi)2 minussumn

j=1(∥∥Bj

∥∥ + ξ∥∥Δj

∥∥)2

times (1 minus 0) ξ isin [0 1]

le2[1 + 2

summi=1 (Ai + Δi)2

][summi=1 (Ai + Δi)2Δi +

sumnj=1(∥∥Bj

∥∥ + ∥∥Δj

∥∥)2∥∥Δj

∥∥]

1 minussummi=1 (Ai + Δi)2 minussumn

j=1(∥∥Bj

∥∥ + ∥∥Δj

∥∥)2= Serr

(237)

Journal of Applied Mathematics 11

3 Numerical Experiments

In this section we use a numerical example to confirm the correctness of Theorem 25 and theprecision of the perturbation bound for the unique positive definite solution XU of (11)

Example 31 Consider the symmetric linear matrix equation

X minusAlowast1XA1 minusAlowast

2XA2 + Blowast1XB1 + Blowast

2XB2 = I (31)

and its perturbed equation

X minus Alowast1XA1 minus Alowast

2XA2 + Blowast1XB1 + Blowast

2XB2 = I (32)

where

A1 =

⎛⎝ 002 minus010 minus002

008 minus010 002minus006 minus012 014

⎞⎠ A2 =

⎛⎝ 008 minus010 minus002

008 minus010 002minus006 minus012 014

⎞⎠

B1 =

⎛⎝ 047 002 004

minus010 036 minus002minus004 001 047

⎞⎠ B2 =

⎛⎝010 010 005

015 0275 0075005 005 0175

⎞⎠

A1 = A1 +

⎛⎝ 05 01 minus02

minus04 02 06minus02 01 minus01

⎞⎠ times 10minusj A2 = A2 +

⎛⎝minus04 01 minus02

05 07 minus1311 09 06

⎞⎠ times 10minusj

B1 = B1 +

⎛⎝ 08 02 005

minus02 012 014minus025 minus02 026

⎞⎠ times 10minusj B2 = B2 +

⎛⎝ 02 02 01

minus03 015 minus01501 minus01 025

⎞⎠ times 10minusj j isin N

(33)

It is easy to verify that the conditions (222)ndash(224) are satisfied then (31) and itsperturbed equation (32) have unique positive definite solutions XU and XU respectivelyFrom Berzig [7] it follows that the sequences Xk and Yk generated by the iterative method

X0 = 0 Y0 = 2I

Xk+1 = I +Alowast1XkA1 +Alowast

2XkA2 minus Blowast1YkB1 minus Blowast

2YkB2

Yk+1 = I +Alowast1YkA1 +Alowast

2YkA2 minus Blowast1XkB1 minus Blowast

2XkB2 k = 0 1 2

(34)

both converge to XU Choose τ = 10 times 10minus15 as the termination scalar that is

R(X) =∥∥X minusAlowast

1XA1 minusAlowast2XA2 + Blowast

1XB1 + Blowast2XB2 minus I

∥∥ le τ = 10 times 10minus15 (35)

By using the iterative method (34) we can get the computed solution Xk of (31) SinceR(Xk) lt 10 times 10minus15 then the computed solution Xk has a very high precision For simplicity

12 Journal of Applied Mathematics

Table 1 Numerical results for the different values of j

j 2 3 4 5 6

XU minusXUXU 213 times 10minus4 616 times 10minus6 523 times 10minus8 437 times 10minus10 845 times 10minus12

SerrXU 342 times 10minus4 713 times 10minus6 663 times 10minus8 512 times 10minus10 977 times 10minus12

we write the computed solution as the unique positive definite solution XU Similarly we canalso get the unique positive definite solution XU of the perturbed equation (32)

Some numerical results on the perturbation bounds for the unique positive definitesolution XU are listed in Table 1

From Table 1 we see that Theorem 25 gives a precise perturbation bound for theunique positive definite solution of (31)

4 Conclusion

In this paper we study the matrix equation (11) which arises in solving some nonlinearmatrix equations and the bilinear control system A new method of perturbation analysisis developed for the matrix equation (11) By making use of the matrix differentiationand its elegant properties we derive a precise perturbation bound for the unique positivedefinite solution of (11) A numerical example is presented to illustrate the sharpness of theperturbation bound

Acknowledgments

This research was supported by the National Natural Science Foundation of China(11101100 11171205 11261014 11226323) the Natural Science Foundation of GuangxiProvince (2012GXNSFBA053006 2011GXNSFA018138) the Key Project of Scientific ResearchInnovation Foundation of Shanghai Municipal Education Commission (13ZZ080) theNatural Science Foundation of Shanghai (11ZR1412500) the PhD Programs Foundation ofMinistry of Education of China (20093108110001) the Discipline Project at the correspondinglevel of Shanghai (A 13-0101-12-005) and Shanghai Leading Academic Discipline Project(J50101) The authors wish to thank Professor Y Zhang and the anonymous referees forproviding very useful suggestions for improving this paper The authors also thank DrXiaoshan Chen for discussing the properties of matrix differentiation

References

[1] L A Sakhnovich Interpolation Theory and Its Applications Mathematics and Its Applications KluwerAcademic Publishers Dordrecht The Netherlands 1997

[2] T Damm ldquoDirect methods and ADI-preconditioned Krylov subspace methods for generalizedLyapunov equationsrdquo Numerical Linear Algebra with Applications vol 15 no 9 pp 853ndash871 2008

[3] L Zhang and J Lam ldquoOn H2 model reduction of bilinear systemsrdquo Automatica vol 38 no 2 pp205ndash216 2002

[4] M C B Reurings Symmetric matrix equations [PhD thesis] Vrije Universiteit Amsterdam TheNetherlands 2003

[5] A C M Ran and M C B Reurings ldquoThe symmetric linear matrix equationrdquo Electronic Journal ofLinear Algebra vol 9 pp 93ndash107 2002

Journal of Applied Mathematics 13

[6] A C M Ran and M C B Reurings ldquoA fixed point theorem in partially ordered sets and someapplications to matrix equationsrdquo Proceedings of the American Mathematical Society vol 132 no 5 pp1435ndash1443 2004

[7] M Berzig ldquoSolving a class of matrix equation via Bhaskar-Lakshmikantham coupled fixed pointtheoremrdquo Applied Mathematics Letters vol 25 no 11 pp 1638ndash1643 2012

[8] J Cai and G Chen ldquoSome investigation on Hermitian positive definite solutions of the matrixequation Xs +AlowastXminustA = Qrdquo Linear Algebra and Its Applications vol 430 no 8-9 pp 2448ndash2456 2009

[9] X Duan and A Liao ldquoOn the existence of Hermitian positive definite solutions of the matrix equationXs +AlowastXminustA = Qrdquo Linear Algebra and its Applications vol 429 no 4 pp 673ndash687 2008

[10] X Duan A Liao and B Tang ldquoOn the nonlinear matrix equation X minus summi=1 A

lowasti X

δiAi = Qrdquo LinearAlgebra and its Applications vol 429 no 1 pp 110ndash121 2008

[11] S M El-Sayed and A C M Ran ldquoOn an iteration method for solving a class of nonlinear matrixequationsrdquo SIAM Journal on Matrix Analysis and Applications vol 23 no 3 pp 632ndash645 2001

[12] Z Gajic and M T J Qureshi Lyapunov Matrix Equation in System Stability and Control Academic PressLondon UK 1995

[13] V I Hasanov ldquoNotes on two perturbation estimates of the extreme solutions to the equations X plusmnAlowastXminus1A = Qrdquo Applied Mathematics and Computation vol 216 no 5 pp 1355ndash1362 2010

[14] Z-Y Peng S M El-Sayed and X-l Zhang ldquoIterative methods for the extremal positive definitesolution of the matrix equation X + AlowastXminusαA = Qrdquo Journal of Computational and Applied Mathematicsvol 200 no 2 pp 520ndash527 2007

[15] B R Fang J D Zhou and Y M Li Matrix Theory Tsinghua University Press Beijing China 2006

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2012 Article ID 483624 11 pagesdoi1011552012483624

Research ArticleRefinements of Kantorovich Inequality forHermitian Matrices

Feixiang Chen

School of Mathematics and Statistics Chongqing Three Gorges University WanzhouChongqing 404000 China

Correspondence should be addressed to Feixiang Chen cfx2002126com

Received 28 August 2012 Accepted 5 November 2012

Academic Editor K C Sivakumar

Copyright q 2012 Feixiang Chen This is an open access article distributed under the CreativeCommons Attribution License which permits unrestricted use distribution and reproduction inany medium provided the original work is properly cited

Some new Kantorovich-type inequalities for Hermitian matrix are proposed in this paper Weconsider what happens to these inequalities when the positive definite matrix is allowed to beinvertible and provides refinements of the classical results

1 Introduction and Preliminaries

We first state the well-known Kantorovich inequality for a positive definite Hermite matrix(see [1 2]) let A isin Mn be a positive definite Hermitian matrix with real eigenvalues λ1 leλ2 le middot middot middot le λn Then

1 le xlowastAxxlowastAminus1x le (λ1 + λn)2

4λ1λn (11)

for any x isin Cn x = 1 where Alowast denotes the conjugate transpose of matrix A A matrix

A isin Mn is Hermitian if A = Alowast An equivalent form of this result is incorporated in

0 le xlowastAxxlowastAminus1x minus 1 le (λn minus λ1)2

4λ1λn (12)

for any x isin Cn x = 1

Attributed to Kantorovich the inequality has built up a considerable literatureThis typically comprises generalizations Examples are [3ndash5] for matrix versions Operator

2 Journal of Applied Mathematics

versions are developed in [6 7] Multivariate versions have been useful in statistics to assessthe robustness of least squares see [8 9] and the references therein

Due to the important applications of the original Kantorovich inequality for matrices[10] in Statistics [8 11 12] and Numerical Analysis [13 14] any new inequality of this typewill have a flow of consequences in the areas of applications

Motivated by the interest in both pure and applied mathematics outlined abovewe establish in this paper some improvements of Kantorovich inequalities The classicalKantorovich-type inequalities are modified to apply not only to positive definite but also toinvertible Hermitian matrices As natural tools in deriving the new results the recent Gruss-type inequalities for vectors in inner product in [6 15ndash19] are utilized

To simplify the proof we first introduce some lemmas

2 Lemmas

Let B be a Hermitian matrix with real eigenvalues μ1 le μ2 le middot middot middot le μn if A minus B is positivesemidefinite we write

A ge B (21)

that is λi ge μi i = 1 2 n On Cn we have the standard inner product defined by 〈x y〉 =sumn

i=1 xiyi where x = (x1 xn)lowast isin C

n and y = (y1 yn)lowast isin C

n

Lemma 21 Let a b c and d be real numbers then one has the following inequality

(a2 minus b2

)(c2 minus d2

)le (ac minus bd)2 (22)

Lemma 22 Let A and B be Hermitian matrices if AB = BA then

AB le (A + B)2

4 (23)

Lemma 23 Let A ge 0 B ge 0 if AB = BA then

AB ge 0 (24)

3 Some Results

The following lemmas can be obtained from [16ndash19] by replacing Hilbert space (H 〈middot middot〉) withinner product spaces C

n so we omit the details

Lemma 31 Let u v and e be vectors in Cn and e = 1 If α β δ and γ are real or complex

numbers such that

Relangβe minus u u minus αe

rang ge 0 Relangδe minus v v minus γe

rang ge 0 (31)

Journal of Applied Mathematics 3

then

|〈u v〉 minus 〈u e〉〈e v〉| le 14(β minus α

)(δ minus γ

) minus [Re

langβe minus u u minus αe

rangRe

langδe minus v v minus γe

rang]12 (32)

Lemma 32 With the assumptions in Lemma 31 one has

|〈u v〉 minus 〈u e〉〈e v〉| le 14(β minus α

)(γ minus δ

) minus∣∣∣∣〈u e〉 minus α + β

2

∣∣∣∣∣∣∣∣〈v e〉 minus γ + δ

2

∣∣∣∣ (33)

Lemma 33 With the assumptions in Lemma 31 if Re(βα) ge 0 Re(δγ) ge 0 one has

|〈u v〉 minus 〈u e〉〈e v〉| le∣∣β minus α

∣∣∣∣δ minus γ∣∣

4[Re

(βα

)Re

(δγ

)]12|〈u e〉〈e v〉| (34)

Lemma 34 With the assumptions in Lemma 33 one has

|〈u v〉 minus 〈u e〉〈e v〉|

le(∣∣α + β

∣∣ minus 2[Re

(βα

)]12)(∣∣δ + γ

∣∣ minus 2[Re

(δγ

)]12)12

[|〈u e〉〈e v〉|]12(35)

4 New Kantorovich Inequalities for Hermitian Matrices

For a Hermitian matrix A as in [6] we define the following transform

C(A) = (λnI minusA)(A minus λ1I) (41)

When A is invertible if λ1λn gt 0 then

C(Aminus1

)=(

1λ1

I minusAminus1)(

Aminus1 minus 1λn

I

) (42)

Otherwise λ1λn lt 0 then

C(Aminus1

)=(

1λk+1

I minusAminus1)(

Aminus1 minus 1λk

I

) (43)

where

λ1 le middot middot middot le λk lt 0 lt λk+1 le middot middot middot le λn (44)

From Lemma 23 we can conclude that C(A) ge 0 and C(Aminus1) ge 0

4 Journal of Applied Mathematics

For two Hermitian matrices A and B and x isin Cn x = 1 we define the following

functional

G(ABx) = 〈AxBx〉 minus 〈Ax x〉〈x Bx〉 (45)

When A = B we denote

G(Ax) = Ax2 minus 〈Ax x〉2 (46)

for x isin Cn x = 1

Lemma 41 With notations above and for x isin Cn x = 1 then

0 le 〈C(A)x x〉 le (λn minus λ1)2

4 (47)

if λnλ1 gt 0

0 lelangC(Aminus1

)x x

rangle (λn minus λ1)

2

4(λnλ1)2 (48)

if λnλ1 lt 0

0 lelangC(Aminus1

)x x

rangle (λk+1 minus λk)

2

4(λk+1λk)2 (49)

Proof From C(A) ge 0 then

〈C(A)x x〉 ge 0 (410)

While from Lemma 22 we can get

C(A) = (λnI minusA)(A minus λ1I) le (λn minus λ1)2

4I (411)

Then 〈C(A)x x〉 le (λn minus λ1)24 is straightforward The proof for C(Aminus1) is similar

Lemma 42 With notations above and for x isin Cn x = 1 then

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le G(Ax)G

(Aminus1x

) (412)

Journal of Applied Mathematics 5

Proof Thus

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2

=∣∣∣xlowast

((xlowastAminus1x

)I minusAminus1

)((xlowastAx)I minusA)x

∣∣∣2

le∥∥∥((xlowastAminus1x

)I minusAminus1

)x∥∥∥2

((xlowastAx)I minusA)x2

(413)

while

((xlowastAx)I minusA)x2 = xlowast((xlowastAx)2I minus 2(xlowastAx)A +A2

)x

= xlowastA2x minus (xlowastAx)2

= Ax2 minus 〈Ax x〉2

= G(Ax)

(414)

Similarly we can get ((xlowastAminus1x)I minusAminus1)x2 = G(Aminus1x) then we complete the proof

Theorem 43 Let A B be two Hermitian matrices and C(A) ge 0 C(B) ge 0 are defined as abovethen

|G(ABx)| le 14(λn minus λ1)

(μn minus μ1

) minus [〈C(A)x x〉〈C(B)x x〉]12

|G(ABx)| le 14(λn minus λ1)

(μn minus μ1

) minus∣∣∣∣lang(

A minus λ1 + λn2

)x x

rang∣∣∣∣∣∣∣∣lang(

B minus μ1 + μn

2

)x x

rang∣∣∣∣(415)

for any x isin Cn x = 1

If λ1λn gt 0 μ1μn gt 0 then

|G(ABx)| le (λn minus λ1)(μn minus μ1

)4[(λnλ1)

(μnμ1

)]12|〈Ax x〉〈Bx x〉|

|G(ABx)| le(

|λ1 + λn| minus 2(λ1λn)12

)(∣∣μ1 + μn

∣∣ minus 2(μ1μn

)12)12

[|〈Ax x〉〈Bx x〉|]12

(416)

for any x isin Cn x = 1

Proof The proof follows by Lemmas 31 32 33 and 34 on choosing u = Ax v = Bx ande = x β = λn α = λ1 δ = μn and γ = μ1 x isin C

n x = 1 respectively

6 Journal of Applied Mathematics

Corollary 44 Let A be a Hermitian matrices and C(A) ge 0 is defined as above then

|G(Ax)| le 14(λn minus λ1)

2 minus 〈C(A)x x〉 (417)

|G(Ax)| le 14(λn minus λ1)

2 minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣2

(418)

for any x isin Cn x = 1

If λ1λn gt 0 then

|G(Ax)| le (λn minus λ1)2

4(λnλ1)|〈Ax x〉|2 (419)

|G(Ax)| le(|λ1 + λn| minus 2

radicλ1λn

)|〈Ax x〉| (420)

for any x isin Cn x = 1

Proof The proof follows by Theorem 43 on choosing A = B respectively

Corollary 45 Let A be a Hermitian matrices and C(Aminus1) ge 0 is defined as above then one has thefollowing

If λ1λn gt 0 then

∣∣∣G(Aminus1x

)∣∣∣ le (λn minus λ1)2

4(λnλ1)2

minuslangC(Aminus1

)x x

rang (421)

∣∣∣G(Aminus1x

)∣∣∣ le (λn minus λ1)2

4(λnλ1)2

minus∣∣∣∣lang(

Aminus1 minus λ1 + λn2λnλ1

I

)x x

rang∣∣∣∣2

(422)

∣∣∣G(Aminus1x

)∣∣∣ le (λn minus λ1)2

4λnλ1

∣∣∣langAminus1x xrang∣∣∣2

(423)

∣∣∣G(Aminus1x

)∣∣∣ le(

|λ1 + λn|λ1λn

minus 21radicλ1λn

)∣∣∣langAminus1x xrang∣∣∣ (424)

for any x isin Cn x = 1

If λ1λn lt 0 then

∣∣∣G(Aminus1x

)∣∣∣ le (λk minus λk+1)2

4(λkλk+1)2

minuslangC(Aminus1

)x x

rang (425)

Journal of Applied Mathematics 7

where C(Aminus1) = ((1λk+1)I minusAminus1)(Aminus1 minus (1λk)I) and

∣∣∣G(Aminus1x

)∣∣∣ le (λk minus λk+1)2

4(λkλk+1)2

minus∣∣∣∣lang(

Aminus1 minus λk+1 + λk2λkλk+1

I

)x x

rang∣∣∣∣2

(426)

for any x isin Cn x = 1

Proof The proof follows by Corollary 44 by replacing A with Aminus1 respectively

Theorem 46 LetA be an ntimesn invertible Hermitian matrix with real eigenvalues λ1 le λ2 le middot middot middot le λnthen one has the following

If λ1λn gt 0 then

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le (λn minus λ1)

2

4(λnλ1)minusradic〈C(A)x x〉langC(Aminus1

)x x

rang (427)

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le (λn minus λ1)

2

4(λnλ1)minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣∣∣∣∣lang(

Aminus1 minus λ1 + λn2λnλ1

I

)x x

rang∣∣∣∣(428)

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le (λn minus λ1)

2

4(λnλ1)2 〈Ax x〉

langAminus1x x

rang (429)

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le

(radic|λn| minus

radic|λ1|

)2

radicλnλ1

radic〈Ax x〉〈Aminus1x x〉 (430)

for any x isin Cn x = 1

If λ1λn lt 0 then

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le (λn minus λ1)(λk+1 minus λk)

4|λkλk+1| minusradic〈C(A)x x〉langC(Aminus1

)x x

rang (431)

where C(Aminus1) = ((1λk+1)I minusAminus1)(Aminus1 minus (1λk)I)

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le (λn minus λ1)(λk+1 minus λk)

4|λkλk+1|

minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣∣∣∣∣lang(

Aminus1 minus λk + λk+1

2λ1λk+1I

)x x

rang∣∣∣∣(432)

Proof Considering

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le G(Ax)G

(Aminus1x

) (433)

8 Journal of Applied Mathematics

When λ1λn gt 0 from (417) and (421) we get

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le

14(λn minus λ1)

2 minus 〈C(A)x x〉

(λn minus λ1)2

4(λnλ1)2

minuslangC(Aminus1

)x x

rang (434)

From (a2 minus b2)(c2 minus d2) le (ac minus bd)2 we have

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le

(λn minus λ1)

2

4(λnλ1)minusradic[〈C(A)x x〉langC(Aminus1

)x x

rang]2

(435)

then the conclusion (427) holdsSimilarly from (418) and (422) we get

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le

14(λn minus λ1)

2 minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣2

times(λn minus λ1)

2

4(λnλ1)2

minus∣∣∣∣lang(

Aminus1 minus λ1 + λn2λnλ1

I

)x x

rang∣∣∣∣2

le(λn minus λ1)

2

4(λnλ1)minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣∣∣∣∣lang(

Aminus1 minus λ1 + λn2λnλ1

I

)x x

rang∣∣∣∣2

(436)

then the conclusion (428) holdsFrom (419) and (423) we get

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le (λn minus λ1)

2

4(λnλ1)|〈Ax x〉|2 (λn minus λ1)

2

4(λnλ1)

∣∣∣langAminus1x xrang∣∣∣2

(437)

then the conclusion (429) holdsFrom (420) and (424) we get

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le

(|λ1 + λn| minus 2

radicλ1λn

)|〈Ax x〉|

(|λ1 + λn|λ1λn

minus 21radicλ1λn

)∣∣∣langAminus1x xrang∣∣∣(438)

then the conclusion (430) holdsWhen λ1λn lt 0 from (417) and (425) we get

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2le

14(λn minus λ1)

2 minus 〈C(A)x x〉

(λk minus λk+1)2

4(λkλk+1)2

minuslangC(Aminus1

)x x

rang (439)

then the conclusion (431) holds

Journal of Applied Mathematics 9

From (418) and (426) we get

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣2 le

14(λn minus λ1)

2 minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣2

times(λk minus λk+1)

2

4(λkλk+1)2

minus∣∣∣∣lang(

Aminus1 minus λk+1 + λk2λkλk+1

I

)x x

rang∣∣∣∣2

le(λn minus λ1)(λk+1 minus λk)

4|λkλk+1|

minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣∣∣∣∣lang(

Aminus1 minus λk + λk+1

2λ1λk+1I

)x x

rang∣∣∣∣2

(440)

then the conclusion (432) holds

Corollary 47 With the notations above for any x isin Cn x = 1 one lets

|G1(Ax)| = 14(λn minus λ1)

2 minus 〈C(A)x x〉

|G2(Ax)| = 14(λn minus λ1)

2 minus∣∣∣∣lang(

A minus λ1 + λn2

I

)x x

rang∣∣∣∣2

(441)

If λ1λn gt 0 one lets

|G3(Ax)| = (λn minus λ1)2

4(λnλ1)|〈Ax x〉|2

|G4(Ax)| =(|λ1 + λn| minus 2

radicλ1λn

)|〈Ax x〉|

(442)

If λ1λn gt 0

∣∣∣G1(Aminus1x

)∣∣∣ = (λn minus λ1)2

4(λnλ1)2

minuslangC(Aminus1

)x x

rang

∣∣∣G2(Aminus1x

)∣∣∣ = (λn minus λ1)2

4(λnλ1)2

minus∣∣∣∣lang(

Aminus1 minus λ1 + λn2λnλ1

I

)x x

rang∣∣∣∣2

∣∣∣G3(Aminus1x

)∣∣∣ = (λn minus λ1)2

4λnλ1

∣∣∣langAminus1x xrang∣∣∣2

∣∣∣G4(Aminus1x

)∣∣∣ =(

|λ1 + λn|λ1λn

minus 21radicλ1λn

)∣∣∣langAminus1x xrang∣∣∣

(443)

10 Journal of Applied Mathematics

If λ1λn lt 0 then

∣∣∣G5(Aminus1x

)∣∣∣ = (λk minus λk+1)2

4(λkλk+1)2

minuslangC(Aminus1

)x x

rang (444)

where C(Aminus1) = ((1λk+1)I minusAminus1)(Aminus1 minus (1λk)I) and

∣∣∣G6(Aminus1x

)∣∣∣ = (λk minus λk+1)2

4(λkλk+1)2

minus∣∣∣∣lang(

Aminus1 minus λk+1 + λk2λkλk+1

I

)x x

rang∣∣∣∣2

(445)

Then one has the followingIf λ1λn gt 0

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le

radicGlowast(Ax)Glowast(Aminus1x

) (446)

where

Glowast(Ax) = minG1(Ax) G2(Ax) G3(Ax) G4(Ax)

Glowast(Ax) = minG1

(Aminus11x

) G2

(Aminus1x

) G3

(Aminus1x

) G4

(Aminus1x

)

(447)

If λ1λn lt 0

∣∣∣xlowastAxxlowastAminus1x minus 1∣∣∣ le

radicG(Ax)Gbull(Aminus1x

) (448)

where

G(Ax) = minG1(Ax) G2(Ax)

Gbull(Ax) = minG5

(Aminus1x

) G6

(Aminus1x

)

(449)

Proof The proof follows from that the conclusions in Corollaries 44 and 45 are independent

Remark 48 It is easy to see that if λ1 gt 0 λn gt 0 our result coincides with the inequalityof operator versions in [6] So we conclude that our results give an improvement of theKantorovich inequality [6] that applies to all invertible Hermite matrices

5 Conclusion

In this paper we introduce some new Kantorovich-type inequalities for the invertibleHermitian matrices Inequalities (427) and (431) are the same as [4] but our proof is simpleIn Theorem 46 if λ1 gt 0 λn gt 0 the results are similar to the well-known Kantorovich-typeinequalities for operators in [6] Moreover for any invertible Hermitian matrix there exists asimilar inequality

Journal of Applied Mathematics 11

Acknowledgments

The authors are grateful to the associate editor and two anonymous referees whose commentshave helped to improve the final version of the present work This work was supported by theNatural Science Foundation of Chongqing Municipal Education Commission (no KJ091104)

References

[1] R A Horn and C R Johnson Matrix Analysis Cambridge University Press Cambridge UK 1985[2] A S Householder The Theory of Matrices in Numerical Analysis Blaisdell New York NY USA 1964[3] S Liu and H Neudecker ldquoSeveral matrix Kantorovich-type inequalitiesrdquo Journal of Mathematical

Analysis and Applications vol 197 no 1 pp 23ndash26 1996[4] A W Marshall and I Olkin ldquoMatrix versions of the Cauchy and Kantorovich inequalitiesrdquo

Aequationes Mathematicae vol 40 no 1 pp 89ndash93 1990[5] J K Baksalary and S Puntanen ldquoGeneralized matrix versions of the Cauchy-Schwarz and

Kantorovich inequalitiesrdquo Aequationes Mathematicae vol 41 no 1 pp 103ndash110 1991[6] S S Dragomir ldquoNew inequalities of the Kantorovich type for bounded linear operators in Hilbert

spacesrdquo Linear Algebra and its Applications vol 428 no 11-12 pp 2750ndash2760 2008[7] P G Spain ldquoOperator versions of the Kantorovich inequalityrdquo Proceedings of the American

Mathematical Society vol 124 no 9 pp 2813ndash2819 1996[8] S Liu and H Neudecker ldquoKantorovich inequalities and efficiency comparisons for several classes of

estimators in linear modelsrdquo Statistica Neerlandica vol 51 no 3 pp 345ndash355 1997[9] P Bloomfield and G S Watson ldquoThe inefficiency of least squaresrdquo Biometrika vol 62 pp 121ndash128

1975[10] L V Kantorovic ldquoFunctional analysis and applied mathematicsrdquo Uspekhi Matematicheskikh Nauk vol

3 no 6(28) pp 89ndash185 1948 (Russian)[11] C G Khatri and C R Rao ldquoSome extensions of the Kantorovich inequality and statistical

applicationsrdquo Journal of Multivariate Analysis vol 11 no 4 pp 498ndash505 1981[12] C T Lin ldquoExtrema of quadratic forms and statistical applicationsrdquo Communications in Statistics A

vol 13 no 12 pp 1517ndash1520 1984[13] A Galantai ldquoA study of Auchmutyrsquos error estimaterdquo Computers amp Mathematics with Applications vol

42 no 8-9 pp 1093ndash1102 2001[14] P D Robinson and A J Wathen ldquoVariational bounds on the entries of the inverse of a matrixrdquo IMA

Journal of Numerical Analysis vol 12 no 4 pp 463ndash486 1992[15] S S Dragomir ldquoA generalization of Grussrsquos inequality in inner product spaces and applicationsrdquo

Journal of Mathematical Analysis and Applications vol 237 no 1 pp 74ndash82 1999[16] S S Dragomir ldquoSome Gruss type inequalities in inner product spacesrdquo Journal of Inequalities in Pure

and Applied Mathematics vol 4 no 2 article 42 2003[17] S S Dragomir ldquoOn Bessel and Gruss inequalities for orthonormal families in inner product spacesrdquo

Bulletin of the Australian Mathematical Society vol 69 no 2 pp 327ndash340 2004[18] S S Dragomir ldquoReverses of Schwarz triangle and Bessel inequalities in inner product spacesrdquo Journal

of Inequalities in Pure and Applied Mathematics vol 5 no 3 article 76 2004[19] S S Dragomir ldquoReverses of the Schwarz inequality generalising a Klamkin-McLenaghan resultrdquo

Bulletin of the Australian Mathematical Society vol 73 no 1 pp 69ndash78 2006[20] Z Liu L Lu and K Wang ldquoOn several matrix Kantorovich-type inequalitiesrdquo Journal of Inequalities

and Applications vol 2010 Article ID 571629 5 pages 2010