n-linear algebra type 1 and its applications, by w. b. vasantha kandasamy & florentin...

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n- LINEAR ALGEBRA OF TYPE I AND ITS APPLICATIONS

W. B. Vasantha Kandasamye-mail: [email protected]

web: http://mat.iitm.ac.in/~wbvwww.vasantha.net

Florentin Smarandachee-mail: [email protected]

INFOLEARNQUESTAnn Arbor

2008



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This book can be ordered in a paper bound reprint from:

Books on DemandProQuest Information & Learning(University of Microfilm International)300 N. Zeeb RoadP.O. Box 1346, Ann Arbor MI 48106-1346, USATel.: 1-800-521-0600 (Customer Service)http://wwwlib.umi.com/bod/

Peer reviewers:Professor Sukanto Bhattacharya, Queensland University,Australia.

Dr.S.Osman, Menofia University, Shebin Elkom, Egypt.

Eng. Marian Popescu and Prof. Florentin Popescu, Craiova,Romania.

Copyright 2008 by InfoLearnQuest and authorsCover Design and Layout by Kama Kandasamy

Many books can be downloaded from the followingDigital Library of Science:http://www.gallup.unm.edu/~smarandache/eBooks-otherformats.htm

ISBN-10: 1-59973-074-XISBN-13: 978-1-59973-074-5EAN: 9781599730745

Standard Address Number: 297-5092Printed in the United States of America



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CONTENTS

Preface 5

Chapter OneBASIC CONCEPTS 7

Chapter Two

n-VECTOR SPACES OF TYPE I AND THEIR PROPERTIES 13

Chapter ThreeAPPLICATIONS OFn-LINEAR ALGEBRA OF TYPE I 81



4

Chapter FourSUGGESTED PROBLEMS 103

FURTHER READING 111

INDEX 116

ABOUT THE AUTHORS 120



5

PREFACE

With the advent of computers one needs algebraic structuresthat can simultaneously work with bulk data. One suchalgebraic structure namely n-linear algebras of type I areintroduced in this book and its applications to n-Markov chainsand n-Leontief models are given. These structures can be

thought of as the generalization of bilinear algebras and bivector spaces. Several interesting n-linear algebra properties are proved.

This book has four chapters. The first chapter justintroduces n-group which is essential for the definition of n-vector spaces and n-linear algebras of type I. Chapter two givesthe notion of n-vector spaces and several related results whichare analogues of the classical linear algebra theorems. In case of n-vector spaces we can define several types of linear transformations.

The notion of n-best approximations can be used for error correction in coding theory. The notion of n-eigen values can beused in deterministic modal superposition principle for undamped structures, which can find its applications in finiteelement analysis of mechanical structures with uncertain

parameters. Further it is suggested that the concept of n-matrices can be used in real world problems which adopts fuzzy

models like Fuzzy Cognitive Maps, Fuzzy Relational Equationsand Bidirectional Associative Memories. The applications of



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these algebraic structures are given in Chapter 3. Chapter four gives some problem to make the subject easily understandable.

The authors deeply acknowledge the unflinching support of Dr.K.Kandasamy, Meena and Kama.

W.B.VASANTHA KANDASAMYFLORENTIN SMARANDACHE



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Chapter One

BASIC CONCEPTS

In this chapter we introduce the notion of n-field, n-groups (n t 2) and illustrate them by examples. Throughout this book F willdenote a field, Q the field of rationals, R the field of reals, C the

field of complex numbers and Z p, p a prime, the finite field of characteristic p. The fields Q, R and C are fields of zerocharacteristic.

Now we proceed on to define the concept of n-groups.

D EFINITION 1.1: Let G = G 1 G2 … Gn (n t 2) whereeach (G i , * i , e i ) is a group with i the binary operation and e i

the identity element, such that G i z G j , if i z j, 1 j, i n. Further G i G j or G j Gi if i z j. Any element x G would berepresented as x = x 1 x2 … xn; where x i Gi , i = 1, 2, …,n. Now the operations on G is described so that G becomes a

group. For x, y G, where x = x 1 x2 … xn and y = y 1 y2 … yn; with x i , yi G i , i = 1, 2, …, n.

x * y = (x1 x2 … xn ) * (y 1 y2 … yn )= (x 1 *1 y1 x2 *2 y2 … xn *n yn ).



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Since each x i *i yi G i we see x * y = (p 1 p2 … pn ) where xi *i yi = p i for i = 1, 2, …, n. Thus G is closed under the binaryoperation *.

Now let e = (e 1 e2 … en ) where e i Gi the identity of G i with respect to the binary operation, * i , i = 1, 2, …, n we seee * x = x * e = x for all x G. e will be known as the identityelement of G under the operation *.

Further for every x = x 1 x2 … xn G; we have1 1 1

1 2 ... n x x x in G such that,1

1 2* ( ... )n x x x x x * 1 1 11 2( ... )n x x x

= 1 1 11 1 1 2 2 2* * ... *n n n x x x x x

= x -1 * x(e1 e2 … en ) = e.

1 1 1 11 2 ... n x x x x

is known as the inverse of x = x 1 x2 … xn. We define (G,*, e) to be the n-group (n t 2). When n = 1 we see it is the

group. n = 2 gives us the bigroup described in [37-38] when n> 2 we have the n-group.

Now we illustrate this by examples before we proceed on torecall more properties about them.

Example 1.1: Let G = G 1 G2 G3 G4 G5 where G 1 = S 3 the symmetric group of degree 3 with

1

1 2 3e

1 2 3

§ ·

¨ ¸© ¹,

G2 = ¢g | g 6 = e 2², the cyclic group of order 6, G 3 = Z 5, the groupunder addition modulo 5 with e 3 = 0, G 4 = D 8 = {a, b | a 2 = b 8 =1; bab = a}, the dihedral group of order 8, e 4 = 1 is the identityelement of G 4 and G 5 = A 4 the alternating subgroup of S 4 with

4

1 2 3 4

e 1 2 3 4

§ ·¨ ¸© ¹.



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Clearly G = S 3 G2 Z5 D8 A4 is a n-group with n = 5.

Any x G would be of the form

2 31 2 3 1 2 3 4x g 4 b

2 1 3 1 3 4 2§ · § · ¨ ¸ ¨ ¸© ¹ © ¹

.

x-1 = 4 51 2 3 1 2 3 4g 1 b .

2 1 3 1 4 2 3§ · § · ¨ ¸ ¨ ¸© ¹ © ¹

The identity element of G is

2

1 2 3 1 2 3 4e 0 1

1 2 3 1 2 3 4§ · § · ¨ ¸ ¨ ¸© ¹ © ¹

= e 1 e2 e3 e4 e5.

Thus G is a 5-group. Clearly the order of G is o(G 1) u o(G 2) u o(G 3) u o(G 4) u o(G 5) = 6 u 6 u 5 u 16 u 12 = 34, 560.

We see o(G) < f . Thus if in the n-group G 1 G2 … Gn, every group G i is of finite order then G is of finite order; 1 d i d n.

Example 1.2: Let G = G 1 G2 G3 where G 1 = Z 10, the groupunder addition modulo 10, G 2 = ¢g | g 5 = 1 ², the cyclic group of order 5 and G 3 = Z the set of integers under +.

Clearly G is a 3-group. We see G is an infinite group for order of G 3 is infinite.

Further it is interesting to observe that every group in the 3-group G is abelian. Thus if G = G 1 G2 … Gn, is a n-group (n t 2), we see G is an abelian n-group if each G i is anabelian group; i = 1, 2, …, n. Even if one of the G i in G is a nonabelian group then we call G to be only a non abelian n-group.

Having seen an example of an abelian and non abelian group wenow proceed on to define the notion of n-subgroup. We need all



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these concepts mainly to define the new notion of linear n-algebra or n-linear algebra and n-vector spaces of type I.

D EFINITION 1.2: Let G = G 1 G2 … Gn , be a n-group, a proper subset H G of the form H = H 1 H 2 … H n with H i z Gi or {e i } but I z H i G i; i = 1, 2, …, n, H i proper subgroup of G i is defined to be the proper n-subgroup of the n- group G. If some of the H i = G i or H i = {e i } or H i = I for somei, then H will not be called as proper n-subgroup but only as m-

subgroup of the n-group, m < n and m of the subgroups H j in G j

are only proper and the rest are either {e j } or I , 1 d j d n.

We illustrate both these situations by the following example.

Examples 1.3: Let G = G 1 G2 G3 G4 be a 4-group whereG1 = S 4, G 2 = Z 10 group under addition modulo 10, G 3 = D 12 thedihedral group with order 12 given by the set {a, b | a 2 = b 6 = 1,

bab = a} and G 4 = Z the set of positive and negative integerswith zero under +.

Consider H = H 1 H2 H3 H4 where H 1 = A 4 the

alternating subgroup of S 4, H 2 = {0, 2, 4, 6, 8} a subgroup of order 5 under addition modulo 10. H 3 = {1, b, b 2, b 3, b 4, b 5}; thesubgroup of D 12 and H 4 = {2n | n Z} a subgroup of Z. ClearlyH is a proper 4-subroup of the 4-group G.

Let K = K 1 K 2 K 3 K 4 G where K 1 = A 4, K 2 = {0,5}, K 3 = D 12 and K 4 = Z. Clearly K is not a proper 4-subgroupof the 4-group G but only a improper 4-subgroup of G.

Let T = T 1 T2 T3 T4 G where T 1 = A 4, T 2 = {0}, T 3 = I and T 4 = {2n | n Z}; clearly T is only a 2-subgroup or

bisubgroup of the 4-group G.

We mainly need in this book n-groups which are only abelian.

Now in this section we define the notion of n-fields.

D EFINITION 1.3: Let F = F 1 F 2 … F n ( n t 2) be suchthat each F i is a field and F i z F j , if i z j and F i F j or F j F i ,

1 d i, j d n. Then we define (F, +, u ) to be a n-field if (F, +) is a



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n-group and F 1 \ {0} F 2 \ {0} … F n \ {0} is a n-groupunder u .

Further

[(a 1 a2 … an ) + (b 1 b2 … bn )]u [(c 1 c2 … cn )]

= (a 1 + b 1 ) u c1 (a 2 + b 2 ) u c2 … (a n + b n ) u cn

and

[(c 1 c2 … cn )] u {[(a 1 a 2 … an )]+ [(b 1 b2 … bn )]}

= 1c u (a 1 b1 ) c2 u (a 2 b2 ) … cn u (a n bn )

for all a i , b i , c i F, i = 1, 2, …, n. Thus (F, +, u ) is a n-field.

We illustrate this by the following example.

Example 1.4: Let F = F 1 F2 F3 F4 where F 1 = Q, F 2 = Z 2,F3 = Z 17 and F 4 = Z 11; F is a 4-field.

Example 1.5: Let F = F 1 F2 F3 F4 F5 F6 where F 1 =Z2, F 2 = Z 3, F 3 = Z 13, F 4 = Z 7, F 5 = Z 19 and F 6 = Z 31, F is a 6-field. Let F = F 1 F2 … Fn, be a n-field where each F i is afield of characteristic zero, 1 d i d n, then F is called as a n-fieldof characteristic zero.

Let F = F 1 F2 … Fm (m t 2) be a m-field if each field F i is of finite characteristic then we call F to be a m-field of finitecharacteristic. Suppose F = F 1 F2 … Fn, n t 2 wheresome F i’s are finite characteristic and some F j’s are zerocharacteristic then alone we say F is a n-field of mixedcharacteristic.

Example 1.6: Let F = F 1 F2 … F5 where F 1 = Q, F 2 = Z 7,

F3 = Z 23 and F 4 = Z 17 and F 5 = Z 2 be a 5-field. F is a 5-field of mixed characteristic.



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Example 1.7: Let F = F 1 F2 … F6 = Z 4 R Z7 Q R Z11 . Clearly F is not a 6 field as F 2 = F 5. We need each fieldF i to be distinct, 1 d i d n.

Note: Clearly F 1 F2 F3 = Q R Z2 is not a 3-field as Q R. Because we need in this case also as in case of bistructuresnon containment of one set in another set.



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Chapter Two

n-VECTOR SPACES OF TYPE IAND THEIR PROPERTIES

In this chapter we introduce the notion of n-vector spaces anddescribe some of their important properties.

Here we define the concept of n-vector spaces over a fieldwhich will be known as the type I n-vector spaces or n-vector spaces of type I. Several interesting properties about them arederived in this chapter.

D EFINITION 2.1: A n-vector space or a n-linear space of type I (n t 2) consists of the following:

1. a field F of scalars2. a set V = V 1 V 2 … V n of objects called n-vectors3. a rule (or operation) called vector addition; which

associates with each pair of n-vectors = 1 2 … n , = 1 2 … n V = V 1 V 2 … V n; +

= ( 1 2 … n ) + ( 1 2 … n ) = ( 1 + E 1 2 + E 2 … n + E n ) V called the sum of and in such

a waya. + = + ; i.e., addition is commutative ( , V).



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b. + ( + ) = ( + ) + , i.e., addition is associative ( , , V).

c. There is a unique n-vector 0 n = 0 0 … 0 V such that + 0 n = for all V, called the zero n-vector of V.

d. For each n-vector = 1 2 … n V, thereexists a unique vector – = – 1 – 2 … – n V

such that + (– ) = 0 n. e. A rule (or operation) called scalar multiplication which

associates with each scalar c in F and a n-vector in V = V 1 V 2 … V n a n-vector c in V called the

product of c and in such a way that

1. 1. = 1. ( 1 2 … n )= 1. 1 1. 2 … 1. n = 1 2 … n =

for every n-vector D in V.

2. (c 1. c2 ). = c1.(c2. ) for all c 1 , c 2 F and V i.e. if 1 2 … n is the n-vector in V we have

(c1. c2 ). = (c1. c2 ) ( 1 2 … n )= c 1 [c 2(( 1 2 … n )]= c 1 [c 2 1 c2 2 … c2 n ]= c 1 [c 2 ].

3. c( + ) = c. + c. for all , V and for all c F i.e., if

1 2 … n and 1 2 … n are n-vectors of V then for any c F we have

c( + ) = c[( 1 2 … n ) + ( 1 2 … n )]= c[ 1 + 1 2 + 2 … n + n ]= (c( 1 + 1 ) c( 2 + 2 ) … c( n + n )]= (c 1 c 2 … c n ) + (c 1 c 2 … c n )= c + c .

4. (c 1 + c 2 ). = c 1 + c 2 for all c 1 , c 2 F and V.



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Just like a vector space which is a composite algebraic structure containing the field a set of vectors which form a group, the n-vector space of type I is a composite of set of n-vectors or n-group and a field F of scalars. V is a linear n-algebra or n-linear algebra if V has a multiplicative closed binary operation “.” which is associative i.e.; if , V, . V, thus if = ( 1 2 … n ) and = ( 1 2 … n ) V then if

. = ( 1 2 … n ) . ( 1 2 … n )= ( 1. 1 2. 2 … n. n ) V

then the linear n-vector space of type I becomes a linear n-algebra of type-I.

Now we make an important mention that all linear n-algebras of type-I are linear n-vector spaces of type-I; however a n-vector space of type-I over F in general need not be a n- linear algebraof type I over F.

We now illustrate this by the following example.

Example 2.1: Let V = V 1 V2 V3 V4 where V 1 = Q[x] thevector space of polynomials over Q. V 2 = Q u Q, the vector space of dimension two over Q,

V3 =a b

a,b,c,d Qc d

- ½§ ·° °® ¾¨ ¸© ¹° °¯ ¿

the vector space of all 2 u 2 matrices with entries from Q and

V4 =a b c

a,b,c,d,e, f R d e f

- ½§ ·° °® ¾¨ ¸© ¹° °¯ ¿

be the vector space of all 2 u 3 matrices with entries from R

over Q. Thus V is a linear 4-vector space over Q of type-I.Clearly V is not a linear 4-algebra of type-I over Q.



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Now we give yet another example of a linear n-vector space of type-I.

Example 2.2: Let V = V 1 V2 V3 V4 V5 be a 5-vector space over Q of type-I, where V 1 = Q[x], the set of all

polynomials with coefficients from Q is a vector space over Q.V2 = Q u R u Q is a vector space over Q,

V3 =

a b c

d e f a,b,c,d,e,f ,g,h,i Q

g h i

- ½§ ·° °¨ ¸® ¾¨ ¸° °¨ ¸© ¹¯ ¿

is a vector space over Q,

V4 =

a 0 0 0

0 b 0 0a, b,c,d, R

0 0 c 0

0 0 0 d

- ½§ ·° °¨ ¸° °¨ ¸® ¾¨ ¸° °¨ ¸° °© ¹¯ ¿

is a vector space over Q and V 5 = R is a vector space over Q.Clearly V = V 1 V2 V3 V4 V5 is a linear 5-vector spaceof type-I over Q. Also V is a linear 5-linear algebra over Q.Thus we have seen from example 2.1 that every vector n-spaceof type-I need not be a linear n-algebra of type-I. Also everylinear n-algebra of type-I is a linear n-vector space of type-I.

Now we can also define the notion of n-vector space of type-I ina very different way.

D EFINITION 2.2: Let V = V 1 V 2 … V n (n t 2) where eachV i is a vector space over the same field F and V i V j , if i jand V i V j and V j V i if i z j, 1 i, j n, then V is defined tobe a n-vector space of type-I over F.

If each of the V i’s are linear algebra over F then we call V to be a linear n-algebra of type-I over F.



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Now we proceed on to define the notion of n-subvector space of the n-vector space of type-I.

D EFINITION 2.3: Let V = V 1 V 2 … V n (n t 2) be a n-vector space of type I over F. Suppose W = W 1 W 2 … W n(n t 2) is a proper subset of V such that each W i is a proper

subspace of the vector space V i over F with W i V i , W i I or (0) such that W i W j or W i W j or W j W i if i j, 1 d i, j d n, thenwe define W to be a n-subspace of type-I over F.

We now illustrate it by the following example.

Example 2.3: Let V = V 1 V2 V3 where V 1 = R u R, a vector space over R and V 2 = R[x] a vector space over R and

V3 =a c

a,b,c,d R d b

- ½§ ·° °® ¾¨ ¸© ¹° °¯ ¿

,

a vector space over R i.e., V is a 3-vector space of type-I over

R. Let W = W 1 W2 W3 V = V 1 V2 V3 where

W1 = R u {0} V1,

W2 =n

2ii i 2

i 0

r x r R V- ½ ® ¾¯ ¿¦ ,

W3 = 3

a 0a,b R V

0 b

- ½§ ·° ° ® ¾¨ ¸© ¹° °¯ ¿

.

Clearly W is a 3 subspace of V of type-I. Suppose

T = R u {0} R a 0

a,b R 0 b

- ½§ ·° °® ¾¨ ¸© ¹° °¯ ¿

V1 V2 V3,

then T is not a 3-subspace of type-I as R u {0} and R are sameor R R u {0}.



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Now we proceed on to define the notion of n-linear dependenceand n-linear independence in the n-vector space V of type-I.

D EFINITION 2.4: Let V = V 1 V 2 … V n be a n-vector space of type-I over F. Any proper n-subset S V would be of the form S = S 1 S 2 … S n V 1 V 2 … V n where I S i contained in V i , 1 d i d n. S i a proper subset of V i. If each of the subsets S i V i is a linearly independent set over F for i = 1,2, …, n then we define S to be a n-linearly independent subset of V. Even if one of the subset S k of V k is not a linearly independent

subset of V k for some 1 d k d n then we call the n-subset of V tobe a n-linearly dependent subset or a linearly dependent n-

subset of V.

Now we illustrate this situation by the following examples.

Example 2.4: Let V = V 1 V2 V3 V4 be a 4- vector spaceover Q, where V 1 = Q[x], V 2 = Q u Q u Q; V 3 = { the set of all 2u 2 matrices with entries from Q} and V 4 = [the set of all 4 u 2

matrices with entries from Q, are all vector spaces over Q. Let S= S 1 S2 S3 S4 be a 4 subset of V,

S1 = {1, x 2, x5, x7, 3x 8},S2 = {(7, 0, 2), (0, 5, 1)},

3

5 1 0 0S ,

0 0 7 3

- ½§ · § ·° °® ¾¨ ¸ ̈ ¸° °© ¹ © ¹¯ ¿

and

4

0 2 1 0 0 0

1 0 0 2 0 0S , , .

0 0 0 0 7 3

3 0 0 1 0 1

- ½ª º ª º ª º° °« » « » « »° °« » « » « »® ¾« » « » « »° °« » « » « »° °¬ ¼ ¬ ¼ ¬ ¼¯ ¿

Clearly we see every subset S i of V i is a linearly independentsubset, for i = 1, 2, 3, 4. Thus S is a 4- linearly independentsubset of V.



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Example 2.5: Let V = V 1 V2 V3 be a 3-vector space over R where V 1 = R[x], V 2 = {set of all 3 u 3 matrices with entriesfrom R} and V 3 = R u R u R u R. Clearly V 1, V 2 and V 3 are allvector spaces over R. Let S = S 1 S2 S3 V1 V2 V3 = V

be a proper 3-subset of V; where

S1 = {x 3, 3x 3 + 7, x 5},

S2 =

6 0 0 0 1 2

0 0 3 , 1 0 1

1 1 0 0 7 0

- ½§ · § ·° °¨ ¸ ¨ ¸® ¾¨ ¸ ¨ ¸° °¨ ¸ ¨ ¸© ¹ © ¹¯ ¿

andS3 = {(3 1 0 0), (0 7 2 1), (5 1 1 1), (0 8 9 1), (2 1 3 0)}.

We see S 1 is a linearly dependent subset of V 1 over R and S 2 is alinearly independent subset over R and S 3 is a linearlydependent subset of V 3 over R. Thus S is a 3-linearly dependentsubset of the 3-vector space V over R.

Now we proceed onto define the notion of n-basis of the n-vector space V over a field F.

D EFINITION 2.5: Let V = V 1 V 2 … V n be a n-vector spaceover a field F. A proper n-subset S = S 1 S 2 … S n of V is

said to be n-basis of V if S is a n-linearly independent set and each S j V j generates V j , i.e., S j is a basis of V j , true for j = 1,2, …, n. Even if one of the S j is not a basis of V j for 1 d j d nthen S is not a n-basis of V.

As in case of vector spaces the n-vector spaces can also havemany basis but the number of base elements in each of the nsubsets is the same.

Now we illustrate this situation by the following example.

Example 2.6 : Let V = V 1 V2 V3 V4 be a 4-vector spaceover Q. V 1 = {all polynomials of degree less than or equal to 5},



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V2 = Q u Q u Q, V 3 = {the set of all 2 u2 matrices with entriesfrom Q} and V 4 = Q u Q u Q u Q u Q are vector spaces over Q.

Now letB = B 1 B2 B3 B4

= {1, x, x 2, x3, x4, x5} {(1 0 0), (0 1 0), (0 2 1)}

0 0 1 0 0 0 0 1, ,

1 0 0 0 0 1 0 0

- ½§ · § · § ·§ ·° °® ¾¨ ¸ ̈ ¸ ̈ ¸¨ ¸° °© ¹ © ¹ © ¹© ¹¯ ¿

{(0 0 0 0 1), (0 0 0

1 0), (0 0 1 0 0), (0 1 0 0 0), (1 0 0 0 0)}V1 V2 V3 V4 = V.

B is a 4-basis of V as each B i is a basis of V i ; i = 1, 2, 3, 4.

Example 2.7: Let V = V 1 V2 V3 V4 V5 be a 5-vector space over Q where V 1 = R, V 2 = Q u Q, V 3 = Q[x], V 4 = R u R u R and V 5 = {set of all 2 u2 matrices with entries from Q}.Clearly V 1, V 2, V 3, V 4 and V 5 are vector spaces over Q. We seesome of the vector spaces V i over Q are finite dimensional i.e.,has finite basis and some of the vector spaces V j have infinite

number of elements in the basis set. We find means to define thenew notion of finite n-dimensional space and infinite n-dimensional space. To be more specific in this example, V 1 is aninfinite dimensional vector space over Q, V 2 and V 3 are finitedimensional vector spaces over Q. V 4 is an infinite dimensionalvector space over Q and V 5 is a finite dimensional vector spaceover Q.

D EFINITION 2.6: Let V = V 1 V 2 … V n be a n-vector spaceof type-I over F. If every vector space V i in V is finitedimensional over F then we say the n-vector space is finite n-dimensional over F. Even if one of the vector space V j in V isinfinite dimensional then we say V is infinite dimensional over

F. We denote the dimension of V by (n 1 , n 2 , …, n n ); n i dimensionof V i , i = 1, 2, …, n.

We illustrate the definition by some examples.



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Example 2.8: Let V = V 1 V2 V3 be a 3-vector space over Q,where V 1 = Q[x], V 2 = {set of all 2 u2 matrices with entries fromQ} and V 3 = Q the one dimensional vector space over Q.Clearly 3-dimension of the 3-vector space over Q is ( f , 4, 1).Thus V is an infinite 3-dimensional space over Q.

Example 2.9: Let V = V 1 V2 V3 V4 be a 4-vector spaceof type-I over Q. Suppose V 1 = {set of all 2 u 2, matrices withentries from Q}; V 2 = Q u Q u Q a vector space over Q,V3 = {All polynomials of degree less than or equal to 7 withcoefficients from Q} and V 4 = {the collection of all 5 u 5,matrices with entries from Q}, we see V 1, V 2, V 3 and V 4 arevector spaces over Q. The 4-dimension of V is (4, 3, 8, 25), so,V is finite 4-dimension 4 vector space over Q of type-I.

Having seen sub n-spaces, n-basis and n-dimension of n-vector spaces of type-I now we proceed on to define the notion of n-transformation of n-vector space of type-I.

D EFINITION 2.7: Let V = V 1 V 2 … V n be a n-vector

space over a field F of type-I and W = W 1 W 2 … W m beanother m-vector space over the same field F of type I, (n d m)(m t 2) and (n t 2). We call T a n-map if T = T 1 T 2 … T n: V o W is defined as T i : V i o W j , 1 d i d n, 1 d j d m for every i. If each T i is a linear transformation from V i to W j , i = 1,2, …, n, 1 d j d n then we call the n-map to be a n-linear transformation from V to W or linear n-transformation from V to W. No two V i’s are mapped on to the same W j , 1 d i d n, 1 d j

d m. Even if one of the T i is not a linear transformation from V ito W j then T is not a n-linear transformation.

We will illustrate this by the simple example.

Example 2.10: Let V = V 1 V2 V3 be a 3-vector space over Q and W = W 1 W2 W3 W4 be a 4-vector space over Q. Vis of finite (3, 2, 4) dimension and W is of finite (4, 3, 2, 4)dimension. T: V o W be a 3-linear transformation defined by T= T 1 T2 T3: V 1 V2 V3 o W1 W2 W3 W4 as



22

T1: V 1 o W1 given byT1

1 1 1 1 1 1 1 1 1 1 11 2 3 1 2 2 3 2 1 2 3(a ,a ,a ) (a a ,a ,a a ,a a a )

T2:V2 o W3 defined by 1 2 2 2 22 1 2 1 2 1T (a ,a ) (a a , a )

and T 3 : V 3 o W4 defined by3 3 3 3 3 3 3 3 3

3 1 2 3 4 2 4 4 1 2T (a ,a ,a ,a ) (a ,a ,a ,a a ) ,

clearly T is a 3-linear transformation or linear 3-transformationor linear 3 transformation of V to W; i.e. from 3-vector space V

to 4-vector space W.

It may so happen that we may have a n-vector space over a fieldF and it would become essential for us to make a linear n-transformation to a m-vector space over F where n>m. In suchsituation we define a linear n-transformation which we call asshrinking linear n-transformation which is as follows.

D EFINITION 2.8: Let V be a n-vector space over F and W a m-vector space over F n > m. The shrinking n-map T from V = V 1

V 2 … V n to W = W 1 W 2 … W m is defined as a map from V to W as follows T = T 1 T 2 … T n with T i : V i o W j; 1 d i d n and 1 d j d m with the condition T j : V j o W k where jmay be equal to k. i.e. the range space as in case of linear n-map may not be distinct.

Now if T i : V i o W j in addition a linear transformation thenwe call, T = T 1 T 2 … T n the shrinking n-map to be a

shrinking linear n-transformation or a shrinking n-linear transformation.

We illustrate this situation by the following example.

Example 2.11: Let V = V 1 V2 V3 V4 V5 be a 5-vector space defined over Q of 5-dimenion (3, 2, 5, 7, 6) and W = W 1

W2 W3 be a 3-vector space defined over Q of 3-dimension

(5, 3, 6). T = T 1 T2 … T5 : V o W can only be ashrinking 5-linear transformation defined by



23

T1 : V 1 o W3,

T2 : V 2 o W1,

T3 : V 3 o W2,

T4

: V4 o W

3

andT5 : V 5 o W1

where

1 1 1 1 1 1 1 1 1 1 1 11 1 2 3 1 2 3 2 2 3 1 3 1T (x ,x , x ) (x x , x ,x , x x ,x x ,x )

where 1 1 11 2 3 1x , x ,x V ,

2 2 2 2 2 2 2 2 22 1 2 1 2 2 1 2 1 2T (x , x ) (x x , x , x x , x , x ),

where 2 21 2 2, x V ,

3 3 3 3 3 3 3 3 3 3 33 1 2 3 4 5 1 2 2 3 4 5T (x ,x ,x ,x ,x ) (x x ,x x ,x x )

where 3 3 3 31 2 3 4x , x , x ,x and 3

5 3x V ,

4 4 4 4 4 4 44 1 2 3 4 5 6 7T (x ,x ,x ,x ,x , x , x )

= 4 4 4 4 4 4 4 4 4 4 4 41 2 2 3 3 4 4 5 5 6 6 7(x x , x x ,x x , x x , x x , x x )

for all 4 4 41 2 7 4x , x ,..., x V and

5 5 5 5 5 5 5 5 5 5 5 55 1 2 3 4 5 6 1 2 3 4 5 6T (x , x , x , x , x ,x ) (x , x , x x , x , x )

for 5 5 51 2 6 5x , x ,..., x V .

Clearly T is a shrinking linear 5 transformation.

Note: It may be sometimes essential for one to define a linear n-transformation from a n-vector space V into a m-vector spaceW, m > n where all the n spaces of the m-vector space may not

be used only a set of r vector spaces from W may be needed r <n < m, in such cases we call the linear n-transformation as aspecial shrinking linear n-transformation of V into W.




24

Example 2.12: Let V = V 1 V2 V3 be a 3-vector space over Q and W = W 1 W2 W3 W4 W5 be a 5-vector space over Q. Suppose V is a finite 3-dimension (3, 5, 4) space and W be afinite 5-dimension (3, 5, 4, 8, 2) space. Let T = T 1 T2 T3 : Vo W be defined by T 1: V 1 o W1, T 2: V 2 o W3, T 3: V 3 o W1 asfollows;

1 1 1 1 1 1 1 11 1 2 3 1 2 2 3 2T (x ,x , x ) (x x , x x ,x )

for all 1 1 11 2 3 1x ,x ,x V ;

2 2 2 2 2 2 2 2 2 22 1 2 3 4 5 2 1 3 5 4T (x ,x , x ,x , x ) (x ,x ,x x , x )

for all 2 2 2 2 21 2 3 4 5 2x , x , x , x , x V ,

3 3 3 3 3 3 3 3 3 33 1 2 3 4 1 2 4 1 2 3T (x ,x ,x ,x ) (x x ,x x ,x x )

for all 3 3 31 2 3x , x , x and 3

4x in V 3.

Thus T:V o W is only a special shrinking linear 3-transformation.

D EFINITION 2.9: Let V be a n-vector space over the field F and W be a n-vector space over the same field F. T = T 1 T 2 … T n is a linear one to one n transformation if each T i is atransformation from V i to W j and for no V k we have T k : V k o W ji.e. no two distinct domain space can have the same range

space. Then we call T to be a one to one vector space preserving linear n-transformation.

We just show this by a simple example.

Example 2.13: Let V = V 1 V2 V3 V4 be a 4- vector spaceover Q and W = W 1 W2 W3 W4 be another 4-vector space over Q. Let V be of (3, 4, 5, 2) finite 4 dimensional spaceand W a (2, 5, 6, 3) finite 4-dimensional space. Let T = T 1 T2

T3 T4: V = V 1 V2 V3 V4 o W1 W2 W3 W4 given by T 1: V 1 o W2 , T2: V 2 o W3 , T 3: V 3 o W4 and T 4: V 4

o W1 where T 1, T 2, T 3 and T 4 are linear transformation. ClearlyT is a linear one to one 4-transformation.



25

Note: In the definition 2.9 it is interesting and important to notethat all T i’s need not be 1-1 linear transformation with dim V i =dim W j if T i: V i o W j i.e., T i’s are not vector spaceisomorphism for i = 1, 2, …, n. Now we give a new name for an-linear transformation T: V o W where T = T 1 T2 … Tn with each T i a vector space isomorphism or T i is 1-1 and ontolinear transformation from V i to W j, 1 d i d n, 1 d j d n.

D EFINITION 2.10: Let V and W be n vector spaces defined over a field F. We say V and W are of same n-dimension if and onlyif n-dimension of V is (n 1 , …, nn ) then the n-dimension of W is

just a permutation of (n 1 , n2 , … , n n ).

Example 2.14: Let V = V 1 V2 V3 V4 V5 be a 5-dimension vector space over R of 5-dimension (7, 2, 3, 4, 5).Suppose W = W 1 W2 W3 W4 W5 is a 5-dimensionvector space over R of 5-dimension (2, 5, 4, 7, 3) then we say Vand W are of same 5-dimension. If X = X 1 X2 X3 X4 X5 is a 5-vector space of 5-dimension (2, 7, 9, 3, 4) then clearly

X and V are not 5-vector spaces of same dimension. So for anyn-dimensional n-vector space V we have only n number of n-vector spaces of same dimension including V.

We just show this by an example.

Example 2.15: Let V = V 1 V2 V3 be a 3-vector space of 3-dimension (7, 5, 3). Then W, X, Y, Z and S of 3-dimension (5,7, 3), (5, 3, 7), (7, 3, 5), (3, 5, 7) and (3, 7, 5) are of samedimension.

In view of this we have the following interesting theorem.

T HEOREM 2.1: Let V be a finite n-dimension n-vector spaceover the field F of n-dimension (n 1 , n 2 , …, nn ), then their exist n finite n-dimension n-vector spaces of same dimension as that

of V including V over F.



26

Proof: Given V is a finite n-vector space of n-dimension (n 1, n 2,…, n n) i.e. each 1 d ni d and i j implies n i n j we know twon-vector spaces V and W are of same dimension if and only if the n-dimension of one (say V) can be got from permuting then-dimension of W, or vice versa. Further from group theory weknow for a set (1, 2, …, n) we have n permutations of the set(1, 2, …, n). Thus we have n n-vector spaces of dimension (n 1,n2, …, n n).

Note: If we have a n-vector space of n-dimension (m 1, m 2, …,mn) with some m i n j, 1 d i d n then we get another set of n n-

vector spaces of n-dimension (m 1, m 2, …, m n) and all its permutations. Clearly this set of m-vector spaces with n-dimension (n 1, n2, …, n n) are distinct from the n-vector spacesof n-dimension (m 1, m 2, …, m n). From this one can conclude wehave infinite number of n-vector spaces of varying dimensions.Only same n-dimension vector spaces can be n-isomorphic.

D EFINITION 2.11: Let V and W be n-vector spaces of samedimension. Let n-dimension of V be (n 1 , n 2 , …, nn ) and that of W be (n 4 , n 2 , n n , …, n 5 ) i.e. let V = V 1 V 2 … V n and W = W 1

W 2 … W n. A linear n-transformation T = T 1 T 2 … T n : V o W is defined to be a n-vector space linear n-isomorphism if and only if T i : V i o W j is such that dim V i =dim W j; 1 d i, j d n.

We illustrate this situation by an example.

Example 2.16: Let V = V 1 V2 V3 V4 V5 and W = W 1 W2 W3 W4 W5 be two 5-vector spaces of same

dimension. Let the 5-dimension of V and W be (3, 2, 5, 4, 6)and (4, 2, 5, 3, 6) respectively. Suppose T = T 1 T2 T3 T4

T5: V o W given by T 1(V 1) = W 4, T 2(V2) = W 2, T 3(V 3) = W 3,T4(V 4) = W 1 and T 5(V5) = W 5; then T is a one to one n-isomorphic, n-linear transformation of V to W (n = 5). SupposeP : V o W where P = P 1 P2 P3 P4 P5 given by P 1:V1

o W2, P2: V 2 o W3, P 3: V 3 o W4, P4: V 4 o W5, and P 5: V 5 o W1 the linear transformation so that P is a 5-linear



27

transformation from V to W. Clearly P is not the one to oneisomorphic 5-linear transformation of V. P is only a one to one5-linear transformation of V.

Now having seen different types of linear n-transformation of an-vector space V to W, W a linear n-space we proceed on todefine the notion of n-kernel of T.

D EFINITION 2.12: Let V = V 1 V 2 … V n be a n-vector space over the field F and W = W 1 W 2 … W m be a m-vector space over the field F. Let T = T 1 T 2 … T n be n-linear transformation of T from V to W defined by T i: V i o W j;

1 d i d n and 1 d j d m such that no two domain spaces aremapped on to the same range space. The n-kernel of T denoted by ker T = ker T 1 kerT 2 … kerT n where

ker T i = {v i V i | T(v i ) = 0 }, i = 1, 2, …, n.Thus

ker T = {(v 1 , v2 , …, vn ) V 1 V 2 … V n | T(v 1 , v2 , …, vn )= T(v 1 ) T(v2 ) … T(vn ) = 0 0 … 0}.

It is easily verified that Ker T is a proper n-subgroup of V.Further Ker T is a n-subspace of V.

We will illustrate this situation by the following example.

Example 2.17: Let V = V 1 V2 V3 be a 3-vector space over Q of 3-dimension (3, 2, 4). Let W = W 1 W2 W3 W4 be a4-vector space over Q of 4-dimension (4, 3, 2, 5). Let T = T 1

T2 T3: V o W be a 3-linear transformation given by

T1:V1 o W4,1 1 1 1 1 1 1 1 1 1

1 1 2 3 1 2 3 1 1 3 2T ( x , x , x ) ( x x , x , x , x x , x )

for all 1 1 11 2 3 1x ,x ,x V ,

1 1 1 1 1 11 1 2 3 1 2 3ker T {(x , x , x ) T(x , x , x ) (0) i.e. 1

3x 0 , 11x 0 ,

12x 0 and

1 11 2x x 0 and

1 11 3x x 0 }



28

Thus ker T 1 = {(0, 0, 0)} is the trivial subspace of V 1,T2: V 2 o W3, 2 2 2 2 2

2 1 2 1 2 1T (x , x ) (x x , x )

for all 2 21 2 2x , x V .

ker T 2 = 2 2 2 21 2 1 2{(x ,x ) T(x , x ) (0)}

i.e. 2 21 2x x 0 and 2

1x 0 which forces 22x 0 . Thus ker T 2 =

{(0 0)}. Now

T3: V 3 o W1

given by3 3 3 3 3 3 3 3 3 3

3 1 2 3 4 1 2 3 4 3 4T (x , x , x , x ) (x x , x , x , x x )

for all3 3 3 31 2 3 4 3x ,x , x ,x V .

Now ker T 3 gives3 31 2x x 0 , 3

3x 0 , 34x 0 , 3 3

3 4x x 0 .

This gives the condition 3 31 2x x and 3 3

3 4x x 0 . Thus

ker T 3 = 3 31 1{(x , x ,0,0)} .

Thus a subspace of V 3. Hence we see the 3-kernel of T is a 1-susbspace of V i.e. ¢{(0 0 0 0) (0 0) 3 3

1 1(x , x ,0,0) }². We

can define kernel for any n-linear transformation T be it a usualn-linear transformation or a one to one n-linear transformation.It is easily verified that for any n-vector space V = V 1 V2 …

Vn and any m-vector space W = W 1 W2 … Wm over the same field F. Suppose T: V o W is any n-linear transformation from V to W then ker T = ker T 1 ker T 2 …

ker T n would be always a t-subspace of V as each ker T i is asubspace of V i , i = 1, 2, …, n. It may so happen that some of the

ker T i may be the zero space in such case we will call thesubspace of V only as a t-subspace of V where 1 d t d n. If allthe subspaces given by ker T i is zero then we call ker T to be then zero subspace of V; i = 1, 2, …, n.

Now we proceed on to give some more results in case of n-vector spaces and their related linear n-transformation.

D EFINITION 2.13: Let V = V 1 V 2 … V n be a n-vector space over a field F of type-I. Let T = T 1 T 2 … T n : V o



29

V be a linear n-transformation of V such that each T i : V i o V i , i= 1, 2, …, n. i.e., each T i is a linear operator on V i then wedefine T = T 1 T 2 … T n to be a n-linear operator on V.Clearly all n-linear transformations need not be n-linear operator on V. Thus T is a n-linear operator on V if and only if each T i is a linear operator from V i to V i , 1d i d n.

This is the marked difference between the linear operator on avector space and a n-linear operator on a n-vector space. All n-linear transformations from the n-vector space V to the same n-vector space V need not always be a n-linear operator.


Example 2.18 : Let V = V 1 V2 V3 V4 be a 4-vector spaceover Q of 4-dimension (5, 4, 2, 3). Let T = T 1 T2 T3 T4 :V o V be a 4-linear transformation given by T 1: V 1 o V2 , T 2:V2 o V3, T 3: V 3 o V4 and T 4: V 4 o V1. Clearly none of thelinear transformation T i’s are linear operators for they havedifferent domain and range spaces; i = 1, 2, 3, 4. So T though is

on the same n-vector space V still T is a linear n-transformationand not a linear n-operator on V, where n = 4.Suppose we define a 4-linear transformation P = P 1 P2

P3 P4 : V o V defined by P 1: V 1 o V1, P 2: V 2 o V2, P 3: V 3 o V3, and P 4: V 4 o V4, clearly the 4-linear transformation P isa 4-linear operator of V.

The above example shows the reader that in general a n-linear transformation of a n-vector space V need not in general be a n-linear operator on V. But of course trivially every n-linear operator on V is a n-linear transformation on V.

We have the following result in case of finite n-dimensional n-vector spaces over the field F.

T HEOREM 2.2: Let V = V 1 V 2 … V n and W = W 1 W 2 … W n be any two n-vector spaces over the field F. Let B =

1 21 1 1 2 2 21 2 1 2 1 2{( , ,..., ) ( , ,..., ) ... ( , ,..., )} n

n n nn n nD D D D D D D D D be a



30

n-basis of V 1 V 2 … V n; i.e., 1 2( , ,..., )i

i i inD D D is a basis of V i

,

i = 1, 2, … , n. Let

1 2

1 1 1 2 2 21 2 1 2 1 2{( , ,..., ) ( , ,..., ) ... ( , ,..., )}

n

n n nn n nC E E E E E E E E E

be any n-vector in W = W 1 W 2 … W n then there is precisely only one linear n-transformation T = T 1 T 2 … T n from V on to W such that i i

j jT D E , j = 1, 2, …, n i , 1< i<n.

Proof: To prove that there is some n-linear transformation Twith T(B) = C, it is enough if we show for the T = T 1 T2 …

Tn we have i ii j jT D E, i = 1, 2, …, n and j = 1, 2, …, n i.

Given ii i i1 2 n( , ,..., )D D D in V i there is a unique n i tuple

i

i i i1 2 n(x , x ,..., x ) such that

i i

i i i i i i i1 1 2 2 n nx x ... xD D D D, for this

vector i we defineT i

i =i i

i i i i i i1 1 2 2 n nx x ... xE E E

true for each i; i = 1, 2, …, n.

Clearly T i is a well defined rule for associating with each vector i in V i a vector T i i in W i. From the definition it is clear that

i ii j jT D E for each j. To see that T i is linear; let Ei =

i i

i i i i i i1 1 2 2 n ny y ... yD D D be in V i and c i be any scalar.

i i i

i i i i i i i ii i 1 1 1 i n n nT(c ) (c x y ) ... (c x y )D E E E.

On the other hand,

i i

i i ic (T ( )) T ( )D E =

i in ni i i i

i j j j j j 1 j 1c x y E E¦ ¦

=in

i i ii j j j

j 1

(c x y ) E¦

and thus T i( c ii + i ) = c i(T i

i) + T ii true for each i; i = 1, 2,

…, n. If U = U 1 U2 … Un is a linear n-transformation



32

T = T( 1 2 … n )= (T 1 T 2 … T n ) ( 1 2 … n )= T 1 1 T 2 2 … T n n = 0

0

…

0.

If the n-vector space V is finite n-dimensional, the n-rank of T isthe n-dimension of the n-range of T and will vary depending onthe nature of the n-linear transformation like if T is a shrinking n-linear transformation, it would be different and so on.

Now we can prove the most important theorem relating the n-rank of T and n-nullity of T for a n-linear transformation only asfor other n-linear transformation like shrinking n-linear

transformation the result in general may not be true.

T HEOREM 2.3: Let V and W be two n-vector space and m-vector space over the field F, m > n and let T is be linear n-transformation i.e. T = T 1 T 2 … T n from V to W is suchthat T i: V i o W j and the W j’s are distinct spaces for each T i , i.e.no two subspaces of V are mapped on to the same subspace inW. Suppose V is (n 1 , n 2 , … , n n ) finite dimensional, then n rank T

+ n nullity T = n dim V.

Proof: Given V = V 1 V2 … Vn is a n-vector space over Fand W = W 1 W2 … Wn is a m-vector space over F (m >n) of dimensions (n 1, n2, …, n n) and (m 1, m 2, …, m n)respectively. T = T 1 T2 … Tn is a n-linear transformationsuch that each T i is a linear transformation from V i to a uniqueW j, i.e. no two vector spaces V i and V k can be mapped to sameW j, if i k; 1 d i, k d n and 1 d j d m . Now n-rank T + n nullityT = n dim Wi.e. n-rank (T 1 T2 … Tn) + n nullity of (T 1 T2 … Tn) = n dim (V 1 V2 … Vn).i.e. rank T 1 rank T 2 … rank T n + nullity T 1 nullity T 2 … nullity T n = (dim V 1, dim V 2, …, dim V n) = (n 1, n2, …, n n).

Suppose N = N 1 N2 … Nn be the p-null space of then-space V; 0 d p d n. Let

= ^ `

i 2 n

1 1 1 2 2 2 n n n

1 2 k 1 2 k 1 2 k , ,..., , ,..., ... , ,...,D D D D D D D D D



33

be a n-basis for N. Here 0 d k i d ni ; i = 1, 2, … , n. If k i = 0 thenthe corresponding null space is the zero space. Now we showthe working for any i; T i: V i o W j. This result which we would

prove is true for all i = 1, 2, …, n.

Let ^ ì i i1 2 k , ,...,D D D be a basis for N i, the null space of T i. There

are vectorsi i

i ik 1 n,...,D D in V i such that ^ ì

i i i1 2 n, ,...,D D D is a basis

for V i; true for each i; i = 1, 2, …, n. We shall now prove that

^ ì ii k 1 i nT ,...,TD D is a basis for the range of T i. The vectorsi i i

i 1 i 2 i nT ,T ,...,TD D Dcertainly span the range of T i and sincei

i jT 0D for j d k i we see thati ii k 1 i nT ,...,TD D span the range, to

see that these vectors are linearly independent, suppose we havescalars c j’s such that

i

i

ni

j i j j k 1

c T ( ) 0

D ¦ .

This says thati

i

ni

i j j j k 1T ( c ) 0 D ¦ and accordingly the vector

i

i

ni i

j j j k 1

c

D D¦

is in the null space of T i. Sincei

i i i1 2 n, ,...,D D Dform a basis for N i

there must be scalarsi

i i i1 2 n, ,...,E E E such that

ik i i i

j j

j 1

D E D¦ . Thus

ik i i j j

j 1

E D¦ – i

i

ni i j j

j k 1 E D¦ = 0.

Sincei

i i i1 2 n, ,...,D D Dare linearly independent we must have i

1 b =

… =i

ik b =

i 1

ik c = … =

i

inc = 0. If r i is the rank of T i the fact that

T ii 1

ik D , . . . , T i

i

inD form a basis, for the range of T i tells us that

r i = n i – k i . Since k i is the nullity of T i and n i is the dimension of



34

V i, we get rank T i + nullity T i = dim. V i. This is true for eachand every i. That is(rank T 1 + nullity T 1) (rank T 2 + nullity T 2) … (rank T n + nullity T n)

= dim (V 1 V2 … Vn)i.e., (rank T 1 rank T 2 … rank T n) + (nullity T 1 nullityT2 … nullity T n)

= dim (V 1 V2 … Vn) that isrank (T 1 T2 … Tn) + nullity (T 1 T2 … Tn)

= (n 1, n2, … , n n).n rank T + n nullity T = n dim V.

Now in the relationn rank T + n nullity T = n dim (V) = (n 1, n2, …, n n).

We assume the n-linear transformation is such that it is notshrinking it is a n-linear transformation given in definition 2.12.Also we see if nullity T i = 0 for some i in such cases we haverank T i = dim V i. Since a p-nullspace in general need not always

be a nontrivial subspace we may have the p-nullspace of the n-vector space be such that p < n.

Now we proceed on to the algebra of n-linear transformations.Let us assume V and W are two n-vector space and m-vector space respectively defined over the field K.

T HEOREM 2.4: Let V and W be any two n-vector space and m-vector space respectively defined over the field F(m > n). Let T and U be n-linear transformations as given in definition from V into W. The n-function (T + U) defined by (T + U) D = T + Uis a n-linear transformation from V into W, if c is any element

from F, the function cT defined by (cT) = cT D is a n-linear transformation from V into W.

The set of all n-linear transformations from V into W withaddition and scalar multiplication defined above is an n-vector

space over the field F.

Proof: Let V = V 1 V2 … Vn be a n-vector space over F

and W = W 1 W2 … Wm (m>n) a m-vector space over F.T = T 1 T2 … Tn a n-linear transformation from V to W.



35

If U = U 1 U2 … Un is a n-linear transformation from Vinto W; define the n-function (T + U) for = 1 2 … n

V by (T + U) = T + U then (T + U) is a n-linear transformation of V into W.

(T + U) (c + )= [(T 1 T2 … Tn) + (U 1 U2 … Un)] [c + ]= [(T 1 + U 1) (T2 + U 2) … (Tn + U n)]

[c( 1 2 … n) + ( 1 2 … n))]= [(T 1 + U 1) (T2 + U 2) … (Tn + U n)]

[(c 1 + 1) (c 2 + 2) … (c n + n)]= [(T 1 + U 1) (c 1 + 1)] [(T 2 + U 2) (c 2 + 2)] …

[(T n + U n) (c n + n)].

Now using the properties of linear transformation on linear vector space we get (T i + U i) (c i + i) = c (T i + U i) ( i) + (T i +U i) ( i) for each i = 1, 2, …, n.

Thus (T + U) (c + ) = {[c(T 1 + U 1) 1 c(T 2 + U 2) 2 … c(T n + U n) n] + (T 1 + U 1) 1 (T1 + U 1) 2 … (Tn + U n)

n} = c(T + U) + (T + U) , which shows (T + U) is a n-linear transformation from V into W.

cT(d + )= c[(T 1 T2 … Tn) [d( 1 2 … n) + ( 1 2

… n)]= c[T 1 T2 … Tn] [(d 1 + 1) (d 2 + 2) …

(d n + n)]

= c[T 1(d 1 + 1) T2(d 2 + 2) … Tn(d n + n)]= T 1(c(d 1 + 1)) T2(c(d 2 + 2)) … Tn(c(d n + n)

(since each cT i is a linear transformation)= T[c(d + )]= c[T(d + )]= d[(cT) ] + (cT) .

This shows cT is a n-linear transformation of V into W.



36

T HEOREM 2.5: Let V be a n-vector space of n-dimension (n 1 , n 2 ,…, n n ) over the field F, and let W be a m dimensional m-vector

space over the same field F with m-dimension (m 1 , m2 , …, mn )(m > n). Then L n(V,W) is the finite dimensional n-space over F

of n-dimension 1 21 2, ,...,ni i i nm n m n m n where L n(V,W) denotes

the space of all n-linear transformations of V into W, 1 d i1 , i2 ,… , in d m.

Proof: LetB = {

1

1 1 11 2 n( , ,..., )D D D

2

2 2 21 2 n( , ,..., )D D D … n

n n n1 2 n( , ,..., )D D D }

be a n-basis for V = V 1 V2 … Vn of the n-vector space of n-dimension (n 1, n2, … , n n).

LetC = {

1

1 1 11 2 m( , ,..., )E E E 2

2 2 21 2 m( , ,..., )E E E … n

m m m1 2 m( , ,..., )E E E }

be a m-basis of the m-vector space W = W 1 W2 … Wm of m-dimension (m 1, m 2, …, m n).

Let L n(V, W) be the set of all n-linear transformation of Vinto W. For every pair of integers (p j, q i), 1 d j d m and 1 d i d n, 1 d p j d m j and 1 d qi d ni, we define a linear transformation

j i p ,qE ; 1 d i d n and k j d m of V i into W j by

j i p ,qE ( it) =

j

i

j i p

0 if t q

if t q

- z°®E °̄

i j j

tq pG E .

By the theorem i j j j i(T )D E their is a unique linear transformation

from V i into W j. We claim that m jni transformation j i

p ,qE form a basis for L i(V i, W j). This is true for each i, i = 1, 2, …, n and theappropriate j, 1 d j d m with no two spaces V i of V mapped intothe same W j. Let T i be a linear transformation from V i into W j, 1d i d n, 1 j m. For each k i k n i. Let A 1k , …,

jm k A be the

coordinates of the vector T iik in the ordered basis

( j

j j j1 2 m, ,...,E E E ) the n-basis of W given in C.



37

T iik =

j

k

m j p p

p 1

A E¦ (1)

We wish to show that

T i = j i j j

j i

j i

m n p ,q p q

p 1 q 1

A E

¦ ¦ (2)

Let U i be the linear transformation in the right hand member of (2). Then for each k

j , i

j i

j i

i p q ii k k p q

p 1 q 1

U A E ( )

D D¦ ¦

= j i i j

j i

j p q kq p

p q

A G E¦ ¦

= j

j j

j

m j

p k p p 1

A E¦

= ii k T D ;

and consequently U i = T i. Now from 2 we see j i p ,qE spans

L i(V i,W j). We must only now show they form a linearlyindependent set. This is very clear from the fact

U i = j, i

j i

j i

p q p q

p q

A E¦ ¦

is the zero transformation, then U iik = 0 for each k, so that

j

j j

j

m j

p k p p 1

A 0E ¦

and thus the independence of the j j pE implies that j p k A = 0 for

every p j and k. Since this is true of every i, i = 1, 2, … , n. Wehave

Ln(V,W) = n1L (V 1,

1iW ) n

2L (V 2,2i

W ) … nnL (V n,

niW );

where i 1, i2, …, i n are distinct elements from the set {1, 2, …,m}and m > n. Hence L n(V,W) is a n-space of dimension

1 2 ni 1 i 2 i n(m n ,m n ,...,m n ) over the same field F. This n-space will



38

be known as the n-space of n-linear transformation of the n-vector space V= V 1 V2 … Vn of n-dimension (n 1, n 2, …,nn) into the m-vector space W = W 1 W2 … Wm of m-dimension (m 1, m 2, …, m n), m > n.

Now having proved that the space of all n-linear transformationsof a n-vector space V into a m-vector space W forms a n-vector space over the same field F, we prove another interestingtheorem.

T HEOREM 2.6: Let V and W be two n-vector spaces of n-dimensions (n 1 , n 2 , … , n n ) and (t 1, t 2 , … , t n ) respectively defined

over the field F. Z be a m-vector space defined over the same field F(m > n). Let T be a n-linear transformation of V into W and U be a n linear transformation from W into Z. Then thecomposed function UT defined by (UT)( ) = U(T( )) is a n-linear transformation from V into Z, D V.

Proof : Given V = V 1 V2 … Vn and W = W 1 W2 …Wn are 2 n-vector spaces over F. Z = Z 1 Z2 … Zm is

given to be a m-vector space over F, m > n. T: V o W is a n-linear transformation; that is T = T 1 T2 … Tn : V o Wwith T i: V i o W j and no two vector spaces in V are mapped intothe same vector space W j, i = 1, 2, …, n and 1 d j d n.

Now U = U 1 … Un: W o Z is a n-linear transformationsuch that U j: W j o Zk , j = 1, 2, … , n and 1 d k d m such that notwo subspaces of W are mapped into the same Z k .

Now(U j T i) (ci + i) = U j [T i (c i + i)

= U j [T i (c i ) + T ( i)]= U j [c T i ( i) + T i ( i)]= U j [c j + j ]

(as T i : V i W j ; j, j W j)= c U j ( j) + U j ( j)= ca k + b k ; ak , bk Zk .



39

Thus U jT i is a n-linear transformation from W j to Z k . Hence theclaim; for the result is true for each i and each j. Thus UT is a n-linear transformation from W to Z.SoU o T = (U 1 U2 … Un) o (T 1 T2 … Tn)

= U 1 1iT U2 2iT … Un ni

T

(i1, i 2, … , i n) is a permutation of 1, 2, 3, … , n. Now we for thenotational convenience recall that if V = V 1 V2 … Vn is an vector space over a field F then V i’s are called as componentsubvector spaces of V. V i is also unknown as the component of V.

Now we proceed on to define the notion of linear n-operator.

D EFINITION 2.15: Let V = V 1 V 2 … V n be a n-vector space over F, a n-linear operator on V is a n-linear transformation T from V to V, such that T = T 1 T 2 … T n with T i:V i o V i for 1 d i d n. Thus in the above theorem not onlyV = W = Z but U and T are such that T i: V i o V i; U i: V i o V i so

that U and T are n-linear operators on the n space V, we seecomposition UT is again a n-linear operator on V.

Thus the n-space L n (V, V) has a multiplication defined ascomposition. In this case the operator TU is also defined. In

general TU UT i.e., UT – TU 0. Now Ln (V, V) would be only a n-vector space of dimension

2 2 21 2( , ,..., ),nn n n n-dimension of V is (n 1 , n 2 , … , n n ).

UT = (U 1 U 2 … U n ) (T 1 T 2 … T n )= (U 1T 1 U 2 T 2 … U nT n ).

TU = (T 1 T 2 … T n ) (U 1 U 2 … U n )= T 1U 1 T 2U 2 … T nU n.

Here T i : V i o V i and U i : V i o V i , i = 1, 2, … , n.

Now only in this case T 2 = TT and in general T n = TT … T; ntimes for n = 1, 2, …, n. We define Tº = I 1 I 2 … I n =

identity n-function of V = V 1 V 2 … V n. It may so happen



40

depending on each V i we will have different power of T i to beapproaching identity for varying linear transformation. i.e. if T : T 1 T 2 … T n on V = V 1 V 2 … V n , such that T i : V i o V i (only) for i = 1, 2, …, n since n-dimension of V is (n 1 , n 2 ,…, nn ). so T o T = T 2 = (T 1 T 2 … T n ) (T 1 T 2 … T n )= 2 2 2

1 2( , ,..., )nT T T . Like wise any power of T. I = I 1 I 2 … I n is the identity function on V i.e. each I i : V i o V i is such that I i( i ) = i for all i V i ; i = 1, 2, …, n. Only under these special conditions we define L n (V, V); elements of Ln(v, v) arecalled special n-linear operators.

L EMMA 2.1: Let V = V 1 V 2 … V n be a n-vector spaceover the field F; let U, T 1 and T 2 be n-linear operators on V; let c be an element of F

a. IU = UI = U where I = I 1 I 2 … I n is the n-identity transformation

b. U (T 1 + T 2 ) = UT 1 + UT 2 (T 1 + T 2 )U = T 1U + T 2U

c. C(U T 1 ) = (C U) T 1 = U (C T 1 ).

Proof: Given V = V 1 V2 … Vn be a n-vector space over F, F a field. I = I 1 I2 … In be the n-identitytransformation of V to V i.e. I j: V j V j; is the identitytransformation of each V j, j = 1, 2, …, n. U = U 1 U2 … Un: V o V such that U i: V i o V i for i = 1, 2, …, n. T i .

i i i1 2 nT T ... T : V o V such that i

jT : V j o V j; j = 1, 2, …, n

and i = 1, 2.U = (U 1 U2 … Un) (I 1 I2 … In)

= U 1 I 1 U2 I 2 Un In = U 1 … Un.

Further IU = (I 1 I2 … In) (U 1 U2 … Un)

= I 1 U 1 I2 U 2 In o U n = U 1 U2 … Un.

Thus IU=UI.



41

U(T 1 + T 2) = UT 1 + UT 2 = [U 1 U2 … Un] [(T 11 T12 … T1n) +

(T21 T22 … T2n)]= [U 1 U2 … Un] (T 11 + T 21)

(T12 + T 22)

… (T1n + T 2n).

We know from the results in linear algebra U (T 1 + T 2) = UT 1 +UT 2 for any U, T 1, T 2 L(V 1 ; V 1) where V 1 is a vector spaceand L(V 1 ,V1) is the collection of all linear operators from V 1 toV1.

Now in U i (T1i + T 2i), U i T1i and T 2i are linear operatorsfrom V i to V i true for each i = 1, 2, …, n. Thus U (T 1 + T 2) =UT 1 + UT 2 and (T 1 + T 2) U = T 1U + T 2U. Further C(UT 1) =(CU) T 1 = U(CT 1) for all U, T 1 Ln(V,V). Let U = U 1 U2 … Un and T 1 = ( 1 1 1

1 2 nT T ... T ) where U i: V i o V i for

each i and 1iT : V i o V i for each i = 1, 2, …, n.

C[(U 1 U2 … Un) 1 1 11 2 n(T T ... T ) ]

= C [ 1 1 11 1 2 2 n nU T U T ... U T ]

= ( 1 1 11 1 1 2 n nCU T CU T ... CU T ).

(CU 1 CU 2 … CU n) (T1

1 T12 … T1

n) = (CU)T 1.

But

C(UT 1) = (CU 1 CU 2 … CU n) (T 11 T1

2 … T1n)

= (U 1 U2 … Un) (CT 1)= (U 1 U2 … Un) ( 1 1 1

1 2 nCT CT ... CT )

= 1 1 11 1 2 2 n nU (CT ) U (CT ) ... U (CT )

= (U 1 U2 … Un) 1 1 11 2 n(CT CT ... CT )

= U(CT 1).

Let us denote the set of all n-linear transformation from V to V;this will also include the set of all n-linear operator T = T 1 T2

… Tn with T i : V i o V i, i = 1, 2, …, n. Let us denote the n-



44

Proof: Suppose first we assume T s is non singular. Let S = S 1 S2 … Sn be a n-linearly independent n-subset of V = V 1 V2 … Vn i.e. S i V i is a linearly independent subset of V i,i = 1, 2, …, n. LetS = { 1

1 1 11 2 k , ,...,D D D 2

2 2 21 2 k , ,...,D D D … n

n n n1 2 k , ,...,D D D }

= S 1 S2 … Sn V1 V2 … Vn. Given T s = T 1 T2 … Tn. Here T i: V i o W j, i

i i ii 1 i 2 i k T ,T ,...,TD D D are

linearly independent for each i for if i i1 i 1C (T )D + … +

i i

i ik i k C (T )D = 0,

then T i (i i1 1C D + … + i i

i ik k C D ) = 0 and since T i is non singular

( i i1 1C D + … +

i i

i ik k C D ) = 0 from which it follows each i

jC = 0, j =

1, 2, …, k i, because S i is an independent set. This is true of eachi, i.e. S = S 1 S2 … Sn is an independent n-set. This showsthe image of S under T s is independent. Suppose T s carriesindependent n-sets onto independent n sets. Let = 1 2 … n be a non zero n vector of V.

Then if S = S 1 S2 … Sn = 1 2 … n with S i = { i}; i = 1, 2, …, n; is independent. The image n-set of S isthe n-row vector T 1 1 T2 2 … Tn n and this set isindependent. Hence T s( ) = T 1 1 T2 2 … Tn n 0

because the set consisting of the zero n-vector alone isdependent. Thus null space of T s is 0 0 … 0.

The following concept of non singular n-linear transformation is little different.

D EFINITION 2.16: Let V and W be two same n-dimension spacesover F i.e. dim V = (n 1 , n 2 , …, nn ) and dim W = 1 2

, ,...,ni i in n n

where (i 1 , i2 , … , in ) is a permutation of ((1, 2, 3, …, n). If T =T 1 T 2 … T n is a special n-linear transformation of V intoW i.e. if T i: V i o W j then dim V i = dim W j = n i for every i. ThenT is n-non singular if each T i is non singular.

In view of this the reader is expected to prove the followingtheorem.



45

T HEOREM 2.9: Let V and W be two n-vector spaces of samedimension defined over the same field F. T is special n-linear transformation from V into W. Then T is n-invertible if and onlyif T is non-singular.

Note : We say the special n-linear transformation is n-invertibleif and only if each T i = T 1 T2 … Tn is invertible for i = 1,2, …, n.

Now we proceed on to define the n-representation of n-transformations by n-matrices. (n 2).

Let V be a n-vector space of n dimension (n 1, n 2, … , n n) and W be a m-vector space of m-dimension (m 1, m 2, … , m n) definedover the same field F. LetB = {

1

1 1 11 2 n( , ,..., )D D D

2

2 2 21 2 n( , ,..., )D D D …

n

n n n1 2 n( , ,..., )D D D }

be a n-ordered n-basis of V. We say the n-basis is an n-orderedn-basis if each of the basis ( i

1, i2, … , i

inD ) of V i is an ordered

basis for i = 1, 2, …, n and

B1 = {(1

1 1 11 2 m, ,...,E E E ) (

2

2 2 21 2 m, ,...,E E E ) … (

m

m m m1 2 m, ,...,E E E )}

be a m-ordered m basis of W. If T is any n-linear transformationfrom V into W i.e. T = T 1 T2 … Tn then each T i: V i o Wk is determined by its action on the vector i

jD ; 1 k m ; true

for each i = 1, 2, …, n and i j n i. Each of the n i vector T i j isuniquely expressible as a linear combination

T ii jD =

k

i

mk

iji 1

A E¦ (1)

1 k m and k iE Wk , the scalars A 1j, A 2j, …,

k m jA being

coordinates of T ii j in the m-ordered m-basis B 1. Accordingly

the transformation T i is determined by the m k ni scalars; A ij viaequation (1). The m k u ni matrix k

iA defined by jiA is called

the submatrix relative to the n-linear transformation T = T 1 T2

… T i … Tn of the pair of ordered basis



46

{i

i i i1 2 n( , ,..., )D D D } and {

k

k k k 1 2 m( , ,..., )E E E }

of V i and W k respectively. This is true for each i and k, 1 i nand 1 k m i.e. the m-matrix of T is given by

i i i1 2 n1 2 n

m m mn n nA A A !

= A (m, n) here 1 2 ni i im , m ,...,m (m 1, m 2, …,nmm ). Clearly

A is only a n-linear transformation map V i o W j and no twoV i’s are mapped onto same W j, 1 i n and 1 j m. Thus if

i is a vector in V i then i =i i

i i i i1 1 n nx a ... x a is a vector in V i

then

T ii = T i

ini i j j

j 1

x§ ·D¨ ¸© ¹¦

=in

i i j j

j 1

x .(T )§ · D¨ ¸© ¹¦

=i k n m

i k j ij j

j 1 i 1

x A

E¦ ¦

=k im n i k

ij j i j 1 i 1

A x

§ ·E¨ ¸© ¹¦ ¦ .

This is true for each i, i = 1, 2, ..., n. If X = X 1 X2 … Xn is the coordinate n-matrix of in the n-basis B then thecomputation above shows that

AX = ( i i i1 2 n

1 2 n

m m mn n nA A A ! ) (X 1 X2 … Xn) is the

coordinate n-matrix of the n-vector T in the ordered basis B1

because the scalars

1i1

nm 1ij j

j 1

A x¦ 2

i2

nm 2ij j

j 1

A x¦ …n

in

nm nij j

j 1

A x¦

is the entry of the i th n-row of the n-column matrix AX. Let usobserve that A is given by the m i u n j, n-matrices over the fieldF, then

T1

1n1 1 j j

j 1x

§ ·D¨ ¸© ¹¦ T2

2n2 2 j j

j 1x

§ ·D¨ ¸© ¹¦ … Tn

nnn n j j

j 1x

§ ·D¨ ¸© ¹¦



47

=i1 1

i1 1

m nm i1ij j i

i 1 j 1

A x

§ ·E¨ ¸© ¹

¦ ¦ i2 2

i2 2

m nm i2ij j i

i 1 j 1

A x

§ ·E¨ ¸© ¹

¦ ¦ …

in nin n

m nm inij j i

i 1 j 1A x

§ ·E¨ ¸© ¹¦ ¦ ,

where (i 1, i2, …, i n) {1, 2, …, m} taken in some order, definesa n-linear transformation T from V into W, the n-matrix of Arelative to the n-basis B and m-basis B 1 which is stated by thefollowing theorem.

T HEOREM 2.10: Let V = V 1 V 2 … V n be a finite n-dimensional i.e., (n 1 , n 2 , …, nn ) n-vector space over the field F

and W = W 1 W 2 … W m , an m-dimensional (m 1 , m2 , …,mn ) vector space over the same field F, (m > n). For each n-linear transformation T from V into W there is a n-mixed rectangular matrices A of orders (m 1u n1 , m2 u nn , …, mn u nn )with entries in F such that > @1 B

T D = A[ ] B for every D V. T o

A is a one to one correspondence between the set of all n-linear

transformations from V into W and the set of all m i u ni , mixed rectangular n-matrices, i = 1, 2, …, n over the field F. Thematrix A = ! ii i n1 2

1 2 n

mm mn n n A A A is the associated n-

matrix with T; the n-linear transformation of V into W relativeto the basis B and B 1.

Several interesting results true for the usual vector spaces can bederived in case of n-vector spaces n t 2 with appropriate

modifications. Now we give the definition of n-inner product on a n-vector

space V.

D EFINITION 2.17: Let F be a field of reals or complex numbersand V = V 1 V 2 … V n a n-vector space over F. An n-inner

product on V is a n-function which assigns to each ordered pair of n-vectors = 1 2 … n and = 1 2 … n in

the n-vector space V a scalar n-tuple from F. ¢ | ² = ¢ 1 2 … n | 1 2 … n ² = ( ¢ 1| 1 ² , ¢ 2| 2 ² , … , ¢ n| n ² ),



48

where ¢ i| i ² is a inner product on V i as i , i V i , this is true for each i, i = 1, 2, …, n; satisfying the following conditions:

a. ¢ + | ² = ¢ | ² + ¢ | ² (where = 1 2 … n , = 1 2 … n and = 1 2 … n where i , i ,

i V i for each i = 1, 2, …, n.) = ( ¢ 1 | 1 ² , ¢ 2 | 2 ² , …, ¢ n |n ² ) + ( ¢ 1 | 1 ² ¢ 2 | 2 ² , …, ¢ n | n ² ) = ( ¢ 1 | 1 ² + ¢ 1 | 1 ² , ¢ 2

| 2 ² + ¢ 2 | 2 ² , …, ¢ n | n ² + ¢ n | n ² )b. ¢ C | ² = C ¢ | ² = (C 1¢ 1 | 1 ² , C 2¢ 2 | 2 ² , …, C n¢ n | n ² )

c. ¢ | ² = D E , the bar denoting the complex conjugation.

d. ¢ | ² > (0, 0, … , 0) if 0 i.e., ( ¢ 1 | 1 ² , ¢ 2 | 2 ² , … ¢ n|n ² ) > (0, 0, … , 0) each i 0 in = 1 2 … n , i =

1, 2, …, n.

On F = 1 2 . . . nnn n F F F there is a n-inner product which we call the n-standard inner product. It is defined on

= ( 1

1 1 11 2, ,..., n x x x ) (

2

2 2 21 2, ,..., n x x x ) … ( 1 2, ,...,

n

n n nn x x x )

and = (

1

1 1 11 2, ,..., n y y y ) (

2

2 2 21 2, ,..., n y y y ) … ( 1 2, ,...,

n

n n nn y y y ) P

by

¢ | ² = ( 1

1 1

1¦

n

j j j

x y ,2

2 2

1¦n

j j j

x y , …,1

¦nn

n n j j

j

x y )

if F is a real field. If F is the field of complex numbers then

¢ | ² = ( 1 11

1¦

n

j j j

x y ,2 22

1¦n

j j j

x y , …,1

¦nn

nn j j

j

y ).

The reader is expected to work out the properties related with n-inner products on the n-vector spaces over the field F.

Now we proceed on to define n-orthogonal sets.

D EFINITION 2.18: Let V = V 1 V 2 … V n be a n-vector space over the field F. We say V is a n-inner product space if onV is defined an n-inner product. Let = ( 1 2 … n ) and



49

= ( 1 2 … n ) V with i , i V i , i = 1, 2, …, n. We say is n-orthogonal to if ¢ | ² = (0, 0, … , 0) = ( ¢ 1| 1 ² , …,¢ n| n ² ) i.e. if each i is orthogonal to i V i i.e. ¢ i| i ² = 0 for i= 1, 2, …, n. This equivalently implies is n-orthogonal to .

Hence we simply say and are orthogonal.

If S = S 1 S 2 … S n V 1 V 2 … V n = V be a n-set of n-vectors in V. S is called an n-orthogonal set provided all pairsof distinct n-vectors in S are orthogonal. An n-orthogonal set iscalled an n-orthonormal set if || || = (1, 1, … , 1) for every inS = S 1 S 2 … S n.

We denote ¢D | E² also by ( D | E ).

T HEOREM 2.11: Let V = V 1 V 2 … V n be a n-vector spacewhich is a n-inner product space defined over the field. Let S =S 1 S 2 … S n be an n-orthogonal set in V. The set of non

zero vectors in S are n-linearly independent.

Proof: Let V = V 1 V2 … Vn be a n-vector space over F.

Let S = S 1 S2 … Sn V = V 1 V2 … Vn be aorthogonal n-set of V. To show the elements in the n-sets are n-orthogonal. Let i i

1 2,D D , …,i

imD S i for i = 1, 2, …, n. i.e.

1 11 2,D D , …,

1

1mD S1,

2 21 2,D D , …,

2

2mD S2 and so on. n n

1 2,D D ,

…,n

nmD Sn. Let i i

1 2,D D , …,i

imD be the distinct set of n-vectors

in S i and that i = i i1 1c D + i i

2 2c D + … +i i

i im mc D . Then

( i | ik D ) =

im

i i i j j k

j 1c |§ ·D D¨ ¸© ¹¦

= i i i j j k

j 1

c ( | )D D¦

= ik c ( i i

k k |D D ).

Since ( i ik k |D D ) 0, i

k c 0z .



50

Thus when i = 0 then each ik c = 0 for each i. So each S i is an

independent set. Hence S = S 1 S2 … Sn is a n-independent set.

Several interesting results including Gram-Schmidt n-orthogonalization process can be derived.We now proceed onto define the notion of n-best

approximation to the n-vector relative to a n-sub-vector spaceW.

D EFINITION 2.19: Let V = V 1 V 2 … V n be a n-inner product n-vector space over the field F. W = W 1 W 2 … W n be a n-subspace of V. Let = 1 1

1 2 ... nn E E E V the n-

best approximation to by n-vectors in W is a n-vector =1 11 2 ... n

nD D D in W such that || – || || – || for every n-vector in W i.e. || i

i – ii|| || i

i – ii|| for every i

i W i and this is true for each i; i = 1, 2, …, n. We know if =

1 11 2 ... n

n E E E and if is a n-linear combination of an n-orthogonal sequence of non zero-vectors 1 , 2 , …, m whereeach i = i

1 i2 … i

n , i = 1,2,…, m, then

= 1 2

1 1 1 2 2 21 2

2 2 21 21 1 1

...

§ ·¨ ¸ ¨ ¸© ¹¦ ¦ ¦

nn n nmm m

k k k k n k k

nk k k k k k

E D D E D D E D D

D D D .

The following theorem is left as an exercise for the reader to prove.

T HEOREM 2.12: Let W = W 1 W 2 … W n be a n-subspaceof an n-inner product space V = V 1 V 2 … V n and =

1 11 2 ... n

n E E E be a n-vector in V

1. The n-vector = 1 11 2 ... n

nD D D in W is a n-best approximation to by the n-vector in W if and only if -

is n-orthogonal to every n-vector in W.2. If a n-best approximation to by n-vectors in W exists,

it is unique.



51

3. If W is n-finite n-dimensional and { 1 1

1 2 ... nnD D D }

is any n-orthonormal n-basis for W then the n-vector

=1 1 1 2 2 21 2

1 2 2 2

( | ) ( | )|| || || ||¦ ¦k k k k

k k k k

E D D E D D D D

2

( | )|| ||

¦!n n nn k k

nk k

E D D D

is the unique n-best approximation to by n-vector in W.

Now we proceed on to define the notion of n-orthogonalcomplement.

D EFINITION 2.20: Let V be a n-inner product n-space and S anyn-set of n vectors in V. The n-orthogonal complement of S is then-set S A of all n-vectors in V which are n-orthogonal to every n-vector in S; where S = S 1 S 2 … S n V = V 1 V 2 … V n and S A = 1 2 ...A A A nS S S V. i.e. each A

iS is theorthogonal complement of S i for every i, i = 1, 2, …, n. We call

to be the n-orthogonal projection of on W. If every n-vector in V has an n-orthogonal projection on W, the n-mapping that assigns to each n-vector in V its n-orthogonal projection on W is called the n-orthogonal projection of V on W.

The reader is expected to prove the following theorems.

T HEOREM 2.13: Let V = V 1 V 2 … V n be a n-inner

product n-vector space defined over the field F. W a finitedimensional n-subspace of V and E the n-orthogonal projectionof V on W, Then the n-mapping o ( – E ) is the n-orthogonal projection of V on W A.

T HEOREM 2.14: Let W = W 1 W 2 … W n V be a finitedimensional n-subspace of the n inner product space V = V 1 V 2 … V n and let E = E 1 E 2 … E n be the n-orthogonal

projection of V on W. Then E is an n-idempotent n-linear transformation of V onto W and W A is the n-null space of E and



52

V = W W A i.e. if V = V 1 V 2 … V n and W = W 1 W 2 … W n and W A = 1 2 ...A A A nW W W then V = (W W A ) =

(W 1 1AW ) (W 2 2

AW ) … (W n AnW ).

T HEOREM 2.15: Under the conditions of the above theorems, I- E is the n-orthogonal projection of V on W A. It is an n-idempotent linear n-transformation of V onto W A , with null

space W.

T HEOREM 2.16: Let { 1 11 2 ... n

nD D D } be an orthogonal n- set of non-zero vectors in an n-inner product space V. If is any

vector in V then k |( | k k D )| 2 / || k k D || 2 || || 2 where =1 11 2 ... n

n E E E V.

It is pertinent to mention here that the notion of linear functionaldual space or adjoints cannot be extended in an analogous wayin case of n-vector spaces of type I.

Now we proceed on to define the notion of n-unitary operatorson n-inner product n vector spaces V over the field F.

D EFINITION 2.21: Let V and W be n-inner product n-vector space and m vector space over the same field F respectively. Let T be a n-linear transformation from V into W. We say that T

preserves n inner products if (T | T ) = ( | ) for all , V i.e. if V = V 1 V 2 … V n and W = W 1 W 2 … W n and T = T 1 T 2 … T n with = 1 2 … n and = 1

2 … n V. T i: V i o W j. with no two V i mapped on to the same W j , then T i i ,T i i W j and (T i i | T i i ) = ( i | i ) for every i,i = 1, 2, …, n. An n-isomorphism of V into W is a n-vector spaceisomorphism T of V onto W which also preserves n-inner

products.

T HEOREM 2.17: Let V and W be n-finite dimensional n-inner vector spaces of same n-dimension i.e. dim V = (n 1 , n 2 , … , nn )

and dim W = 1 2, ,..., ni i in n n where (i 1 , i2 , … , in ) is a



53

permutation of (1, 2, …, n) defined over the same field T. If T =T 1 T 2 … T n is a n-linear transformation from V into W the following are equivalent

1. T preserves inner products i.e., each T i in T preservesinner product i.e. T i: V i o W j; 1 i , j n.

2. T is an n-inner product n- isomorphism3. T carries every n-orthonormal n-basis for V onto an n-

orthogonal n-basis for W.4. T carries some n-orthogonal n-basis for V onto an n-

orthonormal basis for W i.e. T i carries some orthogonal basis of V i into an orthogonal basis for W j.

The reader is expected to prove the following theorems.

T HEOREM 2.18: Let V and W be n-dimensional finite inner product n-spaces over the same field F. Then V = V 1 V 2 …

V n is n-isomorphic with W = W 1 W 2 … W n i.e. each T i:V i o W j is an isomorphism for i = 1, 2, …, n if V and W are of

same n-dimension.

T HEOREM 2.19: Let V and W be two n-inner product spacesover the same field F. Let T = T 1 T 2 … T n be a n-linear transformation from V into W. Then T preserves n-inner product if and only if ||T || = || || i.e. ||(T 1 T 2 … T n )( 1 1

1 2 ... nnD D D )|| = ||T 1( 1

1D ) T 2( 22D ) … T n( n

nD )|| =

(|| 11D ||, || 2

2D ||, … , || nnD ||) for every V i.e. for every i V i ,

i = 1, 2, …, n.

We define the notion of n unitary operator of a n-vector space Vover the field F.

D EFINITION 2.22: A n-unitary operator on an n-inner product space V is a n-isomorphism of V onto itself.

D EFINITION 2.23: If T is a n-linear operator on an n-inner product space V = V

1V

2 … V

n , then we say T = T

1 T

2

… T n has an n-adjoint on V if there exists a n-linear



54

operator T*= 1 2 ... nT T T on V such that (T | ) = ( |

T* ) for all = 1 11 2 ... n

nD D D , = 1 11 2 ... n

n E E E in V

= V 1 V 2 … V n i.e. T i( |i ii iD E ) = ( | *i i

i iT D E ) for each i =1, 2, …, n.

It is easily verified as in case of adjoints the n-adjoints of T notonly depends on T but also on the n-inner product on V.

Interesting results in this direction can be derived for any reader.The following theorems are also left as an exercise for thereader.

T HEOREM 2.20: Let V = V 1 V 2 … V n be a finite n-dimensional n-inner product n-space defined over the field F. If T and U are n-linear operators on V and c is a scalar, then

1. (T + U)* = T* + U* i.e. if T = T 1 T 2 … T n and U = U 1 U 2 … U n then in (T + U)* we have for eachi, (T i + U i )* = *

iT + *iU , i = 1, 2, …, n.

2. (cT)* = cT*

3. (TU)* = T*U*, here also (T iU i )* = *iU *

iT for i = 1, 2,…, n. i.e. (TU)* = (T 1U 1 )* (T 2U 2 )* … (T nU n )* =

*1U *

1T *2U *

2T … *nU *

nT

4. (T*)* = T since ( *iT )* = T i , for each i = 1, 2, …, n.

T HEOREM 2.21: Let U be a n-linear operator on an n-inner product space V, defined over the field F. Then U is n-unitary if and only the n-adjoint, U* of U exists and UU* = U*U = I.

T HEOREM 2.22: Let V = V 1 V 2 … V n be a n-vector spaceof a n-inner vector space of finite dimension and U be a n-linear operator on V. Then U is n-unitary if and only if the n-matrix

related with U in some ordered n-orthonormal n-basis is also an-unitary matrix i.e. if A = A 1 A2 … An is the n-matrix



55

each A i in A is unitary i.e. A i * Ai = I for each i, i.e. A * A = A1*A1 A2*A2 … An*An = I 1 … I n.

Several interesting results can be obtained using appropriate andanalogous proper modifications.

Now we proceed on to define the notion of n-normaloperator or normal n-operators on a n-vector space V. The

principle objective for doing this is we can obtain someinteresting properties about the n-orthonormal n-basis of V = V 1

V2 … Vn.

Let the n-orthonormal n-basis of V be denoted by B =

{( 11 1 11 2, ,..., nD D D ) ( 2

2 2 21 2, ,..., nD D D ) … ( 1 2, ,..., nn n nnD D D )}

where each ( 1 2, ,...,i

i i inD D D ) is a orthogonal basis of V i for i = 1,

2, …, n. Let T = T 1 T 2 … T n the n-linear operator on V be defined by T i

i jD = i i

j jc D for j = 1, 2, …, n i and for each T i , i =

1, 2, …, n. This simply implies that the n-matrix of T (consequently each matrix of T i in the ordered basis( 1 2, ,...,

i

i i inD D D ) is a diagonal matrix with the diagonal entries

( 1 2, ,...,i

i i inc c c ) is a n-diagonal n matrix given by

D =

1

11

12

1

0

0

ª º« »« »« »« »« »

« »¬ ¼

%

n

c

c

c

2

21

22

2

0

0

ª º« »« »« »« »« »

« »¬ ¼

%

n

c

c

c

…

1

2

0

0

ª º« »« »« »« »« »« »¬ ¼

%

n

n

n

nn

c

c

c

.



56

The n-adjoint operator T*= T* 1 T*2 … T*n of T = T 1 T 2 … T n is represented by the n-conjugate transpose nmatrix i.e. once again a n-diagonal n matrix with diagonal

entries1 1

1 2, ,...,

c c i

i

nc ; i = 1, 2, …, n. If V is a real n-vector space over the real field F then of course we have T = T*

D EFINITION 2.24: If V = V 1 V 2 … V n be a n-dimensional n-inner product n-vector space and T a n-linear operator on V be say T is n-normal if it commutes with its n-adjoint T* of T i.e.TT* = T*T.

Now in order to define some more properties we now proceedonto define the notion of n-characteristic values or n-eigenvalues of a n-vector space V and so on.

D EFINITION 2.25: Let V = V 1 V 2 … V n be a n-vector space over the field F of type I. Let T = T 1 T 2 … T n be an-linear operator on V. A n-characteristic value (or equivalently characteristic n-value) of T is a n-tuple of scalars

1 21 2 ... n

nc c c such that their exists a non zero n vector =1 21 2 ... n

nD D D in V with T = c . i.e. ( 1 21 2 ... n

nc c c )

( 1 21 2 ... n

nD D D ) = 1 11 1c D 2 2

2 2c D … n nn nc D = T 1 1

1D

T 2 22D … T n n

nD ; If c = 1 21 2 ... n

nc c c is the n-characteristic value of T then

a. any = 1 21 2 ... n

nD D D such that T = c is called the n-characteristic n-vector of T associated with the n-characteristic value c = 1 2

1 2 ... nnc c c .

b. The collection of all = 1 21 2 ... n

nD D D such that T= c is called the n-characteristic space associated with c. n-characteristic values will also be known as n-eigen values or n-spectral values.

If T is any n-linear operator on the n-vector space V and c any n scalar the set of n-vector in V = V 1 V 2 … V n such that



57

T = c is a n-subspace of V. It is the n-null space of the n-linear transformation (T – cI) = (T 1 T 2 … T n ) – ( 1 2

1 2 ... nnc c c ) (I 1 I 2 … I n )) = (T 1 – 1

1c I 1 ) (T 2 – 22c I 2 ) … (T n –

nnc I n ). We call c the n-characteristic value of

T and if this n-subspace is different from the zero subspace i.e.if (T – cI) = (T 1 – 1

1c I 1 ) … (T n – nnc I n ) fails to be one to one

i.e. each T i – iic I i fails to be one to one. If V is a finite n-

dimension n-vector space, (T – cI) fails to be one to one. Onlywhen the n determinant i.e. det(T – cI) = det(T 1 – 1

1c I 1 ) det(T 2 – 2

2c I 2 ) … det(T n – nnc I n ) (0 0 … 0) i.e. each

det(T i – iic I i ) 0 for i = 1, 2, …, n.

This is made into the following nice theorem

T HEOREM 2.23: Let T be a n-linear operator on a finite n-dimensional n-vector space V = V 1 V 2 … V n and c =

1 21 2 ... n

nc c c be a n scalar then the following are equivalent

a. c = 1 21 2 ... n

nc c c is a n-characteristic value of T =T 1 T 2 … T n i.e. each c i

i is a characteristic valueof T i ; i = 1, 2, …, n.

b. The n-operator (T – cI) = (T 1 – 11c I 1 ) … (T n –

nnc I n ) is non singular (i.e. non invertible) i.e. each ( T i – iic I i ) is non invertible i.e. non singular for each n-vector

spaces, i = 1, 2, …, n.c. det(T – cI) = ( 0 0 … 0) i.e. det ( )i

i i iT c I = 0 for each i = 1, 2, …, n.

Now we give the analogous for n-matrix.

D EFINITION 2.26: Let A = A1 A2 … An be a n-squarematrix where each matrix A i is n i u ni matrix i = 1, 2, …, n; if i

j then n i n j , 1 i, j n over the field F, a n-characteristicvalue of A in F is a n scalar C= 1 2

1 2 ... nnC C C ; i

iC F , i =



58

1, 2, 3, … , n such that the n-matrix (A – CI) = 11 1 1( ) A C I

22 2 2( ) A C I … ( )n

n n n A C I is singular, i.e. C is a n-characteristic value of A if and only if det (A – CI) = 0 0 …

0 or equivalently det (CI – A) = 0 0 … 0, i.e. if det ( )i

i i i A C I = 0 for each and every i, i = 1, 2, …, n we formthe matrix (xI – A) where x = x 1 x2 … xn with polynomial entries and consider the n-polynomial f = det(xI – A) = det(x 1 I 1

– A1 ) det(x 2 I 2 – A2 ) … det(x n I n – An ) = f 1 f 2 … f n inn variables x 1 , x2 , … , xn. Clearly the n-characteristic values of

A in F are just the n-tuple scalars C = 1 21 2 ... n

nC C C in F such that

f(C) = 0 0 … 0= 1 2

1 1 2 2( ) ( ) ... ( )nn n f C f C f C .

For this reason f is called the n-characteristic n-polynomial of A.

It is important to note that f is a n-monic polymonial which hasdegree exactly (n 1, n2, …, n n ) is the n-degree of the n-monic

polynomial f = f 1 f 2 … f n.

We can prove the following simple lemma.

L EMMA 2.2 : Similar n-matrices have same n-characteristic polynomial.

Proof: We just recall if A and B are the mixed square n-

matrices of dimension (n 1, n2, …, n n ) and (n 1, n2, …, n n ) i.e.same dimension i.e. identity permutation of (n 1, n 2, …, n n ) . Wesay A is similar to B or B is similar to A if their exists ainvertible n matrix P of dimension (n 1, n 2, …, n n ) such that if A= A 1 A2 … An, B = B 1 B2 … Bn and P = P 1 P2

… Pn then B = P -1AP i.e. B = B 1 B2 … Bn = 11P A1

P1 12P A2 P2 … 1

nP An Pn; i.e. each A i is similar B i for i= 1, 2, …, n.



59

Suppose A and B are similar n-mixed square matrices of identical dimension i.e. order A i = order B i for i = 1, 2, …, nthen B = P -1AP the

n-det (xI – B) = n-det(xI – P -1 A P)= n-det (P -1(xI –A) P)= n-det P -1 det(xI – A) .det P= 1 1

1 1 1 1 1det P det(x I A )det P 1 2

2 2 2 2 2det P det(x I A )det P … 1 n

n n n n ndet P det(x I A )det P = det(xI – B)

=11 1 1det(x I B )

22 2 2det(x I B ) …

nn n ndet(x I B )

i.e. n-det (xI – B) = n-det (xI – A) .

D EFINITION 2.27 : Let T = T 1 T 2 … T n be a special n-linear operator on a finite dimension n-vector space V = V 1 V 2 … V n. We say T is n diagonalizable if there is a n-basis

for V each n-vector of which is a n-characteristic n-vector of T.

The following two lemmas are left as an exercise to the reader.

L EMMA 2.3 : Suppose T = C where T = T 1 T 2 … T n , = 1 2

1 2 ... nnD D D and C = 1 2

1 2 ... nnC C C . If F = F 1

F 2 … F n is any n-polynomial then f(T) = f(C) i.e.,1 2

1 1 1 2 2 2( ) ( ) ... ( ) nn n n f T f T f T D D D =

1 1 2 21 1 1 2 2 2( ) ( ) ... ( )n n

n n n f C f C f C D D D .

L EMMA 2.4 : Let T = T 1 T 2 … T n be a n-linear operator on a finite (n 1 , n 2 , … , n n ) dimensional n-vector space V = V 1 V 2 … V n. Let {(

1

1 1 11 2, ,..., k C C C ) (

2

2 2 21 2, ,..., k C C C ) …

( 1 2, ,...,n

n n nk C C C ) } be distinct n-characteristic values of T 1 T 2

… T n and let 1

1 1 11 2, ,..., k W W W be the subspaces of V 1

associated with characteristic values1

1 1 11 2, ,..., k C C C respectively,

22 2 2

1 2, , ..., k W W W be the subspaces of V 2 with associated



60

characteristic values2

2 2 21 2, , , k C C C ! respectively; and so on and

let 1 2, ,...,n

n n nk W W W be the subspaces of V n with associated

characteristic values1 2

, ,...,n

n n n

k C C C and if

1

1 1 1 11 2 ... k W W W W

and if

2

2 2 2 21 2 ... k W W W W , …, 1 2 ...

n

n n n nk W W W W

and if W = W 1 W 2 … W n the n-dim W = (dimW 1 , dimW 2 ,…, dimW n ) with dim W j = 1 2 j

j j jk dimW dimW ... dimW for

each j = 1, 2, …, n; and if ii B is an ordered basis of W

i , i = 1,

2, …, k; then ( 1 21 2, ,..., n

n B B B ) is the n-ordered n-basis of W 1 W 2

… W k .

Using these lemmas the reader is expected to prove thefollowing theorem.

T HEOREM 2.24: Let T = T 1 T 2 … T n be a n-linear operator of the finite n-dimensional n vector space V = V 1 V 2

… V n. Let {( 1

1 1 11 2, ,..., k C C C ) , (

2

2 2 21 2, ,..., k C C C ) …

( 1 2, ,...,n

n n nk C C C ) } be the distinct n-characteristic n-values of T

and let (W 1 W 2 … W n ) be the null n-subspace of (T – CI)i.e. W i is a subspace of ( )i

i iT C I for i = 1, 2, …, n. Then the following are equivalent

1. T is n-diagonalizable2. The n-characteristic polynomial for T is

f = f 1 f 2 … f n where f i =1 2

1 2( ) ( ) ...( )ii ik i

i

d d d i i ii i i k x C x C x C for every i = 1, 2, … , n. and

dim W i = d i where 1 2 ... i

i i i ik d d d d for every i = 1, 2, … ,

n. dim W 1 + dim W 2 + … + dimW k = dim V = (n 1 , n 2 , … n n ) i.e.,dim W 1 = dim 1

1W + dim 12W + … + dim

1

11k W n , dimW 2 =



61

dim 21W + dim 2

2W + … + dim2

12k W n and so on and dimW n =

dim 1nW + dim 2

nW + … + dimn

nk nW n .

The proof left as an exercise for the reader.

Now we proceed on to define the notion of n-annihilating polynominals.

Let T = T 1 T2 … Tn be a n-linear operator on a n-vector space V over the field F. If p(x) = p 1(x) p2(x) …

pn(x) be a n-polynominal in x with coefficients from F then p(T)= p 1(T1) p1(T1) … pn(Tn) is again a n-linear operator on

V. If q(x) = q 1(x) q2(x) … qn(x) is another n-polynomialover F then

(p + q) (T) = p(T) + q(T) . pq(T) = p(T) q(T)

= p 1(T) q 1(T) p2(T) q 2(T) … pn(T) q n(T) .

Therefore the collection of n-polynomials p(x) which n-annihilate T in the sense that p(T) = 0, is a n ideal in the n-

polynomial algebra F[x]. Clearly Ln

(V, V) is a n-linear space of dimension ( 2 2 21 2 nn , n ..., n ) where n i is the dimension of the vector

space V i in V = V 1 V2 … Vn. If we take in the n-linear operator T = T 1 T2 … Tn, for each T i a 2

in 1 power of

T i for i = 1, 2, …, n then2i

2i

ni i i 2 i0 1 i 2 i in

C C T C T ... C T 0 for

some scalars i jC not all zero, 1 d j d 2

in .

So the n-ideal of polynomials which n-annihilate T contains

a non zero n-polynomial of n-degree ( 2 2 21 2 nn ,n ,..., n ) or less.

Now we define the notion of n-minimal polynomial for T =T1 T2 … Tn.

D EFINITION 2.28: If T is a n-linear operator on a finitedimensional n-vector space V over the field F. The n-minimal

polynomial for T = T 1 T 2 … T n is the unique n-monic generator of the n-ideals of polynomial over F which n-

annihilate T, i.e., the n-monic generator of the n-ideals of polynomials over F which annihilate each T i for i = 1, 2, …, n.



62

The term n-minimal comes from the fact that the n-generator of a polynomial n-ideal is characterized by being the n-monic

polynomials each of minimum degree that is every ideal in then-ideals; that is the n-minimal polynomial p = p 1 p2 …

pn for the n-linear operator T is uniquely determined by thesethree properties. In p = p 1 p2 … pn , each p i is a monic

polynomial over the scalar field F, which we shortly call as then-monic polynomial over F. p(T) = 0 implies p i(T i ) = 0 for eachi, i = 1, 2, …, n; i.e., p 1(T 1 ) p2(T 2 ) … pn(T n ) = 0 0 …

0. No n-polynomial over F which n-annihilates T has smaller degree than p, i.e., polynomial over F which annihilates T i has

smaller degree than p i for each i = 1, 2, …, n. If A is a n-mixed

square matrix over F i.e., A = A 1 A2 … An is a n-mixed matrix where each A i is a n i u ni matrix over F, we define the n-minimal polynomial for A in an analogous way as unique n-monic generator ideal of all n-polynomials over F which n-annihilate A or annihilates A i for each i, i = 1, 2, …, n.

Similar results which hold good in case of linear vector spacescan be analogously extended to the case of n-vector spaces with

proper and appropriate modifications.The proof of the following interesting theorem can be obtained

by any interested reader.

T HEOREM 2.25: Let T = T 1 T 2 … T n be a n-linear operator on a (n 1 , n 2 , …, nn ) finite dimensional n-vector space[or let A = A 1 A2 … An , a n-mixed square matrix whereeach A i is a n i u ni matrix, i = 1, 2, …, n] then n-characteristicand n-minimal polynomial for T[for A] have the same n-rootsexcept for multiplicities.

The Cayley-Hamilton theorem for n-linear operator T on the n-vector space V is stated, the proof is also left as an exercise for the reader.

T HEOREM 2.26: (CAYLEY HAMILTON THEOREM FOR n-VECTOR

SPACES ) Let T = T 1 T 2 … T n be a n-linear operator on a finite (n 1 , n 2 , …, n n ) dimensional n-vector space V = V 1 V 2



64

… , n. i.e., T(W) is contained in W or this is the same as T i(W i )is contained in W i for i = 1, 2, …, n.

L EMMA 2.5: Let V be a finite (n 1 , n 2 , …, nn ) dimensional n-vector space over the field F. Let T = T 1 T 2 … T n be a n-linear operator on V such that the n-minimal polynomial for T is a product of linear n-factors p = p 1 p2 … pn where p i

= 1

1 ... ; k i

i

r r i i ik j x C x C C F , 1 d j d k i , for i = 1, 2, …, m.

Let W = W 1 W 2 … W n be a proper (W z V) subspace of

V where each 1 ... i

i i ik W W W , i = 1, 2, … , n. which is n-

invariant under T. There exists a vector D = D 1 D 2 … D n in V such that D is not in W;

(T – CI) D = (T 1 – C 1 I 1 ) D 1 (T 2 – C 2 I 2 )D 2 … (T n – C n I n )D n

is in W for some m-characteristic values1

1 1 1 11 2( , ... )k C C C C , 1 d

k 1 d n1;2

2 2 2 21 2( , ,..., )k C C C C and so on.

The proof can be derived without much difficulty; infact verystraight forward, using the working for each T i: V i o V i and

i

i i i1 k W W ... W , 1 d k i d ni. When the result holds for every

component of V and T it is true for the n-vector space and its n-linear operator T which is defined on V.

The following theorem on the n-diagonalizablily of the n-linear operator T on V is given below.

T HEOREM 2.27: Let V = V 1 V 2 … V n be a finite (n 1 , n 2 ,… , nn ) dimensional n-vector space over the field F and let T =T 1 T 2 … T n be a n-linear operator on V. Then T is ndiagonalizable if and only if the n-minimal polynomial for T hasthe form,

p = ^ `1

1 11 ... k x C x C



65

^ `2

2 2 21 2 ... k x C x C x C …

^ `1 2 ... n

n n nk x C x C x C

wherei

i jC are distinct elements of F (i.e.,

1

1 1 11 2, ,..., k C C C , forms a

distinct set in F,2

2 2 21 2, ,..., k C C C forms a distinct set in F, so on

1 2, ,...,n

n n nk C C C forms a distinct set of F) .

Proof: We know if T = T 1 T2 … Tn is n-diagonalizableits n-minimal polynomial is a n-product of distinct linear factors

i.e., each T i: V i o V i (where V i is a component of the n-vector space V = V 1 V2 … Vn and T i is a linear operator of V i and a component of T).

So we can say if p i = i

i i i1 2 k x C x C ... x C the

minimal polynomial associated with the diagonalizable operator T i then the p i is a product of distinct linear factors. This is truefor each i; i = 1, 2, …, n, Hence the claim. So to prove theconverse, let W = W 1 W2 … Wn be the n-subspacespanned by all the n-characteristic n-vectors of T and supposeW z V that is; each W i z V i for i = 1, 2, …, n.

This implies we have a n-vector D= D1 D2 … Dn notin W (i.e., each Di W i for i = 1, 2, …, n.) and a n-characteristic value C = C 1 C2 … Cn of T such that thevector E= (T – CI) Dlies in W i.e., E= E1 E2 … En thenEi = i i

i j iT C I D lies in W i (1 d j d k i) this is true for each i, i =

1, 2, …, n. Since Ei W

iwe have i

i i i i1 2 k ...E E E E(true for

each i, i = 1, 2, …, n) where i i ii j j jT CE E; 1 d j d k i and i = 1, 2,

…, n and hence the vector in W i.

i i

i i i i i i i ii 1 1 k k h (T ) h (C ) ... h (C )E E E is in W i for every

polynomial h i; this is true for each i, i = 1, 2, …, n. Now pi = ( i

jx C ) q i for some polynomial q i alsoi i i

i i j jq q (C ) x (C )h (this is true for each i, i = 1, 2, …, n).



66

We have i i i i i i i ii i i j i i j i iq (T ) q (C ) h (T )(T C I ) h (T )D D D E, 1

d i d n. But i iih (T ) E is in W i (for each i) and since 0 = p i(T i) Di

=

i i

i j i i iT C I q (T ) D, the vector ii iq (T ) D is in W i.

Therefore i ii jq (C ) D is in W i. Since Di is not W i we have

ii jq (C ) = 0 true for every i = 1, 2, …, n. This contradicts the fact

p i has distinct roots for i = 1, 2, …, n. Hence the claim.

How ever we give an illustration of this theorem so that thereader can understand how it is applied in general.

Example 2.19: Let V = V 1 V2 V3 where V 1 = Q u Q, V 2 =Q u Q u Q u Q and V 3 = Q u Q u Q i.e., V a 3-vector space over Q of finite dimension and 3-dimension (2, 4, 3) .Define T : V o V by T = T 1 T2 T3 :V1 V2 V3 o V1 V2 V3 by T 1:V1o V1 defined by the related matrix

1

1 2A

0 2ª º« »¬ ¼

.

T2:V2o V2 defined by the related matrix

2

2 1 1 3

0 1 2 1A

0 0 3 5

0 0 0 4

ª º« »« »« »« »¬ ¼

and T 3:V3o V3 defined by the related matrix

3

5 6 6

A 1 4 2

3 6 4

ª º« » « »« » ¬ ¼

.

The 3-matrix associated with T is given by



69

Proof: Now to prove the converse statement we proceed asfollows; from the basic definition and properties the condition 1to 4 are true, which can be easily verified.

Suppose we have E = E 1 E2 … En where each E i is a

i

i i i1 2 k E , E ,..., E k i number of linear operators of V i, V i a

component of the n-vector space V = V 1 V2 … Vn andwhat we prove for i, is true for i = 1, 2, … , n.

Given E satisfies all the three conditions given in (1) (2) and(3) and if we let i

jW to be range of i jE then certainly V = W 1

… Wn wherei

i i i1 k W W ... W by condition (3) we have

for D= D1 D2 … Dn,

D= (1 1

1 1 1 1 1 11 1 2 2 k k E E ... ED D D)

(2 2

2 2 2 2 2 21 1 2 2 k k E E ... ED D D) …

(n n

n n n n n n1 1 2 2 k k E E ... ED D D)

for each D V j where each I i =i

i i1 k E ... E , i = 1, 2, … , n and

p k i i j jE .E 0 if p z k; 1 d j d k i and i

i i i1 k ...D D Dtrue for i =1, 2, …, n. This is true for each Di V i and hence for each D V and i i

j jE D in W i. This expression for each Di is unique and

hence each D is unique, because if D = (1

1 11 k ...D D)

(2

2 2 21 2 k ...D D D) … (

n

n n n1 2 k ...D D D) is unique with

each Di W i , i.e., i jD i

jW . Suppose i jD = i i

j jE E then from (1)

and (2) we haveik

i i i i j j jk

k 1

E ED D¦

=ik

i i i j k jk

k 1

E E E¦ = i 2 i j j(E ) E

= i i j jE E = i

jD .



70

This is true for every i, i = 1, 2, …, n and every j, j = 1, 2, …, k i.This proves each V i is direct sum of W i, hence V is a direct sum

1

1 11 k W ,...,W , … ,

n

n n n1 2 k W , W ,..., W . Hence the result.

Now we give a sketch of proof of the following theorem.However reader is expected to prove the theorem.

T HEOREM 2.29: Let T = T 1 T 2 … T n be a n-linear operator on the n-space V = V 1 V 2 … V n and let W

1 , …,W n and E 1 , E 2 , …, E n be as in the above theorem. Then anecessary and sufficient condition that each n-subspace W

i tobe n-invariant under T (i.e., each i

jW invariant under T

i ) is that

T commutes with each of the projections E i i.e., TE i = E

iT for i= 1, 2, …, m (i.e., each T i commutes with i

j E i.e., T ii j E = i

j E T i ,

i = 1, 2, …, n and j = 1, 2, …, k i ).

Proof: Suppose T commutes with each i jE i.e., T i commutes

with i jE for j = 1, 2, …, k i. This is true for each T i also. Let D=

D1 D

2 … D

nwith

i i j jWD , then

i i i j j jE D D and for

i i ii j i j jT T (E )D D = i i

j i jE T D (since T i commutes with i jE for j = 1,

2, …, k i and i = 1, 2, …, n) .This shows that T ii jD is in the range

of i jE i.e., i

jW is invariant under T i.

Assume now that each i jW is invariant under T i, 1 < j < k i; i = 1,

2, …, n; we shall show that i ii j j iT E E T for every i, 1 d i d n and

j = 1, 2, …, k i. Leti

iVD

Di =i

i i i i1 k E ... ED D

i

i i i i i1 k T TE ... TED D D.

Since i i jE D is in i

jW which is invariant under T i we must have

T i( i i jE D ) = i i

j jE E for some i jE .

Then i i i j i k E T E D = i i i

j k k E E E



72

5. The range of each i j E is the characteristic space for T i

associated with i jC .

Conversely if there exists (k 1 , k 2 , …, k n ) set of k i distinct n- scalars 1 2, ,...,

i

i i ik C C C , i = 1, 2, …, n and k i distinct linear

operators 1 2, ,...,i

i i ik E E E ; i = 1, 2, …, n which satisfy conditions

(1), (2) and (3) then T i is diagonalizable; hence T = T 1 T 2 … T n is n-diagonalizable. 1 2, ,...,

i

i i ik C C C are distinct

characteristic values of T i for i = 1, 2, …, n and conditions (4)and (5) are satisfied.

Proof: Suppose that T is n-diagonalizable i.e., each T i of T isdiagonalizable with distinct characteristic values (

1

1 1 11 2 k C , C ,..., C )

(2

2 2 21 2 k C ,C ,...,C ) … (

n

n n n1 2 k C ,C ,...,C ), i.e., each set of

(i

i i i1 2 k C , C ,..., C ) are distinct. Let i

jW be the space of

characteristic vectors associated with the characteristic valuesi

jC . As we have seen.

V = (1

1 11 k W ... W ) (

2

2 21 k W ... W ) …

(n

n n1 k W ... W )

where each V i =i

i i1 k W ... W for i = 1, 2, …, n.

Leti

i i i1 2 k E ,E ,...,E be the projections associated with this

decomposition given in theorem. Then (2), (3), (4) and (5) aresatisfied. To verify (1) we proceed as follows for each D= D1 D2 … Dn in V; Di V i; Di =

i

i i1 k E ... ED Dand so

T iDi =i

i i i i1 k TE ... TED D

=i i

i i i i i i1 1 k k C E ... C ED D.

In other words T i =i i

i i i i1 1 k k C E ... C E . Now suppose that we are

given a n-linear operator T = T 1 T2 … Tn along withdistinct n scalars C 1 C2 … Cn = C with scalar i

jC and

non zero operator i jE satisfying (1), (2) and (3) . This is true for



73

each i = 1, 2, …, n and j = 1, 2, …, k i. Since i i j k E .E 0 when j z

k, we multiply both sides of I = I 1 I2 … In

= ( 1

1 1 11 2 k E E ... E ) ( 2

2 2 21 2 k E E ... E ) …

(n

n n n1 2 k E E ... E )

by 1 2 nt t tE E ... E and obtain immediately

1 2 nt t tE , E ,..., E = ( 1

tE )2 ( 2tE )2 … ( n

tE )2.MultiplyingT = (

1 1

1 1 1 11 1 k k C E ... C E ) … (

n n

n n n n1 1 k k C E ... C E )

by 1 2 nt t tE E ... E we have

1 2 n1 t 2 t n tT E T E ... T E = 1 1 2 2 n n

t t t t t tC E C E ... C E which shows that any n-vector in the n range of

1 2 nt t tE E ... E is in the n-null space of (T – CI) =

1 n1 t 1 n t n(T C I ) ... (T C I ) where I = I 1 I2 … In. Since

we have assumed 1 2 nt t tE E ... E z 0 0 … 0, this

proves that there is a nonzero n-vector in the n-null space of (T –CI) = 1 n

1 t 1 n t n(T C I ) ... (T C I )

i.e., that itC is a characteristic value of T i for each i, i = 1, 2, …,

n; for if C i is any scalar then (T i – C iIi) = ( i i1C C ) i

1E + … +

(i

i ik C C )

i

ik E true for i = 1, 2, … , n so if (T i – C iIi)Di = 0, we

must have ( itC – C i) i i

jE D = 0. If Di must be the zero vector theni i jE 0D z for some j so that for this j we have i i

jC C 0 .

Certainly T i is diagonalizable since we have shown thatevery non zero vector in the range of i

jE is a characteristic value

of T i and the fact that I i =i

i i1 k E ... E , shows that these

characteristic vectors span V i. This is true for each i, i = 1, 2, …,n. All that is to be shown is that the n-null space of (T – CI) =( 1

1 k 1T C I ) ( 22 k 2T C I ) … ( n

n k nT C I ) is exactly the n



74

range of 1 nk k E ... E , but this is clear because T D = C D i.e.,

i i ii jT CD D, for each i, i = 1, 2, …, n. Thus

ii k k i i i i j k j

i 1 j 1 (C C )E 0 D ¦* ; i.e.,1 2 nk k k

1 1 1 1 2 2 2 2 n n n n j k j j k j j k j

j 1 j 1 j 1

(C C )E (C C )E ... (C C )E

D D D¦ ¦ ¦

= 0 0 … 0.

Hence ( i i i i j k j(C C )E D = 0 for each j; and each i = 1, 2, …, n

and i i

jE D = 0, k z j; for each i, i = 1, 2, …, n. Since Di =

i

i i i i1 k E ... ED Dfor each i and i i

jE D = 0 for j z k we have Di =i i jE D which proves that Di is the range of i

jE . This is true for

each i hence the claim.

We give the statement of the primary decomposition theoremfor n-vector space V.

T HEOREM 2.31: Let T = T 1 T 2 … T n be a n-linear operator on the finite dimensional n-vector space V = V 1 V 2

… V n over the field F. Let p = p 1 p2 … pn where p i

= 1 21 1 ...

ii ik i

i

r r r k p p p , i = 1, 2, …, n. i.e.,

p =1 21 1 2 2

1 2 1 1 2 2 1 2

1 211 12 1 21 22 2 11 12... ... ... ... nn n

k k k n

n

r r r r r r r r r k k nk p p p p p p p p p

where ik p are distinct irreducible monic polynomials over F and

the i jr are positive integers. Let i

jW be the null space of ( )ir k T

ik p ,

k = 1, 2, …, k i; i = 1, 2, …, n then

1. V = ( 1

1 11 ... k W W ) (

2

2 21 ... k W W ) …

( 1 ... n

n nk W W )

2. each W i is invariant under T ik , i = 1, 2, …, n, 1 d r d k i.



75

3. if T ij is the operator induced on i jW by T i then the

minimal polynomial for T ij isir

ij p , true for j = 1, 2, …, k i

and i = 1, 2, …, n.

Several interesting results can be found in this directionanalogously.

Now we define the notion of n-diagonalizable part and n-nilpotent part of a n-linear operator T.

Given V = V 1 V2 … Vn is a n-vector space over thefield F. T = T 1 T2 … Tn a n-linear operator on V.Suppose the n-minimal polynomial of T is the product of firstdegree polynomials, i.e., the case in which each i

j p is of the

form x – i jC . Now range of i

jE , for each T i in T is the null space

i jW of (

i jr i i

i j i(T C I ) . This is true for each i, i = 1, 2, …, n. Put

D = D 1 D2 … Dn = (

1 1

1 1 1 1 1 11 1 2 2 k k C E C E ... C E )

( 2 22 2 2 2 2 21 1 2 2 k k C E C E ... C E ) …

(n n

n n n n n n1 1 2 2 k k C E C E ... C E ) .

Clearly D is n-diagonalizable operator which we define or call as the n-diagonalizable part of T. Let as consider N = T – D.

NowT = [

1

1 11 1 1 k T E ... T E ] [

2

2 22 1 2 k T E ... T E ] …

[ nn nn 1 n k T E ... T E ]

D = (1 1

1 1 1 1 1 11 1 2 2 k k C E C E ... C E )

(2 2

2 2 2 2 2 21 1 2 2 k k C E C E ... C E ) …

(n n

n n n n n n1 1 2 2 k k C E C E ... C E )

so N = [

1 1 1

1 1 1 1 1 1 1 1 1

1 1 1 1 1 2 2 2 1 k k k (T C I )E (T C I )E ... (T C I )E ]



76

[2 2 2

2 2 2 2 2 2 2 2 22 1 1 1 2 2 2 2 2 k k k (T C I )E (T C I )E ... (T C I )E ]

… [n n

n n n n nn 1 1 1 n k k (T C I )E ... (T C I ) ] .

Clearly

N2 = [1 1 1

1 1 2 1 1 1 2 11 1 1 1 1 k k k (T C I ) E ... (T C I ) E ]

[2 2 2

2 2 2 2 2 2 2 22 1 1 1 2 k k k (T C I ) E ... (T C I ) E ] …

[n n n

n n 2 n n n 2 nn 1 1 1 n k k k (T C I ) E ... (T C I ) E ]

and in general we have

N r = [1 1 1

1 1 r 1 1 1 r 11 1 1 1 1 k k k (T C I ) E ... (T C I ) E ] …

[1 n n

n n r n n n r nn 1 1 1 n k k k (T C I ) E ... (T C I ) E ]

where r t (r 1, r 2, …, r n) i.e., r > r i, i = 1, 2, …, n ( by misuse of notation) we have N r = 0 because the n-operator (T – CI) r will

be (0 0 … 0) i.e., each (T i – i j iC I ) i

jr = 0 where r > i jr for

j = 1, 2, …, k i and i = 1, 2, …, n.

Now we define a nilpotent n-linear operator T.

D EFINITION 2.31: Let N be a n-linear operator on V = V 1 V 2 … V n we say N is n-nilpotent if there exists some positive

integer r, r >r i; i = 1, 2, … , n such that N r = 0.

Note: If N = N 1 N2 … Nn then N i: V i o V i is of

dimension n i, n i z n j if i z j true for i = 1, 2, …, n so we mayhave

ir i N = 0, i = 1, 2, … , n. We may not have r i = r j, if i z j;

hence the claim.

Now we give only a sketch of the proof however the reader isexpected to get the complete the proof using this sketch.

T HEOREM 2.32: Let T = T 1 T 2 … T n be a n-linear operator on a finite dimensional n-vector space V = V 1 V 2



77

… V n over the field F. Suppose the n-minimal polynomial for T decomposes over F in to product of n-linear polynomials, thenthere is a n-diagonalizable n-operator D on V and a n-nilpotent operator N on V such that

I. T = D + N II. DN = ND.

The n-diagonalizable operator D and the n-nilpotent operator N are uniquely determined by (I) and (II) and each of them is a n-

polynomial in T.

Proof: We give only a sketch of the proof. However theinterested reader can find a complete proof using this sketch.

Given V = V 1 V2 … Vn finite (n 1, n2, …, n k )dimensional a n-vector space. T = T 1 T2 … Tn a n-linear operator on T such that T i: V i o V i for each i = 1, 2, …, n. Wecan write each T i = D i + N i, a nilpotent part N i and adiagonalizable part D i; i = 1, 2, …, n.

ThusT = T 1 T2 … Tn

= (N 1 + D 1) (N2 + D 2) … (N n + D n)

= (N 1 N2 … Nn) + (D 1 D2 … Dn)

i.e., T = N + D where N = N 1 N2 … Nn and D = D 1 D2 … Dn. Since each D i and N i not only commute but are

polynomials in T i we see D and N commute and are n- polynomials of T, as the result is true for each i, i = 1, 2, … , n.Suppose we have T = D 1 + N 1, i.e.,

T = T 1 T2 … Tn = ( 1 1

1 1D N ) ( 1 12 2D N ) … ( 1 1

n nD N )= D 1 + N 1

where D 1 is the n-diagonalizable part of T i.e., each 1iD is the

diagonalizable part of T i for i = 1, 2, …, n and N 1 the n-nilpotent part of T i.e., each N i is the nilpotent part of T i for i =1, 2, …, n.



78

Since each 1iD and 1

i N commute for i = 1, 2, …, n we have D 1 and N 1 also n-commute with any n-polynomial in T. Hence in

particular they commute with D and N.

Now we have D + N = D1

+ N1

i.e., D – D1

= N1

– N andthese four n-operator commute with each other. Since D and D 1 are n-diagonalizable they commute and so D – D 1 is also n-diagonalizable.

Since both N and N 1 are n-nilpotent they n-commute andthe operator N – N 1 is also n-nilpotent. Since N – N 1 = D – D 1 and N – N 1 is n-nilpotent we have D – D 1 the n-diagonalizablen-operator is also n-nilpotent.

Such an n-operator can only be a zero operator, for since itis n-nilpotent, the n-minimal polynomial for this n-operator is of the form 1 2 nr r r x x ... x with ir x 0 for appropriate m i t r i,i = 1, 2, …, n. But since the n-operator is n-diagonalizable then-minimal polynomial cannot have repeated n-roots hence eachr i = 1 and the n-minimal polynomial is simple x x … xwhich confirms the operator is zero. Thus we have D = D 1 and

N = N 1.

The interested reader is expected to derive analogous resultswhen F is the field of complex numbers. Now we proceed on to work with n-characteristic values n-

characteristic vectors of a special n-linear n-operator on V.Given V is a n-vector space say of finite dimension, V = V 1

V2 … Vn of dimension (n 1, n1, …, n n) defined over thefield F. Let T = T 1 T2 … Tn be a special n-linear operator on V; i.e., T i: V i o V i for each i, i = 1, 2, …, n.

We say C = (C 1 C2 … Cn) is a n-characteristic valueof T if some n-vector D= D1 D2 … Dn we have T D= C D,i.e., T D= (T 1 T2 … Tn) (D1 D2 … Dn) = (C 1 C2

… Cn) (D1 D2 … Dn) i.e., T o T = T 1D1 T2D2 …TnDn = C 1D1 C2D2 … CnDn, i.e., each T iDi = C iDi for i

= 1, 2, …, n.Here D = D1 D2 … Dn is defined to be the n-

characteristic vector of T. The collection of all Dsuch that T D=CDis called the n-characteristic space associated with C.



79

We shall illustrate the working of the n characteristicvalues, n-characteristic vectors associated with aT.

Example2.20: Let V = V 1 V2 V3 be 3 vector space over Qwhere V 1 = Q u Q u Q, V 2 = Q u Q and V 3 = Q u Q u Q u Q arevector spaces over Q of dimensions 3, 2 and 4 respectively i.e.,V is of (3, 2, 4) dimension. Define T: V o V where the 3-matrix associated with T is given by

A = A 1 A2 A3

=

1 0 2 13 0 2

1 2 0 2 5 00 1 5

0 3 0 0 3 70 0 70 0 0 4

ª ºª º « »ª º« » « » « »« »

« »¬ ¼« » « »¬ ¼ ¬ ¼

.

Now we will determine the 3-characterstic values associatedwith T. The n-characteristic polynomial

p =

x 3 0 2x 1 2

0 x 1 50 x 30 0 x 7

ª º ª º« » « »« » ¬ ¼« »¬ ¼

x 1 0 2 1

0 x 2 5 0

0 0 x 3 0

0 0 0 x 4

ª º« » « »« »« »¬ ¼

= (x – 3) (x – 1) (x –7) (x – 1) (x – 3) (x – 1) (x – 2)(x – 3) (x – 4).

Thus the 3- characteristic values of A = A 1 A2 A3 are {3, 1,7} {1, 3} {1, 2, 3, 4}. One can find the 3-characteristicvalues as in case of usual vector spaces and their set theoreticunion will give 3-row mixed vector, which will be 48 in number as we have 48 choices for the 3-characterstic values as {3} {1} {1}, {3} {1} {2}, {3} {1} {3}, {3} {1} {4} so on and {7} {3} {4}.



80

Now having seen the working of 3-characteristic values we just recall in case of matrices A we say A is orthogonal if AA t =I. Further A is anti orthogonal if AA t = – I.

Now we for the first time define the notion of n-orthogonalmatrices and n-anti orthogonal matrices.

D EFINITION 2.32: Let A = (A 1 A2 … An ) be a n-matrix. At = (A 1 … An ) t = 1 2 ... t t t

n A A A .

AAt = 1 1 2 2 ... t t t n n A A A A A A .

We say A is n-orthogonal if and only if AA t = I 1 I 2 …

I n where I j is the identity matrix, i.e., if A = A 1 A2 … An ismi u ni matrix i = 1, 2, …, n; then AA t = I 1 I 2 … I n is suchthat I j is a m j u m j identity matrix, j = 1, 2, …, n. We say A isanti orthogonal if and only if AA t = (– I 1 ) (–I 2 ) … (– I n )where I j is m j u m j identity matrix i.e., if

I =

ª º« »« »« »« »¬ ¼

1 0 0 0

0 1 0 0

0 0 1 00 0 0 1

then

–I =

ª º« »« »« »« »¬ ¼

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

.

Now we say AA t is n-semi orthogonal if AA t = B 1 B2 … Bn ; some of the B i’s are identity matrices and some are not identity matrices on similar lines we define n-semi antiorthogonal if in AA t = C 1 C 2 … C n some C i’s are –I i and

some are not – I j.

It is not a very difficult task for the reader can easily get

examples of these 4 types of n-matrices.



81

Chapter Three

APPLICATIONS OFn- LINEAR ALGEBRA OF TYPE I

In this chapter we just introduce the applications of the n-linear algebras of type I. We just recall the notion of Markov bichainsand indicate the applications of vector bispaces and linear

bialgebras in Markov bioprocess. For this we have to firstdefine the notion of Markov biprocess and its implications tolinear bialgebra / bivector spaces. We may call it as Markov

biprocess or Markov bichains.Suppose a physical or mathematical system is such that at

any moment it occupies two of the finite number of states(Incase of one of the finite number of states we apply Markov

chains or the Markov process). For example say about aindividuals emotional states like happy, sad etc., suppose asystem move with time from two states or a pair of states toanother pair of states; let us construct a schedule of observationtimes and a record of states of the system at these times. If wefind the transition from one pair of state to another pair of stateis not predetermined but rather can only be specified in terms of certain probabilities depending on the previous history of thesystem then the biprocess is called a stochastic biprocess. If inaddition these transition probabilities depend only on the



82

immediate history of the system; that is if the state of the systemat any observation is dependent only on its state at theimmediately proceeding observations then the process is calledMarkov biprocess or Markov bichain.

The bitransition probability p ij =1 1 2 2

1 2i j i j p p (i, j = 1, 2,…,

k) is the probabilities that if the system is in state j = (j 1, j2) atany observation, it will be in state i = (i 1, i2) at the nextobservation. A transition matrix

P = [p ij] = 1 1 2 2

1 2i j i j p pª º ª º¬ ¼ ¬ ¼

is any square bimatrix with non negative entries for which the bicolumn sum is 1 1. A probability bivector is a column bivector with non negative entries whose sum is 1 1.

The probability bivectors are said to be the state bivectors of the Markov biprocess. If P = P 1 P2 is the transition bimatrixof the Markov biprocess and x n = n n

1 2x x is the state bivector at

the n th observation then x (n+1) = P x (n) and thus (n 1) (n 1)1 2x x =

(n) (n)1 1 2 2P x P x . Thus Markov bichains find all its applications

in bivector spaces and linear bialgebras.

Now we proceed onto define the new notion of Markov n-chainn t 2. Suppose a physical or a mathematical system is such thatat any time it can occupy a finite number of states; when weview them as stochastic biprocess or Markov bichains when wemake an assumption that the system moves with time from onestate to another so that a schedule of observation times keeps thestates of the system at these times. But when we tackle realworld problems, say even for simplicity; emotions of a personmay be very unpredictable depending largely on the situationand the mood of the person and its relation with another so suchstudy cannot come under Markov chains. Even more is thecomplicated situation when the mood of a boss withsubordinates; where mood of a person with a n number of

persons and with varying emotions at a time and in such cases

more than one emotion is experienced by a person and suchstates cannot be included and given as a next set of observation.



83

These changes and several feelings say at least n at a time (n >2) will largely affect the transition n-matrix

P = P1 … P

n

1 1 n n

1 n

i j i j p pª º ª º ¬ ¼ ¬ ¼

!

with non negative entries which we will explain shortly. Weindicate how n-vector spaces and n-linear algebras are used inMarkov n-process (n t 2), when n = 2 the study is termed asMarkov bioprocess. We first define Markov n-process and itsimplications to linear n-algebra and n-vector spaces; which wemay call as Markov n-process and Markov n-chains.

Suppose a physical or a mathematical system is such that atany moment it occupies two or more finite number of states (incase of one of the finite number of states we apply Markovchains or the Markov process; in case of two of the finitenumber of state we apply Markov bichains or Markov

biprocess). For example individual emotional states; happy, sad,cold, angry etc. suppose a system move with time from n statesor a n tuple of states to another n-tuple of states; let us constructa schedule of observation times and a record of states of the

system at these times. If we find the transition from n-tuple of states to another n-tuple of states not predetermined but rather can only be specified in terms of certain probabilities dependingon the previous history of the system then the n-process iscalled a stochastic n-process. If in addition these transition

probabilities depend only on the immediate history of thesystem that is if the state of the system at any observation isdependent only on its state at immediately proceedingobservations then the process is called Markov n-process or Markov n-chain.

The n-transition probability

1 1 2 2 n n

1 2 nij i j i j i j p p p p !

i, j = 1, 2, …, K is the probabilities that if the system is in state j= (j 1, j2, …, j n) at any observation it will be in state i = (i 1, i2, …,in) at the next observation.

A transition matrix associated with it is

1 1 n n

1 nij i j i jP [p ] [p ] ... [p ]



84

is a square n-matrix with non negative entries for all of which n-column sum is (1 … 1). A probability n-vector is a columnn-vector with non negative entries whose sum is 1 … 1.

The probability n-vectors are said to be the state n-vectorsof the Markov n-process. If P = P 1 … Pn is the transition n-matrix of the Markov n-process and m m m

1 nx x x ! is thestate n-vector at the m th observation then x (m+1) = Px (m) and thus

m 1 m 1 (m) (m)1 n 1 1 n nx x P (x ) P x ! ! . Thus Markov n-

chains find all its applications in n-vector spaces and linear n-algebras. (n-linear algebras).

Example 3.1 : (Random Walk): A random walk by n persons onthe real lines i.e. lines parallel to x axis is a Markov n-chainsuch that

1 1 n n

1 n j k j k p p ! = 0 … 0 if k t = j t – 1 or j t + 1, t

= 1, 2, …, n. Transition is possible only to neighbouring statesfrom j to j – 1 and j + 1. Here state n-space is S = S 1 … Sn where S i = { … –3 –2 –1 0 1 2 3 …}; i = 1, 2, …, n.

The following theorem is direct.

T HEOREM 3.1: The Markov n-chain1 1{ ; 0}m X m t …

{ ; 0}nm n X m t is completely determined by the transition n-

matrix P = P 1 … P n and the initial n-distribution

1

1{ } { }n

n K K P P ! defined as

11 0 1 0[ ] [ ]n

n n P X K P X K !

1 n K K p p t! 0 … 0

and

1

1 1 ¦ ¦!

n

n n

k k K S K S

p p = 1 … 1.

The proof is similar to Markov chain.

The n vector 1 n

1 1 n n1 n 1 nu (u u ) (u u ) ! ! ! is called a

probability n-vector if the components are non negative andtheir sum is one.



86

for all m t (1, …, 1) where

t t t

t ti j j p p for t = 1, 2, …, n i.e. all

the rows of each P t is the same then we say P is an n-

independent trial.We can also define the notion of Bernoulli n trials. We justdepict n-random walk with absorbing barriers. Let the possiblen-states be

1 n

1 1 1 n n n0 K 0 1 K (E , E , ,E ) (E ,E , ,E ) ! ! ! .

Consider the n-matrix of transition n-possibilities

1 1 n n

1 ni j i jP P P !

=

1 1 n n

1 1 n n

1 1 n n

1 n

K K K K

1 0 0 0 1 0 0 0

q 0 p 0 q 0 p 0

0 q 0 p 0 0 q 0 p

0 q 0 0 q

0 0 1 0 0 1u u

ª º ª º« » « »« » « »« » « » « » « »« » « »« » « »¬ ¼ ¬ ¼

! !

! !

! #

! !

! !

From each of the interior n states

1 1 n 1

1 1 n n1 K 1 K {E , ,E } {E , , E },

! ! !

n-transmission are possible to the right and left neighbour with

t tt i ,i 1 t(p ) p , t tt i ,i 1 t(p ) q ; t = 1, 2, …, n.

However no n-transition is possible from either 1 n

0 0 0E (E E ) ! and 1 nK K K E [E E ] ! to any other n-

state.This n-system may move from one n-state to another but

once E 0 or E K is reached the n-system stays there permanently.

Now we describe random walk with reflecting barriers.

Let1 1 n n

1 ni j i jP P P ! be a n-matrix with



87

1 1

1 1

11

1 1

1 1

q p 0 0 0 0

q 0 p 0 0 0P 0 q 0 p 0 0 0

0 0 0 q 0 p

0 0 0 0 q p

ª º« »« »« »

« »« »« »¬ ¼

!

!

!

!

…

n n

n n

n n

n n

n n

q p 0 0 0 0

q 0 p 0 0 0

0 q 0 p 0 0 0

0 0 0 q 0 p0 0 0 0 q p

ª º« »« »« »« »« »« »¬ ¼

!

!

!

!

p t and q t for t = 1, 2, …, n is defined by

t t

tt t

t t t t tij n t n 1 t t

p if j i 1

P P (X j |X i ) q if j 0

0 otherwise

- ° ®°¯

true for t = 1, 2, …, n.It may be possible that

( 2)

t t t t

t ti j i j p 0, p 0 but

(3 )

t t

ti j p 0! . We

say the state j t is accessible from state i t if ( n )

t t

ti jP 0! for some n >

0.In notation i t o jt i.e. i t leads to j t. If i t o jt and j t o it then i t

and j t communicate and we denote it by i t l jt, if this happenswe say they n-communicate. If only some of them communicate

and others do not communicate we say the n-system semicommunicates.

Here j

( n )t

t t

t t tt t t

jt ti j t t t

q p for j 0,1,2, , i n 1

P j p for j j n

0 otherwise.

- ° ®°¯

!

The state i t is essential if i t o jt

implies i t m jt i.e. if any state j t is accessible from i t then i t is accessible from that state, true for t



88

= 1, 2, …, n. Let = 1 … n denote the set of allessential n state i.e. each t denotes the set of all essential states,t = 1, 2, …, n. States that not n-essential are called n-inessential.We have semi essential if a few of the t’s are essential. Wehave semi essential state as m-essential state where m < n andonly m out of the n states are essential rest inessential or n – minessential state.

A Markov n-chain is called n-irreducible (or n-ergodic) if there is only one n communicating class i.e. all states n-communicate with each other or every n-state can be reachedfrom every other n-state.

A n-subset c = c 1 … cn of S = S 1 … Sn is said to

be closed (or n-transient) if it is impossible to leave c in one stepi.e. p ij = 0 … 0, i.e.

1 1 n n

1 ni j i j p p ! = 0 … 0 for all i

c i.e. (i 1, …, i n) c1 … cn and all (j 1, …, j n) c for all i t

ct and all j t ct; t = 1, 2, …, n.We say a n-subset c = c 1 … cn of S = S 1 … Sn is

semi n-closed (or semi n-transient) if it is impossible to leave(only m of the) c t’s, 1 d t d n, m < n in one state; i.e.

t t

ti j p 0

for all i t ct, and for all j t ct. We call this also m-closed (m <n) or m-transient, m = 1, 2, …, n –1. If m = n – 1 we call c to behyper n-closed (or hyper n-transient).

A Markov n-chain is n-irreducible if the only n-closed set inS is S itself i.e., there is no n-closed set other than the set all of n states.

We say a Markov n-chain is semi irreducible or m-irreducible (m < n) if the closed sets in S = S 1 … Sn are min number from the n-states {S 1, …, S n}, m < n. If m = n – 1then we say the Markov n-chain is hyper n irreducible.

A single n-state {K 1, …, K n} forming a closed n-set iscalled n-absorbing (n-trapping) i.e., a n-state such that the n-system remains in that state once it enters there. Thus a n-state{K 1, …, K n} is n absorbing if the th th

1 n{K , , K }! rows of thetransition n-matrix P = P 1 … Pn has 1 on the main n-diagonal and 0 else where.



89

Example 3.3 : Let P = P 1 … P4 be a transition 4 matrixgiven by

P =

1 10 0 0 0 02 21 0 0 0 0 0 0

1 20 0 0 0 03 30 0 0 1 0 0 0

51 10 0 0 07 7 75 30 0 0 0 08 8

3 1 10 0 0 05 5 5

ª º« »« »« »« »« »« »« »« »« »« »« »« »¬ ¼

1 10 0 02 21 1 10 0 3 3 3

0 0 1 0 0

1 4 0 0 05 57 20 0 09 9

ª º« »« »« »« »

« »« »« »« »¬ ¼

1 0 0 0

810 09 971

0 08 81 1 104 2 4

ª º« »« »« »« »« »« »¬ ¼

0 0 0 1 0 0

1 10 0 0 02 231 0 0 0 04 4

7 10 0 0 0 8 80 1 0 0 0 0

0 0 0 0 0 1

ª º« »« »

« »« »« »« »« »« »« »¬ ¼

.

Clearly the n-absorbing state is (4, 3, 1, 6).



90

Several interesting results true in case of M C can be proved for M – n – C with appropriate changes and suitable modifications.

Now we briefly describe the method for spectral m-decomposition (m t 2). Let P = P 1 … Pm be a N u N m-matrix with m set of latent roots

1 2 m

1 1 2 2 m m1 N 1 N 1 N, , ,O O O O O O! ! ! ! all distinct and simple i.e.

each set of latent rootst

t t1 N{ }O O! are all distinct and simple for

t = 1, 2, …, m; then

1 1 m m

1 1 m m1 i 1 i m i m i(P I ) U (P I ) UO O! = 0 … 0

for the n-column latent n-vector 1 m

1 m

i iU U ! and

1 1 m m

1 1 m mi 1 i i m iV (P I) V (P I)c cO O! = 0 … 0

for the row latent n-vector 1 m

1 mi iV V ! .

1 m 1 1 m m

1 m 1 1 m mi i i i i iA A U V U Vc c ! !

are called m latent or m-spectral m-matrix associated with

1 m

1 mi i( , , );O O! it = 1, 2, …, N t, t = 1, 2, …, m.

The following properties of 1 m

1 m

i iA A ! are well known

(i)1 m

1 mi iA A ! ’s are m-idempotent i.e.

(1 m

1 mi iA A ! )2 =

1 m

1 mi iA A !

i.e. each t

2tiA =

t

tiA , t = 1, 2, …, m.

(ii) They are n-orthogonal i.e.

1 t

1 ti j t tA .A 0, i j z ; t = 1, 2, …, m.

(iii) They give a spectral n-decomposition

P1 … Pn =1 m

1 1 m m

1 m

N N1 1 m mi i i i

i 1 i 1

A A O O¦ ¦! .

It follows from (i) to (iii), that

1 mK K K 1 mP P P !



91

1 m1 m

1 1 m m

1 m

K K N N

1 1 m mi i i i

i 1 i 1

A A

§ · § ·O O¨ ¸ ¨ ¸¨ ¸ ¨ ¸© ¹ © ¹

¦ ¦!

1 m

1 m1 1 m m

1 m

N NK K 1 mi i i i

i 1 i 1A A

O O¦ ¦!

1 m

1 m

1 1 1 m m m

1 m

N NK K 1 1 m mi i i i i i

i 1 i 1

U V U Vc c

O O¦ ¦! .

Also we know that1 mK K K K 1 1 1 1 m m 1

1 mP UD U U D (U ) U D (U ) !

where 1 m

1 1 m m

1 N 1 NU {U , ,U } {U , , U } ! ! !

andD = D 1 D2 … Dm

=

1 m

1 m1 1

1 m2 2

1 m N N

0 0 0 0

0 0 0 0

0 0 0 0

ª º ª ºO O« » « »O O« » « » « » « »« » « »O O« » « »¬ ¼ ¬ ¼

! !

! !!

# # # # # #

" "

.

Since the n-latent n-vectors are determined uniquely only upto amultiplicative constant, we have chosen them such that

1 1 m m

1 mi i i iU V U Vc c ! = (1 … 1).

One can work for any m-power of P to knowt

tiO ’s and

t

tiA ’s; t

= 1, 2, …, m. Now even if we say 1 mK K K 1 mP P P ! we

work for K = (K 1, …, K m) and when the working with any P t isover that t th component remains as it is and calculations are

performed for the rest of the components of P. With the adventof the appropriate programming using computers simultaneousworking is easy; also one needs to know in the presenttechnologically advanced age one cannot think of computingone by one and also things do not occur like that in manysituations. So under these circumstances only the adaptation of n-matrices plays a vital role by saving both time and economy.Also stage by stage comparison of the simultaneous occurrenceof n-events is possible.



92

Matrix theory has been very successful in describing theinterrelations between prices, outputs and demands in aneconomic model. Here we just discuss some simple models

based on the ideals of the Nobel-laureate Wassily Leontief. Twotypes of models discussed are the closed or input-output modeland the open or production model each of which assumes someeconomic parameter which describe the inter relations betweenthe industries in the economy under considerations. Usingmatrix theory we evaluate certain parameters.

The basic equations of the input-output model are thefollowing:

11 12 1n

21 22 2n

n1 n 2 nn

a a aa a a

a a a

ª º« »« »« »« »¬ ¼

""

# # #

"

1

2

n

p p

p

ª º« »« »« »« »¬ ¼

#=

1

2

n

p p

p

ª º« »« »« »« »¬ ¼

#

each column sum of the coefficient matrix is one

i. p i t 0, i = 1, 2, …, n.ii. aij t 0, i , j = 1, 2, …, n.iii. a ij + a 2j +…+ a nj = 1

for j = 1, 2 , …, n.

p =

1

2

n

p

p

p

ª º« »« »« »

« »¬ ¼

#

are the price vector. A = (a ij) is called the input-output matrix

Ap = p that is, (I – A) p = 0.

Thus A is an exchange matrix, then Ap = p always has anontrivial solution p whose entries are nonnegative. Let A be an

exchange matrix such that for some positive integer m, all of the



94

and the consumption matrix,

C =

11 12 1k

21 22 2k

k1 k 2 kk

V V Vª º« »V V V« »« »

« »V V V¬ ¼

"

"# # #

"

.

By their nature we have

x t 0, d t 0 and C t 0.

From the definition of Vij and x j it can be seen that the quantityVi1 x1 + Vi2 x2 +…+ Vik xk

is the value of the output of the i th industry needed by all k industries to produce a total output specified by the productionvector x.Since this quantity is simply the i th entry of the column vector Cx, we can further say that the i th entry of the column vector x –

Cx is the value of the excess output of the ith

industry availableto satisfy the outside demand. The value of the outside demandfor the output of the i th industry is the i th entry of the demandvector d; consequently; we are led to the following equation:

x – Cx = d or (I – C) x = d

for the demand to be exactly met without any surpluses or shortages. Thus, given C and d, our objective is to find a

production vector x t 0 which satisfies the equation (I – C)x =d.

A consumption matrix C is said to be productive if (1 – C) –1 exists and (1 – C) –1 t 0.

A consumption matrix C is productive if and only if there issome production vector x t 0 such that x ! Cx.

A consumption matrix is productive if each of its row sums

is less than one. A consumption matrix is productive if each of its column sums is less than one.



95

Now we will formulate the Smarandache analogue for this,at the outset we will justify why we need an analogue for thosetwo models.

Clearly, in the Leontief closed Input – Output model, p i = price charged by the i th industry for its total output in realityneed not be always a positive quantity for due to competition tocapture the market the price may be fixed at a loss or thedemand for that product might have fallen down so badly sothat the industry may try to charge very less than its real value

just to market it.

Similarly a ij t 0 may not be always be true. Thus in the

Smarandache Leontief closed (Input – Output) model (S-Leontief closed (Input-Output) model) we do not demand p i t 0,

p i can be negative; also in the matrix A = (a ij),

a1j + a 2j +…+a kj z 1

so that we permit a ij's to be both positive and negative, the onlyadjustment will be we may not have (I – A) p = 0, to have onlyone linearly independent solution, we may have more than oneand we will have to choose only the best solution.

As in this complicated real world problems we may nothave in practicality such nice situation. So we work only for the

best solution.On similar lines we formulate the Smarandache Leontief

open model (S-Leontief open model) by permitting that x t 0 , dt 0 and C t 0 will be allowed to take x d 0 or d d 0 and or C d 0. For in the opinion of the author we may not in reality have the

monetary total output to be always a positive quality for allindustries and similar arguments for di's and C ij's.

When we permit negative values the corresponding productionvector will be redefined as Smarandache production vector (S-

production vector) the demand vector as Smarandache demandvector (S-demand vector) and the consumption matrix as theSmarandache consumption matrix (S-consumption matrix). Sowhen we work out under these assumptions we may havedifferent sets of conditions



96

We say productive if (1 – C) –1 t 0, and non-productive or not up to satisfaction if (1 – C) –1 0.

The reader is expected to construct real models by takingdata's from several industries. Thus one can develop severalother properties in case of different models.

Matrix theory has been very successful in describing theinterrelations between prices outputs and demands.

Now when we use n-matrices in the input – output modelwe can under the same set up study the price vectors of all thegoods manufactured by that industry simultaneously. For in the

present modernized world no industry thrives only in the production one goods. For instance take the Godrej industries it

manufacturers several goods from simple locks to bureau. So if they want to study input output model to each and every goodsit has to work several times with the exchange matrix; but withthe introduction of n-mixed matrices we can use the n-matrix asthe input output n-model to study interrelations between the

prices outputs and demands of each and every goodsmanufactured by that industry. Suppose the industrymanufactures n-goods, n t 2.

Thus A = A 1 … An is an exchange n-matrix whereeach A i is a n i × n i matrix i = 1, 2, …, n. The basic n-equationsof the input – output model is the following

1

1

1

1 1 1 1

1 1 111 12 1n 1

11 2 121 22 2n

1n1 1 1

n 1 n 2 n n

a a a p

a a a

pa a a

ª ºª º« »« »« »« »« »« »« »¬ ¼« »

¬ ¼

!

!#

# # #

!

2

2

2 2 2 2

2 2 211 12 1n

2 2 221 22 2n

2 2 2n 1 n 2 n n

a a a

a a a

a a a

ª º« »« »« »« »« »¬ ¼

!

!

# # #

! 2

21

2n

p

p

ª º« »« »« »¬ ¼

# !



97

n

n

1

n n n n

n n n11 12 1n n

1n n n21 22 2n

nnn n n

n 1 n 2 n n

a a a p

a a a

pa a a

ª ºª º« »« »« »« »« »« »« »¬ ¼« »¬ ¼

!

!#

# # #

!

=

1 2 n

1 2 n1 1 1

1 1 nn n n

p p p

p p p

ª º ª º ª º« » « » « » « » « » « »« » « » « »¬ ¼ ¬ ¼ ¬ ¼

# # ! #

each n column sum of the coefficient n-matrix is (1 … 1)

(i) ti p 0;t t = 1, 2, …, n.

(ii)t t

ti ja 0;t it, j t = 1, 2, …, n t and t = 1, 2, …, n.

(iii)1 t t t

t t tij 2 j n ja a a 1 ! for j t = 1, 2, …, n t and t = 1, 2, …, n.

p = p 1 … pn

1 2 n

1 2 n1 1 1

1 2 n2 2 2

1 2 nn n n

p p p

p p p

p p p

ª º ª º ª º« » « » « »« » « » « » « » « » « »« » « » « »« » « » « »¬ ¼ ¬ ¼ ¬ ¼

!# # #

are the price n-vector of the n-goods.A = A 1 … An =

1 1 n n

1 ni j i j(a ) (a ) !

is called the input-output n-matrix.Ap = p that is (I – A) p = 0 … 0

i.e. 1 1 1 n n n(I A ) p (I A ) p ! 0 … 0.

Thus A is an exchange n-matrix then Ap = p always has anontrivial n-solution p = p 1 … pn, whose entries arenonnegative. Let A be the exchange n-matrix such that for somen-positive integers (m 1, …, m n) all the entries of

1 nm mm1 nA A A ! are positive. Then there is exactly only

one linearly n-independent solution of (I – A)p = 0 … 0



98

and that it may be chosen such that all of its entries are positivein Leontief open production n-model.

Thus the model provides at a time i.e. simultaneously the price n-vector i.e. the price vector of each of the n-goods. Whenn = 1 we see the structure corresponds to the Leontief openmodel. When n = 2 we get the Leontief economic bi models.This n-model is useful when the industry manufactures morethan one goods and it not only saves time and economy but itrenders itself stage by stage comparison of the price n-vector which is given by p = p 1 … pn.

Now we proceed onto describe the S-Leontief open n-modelusing n-matrices.

In reality we may not always have the exchange n-matrix A= A 1 … An =

1 1 n n t t

1 n ti j i j i j(a ) (a ), a 0. t! For it can also be

both positive or negative. Thus in S-Leontief closed (input -output) n-model we do not demand

t t

t ti i p 0, pt can be negative

also in the n-matrix A = A 1 … An =1 1 n n

1 ni j i j(a ) (a ) !

wheret t t

t tij K ja a 1 z! for every t = 1, 2, …, n. i.e. we permit

t t

t

i ja to be both positive and negative, the only adjustment will

be, we may not have (I – A)p = 0 … 0 to have only one n-linearly independent solution, we may have more than one andwe will have to choose only the best solution which will behelpful to the economy of the nation. The best by no meansshould favour in the interrelation high prices but a medium pricewith most satisfactory outputs and best catering to the demandsas it is an economic n-model.

So n-matrices will be highly helpful and out of one set of solution which will have n-components associated with theexchange n-matrix A = A 1 … An, we have to pick up fromthe nontrivial solution p 1 = p 1 … pn the best suited p i’s andonce again find a 1 n p p pc c c ! with the estimated p i’s from

the earlier p remain as zero and choose the best j pc for the

solution p cand so on. The final p = p 1 … pn will be filledwith the best p i’s and p j’s and so on.



99

Thus the solution would be the best suited solution of theeconomic model.

The difference between Leontief closed or input output n-model and the S-Leontief closed or input output economic n-model is that in the Leontief model there is only oneindependent solution where as in the S-Leontief closed inputoutput economic n model we can choose the best solution fromthe set of solutions so that best solution also may vary from

person to person for what is best for one may not be best for theother that too when it describes the interrelations between

prices, outputs and demands in an economic n-model. Now we briefly describe the Leontief open production n-

model. In contrast with Leontief closed n-model here the n-setof or n-tuple of industries say (K 1, …, K n) where output of K i industries are distributed only among themselves the open nmodel attempts to satisfy an outside demand for the n-outputs,true for i = 1, 2, …, n. Portions of these n-outputs may still bedistributed among the (K 1, …, K n) set of industries themselvesto keep them operating, but there is to be some excess some net

production with which to satisfy the outside demand.In the closed n-model the n-outputs of the industries were

fixed and the objective was to determine the n-prices for thesen-outputs so that the equilibrium condition that expendituresequal income was satisfied.

t

tix monetary value of the i t

th industry from the t th unit i.e.

we haveK 1 = industries in the first unit denoted by c 1 K 2 = industries in the second unit denoted by c 2 # K t = industries in the t th unit denoted by c t

and so onK n – industries in the n th unit denoted by c n.

t

tid monetary value of the output of the i t

th industry need

to satisfy the outside demand.

t t

ti jV monetary value of the output of the i t

th industry

needed by the j tth

industry to produce one unit of monetary valueof its own profit.



100

This is true for every t; t = 1, 2, …, n. With these qualities

we define the n-production vector which is a n-vector.

x = x 1 … xn =

1 2 n

1 2 n1 1 1

1 2 nK K K

x x x

,

x x x

ª º ª º ª º« » « » « » « » « » « »« » « » « »¬ ¼ ¬ ¼ ¬ ¼

# # ! #

the n-demand vector which is a n-vector,

d = d 1 … dn

1 n

1 n1 1

1 nK K

d d

d d

ª º ª º« » « » « » « »« » « »¬ ¼ ¬ ¼

# ! #

and the n-consumption matrix which is a n matrix

c = c 1 … cn

=

1

1

1 1 1 1

1 1 111 12 1K

1 1 121 22 2K

1 1 1K K 2 K K

ª ºV V V« »V V V« »

« »« »« »V V V¬ ¼

!!

# # #

!

…

n

n n n n

n n n11 12 1K

n n n21 22 2Kn

n n nK 1 K 2 K K

ª ºV V V« »V V V« »

« »« »V V V« »¬ ¼

!!

# # #

!

.

We have x t 0 … 0 i.e. x = x 1 … xn t 0 0 … 0,d t 0 … 0 i.e. d = d 1 … dn t 0 … 0 and c t 0 … 0 i.e. c = c 1 c2 … cn t 0 … 0.

From the definition of t t t

t ti j jand xV it can be seen that the

quantityt t t t

t t t t ti 1 1 i 2 2 i K x xV V V! is the value of the th

ti

industry of the t th unit needed for all K t industries to produce atotal output specified by the production component vector

t

t t t1 K x x x ! of the n-vector. x = x 1 x2 … xn. This is

true for each t; t = 1, 2, …, n. Since the quantity is simply the



101

thti entry of the t th unit column n vector c txt we can further say

that the thti entry of the column vector x t – c tx

t is the value of the

excess output of the thti industry available to satisfy the outside

demand for t = 1, 2, …, n. Thus the excess n-output of the (i 1,…, i n) industry is given by the n-column vector

1 1 n n1 nx cx x c x x c x ! . The value of the outside

demand for the n output 1 n(i , , i )! is the th1 n(c , ,c )! entry of

the demand vector 1 nd d d ! . Consequently, we are ledto the following equation.

1 1 n n1 n 1 nx cx x c x x c x d d ! !

(I – c) x = d(I1 – c 1) x 1 … (In – c n)xn = d 1 … dn,

for the demand to be exactly met without any surplus or shortages. Thus given c and d our objective is to find a

production n-vector x = x 1 … xn t 0 … 0 whichsatisfies the n-equation (I – c) x = d (I 1 – c 1) x 1 … (In – c n)

xn = d 1 … dn.The consumption n-matrix c = c 1 … cn is said to be n-

productive if (1 – c) -1 = (1 – c 1)-1 … (1 – c n) –1 exists and (1 – c) –1 t 0 … 0. A consumption n-matrix c = c 1 … cn is productive if and only if there is some production n-vector x =x1 … xn t 0 … 0 such that x > cx; x 1 … xn > c 1x1

… cnxn.

A consumption n-matrix is productive if each of the n-rowsums is less than one. A consumption n-matrix is n-productiveif each of its column sum is less than one.

Now we will formulate the Smarandache analogue for this,at the outset we will justify why we need an analogue for theopen or production n-model.

In the Leontief open n-model we may assume also x d 0, or d d 0 and or c d 0. For in the opinion of the author we may notin reality have the monetary total output to be always a positive



102

quantity for all industries and similar arguments for tid ’s and

tijc ‘s.

When we permit negative values the corresponding production n-vector will be redefined as S-production n-vector the demand n-vector as S-demand n-vector and the consumptionn-matrix as S-consumption n-matrix. Under these assumptionswe may have different sets of conditions.

We say n-productive if (1 – c) -1 > 0 and non n-productive or not upto satisfaction if (1 – c) -1 < 0.

Now we have given some application of these n-matrices toindustrial problems.

Finally it has become pertinent here to mention that in theconsumption n matrices a particular industry or many industriescan be used in several or more than one consumption matrix. Soin this situation only the open Leontief n-model will serve it

purpose. Also we can study the performance such industrieswhich is in several groups i.e. in several c i’s. One can alsosimultaneously study the group in which an industry has the

best performance also the group in which it has the worst performance. In such situation only this model is handy.



103

Chapter Four

SUGGESTED PROBLEMS

In this chapter we suggest some problems for the readers.Solving these problems will be a great help to understand thenotions given in this book.

1. Find all p-subspaces of the n-vector space V = V 1 V2 V3 V4 where n = 4 and p d 4 over Q.

V1 =a b e

a, b,c,d,e,f Qc d f

- ½§ ·° °® ¾¨ ¸© ¹° °¯ ¿

,

V2 = (Q u Q u Q u Q) over Q,V3 = {Q[x] contains only polynomials of degree less than or equal to 6 with coefficients from Q} and

V4 =a b c d

a, b,..., g, h Qe f g h

- ½§ ·° °® ¾¨ ¸© ¹° °¯ ¿

.

What is the 4-dimension of V? Find a 4 basis of V.

2. Let V = V 1 V2 V3 and W = W 1 W2 W3 W4 be 3vector space and 4 vector space over the field Q of 3

dimension (3, 2, 4) and 4 dimension (5, 3, 4, 2) respectively.



104

Find a 3 linear transformation from V to W. Also find ashrinking 3 linear transformation from V into W.

3. Let V = V 1 V2 V3 and W = W 1 W2 W3 be 3-vector spaces of dimensions (4, 2, 3) and (3, 5, 4) respectivelydefined over Q. Find the 3 linear transformation from V toW. What is the 3 dimension of the 3-vector space of all 3linear transformation from V into W?

4. Let V = V 1 V2 V3 V4 be a 4-vector space definedover Q of dimension (3, 4, 2, 1). Give a 4 linear operator Ton V.

Verify: 4 rank T + 4 nullity T = n dim V = (3, 4, 2, 1).

5. Define T: V o W be a 4 linear operator where V = V 1 V2 V3 V4 and W = W 1 W2 W3 W4 with 4-

dimension (3, 2, 4, 5) and (4, 3, 5, 2) respectively, such that4 kerT is a 4-dimensional subspace of V. Verify 4 rank T +4 nullily T = 4 dim V = (3, 2, 4, 5).

6. Explicitly describe the n-vector space of n-linear transformations L n (V,W) of V = V 1 V2 V3 into W =W1 W2 W3 W4 over Q of 3-dimension (3, 2, 4) and4-dimension (4, 3, 2, 5) respectively.

7. What is n-dimension of L n (V,W) given in the problem 6?

8. For T = T 1 T2 T3 defined for V and W given in problem 6; T 1 : V 1 o W3, T 1 (x y z) = (x + y, y + z) for allx, y, z V1, T 2 : V 2 o W2 defined by T 2 (x1, y 1) = (x 1 + y 1,2y1, y 1) for all x 1, y 1, V2 and T 3 : V 3 o W4 defined by T 3 (a, b, c, d) = (a + b, b + c, c + d, d + a, a + b + d) for all a, b,c, d V3. Prove 3 rank T + 3 nullity T = dim V = (3, 2, 4).

9. Let V = V 1 V2 V3 V4 be a 4 vector space over Q,where V 1 = Q u Q u Q, V 2 = Q u Q u Q u Q, V 3 = Q u Q,V4 = Q u Q u Q u Q u Q, j j j j j

1 2 3 4T T T T T : V o V



105

jiT : V i o V i; i = 1, 2, 3, 4.

Define two distinct 4 transformations T 1 and T 2 and find T 1 o T 2 and T 2 o T 1.

10. Give an example of a special linear 6-transformation T = T 1 T2 T6 of V into W where V and W are 6 vector

space of same 6 dimension.

11. Let T : V o W where 3-dim V = (3, 7, 8) and 3-dim W =(8, 3, 7). Give an example of T and find T -1. Define T onlyas a 3 linear transformation for which T -1 cannot be found.

12. Derive for a n-vector space the Gram-Schmidt n-orthogonalization process.

13. Prove every finite n-dimensional inner product n-space hasan n-orthonormal basis.

14. Give an example of a 4-orthogonal matrix.

15. Give an example of a 5-anitorthogonal matrix.

16. Give an example of a 7-semi orthogonal matrix.

17. Give an example of a 5-semi antiorthogonal matrix.

18. Is A =

3 1 8

3 1 0 2 1 1 1 1 0 0 1 1

1 1 6 1 2 0 0 2 1 1 0 10 2 0 1 0 1 2 3 4 0 1 4

1 0 5 0 5 6 7 0 12 1 0 1

1 1 0

ª º« »ª º ª º « »« » « »« »« » « » « »« » « » « »« » « » « »¬ ¼ ¬ ¼« »« »¬ ¼

a

3-semi orthogonal 3 matrix?

19. Find the 4-eigen values, 4-eigen vectors of A = A 1 $2

$3 $4 =



106

=

0 1 2 3 63 0 1 0

3 0 1 0 2 1 0 20 4 1 3 5 1

0 1 4 0 0 2 1 1

0 5 0 1 0 10 0 5 0 0 1 0 01 2 1 0

0 0 0 0 5

ª ºª º « »ª º« » « »ª º« »« » « » « »« »« » « »¬ ¼« »« » ¬ ¼ « »¬ ¼ « »¬ ¼

.

Find the 4-minimal polynomial and the 4-characteristic polynomial associated with A. Is A a diagonalizabletransformation? Justify your claim.

20. Give an example of 5-linear transformation on V = V 1 V2 V5 which is not a 5-linear operator on V.

21. Let V = V 1 V2 V3 be a 3-vector space over the field Qof finite (5, 3, 2) dimension over Q. Give a special 3 linear operator on V. Give a 3 linear transformation on V which isnot a special linear operator on V.

22. Define a 3-innerproduct on V given in the above problem

and construct a normal 3 linear operator T on V such thatT*T = TT*.

23. Let V = V 1 V2 V3 V4 be a 4-vector space of (3, 5, 2,4) dimension over Q. Find a 4-linear operator T on V so thatthe 4-minimal polynomial of T is the same as 4-characteristic polynomial of T. Give a 4-linear operator Uon V so that the 4-minimal polynomial is different from the

4-characteristic polynomial.

24. Let V = V 1 V2 V3 V4 V5 be a 5-vector space over Q of (2, 3, 4, 5, 6) dimension over Q. Construct a linear operator T on V so that T is 5-diagonalizable.

25. Let V = V 1 V2 V3 V4 be a 4-vector space over Q.Define a suitable T and find the n-monic generator of the 4-ideals of the polynomials over Q which 4-annihilate T.



107

Prove or disprove every 4-linear operator T on V need not4-annihulate T.

26. State and prove the Cayley Hamilton theorem for n-linear operator on a n-vector space V.

27. Let V = V 1 V2 V3 V4 V5 be a 5-vector space over Q of (2, 4, 6, 3, 5) dimension over Q. Give a 5-basis of V sothat Cayley Hamilton Theorem is true. Is Cayley HamiltonTheorem true for every set of 5-basis of V? Justify your claim.

28. Given V = V 1 V2 V3 V4 is a 4-vector space over Q of dimension (3, 7, 4, 2). Construct a T, a 4 linear operator onV so that V has a 4-subspace 4-invariat under T. Does Vhave any 4-linear operator T and a non-trivial 4-subspace Wso that W is 4-invariant under T? Justify your answer.

29. Let V = V 1 V2 V3 V4 V5 be a 5-vector space of (2,4, 5, 3, 7) dimension over Q. Construct a 5-linear operator V on T so that the 5-minimal polynomial associated with Tis linearly factorizable. Find a T on V so that the 5-minimal

polynomial does not factor linearly over Q.

30. Let V = V 1 V2 V3 be a 3-vector space of (2, 4, 3)dimension over Q. Find L 3 (V, V) the set of all 3-linear transformations on V. Suppose 3

SL (V,V) is the set of allspecial 3-linear transformations on V.

a. Prove 3SL (V, V) L3(V, V).

b. What is the 3-dimension of L 3 (V, V)?c. What is the 3-dimension of 3

SL (V, V)?

d. Find a set of 3-orthogonal 3 basis for 3SL (V, V).

e. Find a set of 3-orthonormal 3-basis for L 3 (V, V)f. Find a T : V o V, T only a 3-linear transformation

which has a nontrivial 3-null space.

g. Find the 3-rank T of that is given in (6)



108

h. Can any T 3SL (V, V) have nontrivial 3-null space?

Justify your answer.i. Define a 3-unitary operator on V.

j. Define a 3-normal operator on V which not 3-unitary.

31. Let V and W be two 6-inner product spaces of samedimension (W z V) defined over the same field F. Define aT linear operator from V into W which preserves inner

products by taking (3, 4, 6, 2, 1, 5) to be the dimension of Vand (6, 5, 4, 2, 3, 1) is the dimension of W.Does every T L6 (V, W) preserve inner product? Justifyyour claim.

32. Given V = V 1 V2 V3 is a (4, 5, 3) dimensional 3-vector space over Q. Give an example of a 3-linear operator T onV which is 3-diagonalizable. Does their exist a 3-linear operator T c on V such that T cis not 3 diagonalizable?Justify your answer.

33. Let V = V 1 V2 V3 V4 V5 be a (3, 4, 5, 2, 6)

dimension 5-vector space over Q. Define a 5 linear operator T on V and decompose it into the 5-nilpotent operator and5-diagonal operator.

a. Does there exist a 5-linear operator T on V such thatthe 5-diagonal part is zero, i.e., the operator T isnilpotent?

b. Does there exist a 5-linear operator P on V such that itis completely 5-diagonal and the 5-nilpotent part of itis zero.

c. Give examples of the above mentioned 5-operator in(1) and (2)

d. What is the form of the 5-minimal polynomial in caseof (1) and (2)?

34. Define for a n-vector space V over a field F the notion of n-independent n-subspaces of V. Give an example when n =

4.



109

35. Let V = V 1 V2 V6 be a 6-vector space over Q.Define a 6-linear operator E on V such that E 2 = E.

36. Let V = V 1 V2 V3 V4 be a 4-vector space over Q of (3, 4, 5, 2) dimension. Suppose V = 1 1

1 2W W

2 2 2 3 3 3 4 41 2 3 1 2 3 1 2W W W W W W W W ,

Define 4-linear operators, 1 2 3 4i j k mE E E E ; i = 1, 2; j = 1,

2, 3; k = 1, 2, 3 and m = 1, 2 such that each i pE is a

projection, i = 1, 2, 3, 4 and

i i p jE E i p

= 0 if p j

=E if p j.

z-°®°̄ .

37. Prove if T is any 4-linear operator on V then i jTE = i

jE T

for i = 1,2, 3,4. j = 1, 2 or 1, 2, 3 or 1, 2, for the V given inthe problem 36.

38. Given V = V 1 V2 V3 V4 V5 to be a 5-vector spaceover Q of (2, 3, 4, 5, 6) dimension. Define T a linear operator on V and find the 5 minimal polynomial for T. Isevery 5-subspace of V related with the 5-minimal

polynomials i.e. the 5-null space of the minimal polynomials invariant under T?Obtain the 5-nilpotent and 5-diagonalizable operator N andD respectively so that T = N+D.Verify ND = DN for the same N and D of T.

39. If T is a 7-linear operator on V = V 1 V2 V7 of (3,2, 5, 1, 6, 4, 7) dimension over Q. Is the generalized CayleyHamilton Theorem true for T?

40. Prove for a 3-vector spaces V = V 1 V2 V3 of (3, 4, 2)dimension over Q and W = W 1 W2 W3 of dimension(4, 5, 3) over Q if T is any 3 linear transformation find the 3matrix associated with T. Find the 3-adjoint of T.



110

41. For any n-linear transformation T of a n vector space V =V1 V2 Vn of dimension (n 1, n2, nn) into a m-vector space W (m>n) of dimension (m 1, m 2, …, m m) over Q. Prove there exists a n-matrix A = (A 1 A2 An)which is related to T. Prove L n (V, W) # {set of all n-matrices A 1 A2 An where each A i is a n i u m j matrix with entries from Q}.

42. If V = V 1 V2 Vn is a n-vector space over the fieldF of (n 1, n 2,…, n n) dimension. If T : V o V is such that T i :V i o V i; i = 1, 2, …, n. Show S

nL (V, V) # {All n-mixedsquare matrices A = (A 1 A2 An) where A i is a n i u ni matrix with entries from F}.

43. Define n-norm on V an inner product space and is it possible to prove the Cauchy Schwarz inequality?

44. Derive Gram-Schmidt orthogonalization process for a n-vector space V with an inner product for a n-set of n-independent vectors in V.

45. Let V be a n-inner product space over F. W a finitedimensional n-subspace of V. Suppose E is a n orthogonal

projection of V on W, with E an n-idempotent n-linear transformation of V onto W. W A the n-null space of E.Prove V = W WA.



111

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113

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116

INDEX

B

Bigroup, 8

C

Cayley Hamilton theorem for n-vector spaces of type I, 62-3Characteristic n-value in type I vector spaces, 56-7

E

Essential n-states 87-8

F

Finite n-dimensional n-vector space, 20

H

Hyper n-irreducible, 88



117

I

Infinite n-dimensional n-vector space, 20

L

Leontief model, 92Leontief open production n-models, 99Linear n-algebra of type I, 13-8Linear n-transformation, 21Linear n-vector space of type I, 13-8

Linearly dependent n-subset, 18

M

Markov bichains, 81-2Markov bioprocess, 81-3Markov chains, 81-2Markov n-chains, 81-4Markov n-process, 82-4m-idempotent, 90m-n-C stochastic n matrix, 85-6m-spectral m –matrix, 90

N

n-adjoints of T in type I n-vector spaces, 55-6

n-annihilating polynomials in n-vector spaces of type I, 61-2n-basis of a n-vector space, 19n-best approximation in n-vector spaces of type I, 50n-characteristic n-polynomial in type I n-vector spaces, 57-8n-characteristic n-vector in type I n-vector spaces, 56-7n-characteristic value in type I n-vector spaces, 56-7n-diagonalizable n-linear operator, 59n-diagonalizable n-linear operator, 71n-eigen value of the stochastic n-matrix, 85-6n-eigen values in type I vector spaces, 56-7



118

n-ergodic, 88n-field of characteristic zero, 11-2n-field of finite characteristic, 11-2n-field of mixed characteristic, 11-2n-field, 7, 10-1, 81-4n-group, 7-10n-independent trial, 85-6n-inner product of n-vector space of type I, 47-48n-invariant under T, 63-4n-irreducible (n-ergodic), 88-9n-kernel of a n-linear transformation, 27n-latent n vectors, 91

n-linear algebra of type I, 13-8n-linear operator of a type I n-vector space, 28-9n-linear transformation, 21n-linearly independent subset, 18n-minimal polynomial, 77-8n-monic polynomial, 61-3n-nilpotent n-linear operator, 76n-normal linear n-operator on type I n-vector spaces, 56n-orthogonal complement of a n-set

in a n-vector space of type I, 51-2n-orthogonal n-vectors, 48-9n-orthogonal, 80n-orthogonal, 90n-projection of n-linear operator, 67-8n-range of a n-linear transformation, 31-2n-row probability n-vector, 85n-semi anti orthogonal, 80

n-semi orthogonal, 80n-subfield, 83n-subgroup, 10n-subspace of type I, 17n-system semi communicates, 87-8n-unitary operator of type I vector space, 53n-vector space linear n-isomorphism, 26n-vector space of type I, 13-8



119

O

One to one n-linear transformation, 24

P

Probability n-vector, 85

R

Random walk with reflecting barriers, 86-7Random walk, 84

S

Same n-dimension n-vector space, 25Shrinking n-linear transformation, 22Shrinking n-map, 22S-Leontief n-closed n-model, 98S-Leontief n-open models, 98S-Leontief open model, 95Special n-linear operators, 39-40Special n-shrinking transformation, 23-4Special shrinking n-transformation, 23-4Spectral n decomposition, 90

T

Transition matrix, 83-4

Transition n-matrix, 84-5



ABOUT THE AUTHORS

Dr.W.B.Vasantha Kandasamy is an Associate Professor in theDepartment of Mathematics, Indian Institute of TechnologyMadras, Chennai. In the past decade she has guided 12 Ph.D.scholars in the different fields of non-associative algebras,algebraic coding theory, transportation theory, fuzzy groups, andapplications of fuzzy theory of the problems faced in chemicalindustries and cement industries.

She has to her credit 646 research papers. She has guidedover 68 M.Sc. and M.Tech. projects. She has worked incollaboration projects with the Indian Space Research

Organization and with the Tamil Nadu State AIDS Control Society.This is her 37 th book.

On India's 60th Independence Day, Dr.Vasantha wasconferred the Kalpana Chawla Award for Courage and DaringEnterprise by the State Government of Tamil Nadu in recognitionof her sustained fight for social justice in the Indian Institute of Technology (IIT) Madras and for her contribution to mathematics.(The award, instituted in the memory of Indian-Americanastronaut Kalpana Chawla who died aboard Space ShuttleColumbia). The award carried a cash prize of five lakh rupees (thehighest prize-money for any Indian award) and a gold medal.She can be contacted at [email protected] can visit her on the web at: http://mat.iitm.ac.in/~wbv

Dr. Florentin Smarandache is a Professor of Mathematics andChair of Math & Sciences Department at the University of New

n-linear algebra type 1 and its applications, by w. b. vasantha kandasamy & florentin...

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