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C{ONTEI!}ISICIIAPIER ONI'
SYSTEUS OF LTNEAR EgUArIOIvS
Articles Pagles
1.1 Introduction to systems of linear equations I1.2 Degenerate and non-degenerate linear equations 41.3 SoliUon of a non_homogeneous system of linear equtions 4
L.4 Solution of a system of homogeneous linear equations 20
ETERCISES .ICHAPTER, fiTO
DEf,ERMINANTS
2.1 Introduction2.2 Deflnition of determinant2.3 Sarms diagrams for determinants2.3. f Sarms diagram for determinant of order 22.g.2 Sarms diagram for determinant of order 32.4 Minors and cofactors2.5 Fundamental properties of determinants2.6 Expansion of determinant (kplace's expansion)
2.7' Application of determinant to linear equations(Crammer's rule)
Multiplication of two determinants of finite order
Multiplication of two determinants of the same order
Multiplication of two determinants of different orders
MultiplicationtheQrgm.:' .
EtrEEQIqpp - ' (Al
2.9 AdJointdeterminant2.lO Inverse or reciprocal determinant2.ll Symmetric and skew-symmetrlc determinants
2.1 1. I Froperties of symmetric determinant2.11.2 Ortho-symmetric determinant2. 1 1,3 Properties of skew-symmetric determinant2.12 Skew-determinant2.13 Differentiation of a deterlnina$t " . ,'
Llst of books mltten bY the author
Professor Md. Abdur Rahaman'
1. College Linear Algebra2. College Modern Algebra3. College Higher Algebra
@asiI agJura & Fundamentals of Mathematics)
4. b-oU.g" frathematical Methods [Volume one)
(Spec."ial Functions & Vector Analysis)
5. boU"g" Mathematical Methods [Volume two)
(Integral Transforms & Boundary Value Problems)
6. +rqq"EEE< aq"rfalg",, g*W'm* lffitH HHH##P*'*")8- Ideal Solution of Cotlege Linear Algebra
; rrrrsi Ers< Aq{Flrc< c"t{ rTfl{rq t
ib. E-;"F* qilfie e +imi-fiq< qm{'Tflttq
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2.82.8.r2.8.22.8.3
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EXERCIIIETT - 2 EI
CIIAPTER THREEMATRIX AI.GEBRA
Artlcles : : **;;3. f Introduction .o3.2 Definition of matrix r!)
5:5 aJaiiiot "rra-"".l"t multiplication of matrices 95
5:; M;irit multiPlication 393.5 TransPose of a matrix ,3:6 Compiex con3rrgate (or-conjugate) of a matrix 97g'.; c""jig"t" tti.rEfo*i of a coriplex matrix 98
5,8 Sp.tii types of 'matrices wlth-examples ,983.9 Theorems o" ii"""pose.matrix' 1O3
3.10 Theorems o, **pi"" conjugate of a matrix 104
3.I1 Theorems on the 6on3ugat-e tlanspose of a
comPlex matrix - 106
3.12 Theorems on syrm.ttic and skew-symmetric matrices !O73.13 Theorems o"'li"i*iuuo and skew-Hermitian matrices lto3.I4 Theorems on idempotent matrices 113
3.15 Singular urrJ ttott-"ingular matrices ll43.16 Inverse matrix 115
3.17 Adjoint or adjugate matrix 1'65. iA Pr6cess of maiirg the inverse of a square matrix ::'-3.19 Theorems on inverse matrix I' r /
3.2O Theorems on orthogonal matrices I19g.2l Theorems on Unitaiy matrices l2O
3.22 Theorems on i,,"ol'iorY matrix 122
ilrg sor.:uon or ii.r.* .qr.jtio.r" by applying matrices 122
ETE,RCISES - 3CHAPTER FOITR
RANT OF A MATNtrT:
Necessary definitions applied itt tTFTrace of i matrx and properties of traceDeflnitton of rank of a matrixn"J"-"Uo" of a matrix to the normal formSvlvester's lawC6"aiu"" f"r consistency of a System of linear equations
Sweep out method and its applications
4.L4.24.34.44.54.64.7
r42
153154r58162167t67r82
184urDRcrsDs - 4
5.15.25,35.4
5.5&6
5.75.85.9.5.105.115.L25.135.14
.:jP'
Pages1951951s6
19619819819920020t202203203
CHAPTER FIVEVECTORS IN IRN AND CN
ArtlclesIntroduction to vectorsE;iid;r, n-space IRn
Vector addition andscalar multiplication.in If;;";;r;;oerties of the vectors in IR' under
;;;;';deition and scalar multiplication ,';;;; scatar produet of lro veetors in IR-
Oi"t^t "" and horm in IRI - - -*n
S;src properties of dot product in tR-CauctrY- S-ctr*ar1 ineq-UafitYMinkowski' s -inequatilY'Curves in q{^iit
"" in IRn --B
Spatial vectors and f, ,f, li' notationt T*'*iro"" Product of two vectors in IR"
Qpmplex numbersVectors in C"
EIGRCISES-5"''
rmcIgR SPACT,S -
i ':,.
,,
Binary operau-ffifriforry1ositior) on a 9et ''O"ntiti.i, of group with o<ainplEs' , .
Definition of ring with examPles'
2042A4205
6.16.26.3
210 :; l
213?132t+2t52152r6217224225233238244,t4s
6.4 Definition of ffeld with examples
6.5 Field properties 9f real nqmbers.
6.6 Definltion of a vector Spaie (or iineat' siiace)
.--.!l
i6*
6.7 ExamPles of vector sPace
6.8 SubsPaces of a veitor sPace'^ / r-^ ^f ^..I^^*aaac -.1-
6.9 - -WarrrPles of subsPaces
-Odd w.", ""*a'"ett"" 't t
ofr
CTIAPTER ONE
SYSTEMS OF LTNEAR EguATroNS
l.llntrodtrctiontosystemsoflirrearoqrratlonsMatrlx theory l{i rnore generally knovm as Ilneer Alglebra
which is orl$i,atecl i, .he st.udy of systems of linear
equatlons in se'reral rrnhnowns (or variatrles) and in the
attempt to find general methocls for their solutions' An
equation in two or'rnore variables (or unkno$rns) is llnear if
it contains no terrns of s;c'r:ond degree or greater' that is' if it
contalns no protlucts or po-lvers <lf the variablcs or roclts of the
variables. Ail variables occur oniy to the ii.sr power- and do
not appear as arglrrnents for tri$onometric' logarithmic' or
exponent"ial f'unctions'
A straight line in the cartesion ly-plane can be
represented al$ebratcally by an equation of the form
ax+ by = c where a' b and c are real constants (or nunrbcrs) and
x and y are variahles' An equation of this kind is called a
linear equation in 'the variables x autl y' Sirnilariy'
ax+by +c:t,+d=oisalinear equatlon in three variables
x, y and z which represents a plane in three dimensional
$pace. In general, an equation is called llnear if lt is of the
inrrna-rx, +a2x2+..' "' *o., lh=b(l)whereavaz"' "' '&''
which are to be determined'
lfi:=o'then{lJiscalleclahornogeneotls'linearequatlonand if b * o then {1) is calied a non-homogeneout llnear
equation.ExarnPles of linear equations'(i) Y*mx = o which is a homogienefius linear equatiort
representing the sllaight line passing through the
origin. " :
linear Algebra-l
COLLEGE LINEAR AT,GEBRA.i
(ll) 2x + 3y = 5 which is a non-homogeneous lineareiluation'iepre.denting a strai$ht line not passing
through,the origin.(llU ,i,+ 2y + 5z = 2O which is a non-homogeneous linear
r eeuation representing a Plane.(iv) \ - k - ax,-2xu= 3 (Non-homogeneous)
(v) xt + k+ ...'.. + r; = I (Non-homogeneous)'
Examples of non-Ilnear equations(i) 2fi +3Y=1(ii) x- xy = 2(iiil * nY'+4x+4=4(M a-# +2hry + bY2 =oM 6l + 13.rY+ 6f -5x-5Y+ I =oEquatlon(t) represents a Parabola'(it) represents a hYPerbola,(tii) represents a circle,(iv) represents a pair of straight lines passing through
the orlgin and ;
(v) represents a pair of straight lines not passingtlrrough the origin.
I.et'IR be the set of real numbers. 'I'hen a sotutlon of the
linear equation alxi + a2x2 + ... ... + g',-rh = b is any n-tuple(o.d2, on) of elements of IR such that the equation is
saUsliea when we substitute x, = ar, h, = 02, ,,. "' ' .rn = s,r' The
set of all such solutions of this linear equation is called the
solution set.Now we conslder the following two iinear equations :
4r x+-brY = crl /arx+bry=czlIf we interpret x, y as coordinates in the 'qr-plane, then
cmir of the al:ove two linear equations represents a straightlgme and {o,g) is a solution of the above two equations if and
orrly if flne point P with coordinates c[,P lies on both lines.
Hence there are three possible cases :
i.
SYSTEMS OF LINEAII EOUATIONS 3
Ca6€ I No solution if the lines are parallel
GLee tr l'rccisely one solution if they intcrsect
case m Inlinitely nlanv s;olt'ttions if they coincide'
These cases are illrrstrat.ed bY the following exarnples :
Exampl* l. The linctri si:i1eln i I I ; 3) nt" no solution'
since the lit:es representccl by tbese two linear equations are
parallel.
Example 2. The linear system |i'r==tr) has only one
solution, since the lines represented by these two equationsintersect aL(2, l).
v(o,2)
COLLEGE LINEAR ALGEBRA
Exarnple 3.'Ihe linear svstem ;:l;r'=u) n""
rn.rny sloltrtiotrs, sincc the lirres regrresented
ecluatiorrs coincide.
(o,3)
*-*-*4 x
.\-6
SYSTEMS OF LINEAR EOUATIONS
3r rxr + anh, + "' "' + dt,''\' = P'''l
^r,"i * oi+ * "' "' +'l:..o'..] o'l
,r,
ar,iXI + Ar..,2xz+ "' "' + z1-"''r;=b- J
#;t; thecoeflicients au' i = l'2' "" tnj = 1,2, ..., n
of the variables and the free terms bt' i = l'2' "':m are real
numbers taken f'o* m' the set of real nurnbers' If the b1 are all
zero, then the systenr (1) is called a homogeneous sJrstem' If at
least one bl is not zerc!' then the system (1) is called a non'
homogeneor" "y"tlm' o ""ot"""e
of numbers crl' 0b' "" crn is
called solutlon of the system of linear equations given by (1) if
x1 = (t1, h=%, ' "' ; = cln ls a solution of every equation in
the system (1)'
A solution vector of (1) is a vector x.whose components
constitute a solution of (1)' If (1) is homogeneous' it has at
leastthetrivial(orzero)solutlonXl=o,.xz=a,...',i'=o.lf x =(or, cz""'or,) is a solution of the homogeneous
system and at least one s' + o' then it is called a rron-zero or
non-triviat solutlon of tl:e homogeneous system'
A system of linear equations is called conslstent if it has
at least one solution and lnconslstent if it has no soluUon' A
consistent system is called determlnate if it has a uniq'ue
solution .,a itaJuEli-tf, ir it r'""-ilo[ thau one solution"
An indeterminate system of linear equations always has an
i"n"n" number of solutions'
Two systems of linear equations are called equlvalent if
every solution of the llrst system is a solution of the second
and converselY (vice versa)'
';Li.{{-I
i
infinitely
by these
(3,o)-3 '..Y-\'x-.
:;.
L.2 $egenerate and non-degienerate Hnear eqrrations'
'fhe ijeneral linear eqrration arxl + a2x2 + "' "' + q/i = b is
also callt:d non-degenerate llnear eguation'
Alirrearequationissaidtclbedegenerateifit}rastlrefr:rrnoxl +oJd2 + ' ' '" +'o'rcr' = b' Tltat is' if every coefficient of the
variable is equal ks zero.'Ihe solution of such a degenerate
linear equation is as "follows :
(i) If the constant l: * o, then t'he abol'e equation has no
solution.(ii) If the constant Ll = o, then every vect.ttr u = (crt ' %' .'"" c'/J
is a solui-ion of the allove equaUon"
1.3 $tolution of a non-hcmogeneous system of linear
eguatiouo: .
A system of linear equations tor a set cf :l sis:riltatreous
linear equations) ln n variables [or urlknr:wns) xt'xt' "" "" 4'
is a s.e[ of equations of the form
{IEllff&$
ir#\
I
tJ
INIl
I
System of linear equations
Inconsistent
Uniquesolution
':l
7COI,LBGE'I,IN EAR ALGEBRA
A slraighllbr',vard method of solution, known as Gausslan
elimlnalion involves a successir,'e "Whitling away" of the
var.iatlles irr order to isolate their values. This method is
birsed on l itc following three eiementary operations which
alter the Ibrm of the equations' but not the solutions :
(i) Interchange a pair of equations.
(ii) Multiplying an equationthrough bya non-zero number'
(iii) Adding a multiple of one equation to another equation
er, equivalently, if we consider a system of m llnear
equattons tn the n unknowns xr.&, ...., .r1 given by (l)' we can
reduce lt to a slmpler system as follows :
Proocsg I(i) Interchan$e equations so that the first unknown x, has
a non-zero co-efllcient in the first equation i' e, ar, * 0'
(ii) For each Dl, apply the operation
Ir -+ -a,r L, + ar r\That is, replacc the linear equation Lr by the equation
,gbtained by multiplying the first equation L, by - air ' and the
ith equaticin L, by a,, ancl then adding' We lhen obtain the
following system which is cquivalent to the system (i)'
SYSTEMS OF LINBAR EgUATIONS
3t tXt + atzxz + ... ." * atrr 4, = bl I^''jr\r+
a'21r+L tr*r + "'+ dz'4' =o''l e)
I
d' n.z\z+ a'-ir*t \r*t + "' * 2'-r,4. - b'*J
whJre"ar, *b.ri t, denotes the first unknown with
a'2Jz* in an equauon excJpt ttre first' Here D I and so {2 * xr '
_ It is to be noted that the system (2) of equations excluding
the first equation, form a subsystem which has fewer
equations and fewer unknowns than the original system (1)'
Repeating the above process rvith each new smhller subsystern
we obtain by induction ttrat' the system (1) is either
inconslstent or is reducible to an equivalent system of the
following form w'hich is known as echelon forrn :&rrxr * anh +..' "' + ar, {, = Pllu'or\r* u''Jr*' \r*t+ "'+ dt, x,- o''!
B)I
'^;
or\, * d r r*r{,.'r' * ... + a'* )i, - b',
J
where ,. rr,...1, t.d the leading co-e{ficients are not zero
i. e, arr *O,drrr*O, ..-,4:, *0'
Deflnltlon : In reduced echelon form the unknowns x'
which do not appear at the beginning of any equation (i * 1' J2'
... j.) are known as free variables' We also note that
(i) if an equation 0x, t 0.rA * "' + O{, = b' b* O occurs' tlren
the system is lnconslstent and has no solution'
(ii) if an equation Ox, + O& * "' + O-r1= O occurs' then the
equation can be cleleted without affecting the solution'
Process 2 : Consider the following system of m linear
equations (or set of m simultaneous linear equations) in n
unknowns \, x2,...,x"
4*r;l,l
Ii!;\il.
x.i!
t'
9;rzczzcqz
czz
ili!,l
I
1
1
{.J
I
t,l
d-rx, *.\rt-f: -i' ... + ilrr,{.,= brr.,
We reduce the s.r'stem (1) to a simpler slrsterl as follows :
Flrst step Dilminations of xl from the second' third' ""mth equations. We may assume that the order (rute) of the
equations and tlle order (rule) of the unknowns in each
equation such that art * O' The variable xl can then be
eliminated from the seconcl, third,...,mth equatlons by
subtracting
9,t*"" the lirst equation from the second equation8rr
$-t,i*"" the first equation from the third equation8rt
1* ti*"" the tirst eqrration from the mth equationottThts glves a ncw systenr of equittiotls of thc formallxl + anXZ + ..'* flI,,'t'r, =.1).t I
cz*z + '..+('a,-\,=D z I {2)
Jr* x2 + ... * crrrxn = b* n,l
Any solutiolr of lhe system (l) is a solution of the system
(2) and converselY.
SecondstepEliminationof.rifromthethird,fourth'......'mth equations in the system (2)' If the co-efficients err' czs ""c*r,, h the system (21are not all zero' we may assume that the
order (rule) of the equations and the unknowns such that
c22* O. Then we nray eliminate x, from the third' fourth' ""mttr, equations of the system (2) by subtracting
SYSTEMS OF LINEAR EOUATIONS
times the second equation {iom lhe third equation
times the second equation form the fourth equation
S!-'2- times thc second equatlon form the mth cquation'
ii'. ,rr,n"r steps are now obvious' In the third step we
eliminate xr, in the fourth step we eliminate xn etc'
This process will terminate only when no equations are
left or when the co-efficients of all the unknowns in the
remaining equations are zeto' We have a system of the form'arrxrl atzk +...+arr,)i' =.b., I
"r.r.? * "'*3o .i'l
c- + +.'. + k*); =b. I B)_a
o= br+t I
I-1o= b,r, )where r s m. We see that there are threc possible cases
(i) No sotution if r < m and one of the nunll;ct:-ifi, ' t ' " i:* is
not zero.
(ii) Precisely one sotrution if r = n nnd b,*t..... i:- if 1:r'esent
are zero.This solution is obtained by solving tlie nth ctiualion of
the system (3) for.r1, then the (n-1) th equation frrr "r;',-t and so
on up to the line.
(iii) Infinitely rnany solutions if r < n and b, * r. ..., bm if
present, are zero, Then any of these solutions is obtained by
ihoosing values at pleasure for the unknowns xr+l' "" 4r
solving the rth equation for x, then (r-l) th equation for xr-,'
and so on uP to the line.
dttxt * a:I2)c)'+'..".a2txl + azt )Y 'i-"'
COLL}] G E LINEAR AI,G EI]RA
'ct,r{, *l',rr i{2, {, = bZ
i (1,
12 COLLEGE LINEAR AT,GEBRA
'l'hus we obtain the equivalent system (with the same
rrolutlous as the sYstem (l).xr+2xr-xs=2)
-r2 -)ca=l I Qlb-xe=l)
Since the second and the
systey (2) are i49-tt-tlggLwe can-Hence we can simPlY wrtte.
third equations of the new
oilggera a-ny_93_9 o[_them.'\. <v '\_/*'n'Z-i
=?) .u
Thlssystem(3)isinechelonformandhastwoequationsin three unknowns and so it has 3-2 = 1 free variable which is
.q and hence it has an infinite mimber of solutions'
Letx3 = a {where ais arbitraryrealnttmber); then k=|+aand x, - -a,I'hus the general solutlonis x, = -a,x2 = I +a'
)h = a, whcre a is anY real number, Now a particular solution
can be cll,rli,rinr:tl l.)"'/ .tlj\Iiilg any value for a, Irt a =1, then xt= - 1'
\-2..t:,, '. i t'1. in other worcls, the S-tuple (-1'2' 1) is a
particr.rli-tt' soltttion of the given system.
E)rample 3. solve the following system of linear equations:3xr-xr+x" --21x, +1xr+2xr=6 I2xr+3x, +)b =O )
Solutlon : Reduce the system to echelon form by the
elemeptary operations. Interchange first and second
equations. Then we obtain the equivalent system'xr+1xr+2x" = 613xr-xr+x* --212xr+1xr+h= OJ
We multiply first equation by 3 and 2 and then subtract
from the second and third equations respectively. Then we get
the equivalent sYstem.xr+1xr+2x, = 61-l6n-5x. =-2O I-7xr-Sxr=-12)
SYSTEMS OF LINEAR EOUATIONS 13
We multiply second equation by - A*d then ad<l with the
third equation, Then we obtain'Equivalent system.xr+5.xr+2x"= 6
I
-l6h-Src" =-20 |13 13t- 16'rt =- 4 J
This system is in echelon form and has three equations in
three unknowns. So the system has a unique solution. From
the third equation we have \ = 4' Putting xe = 4 in the second
equation, we get xz = O,Again putting xz = O and 4 = 4 tn the
first equaUorl, we $et x, = -2.Thus x, ='-2, h = O, lt = 4 or, in other r"'ords' the 3-tuple
(-2, O, a) is l44Nque solutlon of the giver] systeln'
Eramff 4.JJProve that the ftrllow'ing system of linearequations iMnsistent :
?;i3;:?*,;;'I 6y@3\-xz+2d=z J
Proof : Reduce the system to echelon form by elernentary
operations. We multiply tirst equation by 5 and 3 and then
subtract from the second and third equations respectively.
Then we obtain the equivalent system.xr+2xr-3x" =- 1ln-7xz+ll:;. 7 I
/ )m + tl.rc, = 1O )
V/'6" srbtract seconcl equationt'fien we get the equivalent system.
xr+2xr-3x. --11-7xr+ ll.rco = 7 |
O+O = Il )
from the third equation.
or, equivalentll,, "r;?; i&:0 =
-- 11
=713J
l4
-has no solution.,I
Etample 5. solve the following system of linear equations:Zxr+3xr+Sxu+xn -S .l
3x, +4xr+2\+3xn=-21xr+2xr+8:6-xn -8 f tU7xr+9xr+.rt+Bxn =g )
COLLEGE LINEAR ALGEtsRA
't-h"" th: Si"." "y"t.- @chelonO = 3 (which is not
true) Flence th.e stem is incons_is!9n_(. !:_e,t!e system
Solutlon : Reduce the system to echelon form by theelementary operations. Interchange the first and third linearequations of the system (1). Then we have the equivalentsystem
xr+ 2x, + 8x, -xn = g I3x, +4xr+2x"+Sxo=-21zxi + sxl + s.rl + xn- =s I Q)7xr+9xr+\+8xo =g )
I.et us represent the four hnear equations of the system (2)by Lr, Ie, k and Lo respecUvely. Reduce the system to echelonfonn by the elementary operations. Eliminate x, from thesecond' thtrd and fourth linear equations by the operations
l, -+Lt - SLt, h -+f: -2L, andLn -+Ln -ZLr.Iz :Bx,+ 4x, +2xr+Bxn =_)
- 3L, : -Jx, - 6x, - 24\ + 3xn = _24Iz-SL, : 1\-22xr+ 6xo =-26l*:2xr+Sxr+5\*&
SYSTEMS OF LTNEAR EgUAnONS 16,
l'hus we obtain the equivalent systemx|+2tri+8x"-xn =$ I-2*r -22x" + 6xn = -26 Lo,-x2- llxr+3xn -- 13[ t
-1xr- 55.16 + lS.xn = -56 J
Divide the second linear equation of the system (3) by -2.Then we have the equivalent system
xr+2xr+8x"-xn =$ lx2 + Llx"- Sxn = 13 I
-xr-11.r6+3xn =- 13f (4)
-?tr- 55.t6 + lSxo = - 56 J
kt us respresent the four linear equations of the system(41 by Y r, U r, U
" and U n respecUvely.
Apply the operatlons U. -+ Li + L', "dU: +Vn * W,Us: -k - I l.r. + 3x, = -- 13
L'z:4,+11.16-3x, - 13
U"+Ur:0+O+Oi. e.
=QO=O
Ln : -5xr-5516 + 15xn =-56\Ur:1xr+55x, -75xn =65
- 2L, : -2x, - 4x, - L6x" + 2xn =- 16l**2Lt: -x2-Llx"+Sxn =-15
.Uq + 51, : 0+O+O =9.i.e. O =9. ,. ., i , ,
Thus we obtain tJre equivalent system :
xrt 2tc, + 8.rc, -xn - 8 Ixr+llr,r-3?==lr, lU,
Q= 9 )
Divide the fourth equatiort of the systenr (5) by 9 and
interchange it with the third equation we get the new sygtem.xr+2xr+8.r6-xn -Bl i
h, + ttx, - 3x*_= 1t | (6)V_ I I
O='O l
Ln :7x, + 9x, + x" + 8xn =Q- 7L, : -7x, - l4x, - S6x, + Zxn = - 56
L4-ZLr, -5h--55.r6 +l5,xn =-56
t0 COLLEGE LINBAR AI,GEBRA
Theorem : Given the following system in echelon form :
al rxt + apxz + arslql + ... + ah.r; = bl%r 4r+ a4z * 1\2 *t + "' +arr'ri = b,
a4, t. nH. *r {. *r +"'+ qrr4=b.where I < jz <....j. ,rd all * O, aar*O,...,\. +O
Then the solution of the given system is as follows :
There are hvo cases(i) if r = n, i. e. if there are as many equations as unknowns
ttren the system has a unique solution.(ii) if r < n i. e . if ttrere are f,ewer equations than unknowns
then we can arbitrarily assign values to n-r free variables andobtain a solution of the system.
Proof : The proof is done by induction orl the number ofequaUons in the system. If r =l then we get the single linearequatlon alxl + %12 +.., + q,.ri., = b. where a, * 0.
The free variables are jrz, ..., xr,. I-et the free variables be
assigned arbltrary values, say x2 = (b ri = q3, ... , 4, = 0!r.
Putttng these values into t-he above equation and solvingfor x, u,e &t
I,, =i |E--uror.-ilr,rs " .,.-q,qJ
These values give a solution of lJ:e e<1uation. Since puttingthese values we get.
", [+ {b - az {x2 - ... -- ,,,,r,,.)f +'azdz"i- ... * a,.ft, ',, b
or, b = b which is a true statement.
Furthermore, if r - n = l, then we have ax= b where a* O.
cuntafirs an €Ouatiorr of the fo'rn O = 1 which ic ne,t-tnde;-tletTcgthe given system is inconsist€rrt i:e th€ system has Io
v"
MS OF LINEAR EgUATIONS 17
It is to be noted that x= | ," "
solution. because " (i) = b is
true. Mot'eover if x= ct is a solution, i. e' ao = b, then * = 3'Thus the equation has a unique solution as desired'
New let us assume that r > 1 and the tJreorem is true for a
system of (r-1) equations.
. a2J2\r* %, *112+r + "' +ar,'r; = b,
arj,x1,, +41, *rt. *l *." +a-x" =br'as a system in the unknowns {r' ..', {,'Now the system is in echelon fonn. By induction we can
arbitr.arily assign values to the (n - jz+ 1) - (r-I) free variables
irr the reduced system to obtain a solution(saY {z= %2,..., r(-, = cr,,)
As in case r = l, these values and arbitrary values for the
additional y- 2free variables (say x2 = d2' ..,, \z-r= '0:r-r)' Yreld
a solution of the equation withtxr=;} (br - arzGz -... -ar,or,)
(Note that there are (n-j, + 1)-(r-l) + Uz-21 = n-r free
variables). Furthermore, these values for x1 , x2, """,'rqnalsosatisfy the other equations, slnce in these equations' theco-efficients of x1 , ..., \-r are zero.
Now if r = n then j2 = 2. Thus by induction we obtain a
unique solution of the subsystem and then a unique solution
of the entire system. Hence the theorem ls proved.
xample 6. Express the following system of linear
(t)
18 CoLLEGE LINEARALGDBM
Solutlpn :. I*I qq {Eitresent the four liqear equations of the
;',iven system (I) by Lt,Lz, L3 ,and L4 respectively. Reduce the
systern to echelon form by the elementary operations.
Eliminate xl from the second, third and fourth linear
equations by the operationsLa- La- 2Lr,Ia + Ia - 3L, and La -+ La- I-1 respectively.
la:2x1- x2*3x3-ZLtt-2x1 + 2x2-2x3 +2xa-2.x5 ,1=2
- -n:--
L2-2L11 X2+ xs42xa+2x5 = Q
L3:3x1 -2x2+24+xa+xg =I.
:4I;.,,-3",*3? *,3{. *q*n-3"s :9_ .
t,g-3l,ti xz-xs+ 4xa'2x5 =-lL+-Lt:x2+O+3X1 +O =- l..l'hus we obtain ihe following equivalent svstem ((Witli the'same
solutions rs the system ( I ) ).
-Yt-Xz *X3-X4 i')6 =l Ix2 * i3 + 2xa + 2.'--5 =Q L ,.,, ..x2 -.tb + 4xa -2.r5 :-2
1; * -*s'{ =-1J
* kt us represent the four linear equations of the systern (2)'l
byL{,lat, L3', artd La'respectively. Eliminate x2from the
third ariil the fourth linear equations by the opierations
14' -+Lg'- l-z' ?fftd L'4 '+ L '4- L'z :'Y't1 L'2:-2xg+2x4 -4xs =-)L'q- L'2, -x3 +xa -2x5 =- ITtrus the sy.-stern (2) reriuces to,
:. :,-:i,.:;;^3riJ=il (er
*iz"-:,-;::?l t'r'
sYs'r'EMS OF LINEAR EOU.{TIONS 19.
Dlvldlng the third equation of the system (3) by 2 we get
-.ys +& -2x7,= - l. which is identical with the fourth equation.
So wt ('an clisrcgard one of them.
tlen<'e thc system (3) reduces toXt-X2 *Xs-xq+4 =l Ixz*g+2xa+24=g I 6)-xs*x4-2xs --lJ ','
Multiplying the third equation of the system (4) by Iwe get
Xr-xz*xs-&+)6 =1 Ix2*xs+2xa'r2k=O I tO
x.g -xq+2'x5 -l )
Now t& bystem is in echelon foim and there are on\rthree equations in the five unknowns: hence the srystem has
an irrfinite number of solutions and 5 - 3 = 2 free Variables.
Since the three equations begin with the three unkno\MrlS x1,
x2 and *3 repectively, the other two unknowns xa and x5 ore
free variables which may have any real values desired. To
find the general solution 'let us say x+ = a, 'and xs = b where a
and b are any real ndmbers. Putting these values in the thirdequation of the system (5) we get xs = 1 + a - 2b. Puttiqg the
values of x3, xaand )6 in the second equation we get
4+ l'+a-2b+2a,+2b=Oar, x2= - (1 + 3a)
Again, putting the values ot'x2, xs, x4and x5 in the firstccluation of the systemJS)we get
x1 +l+3a+1+a-2b-a+b=lor'Xt=-l-3a+b'
l{ence the general solution is xr= - I - 3a * b, 2= - (1 + 3a),
)6= 1 + a - 2b, h = a,4= b where a and b q5e any real n r^Arorr/:
COIIECE LINEARN.GEBRA2A
1.4 Solution of a systcrn of hooogleoeors tlaoar cqtilrfoos',
, .$ system of linear equations is said to be homo$gneons if
all the constant terms br'b2"" b- of the non-hornogeneous
system are zrro; ttrat ls' the system has tlre form'
a11X1 + 412X2 +"'+arn)h =:lar21x1 +a22X2 +"'*3rrr4r ='|
,..(I)I ...rry
'r;', *t*, x2 + "'+ a,nrrxn= o J
Every homegeneous system of linear'equations is conslstent
Sincexl =O, k=O, "' ' ';h = o is always a solution of the system' This
solution is called the trivial, solutlon. If the other solutions edst'
they are called the noa-trlvlal solutloqs' Thus the above
io*o*un.ous system can always be reduced to an equlvalent
homogeneous system in ebhelon forrr:
a11x1 + a1f02 + arsxs + "' +a rn+r -1Ol;;rir:.':n"'n.:.^+"'+ a..2.,4' I I s
air\r*rdr* 1x].* l*"':t* )h =O J
Hencewehavethefollowtr4ltwoposslbilities:(i) if r = o I' e' the number of equatlons ls equal to the
number of unknowns then the system has only ttre zero
solutlon.
(ii) if r <n i' e' the number of equaUons ls less than the number of
unknowns, then tfre system has non-zero solutlon'
-'Excrrplc 7' find the non-trivial solution of the following system
i
t
II
I
of homogeneous llnear equations:x1 + x2 +2xs
h, +xs-2x1 +3x2 + xs
Solution: Reduce the system to echelon form by the
elementary operations' Ipterchange second and third
=0..l=o I (r)
=oJ
S}nsflTMS OF LINEAR EgUAfiONS 2I
cgllruoni of the glven
8y.tcm.
system (1). Then we get the equivalent
xt +h + 2x3 =-O|
-2xt-+bcz+xs =91 Alx2 + x3 =QJ
I.ct us represent ttre three linear equauons of the system
(2) ty Lt.Leand L3respectivellr' Reduce the system to an
echc)on form by the elementary operations' Eliminate *1
from thc secrnd equation by the operation' la'-+la + 2L1
lo z - 2x1 +3x2+ xs = O
ZLrs2rl+2.x2+4xs =O
la+zl,teS:q + S{3 =O
Thus we obtain the equivalent systemxt+ h. +2 xs =-Ol5x2+5x^ =O t Exz+xs =OJ
Dffilngthesecondequationoft}resysterr(3}B5.wegettlre equivalent qrstem
xt_ xz+2 x. = o 1
tii:;3i *Sirlce t]le seeond and third eqrraUons are ldentical we can
dtsregard one of them' Hence we have the equivalent system
*'*?o:;:=3 )rn'"n is in echeron rorm'
In this echelon form there are only two equations in three
unknowns, henee the system has an infinite number of non-
zero solutons. The systenr has 3 - 2 = I free variable wl:tch is
x3. [,et xs= a.Thus the general solution is x1 = - a' x2= - o" xs-- d
or (- a. - a, a), where a is any real number'
F'or particular solution' Iet a = l' then x1 = - l' x2- - 1'
xl. l, or (- l, - I. l) is a particular solution ofthe system'
22 COLLEGE LINEARAI,GEBRA
Example 8. Find the solution of the- following system ol
hornogeneous linear equations:
Solutlon: Let us represent the five linear equations of the
system (1) by L1 ,L2,L3,La, and L5 respectively. Reduce the system to
echelon form by the elementary operations. Elimlnate x1 from the
second,, third; foirrth and fifth equations by the operations
L2 -+ L2 - Lr , I+ -+ 14 -3L1 , La -+ La - 2L, and'
la"- Ls - 6L1 resPectivelY.
la -Iit: 4x2+O +2xa
Ia: 3x1 - 7x2- xe-6& =Q
- 3L1 : - 3x1 + 3x2+ 3x3+ 3:q : L
I,3-3Lr: -4x2+24*Sxa -Ola: 2x1+2x2-24 -O
-2L,y: -2x1 +2xy+24 +24 =O
L4-2Lr : 4x2+O +2xa=QI4 : 6x1- 2x2 - 4xs- 5xa = Q
-6L1 : -6x1+6x2+6xs+6& =O
k-6l-r: 4x2+2xs+& =O
Thus we obtaln the equivalen:;f^fl
4 xz +2&-4x2 +2xg- 3xa
4xz + 2xa
4x2 +2xg + x4
xt- x2 - x3 - x4 =O-lx1 +3x2-xs+)Q -Ol3x1 '7x2-xs-6x4 =Q | (1)
2x1 +2x2-2x3 =Q I6x1-2x2-4 xs - 5x+ =0 )
& =01=ol=o> a)=Ol=o _,l
xt- x2-x3 - X4 =q4.u.2 +2 x4 =O I
-4x2+24-3xa =0 IAxz + 2-r3 + xa =O .1
-o
In the system (2) the second and fourthidentical we can disregard one of them. So,
reduces to
equations arethe system (2)
SYSTEMS OF I,INEAR QQUATIONS 23
kt us replesent the forrr linear equations of the sy5tem (3)
lty l,'1,L'z,L'3, and La' respectively' Eliminate x2 from the
third arrd lburth cquations by the operations L i3 - L '3 * L;'2,'and L'4 4 I-'q - l,'2
L'3 * L'21 2x3- xo =gL'+ - L'2 :2xs - &=O
Tttus tle systeln (3) r-educes toxt-xz-xs - & -01
4x2 +24 =0 I . ..1- 2lo-i =ol (4)
2r, -xq =O )
In the system (4) the third and fourth equations are
identical we can disregard one of thern. Thus we obtain the
equivalent systemixr-xz-xs - &=Ol45 +2\ =0 l24 -x4 =o ]
(5
In this echelon fomt there are only three equations in fourunknowns, hence the system has an infinite nurnber ofsolutions anird 4 - $ = I free variable which is x4. L,et xa= 4,
where a is any real number. Then h=t, x2= -| ana xr = a.
'Itrus the general solution is x1 = a, x2 =-|, x3= 3, &= a.
( aa \or, [a'-;,i,"tFor particular sclution, let a = 2. Then xt= 2, h= - l, xs = 1,
Xq= 2 or, (2 - 1, 1, 2) is a particular solution of the given
system.Example 9. Find the solution space of tlrc following
homogeneous system of linear equations:
El. P. 1982.(3)
x1 +2x2 +3.rq +15 =q2x1 +3x2 +3xa +:6 =O i
\ +x2 +xS+2X+ +.rB =O I3x1 +5x2 +6xa +24 =O I
2x1 +3x2 +2x3 +5x+ +24=O '
24 COLLEGE LINEARAI-GEBRA
Solution : Reduce the system to echelon form by the
elementary operations' We rnultiply lst equation by 2' l' 3
and 2 and then subtract from 2nd, 3rd, 4th and 5th equations
respectively..Then we have the equivalent system
^ lT-.,7-;*;;l-x2 + xg -x4 =O f
:x :*, i:: )
We subtract 2nd equation from 3rd' 4th and 5th equations'
'Ll:cn we irave the equivalent systemx1 +2X2 +SXa + 15 =\
-xz -Sr<c - x5 =o IxS+Dc4 +x5 =O fO+O+O =o I' 2xg+24 + x5 =o )
x1 +2x2 + 3.ta + -16, =-q-x2 - 3x+ -x5 =-Ol) xs+2xa +)6 =t)1
2xs+2xa +x5 =Q )
We nrultipfy 3jd equailon by 2 and then subtract from 4th
equauon. then we have the equivatent system
x1 +2x2 + 3x4 +)6 =O1
-x2 - Sxa -)6 =Otxt+2xa +x5 =OI-2xa'- xr =0 'l
f'his system is in echelon form ha.Yrng four equations in
five urrknowhs. So the system has 5 - Q = I free variable
which is x5 and it has non-zero solution'
l-et x5= 2a where a is any real number' Putting x5= 2a 7rr
tlre 4tlr equation we get - 2x+- 2a = O' that is' x1= - a' Puttin$
& = - a and x5 = 2a inthe 2ncl & 3rd equations we get xz= 'a and
is= O. trinally putting xz= d, x4= - a and xs= 2a in the lst" eguation. we get
h+ 2a- 3a + ?a=O, i'e X1 = - a'
SYSTEMS OF LINEAR E9U'aTIONS 25
Hence the solution space of the given syst€m is
W = {F a, a, o, - a,2alla€ IRI'
Exanple 10 Determine the values of tr so that' the
following linear system in three variables x' y and z has (i) a
unklue soloution (ii) rnore than one solution (iii) no solution:x+Y-z - l'l
2x +3y +)'z =3 I tO. U. IL lSAx+)'Y +32 =2 ]
Solutlon : Reduce the system bo echelon form by
elementarlr operaUons' We multiply flrst' eglation by 2 and 1
arrd then substract from the second ard tfle ttrird equations
respectively' Then we obtairn the equivaleet Erst€m'x+Y-z =l Iy+()"+28 = 1l()t-t1Y +42 =rJ
We multiply second equation by (X *U sd then substract from
the thir<t equaUon. Then rve obtain the equirraleot system'
x +Y-z =l )y+()'+2)z =l i
{4-a:lxl +2)V =2-)\")
This system is in echelon form' It has a unlque solution if
tlre coefficient of z in the thrid equation is llon-zero i' e' if
)"*2andr! + -3' In case)' =2 third equation is O = Orvhich is
true and the system has more than one solution'
Or,
Clr.
x+Y-z =l 1y+Q'+2lz =l l(6-l -12) =2-),)x+Y-z =l I
y+(r'+2lz =l f(3 + l.) (2-Nz=2- A )
26 COLLEGE IINEARALGEBRA
In case i = - 3. the thirri equation is O = 5 which is not true'
andrlrence the systenr has no solution.
rt[*r nte 11. For what values of .]. arid p the following
Gfstem of linear'equations has (i) no solution'
(ii) more than one solutiotr (iii) a unique solution :
lD. U. H 198q D. U. H (Stat) 19821
Solutlon: Reduce the given system to echelon forrn by the
elernentary operatiorrs. We subtract lst equation frorn 2nd
and 3rd equations. Then we have the'equivalent system.
x+y+z =6 IY+ 2z=4 iy+Q,,-llz =P-6 )
We subtract 2nd equation from 3rd equation. Then we
have the equivalent system
Now from the last sYstem
eases;
(*)(*') we have the following three
(i) If 1= 3 and p + l0 then the 3rd equation of (*) is of the
formo = a wherea= p- 10 + Owhichimpliesthatoisetlual
to a non-zero real number which is not true. Thus we conclude
that for l" =;3 and p + 10, the given system has no solution.
x+Y+z =6 Ix+Zy +32 = lO Ix+2y+)z =$ )
x+Y+z =6 Iy+ 2z =l I().-3)z = pt-10 )
SYSTEMS OF LINBAR EOUATIONS 27
(ii) lf l =3 and $ = IO,. then the 3rd equation of (*') vanishes
& the system will be in echelon form having two e{uations in
three unknows. so it'has 3 - 2 = I free variable which is z and
hence for I = 3 and F = 10, the given systerh will.have more
than. one soltltion.(iii) For a unique solution the coefficient of z in the Srd
equation must be non-zero i.e )" *3. and p may have any
value. Thus for l,*3 and p arhitrary the given system have a
unique solutlon.
Exanple 12. For what values of l. , the following linear
equations have a solution and solve them completely iri each
case:x+y+z =l I. x+2Y + 4" =\^l
x+4Y+LOz =)'2)
Solution: The giiien system of linear equations is
x+Y+ z =l I+2Y *n :! I
x+ 4y + \Oz =)'2 )
Reduce the given system to echelon form by the
elementary operations. We subtract'lst equation from 2nd &
Srd equations respectively. Then we have the equivalent
system.x+!.+z = I Iy +'32 =I-I I
3y+92 =)'2 -1,j
We multiply 2n<i equation by 3 and then strbtract from the
3rd equation. Then we have the equivalent system
. x+!*z=l I x+Y+z =L Iy+32 =,t-l f =Y*9, =tr-I I (-)
o+o =L2'37.+21 o = ?&-3X+2)
2g coLLtsGE LINEARALGEBRA
lf )\2 - 3)," + 2 * o, the given system will be inconsistent ancl
'if ),2 - Nt+ 2 = o, the'above systern will be in echel'on form
having two eguatlons in three unknotrrns. So the system has
3 - 2 = I free varlable whtch is z' So the system has non - Trlro
soluflonfcx)'2-3I+2=Othat ls, (L l) ( X-21=O
=+I=l or'I=2Tfrus the $ven systerr is consistent fur 2' = I and ]" = 2'
Cas I uihrn).= 1
The qrstem ({ will bex+Y+z= lI
y + Zz=OI
\rrhH ls in echctron form where z is afree vartable.
Iretz = a xrtr€tt a is arbitmry real number.
.'.y=-3a.x = l+%Henc-t(lthegiyensystemoflinearequatlorrshisinfinite
number of solutions for I =1 In particular, }et a =' 1, tlren x = 3'
. y = - 3, z = 1. ls a parffcular solution of t]re $iven qlstem'
@ E Whl'=2The system ({') wiU be
x+Y+'=lly+Sz=ll
which.is in echelon form where z is a free variable'
LeLz=bwhere b is anyreal number. "'
y = 1 - 3b, x=2b'
Hence the system hers infinite number of solutions for I = 2.
In particular, let b = - l, then x = 4,Y = - 2, z = - | is a
particular solution of the $iven s5r5[s1n.
SYSTEMS OF LINEAR EgUATIONS 29
E:XERCISEA - 1
l. Which of the followtng systems of linear equations are
inconsistent :
xl+)r2=l)x1+2x2=31
x1- x2 =) )
x1 +3x2 +Zxs = 7 )2x1 + x2+3xs = 8 [3x1 + 4x2 +6xs = 16 [
6x1 + 9xz + llx" =29 1
(i) ';::ffi 1) (ii)
(iii)
'Aaswers: Ineonsistent.(i) Inconsistent' (il) Inconsistenl' (iii)
(iv) Inconsistent
2. Which of the following systems of lI:ear equations are
conslstent? Find all sohJtions of the consistent system'
,, ir i'*:W"=. i l @ I - r1l,'S:I ]AnswErs : (i) xr = L, x2 =2, xs =4
{it) x, =- 3r -L+. =-2a-92x3 =a
where a is anY real number'
3. Express the following systems of linear equations ln
xr* xz+ xs=1.|Zxr+2toz+2xs=1YSxi + 3xz +3x, =21
echelon form and solve them :
x1' xr+2x.+ 4 =-Ol... -*i + 3.xr +2x4 = 2
Ltr,2xi+ x2 - xo=lf2x., + 2x2 + x" + 3& =L4 )
xr+2xr+ xo =-116x, + x2+ )ca =-4 I
(ii)2xr-3*r- *, = O f- xr 7 x2-2xs= 7 txrh, = l' )
Aasrrers, (!-*, =1,&=2, 4=-l'*=)t@ "' =- l' x2=-2' xs:7
30
4.
COLLEGE LINEAR AI,GEBRA
Solve eactr of the following linear systems of equations :
2x1-3x2--212x1+ x2 = 1l3x1+2x2= Il3x1+2x2- xs =-155x1 + 3x2 +.P4 = Q
3x1 + x2*3x-r.= tlLlx1 +7x2 =-3O
4x1-Bx2=12ll,Ii) 3x1 -6x2- 9f
-2x1 +4x, =4 1
d,uf,,,,,
WW
x+W-32= 6l2x- y+42= 2 I4x+ 3y -22= 14 )
p.u.s. 1980, Bl@ atrs#ers:
: (i) Inconsistent
. Cu).xi = 3 + 2a, x2 = 4, where a is any real number.
fiii) xt * - 4. xz = 2, xs =7.
. . M x= 2-a,y =2 +2a,2=awhere a is any real nurnber.
I 5. Solrra the following systems of ]inear equaUons :
/r,,n\3:::i?)ix: =31' V x1 -3x3 -24=91\ - h*xs*x4 =x1+2x2-XZ-X4 =(rr, 24- 2x2 * xs - x4. =
-3x2 + x3 - xa =-Answers .: ti) xr = ?,- x2 = &, xs = - 2
'!.
I where a is. any'real numbero .''1 l I
(ii) x1 =f g + !i. x2 =-ia + 3, xu = - 3a. .rq - a
where a is any.real number. '.
6.:find the solution sets of the following systems of linear
\
equatfons .: ,,x1 i2x2'+ xs+ X4X1- X2*X3- X4
' (1, x1 +8x2+ xs+54'2x1 +7:x2+24s+4xa
?l9,J
-61- o['= z5l= 2Ci)
SYS'TEMS OIT LINEAR' EQU/\TIONS
Anrwers:ll
(r) x, = j (2 +b- 3a), ri =i (8 - 2b), 4 = B, )e =b
where a anrl b are arbitraql real numbers.q I 11 I
[ii) x1 =r1. xz=-ll, xs=-1T,4=-11,
7. Which of the following systems of linear equations areconsistent? Find ail solutions of the consistent s5lstem :
x1 -5x2 +4x3 +4=4 3x1 + 2x2-x3 +4)E =61
(i) ,T,-.n?,; ix.-;fi . if r,,r -i": ;,:..ru"'..r; :3 IX] *x2*x3 *4=l) 5x1 +x2-Sxs-24=l)
.{nsrrrers :
64 -69 28 8(i) xr =85, xz= EE, x,r =-SEi,rQ =B
94 685 30 79(iil x1 =679, h=62g, 4 =- W, 4 =W.8. [i) Express the following system of linear equations,in
cchelon form and solve it :
x+ 1r --3 I2.x-2y -z =-B I4x *z=-I+l ,,[DU,t&,,rl.FQx-3Y-z =-5 )
(ii) Reduce the following systern of linear-equatious to an
xt-x2+2x3=512x1 +x2*xs=2 | tD.U.P.lgqu ';2x1 - x2 : xs = 4 | tD. U. s. 1981, 198u1x1'+ 34 +2i" = 1) :
Ahswers : (ij x = !nr"- Le, y =ff_. z= a :
.i
where a is any real number.(ii) xr - 2, &, =- 1. xs = l. ' 1 . *:: ';:1: i .i
3l
x1 + 2.x2 + 3xs''+ 44 = q1l.r', +llx, +4xr" =l I(ll) i!x'1 +4x2 +& =21lx, +x3 + 2* =3)
_ ." ..",1,
92 COLLEGE LINEARAI,GEBRA
. 9. (il Reduce the system of linear equations
5x1+2x2-7x3=117xt - *, * i* '- 9l P' u' s' 1s2l2xi+1xr- xs=5J
into echelor-r form and hence solve it'(ii) Use echelon method to solve the following system of
linear equations :
x1+x2+3 =Ol2xt - 2xz - xc + 8 =^o I lD. u. s. r9s4l4xi'xs+r4 - =9 I
x1'-3x2-xg +5 =O J
An$rers : (i)xr =m,*='ffi,*"=#'(ii) xr =Lf ! , * =7, X3 = a where a is any real number'
10. (il Solve the following linear equations :
' xr + x2 + 2xs *]'n = 1]3x1 + 2x2 - xs + ll& =-?l4xi + 3x2 + x" + !4 =lll2xi+x2-3x3+24=l)
(ii) Find aII rational solutions of the following system ol
linear equations :
xt+h -x3 -_&34+4x2-xs-2xa\+2x2+x3
Aosscrs : (i) xr = 5a-b - 4' xe = -7a+ 9' x3 = a' & ?b
where a and b are arbitrary real numbers'
{ii) x, = &'+2b-7,'xz=-Za-b+6'x3 =a'&--bwhere a and b are arbitrary real numbers'
'Qd art rmine whether each system has a non-zerc
solution. If exists, tind the non-zero solutions'
x1 +x2+xr]=gl l', +2x2+ X3 =91(i) xt.-xz*"i=of (ii) 7xi-xz+2L4 =ol
xt + h-*; =Ol xi +12x2 -374=91
AnswEfe: (i) xr =O, h. ='O' 16 =0'
= -Il
= 5)
SYSTEMS OF33l;
JI
(ttl x;-ff ^,
* '=ff ^',xr. .= awhere- a !s anY real number'
Vd.T'ina the non-zero solutions of the following systems
of l-inear homogeneous equatlons :
x1 +3x2+2x3 =ql xr -Sxz-2x3 =91
(i) ri_:__{;"i".i":[ ],,,, *,:**rr, : B I
x1 + l7t tD.u.&,lgeffI''
2x1+x2*xs-5x4=9(iii) xl,+f;*-"*=31
2xt - .x2 - xs +Jq =91(iv) xi + 2i - 3x3 +-3xa =P
' -xr - x2 +2xs-2x+ =O '
t\
Aagwere: (t) ,t =-* a' x2==|' x3=a
where a is anY real number'(ii) Xl = - a, x2 *- &1 X3 =2
. wht" * i" an albitrary real number'
j : -'--' *t a'" a and b 4re arbitrarlr real numbers'
Aursr: 0t *, ='?;E =?' )b=a
where a lc arrirea!number'Llnear AlScbra-8
-rt5
(iv) xt '= a - b' xz = a-b' )6 = a' x4 =b*t"*. *a b are arbitraryreal numb€rs'
In particular, leta =2,-b-= I thenXt = 1' & = 1r *s = 2'{+ = I
'" W# il':T:iiffiii"-lili:"*" ", rhe ro,,owing
34tt tor,iuBb ur.rBnn ar,oBsRA
I':r:i ir (ii) xr =-ia,x2=-2'xs=ra'xa=ai i:
where a is anY real number'
(iii) *, =ai, *r=?i, xs = a'-& =a
where a is anY real number'
14. Solve the following homogeneous systems of linear
eQuatlons :
xt-Zth+2x3=O1(t) 2xi+ x2-2xs=Olr''
3xr + +x2-oxr=o13ix1 +4x2' +4 +2xa +34 =9
(ii) ui;**'*,**#,:?Jff #t u;.
i'*i* 10i2+ xs+6xa+5'c5 =o J
x1 + 2x7 + 2x3 -x4 + 3)cg' = q(iii) x1 + 2x2 + 3x3 * x4 +
-16 =.o ! tD' U' H' 19861
3x1 + 6x2+8;; *xa-+S'{ =o J
x1 +2x2 -2x3 +2!n * F =Ol(iv) x1 +2x2--; . #- -ziu =9[ o'u' H' 1ee8l" 2xt+4x2-7x3+ &+ .l6=u)
Arsrart': $l xt=7, *r=T'x5=z
where a is anY real number'
(it) xl, = - 3a- 5b' 4 =2a+ 3b' JF3 = 4 h= O' :6=b
*t"r" a and b are arbitrary real numbers'
'tiii) xr =-2a+ 5b- 7c,tt'z=o"x3= -2b+2c'xa =fo'x5 = c"
where a' b and c tire arbitrary real numbers' l
(M & =-%.-4b,ry=u'xs.i, b:3 =b';6=Owherea and b are arbitrary;real numbers'
lfilFind a general soluticn and a ilticylar solution of
the following system of homogbneous linear equations :
", - 3x2' .t ' ' +**- =9) :. i :qa'lrli;
2x',+ xq+3xs- &=01Sxi -lZi,*;6ft3'+:7-f,ard=OJ " :,,ti'.,.
S'STEMS oF LINEAR EQUATIoNS :r | :i 35
Annrorr:General solutlon : x1
a2a= o, x2=;, *, =!,4 =a
cqrtallrlns :
J(1 -3x2++*"- -'q =9.)3xi + tt'2+2xs+ x+ =9[2xv-4x2+6x3+xa =^ufixi+zi +24 =91Z*i - a*i * e*" =o )
wherc a ls arrY rcal number'
Partlctrlar solution i Xl = A' xb = 3' x3 = 2'/Y. =9'
(ll) Solve the following homogeneous system:"of,''llnear
Aosycf, ! X1 - - a, )h, = x3 = &, )Q =O rrr
where a is an arbttrary real ntrnber'
0ll) Solve the following system of linear equauons :
xr * x2 +2X3 + .lq= .5^l2x1+3x2+ Xs- &=lPl
2I'2;Xe:&=,;) ...,r;-/"!A;
Arlgwer ? x1 = !' h '= 3' x^'= l' 4 = - 2'
t{gdReduce the following system of linear equations to
an echelon form alr0 solve it : '' : ir " '' : 'r'' ' 'l"r;
x+2Y-32= 4 )x+3y* z=lli E).U,s.tg8ol
2x*'5Y-47=13 | '
2x+61*4':22) ' . i ,i',. ,-
(u) Solve the folioruing system "U'""T,-"ffit"ft,'x+ 2y - P,= 6- ] . i, , [D.,Ui;$, 198O 1
2x- y+42\;21 (tmlrovemcnt)
uslng echelon form or otherwise'
36 COLLEGE LINEAR +I'GEBRA
bllowlng linear equations by using echelon
form :
x+2Y +22 = 23x-2Y- z = 52x-5v +32 =-4x+iY+62 = O
ID.u.s. 198ll
p. u. 3. 1e8o I
llasircrB:,(i) x = L,Y =3,2= L
(li) x'r= ? - ^, y = 2a + 2, z = awlrere a is any rea-l number'
1,
17. Find the general solution and also a particular
solution of the following system of llnear equatlons :
2xt- x2- x3'+ 3xa=1'l4x1-2x2- xs+ &=: I6x1 - 3x2 - xs - Jq =-v- I
2i - x2 +24 -12& =lO,
(lt xt =ia+b+2
Aaswers : General solutton , Wr] 3o * uLxa =b
where a and b are any two real.npmbes'
Partlcular solutlon ixl = 2'xz=2'xs=-2'4=-L'
,:*1{. fiotyethe following system of 'llnear equatlons :
X1 t x2 +2x+ + 3'rs = '5' )24 +4x2- x3 +1& +1xs =- I Lxi +3& +5'rq +2x5 =-?^f
Sxi * 7.r2 -3x, + 9xa + 2x'' = -i6J,*, *q.v, - 4x3 + 2x4 -r 7'tq = -, ., ':
. .m-: ,0 = 2, x2 =- 3, xs = l' &.= o'4=2' i"'. I' .:":t-;.r,;\-
.": i -..o- rr*-:*i, ..:]i i,
SYSTEMS OF LINEAR ESUATloNs 37
lg.Sotvethefollowingsystemofhneareqtrations:'x"+iv+32 =-l l'
n /Zx-v -z' = 5 I
^{Y ';".'3u;; =-2 I
d/ ax+5v+42=3 I\tt -;;;iv * +" = - 10 J
A5rer i x=2,y=-3,2=1.2O. Solve the following systems of linear equations :
fi - #,i'#,=* #i:Y:# : f )'""'1esux1 +x2+ xs = il 'Xt-Xc+2x3=-31(iii) Bn - i, + 1xs = -.21
zxi-n- x3= 4 )
Answcra : (i) Inconsistent (ii) Inconsistent (fli) Inconsistent'
21. Find all solutions of the following system of linear
equations :
#,: !r;'{.2:#.:Er =i}S"r * 2i2 - 4x, - 3x+ -9x5' = I
Aogtper 3 x1 = t, &, = 2a' xs = d' x4 = -3b' x5 =5
where a and b are arbitrary real numbers'
22. Solve the follorvin$ linear equations :
x1 +2x2 - 3x3 + 1*, =^1]
z,xi + 5i2 - 54 + 6& -* I { D. u. H" 1s871x, +4xc- xs =O I
2xy + 3x2 - 7xs +lO4= | 1
Answer ! x1 =- I + 5a -Bb' n= I -a +2b' xs=a' & =b
where a and b are arbitrary real numbers'
23. Find all solutions of the following homogelleous
ID.U.IL r9@l
syslt'nts of linear equations :
x+2Y-32=OlItl 2x+5v+22=O I\" 3x- Y-+z=o )
tD.u.H. re86l
38 COLLtrGE I,[NEAI?. ALGT'BRA
x1 +4x2 +Sx:i +3& =9'l2xi +Si + 5x3 + xa =O L(ii) 3x, + 2$ +5x3 - x+ =9 [
4xi+x2+5x3-3-ta =OJ
x1- x2- X3- &=9xi +Sx2- xs*-&=9
(iiil 'lxi -Zxi - x3 - 6xr =!2x1 +2i2-24 =Ooxi - zxi'- 4xs - 5xo= I
ID.U.P, 1e851
lD. u. P. 19881
2x1 +2x2 - xs - +x5 =9]-xr - xi+2xs-3xa+rb=UL
{iv) xi * xz -2xs - rq =9 I
xs +.q+.rc5=0 I
.&rswers : (i) x= Y = z =O' No non-zero solution'
(ii) x1 = b -a, x2= - (a + b)' x3 = aarrd'q = tr-
where a and b are arbitrary real numbers'
(iii) xr = 2,a'x2=-?'x3 =a41%. where a is arbitrary real number'
(M xr = - o.-b, xz = &' x3 = - b' xr = O' 'rj; =tl
where a and b are arbitrary real numtrers''" 24."Reduce the following system of linear equations
echelon form and solve it :
xr * x2* x3+24+?;g =Ll- 2xr - 2x2 + x3 - 3xa -3xs =,P !3x1 +2x2 +& +cxs=t-Ylx1*x2+2xs- &+9x5=I8J
,,eryy"' i x1 = g -'' "#,:';,::# ;,1;f..",*.
25. Solve the following system of linear equatiorrs :
x1 +2x2 + 3x3 + 4x1 = 5l2xi + x2+Axs*-x, =?'!3xr +4x2+ x3 +5x1 =91i*i +3i2 +5x3+2xa=31
15 -tl io. 83
, Ar"y:"i lr= )1,x2=-11' xs= zl' xr= 7t'
to
SYSTEMS OF LINEAR EOUATIONS 39
26. Find a general solution and also a particular solution
of the followtng system of linear equations :
2x1 + 3xz -x3 + lxa - 24 = 4, lxr* 2x2+ x3+3xa- -4 =-ll2x1 + x2-O4 +74+-r3* = II5xi + I Lxl +7x^ +124 -l0xs = 4)
Ansrrers ! xr = - M-21a- 24b'& = 39 + lla+ 15b'
xs = - 15 - 4a- 5b' r+ = 3' xs =b'
where a and b are arbitrary red numbers'
In particular xl = - 85' x2= 50' xs=- 19' & = L')6 =o
27. Reduce the following system of linear equations into
echelon form and solve it :
x1 * 2x2 +' 3x3 + 4x4 + 55 = t,lz"i + 3i2 + 1i3 + \xa + 9o" =:t[3xi +5x2+ 6x3+ 7+*-*=?l-axi +7x2 + t0x3 +I-3xa + lPxs = I I
icxi+Si+ 9x3+l0xa+ S'rs= 3)
Answer i xl = *7 * 4b' h' ='7 + a + lob' x3 = - 2 -2a- 7b'
xA = a.xs = b where a and b are arbitrar)' real numbers'
]$-Reduce the following system of linear equations into
echelon form and solve it :
4x1 + 2xz + 5x3 + 7x4 + -h =?)X7 * X2* x:t+ 4 *!'q =ll
2n +3x2 + 4xs +5& +6:tb = I I3x1+9xz+7x3 + & +8x5=?l5x1 + h, + xs +6xa + .16 =UJ
Answer: xl = l, xz =O, '6 = 1,;8 =- l'xs =O'
2g.Determinethevaluesof.},suchthatthefollowingsystem in unknowns x Y and zhas
{i) a unique solution' (ii) no solution' (iii) more than one
solution :
lx+ Y* z=L Ix+)'Y+ z=l lx+ Y +)'z=l I
B.U.H 1S5l
Arcw?rs : (i) tr* I and L*-2,(u) i'=-2' {iii} }' = 1'
4o t ir.rnan ercbgRA,y'\
,( [9 Determin'e the values of ]...such that the following
systEm in unknowns x y and z has (i) a unique solution, (ii) no
solution (iti) morb than one solution :
^ x -32=-3 )
)zx+xy- z---2I1 x+2y +Az= I J
I
AttJw.*: (i) tr+2and l"+-5'(li) l'=-5,(iit) l'=2'31. Deterinihe the values oI )' such that the following
system of linear equations in unknowns x, y and z has {i) a
.r.riqr" soluilon (ii) no solution (iii) more than one solution :
- x+ Y+)'z=L )x+l"Y * z= ?tlIx+y + z=)tz )
Answers: (i) )' * -2, )"* f (fi) ?t=-2 (iii) lt = I
32. Find out the conditions on o, p and y so that the
followtng systems of 'non-homogeneous linear equations has
a solution :
x+ 2Y- 3z= cr I x+W- 9'= g I(i) 3x- Y+ 2z= Pi (ii) 2x+6Y-tt'=-9 I
2x- IOY + 16z=21 ) 2x-4Y + l4z=21 )
Answer:(i)2cr=P-Y (ii) scr =29 +Y.
33. Find out the condltlons on a, b and c so that the
of non-homogeneous linear equations is
consistent and also solve the system for a = l, b = l "'d " ='- 2'2x+Y+ z=alx-2Y +
-z =b Ix+ Y''22=c')Answer : Condition for consistent is a + b + c = O and
.-the,Emerul solution is x= cr- l,Y = d' - l'z=awhere a iS arbitrary real number. :.. , . :,,,:
A particular solution,is x= l, Y = l, z = 2'
XifiBEAICPnnA&1 IatrpfrcfuJ. J. $y{,tprter was t}rc ffrst man sho intnoduced the uprd
'metrf in 1850 and later on in 1858.fr&ur Crytq, de\rclopadthe theory of matrices in a systemaflc -way. Matrjx ls Apowerful tool of modsn, mathematics wh{h ie origtuated inthe study of llnear equa$ons, and it haq wlde appliqa,sons itrevery branch of sclence and especlally tn Physic+ Che-rnistry,Mathematics, Statistics, Economics, and Engineertngp.
3.2 Dcfri6oa orf nair|rA metrir is a rectangular array of nunrbers (real or
complex) enclosed,by a palr of brackets (or double verticalrolls) and t}re numbers in the array are called the enfies orthe clemcntc of the matrix. that i1 a rectangular artay of (realor complex) numbers of ttre form
is callqd a natfk. The number.s d1r,ilre, ... ..., amn arecalled the entrlec or the etreroeuts of the matrix. The abovematrix has m rows and n bolumns and ls called an (m xn|mat1k {read "m by n matrix"). The matrix of m rows and ncolunrns is said to be of order "m by n" or ra x n. The above
matrix is also denoted by Iary], .,';rl'3: ..']Tlre m horizontal n-tuples (arr, are, ..., ar.,), tazt, d.zz, ...,
az.,), ... ..., (ornr , &m2,..., 2*r) are the m rows of the matrix andlhe n vertical m-tuples ::
.COH,.EGE LINEAR ALGEBRA
I- irii'l'f arz I [ri" Ilorrll^rrl l^r, I1,1.1,,...,1,1L ;-,1 L .,", I L a,,,, r
are its n columns. The element aU, called the ij-entry orii+omponeat, appears in the i th row and j th column.
A matrix consisting ol'a slngle row is called a row matrix(or now vcctor) and a matrix consistir:g of a single colunrn is
called a column matrir (or cotumn vertor).
Matrices are generally denoted by capital letters A, B, X, Y
etc. Square brackets [ ], or, curwed bra'ckets ( ), or. TWo pairs
of parallel lines I I I I ar. used for the mathematical notations
of matrices. In this book we will use the notation [ ].
Emmples of mstriees
ymprc ,.o = [; -3 -?l i1 a matrix of order 2 ;i'il;
tlle real ffeld IR and also over the complex flcld C.
The rows of A are (l; O, - 5) and (2, -3, z) and its columns
*. (;) LS) ,"o ft)r2 0 i-1
D"ramfle2. B=l -i 1 4llsamatrixof'order3x3I r+i -s sJ
over tJ:e complex lield C.
The rows of B are (2, O, i), (-i, 1, 4) 4nd (1 + i, - 5, 3) and,.its:/ 2 . , Or zir
columns are [ -i l, I r I ana [+ l.
I t*i/(.-5) \3)TWo matrices A = [ a1 I and B = [ bu I are equal ifand only if
the5r are identical i.e ,if and only if they contain the same
number of rows and the same grmber of colurnns and a1, = b,for all values of i and j.
MATRIX ALGEBRA 95
3.3 Addttton ""n "of"t
nunfpUolUon of matrlces :
Addltion of matrices is deflned onl5l for the matriceshaving same number of rows and the same number of
columns. kt A and B be the two matrices having m rows and ncolumns.
[-a1 1 +b11 ap+bp 31n *b1n Ia+g=l a21+b21 a22+b22 z2n+b2n
II ... I
[-ar,1 + b-1 a.,,2 *b,,1 ?inn * b-r J
The multiplication sf matrix hy numbers (scalars) isdefined as follows : The product of aq (m,r n) matdx A by anumber k is denoted by kA or Ak and is the ( m x n) matrixobtained by multiplying every element of A by k, that is,
[- karr ka, lrarr.lta = l ka'' kazz u""
II ... I
L k.-, ka-z ka*,, JWe also define -A = {- 1) Aand A- B =A + (- B).
If the matrices A, B,'C:are conformable for addition and ifk is any scalar, then we can state that
(i) A+B= B+A(Commutatlvelaw)(ii) (A+B) +C =A+ (B +C)(Associativelaw)(iii) A+O=O+A=A .t: .. a'-'(iv) k(A+B)=kA+kB={A+B)k. : ".-
where O is the zero matrix of the same;order.l r ri$';rill:.,,.ir'. ,; 'r,i
fl
:l
I
[- a,r dn. oh Ii. e. A = 13,,1 = |
a2r dzz az,- I ,rra, L;, ; ::: ;;" J
I brr brz br., In=rhr= | ::, i.: .:: ?"
I
L.:b*r bse b-" J
Then the sum of A and B is
,d
g6, coLIEqE UNEAS, AITEBRA
For examph, ,r^= [ I -? ] *"= [-3 'nJr-
,rs,^*",;Ill,uu, n'J;{ }=[-i 3]
^=l?iL ?.u? I= [8 To ]ena e-"= [:?-u -?--'ul
= [-l i ]g.4 llatrtx h$!fifonttonTWo rratrlces A and B are conformable for multiplication
if the number of columns iu A is equal to the numder of rows
in B. R,
Ib' I' I.etA =,tir az .... a.l and
" =L j lrn I
The AB = erbr *,adb2r ... + a,,hJ =L P=r*o, -.,.,Jr
Alaln, lOt tle rn x p matSlx 6 = [agl and the p xn mptrixp = I h l. thenAB i*&e m x n *"T C = [ q"l wtrere
cs=anbr1 *"oba +... +fu h ?3ib5
ti= | .l', ',r' If the matrices A, B, C are conformable for the in&cited
sums and products, we havg the foIlowing properties :
, , (it (AB)c = A(BC) (Awcl,ntive law)(,) A(B +
"' :f" lgSl (Eistribudve lars)(ii0 (A+BlC=
(iv) k(AB) = (kA)E =
A[kBlwhere k is any scalar.
RemarLs : In the matrix product AB, the matrix A is called
lhe pre-rntltlpllcr (or Pre-factor) ancl B is called the
postmuttlpltcc {or p6t factor) i
For examples, let " = [ ;
-a
MATRIX AI,GEBRA 97
-1 I4lol-1 l*0"=[-i
thenAB= [] I;5:l-,1-?;'i:3 L 'ti:';:Lf,1,?'. I
= t;13:'f -'r-*'Zl3 I = t?? -::l
uo =[ -l il t; -B -ilL3 0lr 1' 1 + Ftl'z t' (-3) + (-l)'o l^. I * (-1) (-11 I
=l i-;r[; i i-it-&)++ o ?)?* 4 q']. I=L
b-: rio- z s Fs) +'o o ' u,l o' (-r) Ir-1 -3 6l
=l 6 6 -I4 l.'.AB+BA-L s -e 151
3;5 TxansPose of a matrirIf A is an m x n matrix over the real field IR ' then the
n x m matrix obtained from the matrix O OI yTt"U lo-:o*tas columns and its columns as rows is called the transpose of
A and is denoted by the symbol ,s' that is' if A = I aij I is an
m x n matrix then 'S' = [aJi Iis an n x m matrix'
Forexamples, leto=[] 3 ? -[l'
3.6 Compler co4iugate (or conjugatc) of a matrix
The conJugate of' a complex nurnber z' = x.r iy is the
cornplex number z = x- iy' If A is an m x n matt"ix over tire
r:ornplex field, then we say ttral. Lhe conJugate o[ a rnatdx A is
i!tc nrll'[rix A whos* fi:lttTteljJ-si n.,..11 11.1.11,;pf:Ctivciy l'hc <:nrtjui':l1cs
r-,il.lre:elernentsofA.'l'ha1 is'il'A=[n'i ]'tlren A =[Ziii l' t'
1.irtr".ri- lUgebra-7
r1
T 1 21
ur".,N=l g i I
| -z 6.1
98 COLLEGE LINEAR ALGEBRA
ifl. if A='l -iLs
rl -i-tA=li 2
t-- ,t 52
ForZ.examples
l+ i I2+3i I
-5 J1-i r2-sL I ._t-5 l
I,
1+2i
L-2i
/ A matrix A-lq_-c,allecl real provirled it satislles the relation
A=ADefinition Imaginily matrixA matrix A is called imaginary Provided it satisfies the
relationA=*A@.:IEEIe*r,I1)re conjugate of the transpose of a given complex matrix
A is said to be the conjugate transpqse of A and is generally
denoted hy the syrnbol A*. That is,
if A = [aq ] is a comPle-x matrix, then
* * 1(e= IQil.
.o,.*rffi o=l--l i, 'rl;, I1' I s t+2i -5 I
4,,r.=[-1 i 1-r, lL r-i 2-3i -5 -l
/*raif B = lt'*'n" ;f;, ?;z:/*"r,8*=['-* ,:i1
-l
[2+5i 3-2iiS/Gpecial types of matrices v{th*exarnplesJ
Squite rnatrix : A gglru< with t hc _same-
nr:mbet :1l9r""3"dcolunrrrs js t'alled a square matrix.;,=.ffi l"r
' s lor. square tnatrices.trr J
reetangulas.maEix:
-,.furexamples, I -L
MATRIX ALGEB.RA 9g
-# igslan8lular matrlx : The number of rows and coluntns of a-,--_.
er of rows and qoltlmns qf ths
,rrrl =.. :* =aral, then the matrix js known as the
-i 21 3I2 o s lrrxl_
are rectangular matrices'
Dj1g9".1 m"lg : Arytelatrix whose element: :iL: O
wry"s$511*Forexampre",Ii 3l*.LB 3 ?]
are diaeonar matrices'
matrices of order 3 and 4 respectively'
fr zero matrix or null matrix-: A qattl']5lnJvhich--srcry
Lt"*il--i" ,ero is called a rllrllJnatrix 61 a zero rn?trix'
;ilr.".[Tl 3] ",otg [[] arezeromatrices
Upper and lower triangular matrices :
A square matrix whose elements aU = O for D j is called an
upper triangular matrix and a square matrix whose elemcnts
a,j = O for i < j is called a lowgr triangutar matrix' i
-l ?6lyli
roo
r 5 o o.l I I 3 3 3].*lowertriangurar"'dl -r2ol*rol-Li -i;l matrices.
L sztj L-A; -i;lSymmetric matrlx : A matrix equat to its transpose i'e q
square matrix such that a1; = oii for I s i' j " t' is said to be
symmetric. In short we can say a square matrlx A will be
symmetric if .Ar =.{.t-ahgl l- 12 -3 I
exampres,"=L! i:landB=L_A ? ,r)
are sJrnmetric matrices.LA and l.B are also symmetric if l, is a scalar'
Skbw-symnetrlc matrk : A matrix equal to the negative
of its transpose i,e a square matrix such that 2U = - a;i and in
which therefore, du = O is said to be skew-sSrmmetric'ro hgl Io t-21
Fore:<amples, A=l-h O f land g=l -l O 3 |- L-e-fol Lz-s oi
. are skew-sYmmetric matrices.
Hermitian matrix : If A = [ ai.1 I is a square matrix over the
complexfield and A* = (Ar)=A i'ea1= aji fori' j= l' 2""' u;
then A is called a Hermitian matrix.
i'rrt. ex;irnples,,' 12 2-3i
A,= | 2-rlji 5I ':I .J
MATRXATGEBRA }OI
In this case dlagonal elements of the matrices will be real
numbers.
Skew-Hcrmltlan matrlx : If A = [ a,J I is a square matrix
overthecomplex{Ield, andA*=( 6t;=-A i'e 3u=- 6ir for
ij = l. 2, ..., nthen A is called a skcw-Hemltlan matrix'f 21 2-31 3l
For exampres, A = L-1-8, _r:i rt .] "",
r i l-1 5ln= l-r-i 2i i Iare skew-HermiUan matrices.
L-s i olIn this case diagonal elements of the matrix will be either
zero or wholly comPlex number.
Tdhogoaal matrk , A t@is said to be
orthogonal tf AAT =AT A = I.
./
For examPles,
COLLEGE LINEAR AI,GEBRA
Forexamp'.", [Lii] *,.[B iare upper triangular matrices.
2 3li 5l3-2 I
o 7)
ttrat is, if.S,/.,
= A-l (the inverse of the rnatrix A)r l 8 41ls o -0 II a, 4 z lA=l O -O -O l*dlq -L 4tL9 9 9J
r | 2 2-1t_-tl'55 3lI zt 2lB=l ;; -A lareorthogonalmatrices.I zz r II _* t
L-s s -3J
.a-
I .+i | ,r.,rl0j -414l
-3 l
l:j
itv{\
idempotent ry!!"_if A14 'Forexamples, I I -3 -s lanal-t 3' [-t 3 5] L I -2
are idemPotent matrices.
I --i: _-;rr
;,1 -.1 . il- t *zi
J
li jrli'r: aii.'r'S;
LA2 COLI-EGE LINEAR ALGEBRA
Nilpotent matrix : A square matrix A is called a nilpotentmatrlx of ordern if A'r = O and An-l + O where n is a positive
integer and O is the null matrix
3.9 Ttreorems on transpose matrix
Theorem I If A and B are comparable matrices and Al and
Br are the transpose matrices of A and B respectively.
then (i) (Ar)r = A (ii) (A + B)r =;S + Br.
(iii) (AB)r = ISAr (iv) (crA)r = cr AJ. where n is a scalar'
Proof : (i) rrt A = [ au l, l=rl;1'...,'7then bY definition ,{' = [a; lr = [Ei I
Now (AI )r = [qi] r = [ a,:l = A .'. (,$)r =,{.(ii) t€tA = [ air ] and B = [ bu ] whereS ill f'...: l'then C =A+B is deflned and [ -qL I = [aiJ I + [ b1 I.
Now by the deflnltion ol'the transpose of C
we have (A -t B)1 = Cl' = [glr = [qrl = [a1i ] + tQi I
= { qf+{\ lr =.S + sT
.'. (A + B)r = Ar + Bf.
(iii) LetA= [au I i= 1,2,..., m; j = l, 2, ...,n
Then AI = [aiJ ]T = [ ajil is arr n x m Inatrix
gr = [h. l, = IbrE I is a p x n matrix.
ThusABisamxpmatrixsothat(AB)r is a p x m matrix.
Also BrAT is a p x m matrix. Therefore, (AB)r and BrAr ltave
same dimensions.
NowAB = [ ct ]where (i, k) th element of AB is
3 ; where i = l,2, ...' mcu. =i ,
?U DJr, k = l, 2, ..., p.J=l
MATRIX ALGEBRA r03
Periodlc matrii : A square matrix A is called periodlc ifAm+l - A where m is a positive integer.
If m is the least positive integer for which Am+l = A, then A
is said to be of 1rcrlod -j ,r I -2 -6-t
period 2.
-]horotutory matrlx : A square matrix A is called anqffiivolutory matrix f A2 =1.
For exampl., o = [-3 -? I ," an involutory matrix.
Unltary matrlx : tet A be a complex square matrix, then A
is called a unltary matrlx ,l** = A*A = I or equivalenfly
A* = A-t.where A*=.:fi'r= 61).
For exampt.", e = [3 ] I ,'n ,=[-i 3:
are nilpotent matrices of order 2.
Normal matrlx : [rt A be a complex square matrix, then A
is called a normal matrix if A*A = AA* where A* is the
conjugate transpose of A.
no, example, [ = [ '.ui ,,.L,] is a normal matrix.
3tsl
$
,'t":
104 COLLEGE LINEAR AI,GEBRA
"lherefore, trre' (t;iith''i:lement of (ABlr
nn=f,' alib;x =I ajrt b$r,.:i=I - j=I
n
j=l
= (k,i)th element of BrAr
Hence (A6;T =g1O'
0v) LetA=[aij I .'. $ = [a1 lr = lajrl
Now (ctA)r = [cxaijlr = [craii l= c taril = aAI
"' (ae)r = crAr'
3.1O Theoremson complex conJugate of amatrixTheorem 1. IfA = 1 a1; i is any m x n complex matrix' then
A=A.Proof : By definition of complex conjugate of a complex
matrix, wehaveA= taulforlll I <i<m' I SjSnand ao isttre
complex conjugate of a1; i'e iu is ttre entry in the i th row and
j thcolumnof A. Again.E= Iiul for { I < i< m' I < j ( n and
E. i" th" complex conjugate of l.i i'e 5'o = ai1 (the entry in the i
# -," ""a i th colum, of tnt mitrx A)' Therefore' the entries
of the complex conjugate of A = tJre corresponding entries of A'
Also A and .E are matrices of the same order' Hence A = A'
T.heorem 2. lf A and B are two complex matrices
conformable to addition, then ffi=[ + B'
Proof : Let A = [ai1 I and B = [br] be any two matrices of order
m xn. Thenwe tar.e + E! = [ au + biil for alt 1 < i< m, 1 Sj S n'
Also by rlefinitioll, we have A = [tul and 3 = [biil
. A *e = 6,il * trul = t a'F5rr,
=, = *,
rJ,=,
= r,
Again 6- +B
Therefore, ATB = [ fu;b;] r"tru I sism, 1 sJ <n' (2
Since the corresponding entries of A + B and AJ-B are
equal and also both f + E and A;E are matrlces of order
m xn, from (1) and (2)' we conclude that A;B ='6' +g'
Theorcm g. I-etA= [aljlbe anymxnmatrixand B = [bplbe
any n xp matrix over the complex fleld i'e A and B are
conformable to the product AB' then ;E = A eProof : Since A and B are conformable to the product AB'
"oeg=[ai1lx Ib1r.l=[cL lwherecft = atlQ1 for all I < i J m'
lsk <pand 1Sj<n.
Rtso A=1ir;forall 1<i3m'lsj sn
and 6=l h* J forall 13jSn, I 5 kcSP'
NowABis defrnedandwehave AB= tqt x [bEl = F" t ttt
wtrere a,*=- aq+.;x firrd-+<i5-m' 1<k<<pand I<i<n'Again ffi = comPlex conJugate of AB
or. AB = [cr lwhene cil = qJ h.- f a,r brk I = 1i, u-,r t- [ -u "Jl( I - r-u -J^
= [d*l (2 since "{'" = ZrZ,
for any two complex numberd z1 artdza '
Sinced,k=irEuforall 1<ism, t=Oi:pand 1 (jsn'
Combining (1) & (2) we conclude that AB = A B'
Theorem 4. L;et6 = [a11] be any m xn conrplex matrix' then
),A = tr A. I ueing any complex number'
= complo( conjugate of lary + b1]
= complex conjugate of c,i where -Q; = ar1 + b,
=bul=[rFo,,]r"t"u1<i sm'I <J sn'
MATRX AI,GEBRA
i 1 +22 = {Tz2where z1 and z2
105
are arry two comPlexsince
numbers.
TL.Jl
t06 COLLEGE LINEAR ALGEBRA
Proof : By definition of the conjugate of a matrix, we have
A =t;ul foralll<i<m, 1<jsnand 2ii isthecomplexconjugate of atj.
atso 1.e- = [ ^;;l=t ^ "11l trtforall I <i<m, I sj<n'
slnceilz'2 =;tr2where \ andz2 are any two complex
numbers.
Again 1 A, = h1l*t.t bl= Ifu forall I si<m' I5j Sn'
=t l4lforall'l<ism, 1<jsn (21
Thus from (1) & (2)' we conclude that the corresponding
entries of l'A and 1 e are equal and they are of same order'
Hence 1.A = )' A .
3.11 Theorers on the conJugate transpose ofa cotnPlex
matrix., Theorem 1. kt A*and B* L,e the conjugate transpose of A
and B respectivelY, then!**(i) (A):A
(ii) (A + B)* = A* + B*, A and B being conforrnable foraddition
(iii) (AB)* = B* An, A and B being conformable formultiPlication.
(iv) (kA)* = [ A*' k being a complex number'
Prqof : (il l.et A*= B, then B = ( Ar) = (A)'
and Br = (t At')'= A
Again (E)t = 1fr = F= e
Therefore, B* = A, since B* = ( E)t = (B')
Hence (C)* = A, since B = A*'
MATRIX ALGEBRA
(ii) BY de{inition' *: h^'" - ".
(A+B)*= 1A+n)'=14+B)'
=(A)r * tB' since(c+P)t=cI+Dir'
= A* + B*, slnce A* = ([)''
(iii) (A8)* = (ABf = (AB)t
= F)t(A)r since (CD)r =ilCI'
= B*A*-
(iv) GA*= o,Af = G[)t =[ (R)" =fA*
Corolla y : For any complex matrix A' (IA)*= IA*
where l, ls a sealar'
9.12 Theorems on sYmmetrlc
matrices.
and shew-sYmmetrlc
r07
be uniquelYand a skew-Theorem 1' Every square matrix can
expressed as the ""* of a symmetric matrix
symmetric matrix'Proof : t et A be a square matrix' and 'S be its transpose'
Then we have
o= i B * f) *; te-'s) =B +c (sav)
where u = i 6 +.tr) and c =| te - A1)
(1)
(2)
Now Br =*jo+ F)r =| o+ * ffi
and cr =i,ro- ,{)r
=ir* *\= n
= j r+ - (s)r)
=jr+-ot= -Lro- ,$) = - c'
r08 COLLEGE LINEAR AI,GEBRA
Thus B is a symmetric matrix and C is a skew-symmetric
matrix. Hence from (1) we conclude that a square matrix can
be expressed as the sum of a symmetric matrix and a skew-
symmetric matrix.To prove the unlqueness of the representation of (l), let if
possibleA=P+O (3)
where P is a sJrmmetric matrix and O ls a skew-symmetric
matrix so that PT =PandOr=-OTtren,S = (P+QIr =Fr+ S=P-O (4)
Adding (3) and (4) we get
A+Ar =2p ...f=2(f +A)
Again subtracting (4) from (3), we get A -,$ =2OI
...8=;(A-s)This establishes the uniqueness of (l).Hence the theorem is Proved.Thcorcm 2. lf A is a square matr'rx, then A + Ar is
symmetric and A - af is skew-s5nnmetric'
Ploof : Eirst PortionIf A is a square matrix, then
6 +.$)T =,$ + (S)T =Ar +A =A+'qrHence by definition A +,S is symmetric'
Second Portion(A - Ar)r = d - (,s)t =.S - (.{r)r =Ar - a = - (A -'S)Hence by delinition A - Af is skew-symmetric'
Theorem 3. If A is a skew-symmetric matrix then
AAr =,SA and A2 is symmetric.
Proof : Flrst PortionSince A is skew-s5rmmetric, Ar = -A
...A,4r=A{-A)=-A2 (i)
andAlA=(-A)A=-A2 (ii)
From (i) and (ii), we get AAr = Ar A.
MIITRIX AI,GEBRA r09
Second Pordon(a.6t;r = (N')r Ar = AS Since (,s)r= A'
and (NA)r = Ar (N)r = 'SA
... Iu\T and .qT A are both symmetric matrices. Therefore,
-A2isasymmetricmatrix.Againsince_lisascalar,A2 is a sYmmetric matrix'
Note : If A is a symmetric matrix' than kA is also a
s5rmmetric matrix'where k is any scalar'
Theorem 4. If A and B are both skew-symmetric matrices
of same order such that AB = BA' then AB is s5rmmetric'
Proof :.If A and B are both skew-symmetric matrices' then
AT = -Aand BT =-Bi'e A=-N andB=-Bf
Also given AB = BA = (-Br) (-AI) = BrAr = (AB)r
"' (AB;r'=66'
Thus AB is a sYmmetric matrix'
Theorems.IfAandBaren-squaresymmetricmatrices,then AB is symmetric if and on{ if A and B commute
(i.e AB = BA).
Proof : Since'A and B are spnmetric' AI =Aand Bir =B'
Now (ABlr = (BA)r = NyriP.'. (AB)r =AB. ./
Thus AB is a sYrlu{retric matrix'
ConverselY , t/= (rrB)'i - IfAr= BA
S;inc'd llr'= IJ a-nc1 Af = i\'
I-lenee A ancl lJ cutnrtrr-tte
Retnrtrk:.[.lrllrttittrixl}1.A1]issYrrriltr:tri<li'ii"Sk(.r1.\j-l]yrtlrTlctri(:irr'cortlirl$il:iAisjsyt::.trtt,:[i:ic1i1.5[,;111.,,-l:i.ytnrrretrit''
tllMATRIX ALGEBRA
tIo3.13 Theorems on Hermitian and skeu'-Hermitlan
matricesTheorem l. In a complex field every square matrix can be
expressed uniquely as the sum of a Herrnitian matrix and a
skew-Hermitian matrix'
Proof : t,et A be a square matrix of order n and A* be the
conjugate transpose of A' Then we can writ€*;=j;.;. L*o-A*)=P+Q(sau (r)
where, =L,W+ A*) and g = j e -a*t
Now P* = i ,o + A*1*= i {t + (e*)*) :
=){a"*A) = P
g* =i,(A - A*)* =f, ra* - (A*)*)
=jta*-nt1*
= -)o- A*) = - o.
Thus P is a l{ermitian and Q is a skew-Hermitiari rnatrix'
Hence from (1) we see that a square matrix A can be expressed
as tlte sum of a Hermitian matrix P and a skew-Hermitian
*"1T $rorr" the uniqueness of. represe.ntalion^ (1.
l."i' o
possible, A be also e"ft"""iUfe in th.e.fbrm A = R + S (2) where
R is Hermitian and Sls skew-Hermitian such that
R*=RandS*=-S'Norv A* = (R + S)* = R* + S*= R- S (3)
(3). we Set R = )W. A*) = P
Subtracting (3) Iiorn (2). lve $et S = )W-A*) = 8'
which establishes the uniqueness of (1)'
Hence the ttreorem is Proved'
Theorem 2. rf A and B T:.T'*"t11 *.T:l"es then
AB + BA is Hermitia" t'la AB - BA is skew-Hermitian'
Proof : Since A and B are Hermitian matrices' 'we have
o.=-o1]ro';{; *i"t" ".
and B* are the complex'coniugate
,r.rr"ro""" of A and ' t"ll.t"tT-"1'r;
Now (AB + BA)" = (AB)'' + (E}.|) -
= B*A* + A*B*
= BA +AB =AB + BA'
Hence nb + BA is a Hermitian matrix'
Again (AB - BA)*= B',I_ f.?-
= tsA-AB
=-(A8-BA)Hence AB - BA is a skew-Hermitian matrix'
Theorem 3' lf A and B are Hermitian' then AB is
. Hermitian if and only if A and I commute'
Proof : Since A and B are Hermitian'
A*= A and B*= B (1) \No*'p1gi* = B*A* = BA = AB' since AB =
BA' \
Thus'by definition' AB is He:mitian' 1
Ag;ain "o""'""'''; = 9u)" = 9*6*= BA
Since A and B ale Hermitian
H::l#t?:'ffi$"square matrix can be uniquerv
expressed as P + i8 *t'ol p and g are Hermitian' '
Proof : kt A be the square matrix of order n and A* be its
conjugate transPose'Then we can write * 1 - {.,
;=i;.^.' ;;;; -a*)=.1>(a+e*) *i le-a*t *
= I) + iQ (savt (r)"wrrere P = |fo *o*t and Q = | (A - a-)
Now P* =lrrn* A*)* = | {e* * (A*)*}
COLLEGE LINEAR ALGEBRA
tL2 COLLEGE LINEAR ALGEBRA
=;d+A)=in*A*) =P
n. = L\ 1o - o*t)*=# 16* - 1a*)*l
=-fita* -A=fio-A*)=o
Therefore, P and Q are Hermltian matrices. Thus from (1)'
we see that A can be expressed in the form P + rO where P and Q
are Hermitian matrices.To prove the uniqueness of the representation (1) let if
possible A = R + i S (2) where R and S are Hermitian' that is'-
R* = R and S*= S.
NowA* = (R+ i S)* = R* -iS* = R-iS (3)
(z) + (3) gives A + A*= 2RI
.'. R=; (A + A*) = P
(2)- (3) gives A - A* = 2i S
...s=*1e-A*)=owhichestablishestheuniquenessof(t).Hencethe
theorem is Proved.Remarks : (f) If A is any n-sguare complex matrix then
A + A* is Hermil"ian and A - Al is skew-Hermitian'
(2) If Ais any n-square complex matrix then AA* and A*A
are both Herrnilian.(3) If A is l-lermitian matrix, ttaen i A is a skew-Hermitian
matrix.(4) If A is a skew-I{ermitian matrix' t-hcn i A is a
Herrnitian malI I-\.
(5)Aisl-lcr.rrtil'iariiIirrtr]oniyif.Ai.stlerrnitiirrr'
t6) .A is sltrw-i'lct'tltiiialr if :r'nii ';tlh' ii A lSr sl'ir'"iv
Ilerrtril i;tn.
ll3MAf,RD( AI,GEBRA
O,ta Scorrammi'lqofg't'otrnrtrlcca r"rii'r'i
tlicarcm f : f iJ e *" idempotent rnatrlces ttrpn AB is
ra"mit rrt 663 = BA'
Proo(: Slrnce A; B are idernpotent matrlces
'. A2 =Aand BF = B, Also gPen 63 = BA'
rro* gos;i = AB BB) = A (BA) B = AjaE) B
= [A'A] (B'B] = tP * = ltB' ! '
Hencc AB is an ldemPotent matrix'
trkor@ g : If a;tB are idempotent matrices' then A + B
*iU il Ue*eot"nt tf and only if AB = BA = O'
P.imf I Sirtee A and B areidunpotent matrlces
"/,}=AardFF=B' :
Nor O* = ;; = 0' ttren (A + B)2= 6 + B) 6 + Bl
:X:#J#ii,;il=o= A + E slnce A? ='&'E?'=B'
HenceA+BisanidemPotentmatrix'' :' :'
Agairr if [A + B) ts an idempotgnt matrix' then: " '
[A+BP-A+E '; i:';': 'r]
or, A2 +AB+BA+BP'=A+8" I '
or, A+AB + EiA+ B =A+ B
or. AB+.84r-0" - :'' '' 'r ' :"-'r:r':' lii r:'
or, AB=-AA (11 . -'^.'' ^"0'b-.44-'*aOu'6FqA'=-(ABIA
"'':.-.t.t;FAts 7,EA?'t-"Be ' !' r
Addkls f rf Jtzl' we get aa + AB; - BA + BA: o ' I
i. e. 2AB*o, .:.aP,t.o+-BA.'.- .:' i_ -^* .,
Hence A + B'is'ide.inpotent if and only lf AB = o = EiA'
Theorem 3. If A is an iclempotent-rnatrix,, therr 13 = I *A is
idemPotentandAB.=O=.E$'' -'' :'
l:\1 coLlFrQPrr,rN.p4rrncEBRA
hoof : We know.t*ratan = lA = A
:'Al$o:since. A is idempotent, A.2 = AAgain since I and A are square matrices, so I - A is also a
squzrre matrix.Now ( l- A)? = (I. A1(I-A) = I _ tA_AI +A2
=I-A_A+A=tr_A.'So by definition I-A is an ldempotent matrix.
Again AE! = A (I-A) = AI - A2 =A-A = O,'. AEI = O
BA = (I -A) A = [A - A2 =A-A= O
... EIA = O.
Theorem 4.If A and B are squ.rre matrices of order n suchthat.AE! = A arld BA = B, then A and B are idempotent.
Proof c ABA = (AB) A = AA = A2 (l) since AB = AAlSo AEA = A (BAl = AEl.= A (2) since BA = B & AEt = A.
, F.tom (I) & (2), we get A2 = fi ... A is idempotent.Similarly, we have , . :
BAB = B tAB) = EIA= B (S) strrqeAB =A & EI{ = B. AIso BAE| = [BA)B = BB = 82 {4) slnce EIA * BFtom (3) and (4) we get B? =B.
... B is idempotent. ,, l
&16. Silngular end non&3iular ntrdGor ' - .
I,et D be the determrnant of the sqtlare matr!( A, then ifD = O, the matrix A {qcalled,the sl4gulif nirtrlr,and tf D * O,the.matrtx A ts call,ed the loa-flng& rnetrt.
As for e<amples
^=I; 11,"=l_i i;]* .=li 2 3']5 6l8 eJ
are singular matrices,
sinceDl = h[=O,D2= lB l=OanODs= lC l=O. :
MATRIX ALGEBRA
12r3l134-11Asain
" = Ll _; ?]' " =Ll _? ?l
are non-singular matrlces']ri;;' = tit=-6+oandD2= Bl=18*o'3.16 Inversematrlx i I ,' ' ' ''A square matrix A is said io be lnvertible if there exists a
unique matrix B such that AB = BA = I where I is the unit
matrix. We call such a matrix B the lnverse of A & is generally
;i|"'u o, ot Here we have to note that if B ts the inverse of
A. then A is the inverse of B'
*,rampre 1. I€t " = ta ll * 'u
= [-3 -]l
*."*= [3 ll I-3 -"1
=[3-3 -?r**'nl=[l ?I='uo= [-3
tll E ]l= [-3;3 3-.11 =[i ?l =i
Therefore, A and g are tnvertble and are inverses of each
other. That is, .[' = B ard d =[: ',o
r'-o'ndcr.,.,o=[i ? -il"u'=f-'3 i i]
i
^-Loi aJ- L 5-2 2J
r15
12 O -lll- 3 -l llp""*=l3 I gJ[,9 _; -Z)
r 6+ O- 5 -2+A+2 2+O,-2-l=l rs-rs* o -5+6+o 3:3131L o-15+15 o+6-6rl o o1
=l o I ol=I. Lo o rJ13 -l I 1r2 O -ll
=L-!u -8 -z llr I 3l
116 COLLEGE LINEARAITEBRA
F 6:5+O O-l+l -3+O+ 3l flOOl-l-so+3o+o o+6-5 l5+o-15 l=lo I o l=IL ro-ro+o o-2+2 -b+o+ oJ Loo tJ
Therefor.e. A and B are invertible and are inverses of each
other. That is, .[l = B and d = A.
3.f76dJoht or adJugPtc natrlrf-arr arz ah I
, I azr azz 42n I
Let A=l .-. :.. ... ... II ... t
L"", q, a." J
Irt4{ $1 ,,?,$ .::, nandJ = l:2,3: ...,rt '
Forrr .the matrix [&11. Ttren the,trylspose of the matrix
I&l k called the adJolnt or the tgjugat! nat;k of the matrixAarrdb{Bnenallydenotedby,AS,A., ' , . t. .:- i.., .::
[-Arr Arz Ar' l- l'Arr Azt A"r -l
I Az, Azz Az,,' l' I Arz l,1zz Anz I
I "' :" : I :' '1 "'' "' I
tA"r. loz ,-. Aoo J ', Lt$,+,, &, An,, J
L€t D be the deterrninBnt of the matrlxA,a,i l"' r ]atz "' ral" I 'lazr 4zz . ..i azn I
then D= la l= 1... ..f I ,l ,
l;t * ,"' t """ I
AS fOf e1alnDleg ,; :i r ;,i I t'. '; , i i ' "''; '. "
firi.gf r 7.;r ';;.-,i
Let R=l I 3 -l luren i 1j '' iLz I oJ iai !'' i "''r l _2 _E.rr r l . & ':_i, i ,,,''
AdiA=l 3 -o sl=['.2,'-i +l; :" L-rl 4 lJ L-s fl : lJ r, ,'1i
' r it
I-ar r iltz aln
- la" a22 azn
I-etthematrix A=1...Lt'' z1e "' Bnn
I-et D be the determinant of the matrix A' Evaluate the
determinant p; if D = 0' thi roatrix Atg atngElsf, and:it fuas no
inverse, if D * O the matrix'A ls noa-eingufar and A-l erdsts'
Find the adjolnt matrix AdJ A of ttre matrix A then
r AdiA'a t =DAdjA=-m
ArtEIAnET
ArtEI
Theo,r.n I : If A-l is the inverse of the n-squarex$atr,( A'
then A^FI = A-rA = I' where I ts the unit matfix of the ""*order.
,'. rfl =
AztEIAzzEl .:AznE]
3.19 Theorems on lnverse matrlx
rt7
squarc matdxMATRIX N.GEBRA
3.18 Irnoccc! offrdng t.hc lnverrce of a
'D is
the
I .'
,, . ,,,r'1.
then,A;-r = ff'*1;(D * O) and Ag are
AntEIAnzEI
t,2,...,nmatrix A
Hl;;. l
Azr
!:'Azrt
I"'i t&l
]H:
Eoof : lrt A = Iarl i'J =
the determinant of the
co-factors of D' rlAA-r = Iaylxo I&l = D
[-ar r ate zln
1I "' ?22 azn
=Dl "'I ...
L"r, a,,2 ann
ll8 COLLEGE LINEAR AI]GEBRA
Similarly, we can show that A-I A = IThusAA_r =A_l A=I.Theorem 2. If A and B arenon-slngular matrices,(AB)-t =B:r A-1. Also O-,f1 =Aand (A-r)r = (.,Sft.
hoof : Since, (AB) 1B-t A-r) =A(BB-t) A-r = AIA-r- AA-r =Iand (B-rA-r) (AB) - Br (A-I A) B = Et-rIB = FrB = L
Thus E)-t A-l is the Arverse of AB i.e (AB)-r = B-r A-lIf A and B are two matrices such that AB = I, then A = B-l
and B =A-1.Therefore, A = B-l = (A-l)-r ... A = 01;l)-1.
Ar (A-r)r = 0rrA)r = (Ilr = I shours that pt-t;r = (AI)-r.
Corotlary : If A1 P*r, ....A., are non-slngular matrlces of the
s.rme order, their product is non-stngular and(ArAz .... &) -t = &-l .... Ar-l 6r-t.
Theorem 3. A matrix has an inverse if and only if it isnon-slngular. A non-slngular matrlx has only one lnverse.
Proof : Suppose that A is of the type (m, n) and that m <n. IfA has an lnverse A-r the products AA-l and A-t A are bothdefined and hence A-r must be of the form (n, m). It follows
that A-l A = Ir,. IfA has rank r, tlten we have r < m < n s r, since
tl:e rank of I, is n and the rank of the product of two matrlces
ls less than or equal to the rank of either factor. Hence m = n
= r dDd A is non-singular. If n <m the result follows sirpilarly,
using AA-l = Im.
Now suppose that 6 = [arJl is non-singular
o o olI O Ot
l=rIo o ll
fi$
,i,I
I{i
r19MAi!'EIX AICEBRA
Since lA l+ O, we can let b11 =ArrEI
(i, J = t, 2' ..., n), where &J are the co-factors of ag in A'
Irt B be the matrix [bryI. Then
*=[i=, ",- ",]= [a, ""ry]=[+r]=,
simirarry, ,o = [i= t h* *]=[L, ffi* ]=,
'
Hence B is an tnverse of A and every non-singular matrixiA has an inverse.
Finally, we have to show that the inverse of a non-
singular matrix is unlque. Suppose that A is non-singular and
A-l-is its inverse. If EIis another inverses of A, then we have
AEI = I Multiply each side of this equation on th9 left by A-t''
then we O-t 6[i) = A-l I =,fl.or, 1e-rd B =A-lor, IB = A-l
or, B =A-l , l:
Hence A has onlY one inverse A l ,
[The rank of a matrix A is the maximum number of
linearly independent rour (or column) vectors of A'l
8.2O Ttcerclrc on otthogoral patrlccs'
Theorem l. If A and B are orthogonal matrices' each of
order n then the matrices AB and BA are also orthqgonql.
Proof : Since A and B are n-roived orthogonal *"ll"-ti-'
The matrix product AB is also a square matrtx o{ o,,i,,dSr naud{AE)r(AB) = (Br S) GB).
=ntr (l\relBBir I,rB
r2O COLLEGE LINEARAI-GEBRA
Itrus AB is an orthogonal matrix of order n.Similarly, BA {BAir = (BA) (ArBr) = B (A.{r)Br
= BITTBT = BE|I = Ir,.
Hence BA is an orthogonal matrix of order n.
Theorcm 2. $ Ais an orthogonal matrtx, then ^[l is uhoorthogonal.
Proof : If A ls orthogonal. we have
fu{r - Ar A = I where I is the identity matrlx.or, (AAr) l= (l{ta) '= ['- t
or, (At) I d 5 .a' larf'=1-l -l -r -ror, ( allr A'= A'( C')t =I,sirrce (.{Tf'=6-'f
Hence A-I is orthogonal by deftnition. That is, inverse ofan orthogonal matrix is also orthogonal.
Theorem 3. Transpose of an orthogonal matrix is alsoorthogonal.
Proof : Here we have to show that if A is an orthogonalmatrix, then Ar is also orthogonal. By deflnition if A ts angrthogonal matrix, then we have AAr =Ar A= I.
or, (a6rF=(AfA)r=F=Ior. (Arfr .S =,S (AT = IHence AT ls orthogonal by deffnition.
i' Redart : For any square matrix A, we have if ArS = I, therrArrr= I.: $lgt Theorcms oa rmttarymatrlccs - . i
Theorem f . If A and B are unitar5r matrices then AB is alsoa unitary matrrx.
Proof:IfAandwe have
AA
and BB*
B are unitary matrices, then by definition
MATRIX AI,GEBRA
Now (AB) (ABlt= (Rg)'(n*a?)r '
= A (BB*) Ar
t2l
= AIA*
-AAt=lSimilarlY, we have
.^s)*!q,):::l'ffi
=BtB=IHence AB is a unitarY matrlx-
Theorcm 2. IfA ts an unitary matrtx' tfren .il ls also unitary '
Proof : By dellnttion if A is an unitary matrix, then we
have AA* = A*A = I.
1ee+;-t=(A*A)-l= i'=Ior, (e*l-'A-r =A-r n*il= I '
or, ( C') * .Cl = A-' (C')*= I Since g1*f' = (A-ll*
.'. C' i" an unitary matrix by delinition
Theoreio 3. Tlanspose of an unitary matrix is also
unitary.Proof : By delinition if A is an unitary matrix, then
AA* .=, el* =.t where I is the unlt'matrlx.i :' " : ': ' :
(aa+;t=1A*6;r=F=I
or, (A*)r At =.$ {A*)'=I
or, 1ar1t6r =4t {Ar)t = I Since (A*)r= (4,}*
Hence by delinition.S is an unitary matrix'
nemart : For any square matrix A"if AAf= I, then A*A= I'
-t{d.1s
*=A*A=I
=BnB=I
(U
121 1
L22 CoLLEGE LINEAR AI.GEBRA
3.22 Tlreorems on lnvolutory matrL.Theorem 1. A matrix A is iavolutory if and only if
(I -A) (I +A) = Q.
Proof : Let (I -A) (I + A) = O + I -A2 =O
=+A2 =1.
=+ A is involutoryConversely, L€t A be involutory. Then by definition,
we have A2 =l.'. (I -A) (I + A) =l - A2 = I - I = O.
' Hence the theorem is proved.
Theorem 2. lf A is an involutory matrix,II
Then i 0 + A) urd; (t - A) are idempotent.
Pnoor, G ,, + a)l'z =|a, + 2A + N)I
=fi (l + 2,4. + I), since ,{2 = I & P = III
= j 0 +A). Thus2 tt + A) is idempotent.
Again {} u - o,}'=i (P - 2A + NlI
=;(l -2A + I) since A2 = I & P = Ill
=i0-A).Thusi 0-A) is idempotent.
Hence the theorem is proved.
3.23 Solutton of ilndar equations by applldng matrices(i) When the linear system has n linear equations in n
unknowns
MATRIX ALGEBRA 123
Suppose that
[",, a,rz ar' I [",] [l,lR=l
azr d2z 1* l, *= l:: lr"al= l'i l
L;;, ;* ::: ;;" I L*"J Li"l
Then equation no (2) reduces to A)( = L (3)
L,et D. be the determinant of the matrix A. Evaluate the
determtnant, if D = O, A is slagular, so A-l does not odst and
hence ttre system has no solution. If D * O, A is non-slngular
So A-l exists and hence the system has a solution' Now
multiply both sides of (3) by A-1, then we have
A-IAX = A-lLfsince A-IA = I
or, IX-A-r L tX=Xor, X = A-lL
Ar r AzrEI 8I...Arz AzzE] -[al "'
At., AznEI EI..
Then xl = (rll , xz =trr2...., 4n = mn is a solution of the given
system of n linear equations.
{t is to be noted that the solution of the system of
equations can also be found by reducing the augmented
matrix of the given system to reduced row echelon form'
matrix-form as
I-arr a.tz aln I fxl| ^r, dzz a2n I I x2
L;;, ;*:: ;"" 1 L*(2)
It]
That is,
X1
X2
xn
4rtlAlAnzEI
IIll
t}J2
ffin
lr
12
1,,
(say)
AertEI
(1)
=lr I=tz L
=rl
+ Om)Qr* d21,X1,
. * arrrr.lc.,
++
+
* O.px2+ azzh
+ d1pX2
txtrxi
rxr
A1
2.2
an
l,€t
The above linear equations can be written in
MATRD( AI,GEBRA l2s
I24 COLLEGE UNEAR AI,GEBRA
(tt) Itrhcn the cyster has m linear equatlons in' n
rrrtnotmga[dm<D"constder the followlng system of m linear equatlons ln n
unknowns xt, h, ...,4tO11X1 * 312X2 + ... + arrr{n = ll lA21X1 +a2fc2+...*a26X,.=t2 l tfta-rxr i'^r*a"*... i"-r'q" = l" J
inwtrich m<l and fu afrd [ ' i =1, 2, "',m,j = 1,2,.'., r\ are constants (scalars)'
The gtven system (l) can be written in matrix-form as
[a,, a'tz ar" I lxrl [], I1", azz d2n
I lI' l= [ ? I t2l
L;;, ; :: ;;" J L,"J Lu,JNow the augmented matrix of the system is
sothatA Iil=,L*J
Multtplylng both stdes of thts equation by 6-t on the left
u,esetAr ^[i]=*'o
"" Ii ]= *'o srnce FrA = I'
Herrcex=A-Ibistheuniquesolutlonofthegtvensystemoflinear equatlons Ax = b.
-/ Worted out eramliles.fuilct.G1!*, -il *r=E 'r-ilEtnd the matrices 2A. A + B and A* B' r ..''.. '''"
,,;;=-[l i tl *"=fi i-rl^-I3:.!, 3:1 ",,T,l - [ fr i -Zl'
i ' . ,',i{
A+B= Il -? -fl.fi i il= [ll i*r'];',*,'.t-i,l . [3
4ln : lrl
"^ ,-l? I..-I
arr,, : l-JWeshallapplyt}teelementaryrowtransformationson
the above augmented matrtx to reduce it to the redueed row
ccheloa fom.The reduced form of the augmented matrix will either give
a solution of the given system or will indicate that the system
is inconsistent.Theorem 8.12 If A is an invertible n x.n matrix, the system
of linear equations Ax = b has a unique solution given by
x= A-lb.Proof : SinceA (A-tb) = (Au{-t)b=Ib=b,
,
.'. x=^&lb is a solution. i
" f L-z 3l ,f-2 -.3 -5Ie-B-LE l-1Jl[-,1 -4 2l
I5
r, t ,,
- =l f,rl
rc " *t,,uor,,
L*J
= 1i..[3ir?l iilfil';t'l *fl:'3 - f,], .'"''
*"-* z.rrtar [l 8l a/rirE* t; ?l ,, , ,, .,Compute the rnatfipt n""te.es3n4 BA" ,:; ' : ;Conversely, suPPose
126 COLLEGE LINEAR AICEBRA
sorution,*= [; 3] I; ?J =[3]?r%..%]=[?. 3J
uo= [; ?] t6 3l trtS Slgl = [i g]
Sowe see thatAB * BA
*^rgfi*=[;3 3]1Calculate tl:e product AB.
*= [; |'Jlii _+
3-'lslsJ
-l-2
r-2 2; -3-la.ra=12 I -61 (
and B =li Z -',
?llz r -r s.l
t0+9 I15+ 15 Ji
4t7
-4
_Jr+z+o 2+6+S- 12+ 3+ lO 4+9+5
t9 n o 2zl .=lrs t8 -t 261It is to be noted rhat the pr@uct BA ig not delined.
+4-3 3++6-5 6+
' ,,.,ii
ProvethatAs +N41A -54I=O ,
where I is the tdfnUty matrix of brder 3 x 3.r-2 2 -31
Pr,oof:A=l 2 I -61L-r -2 oJ
r-2 2 -sl r-2 2 -3162=al{=l 2 I -61 I 2 I -61L-r -2 oJ, L-r -2 oJf 4+4+3 4+2+6 6-12+O I f lf
-14+2+6 4+t+t2 -6-6+0 l=l 4I z-++o -2-2+o B+ 12+o J L-z
r lI 4 -61r-2 2 13']6o=62f,=l 4 tz -tzl| 2 I -6I
L-2 -4 ts J L-l -2 oJ7-22+8+6 22+4+12 -33-%l +O I
=f -8+Stl+12 8+17+24 *12-tO2+O I
L +-8-15 -4-4-go 6+24 +o J
-6:'l-r2ll5J
MATRX AI,GEBRA 127
r-8 38 -57 1
=l 38 4s -1141[-rs - ss so J
.'. A3 + N -21A- 451
r-8 38 -57 1 rll 4 -61=l B8 4s -rr4 l+l 4 tz -nlL-le - 38 30 J L-2 -4 15 J
r-2 2 -3-'l 11 0 0-r
-2t12 l -6l-+slo r olL-t -z o_J Lo o rJ
r-8 38 -57 a r ll 4 -61=l s8 4s -l ra l+l 4 n -nlL-rg - ss 30 J l-z -4 rs Jr -42 42 -63 -'l f 45 0 01
-l 42 2t -1261-l o 4s o I
L-zt 42 o I Lo o 4s)
[-8+ll+42-45 38+4-42-O -574+ 63-O I=l 38+4 -42-O 49+ 17 -21-45 -ll4- 12+ 126-O I
L-rg-2+2t-o -88-4+424 so+ts-o-4s J
rO O O-l
=l O O 0 l= O. HenceAa + A2 -21A-451=O.LoooJ
"=l:,,1 i]* "=l-ithen prove that (ABP =E[rAr.
f l2 3'r rl-221*:Gtuen
"=L-r, s ?J,. *=13 _? tlr-l 5 31 t-=l 7
GtuenB=l 7 -2 tl....gr=l 5 -z. Lz o-sJ Ls I
r-l 7 21 rl -2 21NowBrAr=l 5 -2 ollz b sl
L3 r -3J13 -l 4J
5 3']-2 1lo -31
2'l
sl
129MATRIX AI,GEBRA
[s?I "{iqilSotutlo I GtrrcnA=
4il
:;l
Lil{
3 3l1 llI -5 I
-5 1-l
The sYmmetric Part of o = l' 14 + '{)
=*[[l ?il.li q?]
=*[iii ?i?i?l =+Vt fl=L;
Ttre skevr-symmetrtc part of A = | A- St
4
fl[i qil_lrtL ?-z1Le s
=;[l - ii -2, i-|.=;[-? i -]l=
P::",;1""Ti*'n*trix"=ruli i j ilr"o'tr'oso""lLs 1-b
Proor:G'en^=*11 i j ilnr=nli i j i]
n'we*=* [i'i j i] [l iUnear Atspbra'-g
UNEAR AI.GEBRA
(-1)'F2l +7'5 +2'l-Ll5. 1-21+ F21.5 + o '(-U3. {-2) + 1'5 + t-31 '{-1)
I28 COLLEGE
fFl)'r+7'2+2'3=l s.t +(-21'2+o3Ls.t*1'2 +(-3)'3
(-1)'2 + 7'3 +2'4 I5.2+F2l'3+04 I
3'2+t's+(-.3}'al-2 +21+8 11o-6+O I
6+3-12 I2+#-2
-lO- lO+O-6+5+3
5-4 +O
'lO- IO+01O-6+O
1.5+2'F-2)+3'O(_2).5+E.t-2)+Fl).p2:S+S.F2)+4.0
(u
r-1+14+6=l s-4+oI s*z-srtg 35 271
=l r -2o +lL+ z -sl
r-1+14+6-l z+BS-2[-e+zt *sr19 I
=l ss -zoLn4
5 3'l-z tlo -sl
2 3'r r-1s-rllzs al L z3.27 + l-Ll: 24.2
rl' {-l) + 2' 7 +
=l {-zl ' (-1) + 5'L2' (-l) + 3' 7 +
r.3+2'l+3'(-9) I(-2)' 3 + 5' I + (-l)' (-3)
|
2.3+3'l+4't-31 J3+2-9 1
-6+5+8 | '' i"'' 6+3 -12 J
Asain* =[-i
-4 Izl-sl i
130 COLI,EGE IJNEAR AI'EBRA
.'. AAr = I.
MATRD( ALGEBRA 131
. [9+9+f+9 9-15+3+3 9+3+g-15 9+3_15+3r_ I I9:I5+.3+39+25+I+t 9-5+t- 5 9_b_5 + I I-3619*3+3-159-5+l-5 9+t+t+25 g +I-b-S IL9+3-15+39-5-5+l 9+l-S- b 9+t+25+l )
2-31 3+5i I3 il-i 5 -l
l-360 0 0tr lo s6 o o I-3610 0 36 0 I
Lo o o s6lr1 0 0,otlo l o ol=lo o l'ol=ILo o, o- rJ
ls Hermitlan.12 2+3i 3-5il
pnoof: [=lZ-Si 3 fi I
[3+5i J
*,. AFi,.i :
12 2-3i 3+5i1Ao==l 2+3i 3 i l=.S .. :j
Ls-si -i 5 I
similarry, "'" =* [j
, [l=lo
Lo
3 s 3l[3 3 3-5 1 rlls-5 I1 l-5 ll,s I II -5 rlls r -5
?r?l
'-'. A* :'A. Herice A is Hermitlan'
Exampti io.'prov6 ueii'tt" *it iiA = # [:, , il
proof : Given ^= i [], -l]
o*= t [1, _l]"
No* aet =,1 [],, _ll, [],
=+ [-il i i;i] =;Y, gl =
Similarly. A*A = I.
Thus AA*.=AilA=I : 'i:
"1
Hence A is'an unitary niatrx' i:
Exampleff.Showthatthernatrx4=l 3 5 O I""--'*-- '' L 7"'2''j'lil"'
, :r ' . "" inwlutol-v" t '
[-5 -B 0l f-5 -8 olFroof .A2=axA=i 3 5 o l"l 3"5 o
I'rvvr Ll 2 -,tl Lt 2 -tJ i
o o otl o olo I o l=I.o o rJ
. is unit4ry.
' g,u.Ltrl
.Theiefore, aalgra=:.Hence A is orthoponal. :iI
[3 ?l= it *r"B.rrA= [],.r, '.iJ ,r,as =l?;t tIthen prove that e+B =a.+ g.
Pr,oor: I= [l * r, 'lJ and e= [?;1, -;J
a*e=[3]ri l:'il i,, (1)
AsairrA+B = [3;rl li'lJ,A*rr = [3 ] 4, i:
riJ et
Thus from,{1) and (2), we;get i
-.,\+R = A+B.
COLLEGE LINEAR AICEBRA l-rA21 =(-1)l 3
lz
^r. = t-I) [a
lze.s = Fl) la
r 3-11 L2
rherefore, Adj A = l'l - r7
-'^L
rao,plg/G. Frnd the lnverse of the matrix "
= [? 3l
solutron'I^'r=[? 3 i 3 ? l::::1"]ffS:'*rstand
- Il 3 i ! I I H:r*l'E'J"ii:H"H'3{""#j"*
tl 3 i o I I X,".H:X'lI,i'#&u.il?"i'lJ'?'-'*[o-t i1 -2
- tl -? ',
"' -Zlw; *"*nlv second row 'bv
(-1)'
-te?: =g'A-rr
Hence A is invertible and o'' = [-? -?l
1MATRD( AI,GEBRA
lp?
182 1\ =,,' *=\'rB\ =-u'
-l\ =-n,o., = \-l -',\ = ''
-3r\ = ,n,o*= \? -l\=n'
r25 -24+O 4O-4O+O O-+Q+Q1
=l-i; *ii*o -24+25+o Q+Q+o I=L-i"*-4"-'i -e*ro -2 o+o+lJrt O 01
=LB I ?l=' A2=I
Hence the given matrix A is involutory'
_@r] r, . rrynd the lnverse of the rnatrrx" = u ?l
^ Slolutlon : Irt D be the determinant of the matrDq then
I r=
I ] ?l = 4 - 6 =' 2 *o' so the matrtxA is non-sin$ular and
hence ,t-l dsts'Nourthe cofactors of D are
Atr=4, A12=-3I9t=-2, t"zz= L
rhenAdJ o=I-t -?f = [-3 -?l
;. x, ={eay,=} Li -11=i ffi -',1
*-fv/6(b)' zurd *".i;*T ".T rnatrix
a=14 o -t l.Ls 3 2)
e-'=*AdjA=*,Fie ld
Tfr3 11 tlI =l-tt -5 141
I Ltz-e 4J
1-1141.+)
Eolutlnn I Ict D be the determlnant of the matrlx;
lz -I 3 l =2(O+3)+ 1(8+3)+3(12-O)tfrerrp = 14 O -t I'"*'"- 16 3 i [=o+11+!fi=s3*o''lio the matrin A ls non-singular and A-l eJdsts' Now tlre
ohctors orD are ^,,
= i3 -11 =t'
.i|,r!li.t.'
la- t4,, = (-U l3 2
l+ ol
T rr.e'. = l6 5l = tz'
-; IV{ATRIX ALCEARA r35134 COLLEGE LINEAR ALGEBRA
[-r 2 -3rElampler+.tfR=l 2 t O lanal+ -2 sl
f2 I -l'tn=l O 2 t I nrra a-,n. p.U. p. rgg5l
Ls 2 -slsolutron: Giveno= [-i i -B
Il+ -2 s_l
l_r 2 -sl=- 1(5+0)-2(ro-o)- g(4-41rrtD=hl=l 2 I oi
l4 -2 5l=-5-2o+24=-l*oSo A is non-singular and hence A-l odsts.
Cof;actorsof- I =or, = l-; g
2=A,=Fl) l? 3l=-rolz rl
-3=e13=14 _tl =-t
2 = Azr= Fttl-|l-r -at=Pw=l 4 5
" O=A23=pl)
lz* 4=Asr = I I
l_r- -2=4o=(-l) | 2I
r I r5 4 3l r-5 4 -31e-, =* Adr A = i.=L_lo '" -g l= L lt i g l
r-5 4 -31f2 1 -llrhus A-,u=L
3o i g lL8 3 _i J
r-IO+O-15 - 5 + 8 - 6 5 +4+ 9 I=l 2o+o+3o 10-14+ 12 -10-7 - LP
I
L to* o+25 8- 12+ 1o -8-6- 15 Ir-25 -3 18 I
=l so 8 -sE I.
L+r 6 -2e)exampte 15. Ft;d tne invS;7f the following matrix by
using row canonlcal form' o =l i i -ll t
lrvrrrr "-L? u -+)Solution:
13 4 -1 i 1 9 9ltnte.ctangefirstandrer,r=[r 3 :, 3 I !]#;;aro#s.
rl O 3 : O I OlWemultiplyfrstrow-by3.3nd-lA 4 -i ; I o o lz"nathensubtractfromthe
l; ; i : o o i lsecona and third rows respectlvely'
. . "fl O 3 : O I 9 l'srUtr"ctthird rowfrom' -Lg 3:13 i L -3 ?1,;;;;;r;;; -
. rl O 3 iO r Ol- I o -i o i r -t -r lnautuplv the second row bv (-t)'
Lo E-loio-2 IJTl o 3 : o I 9lw"*uliiptvsecondrowbv5and
-Lgl_?,;-J-;11,ffi'ffiHJ,il;;'il,iiii,i.*r
- l-l ! 3 : -i i ? lwe murtiplv *rird row t, (- +r)Loo-toi 5-7-4)L
=$
-31=-n
=!2
-2
=$
-3q
-8-Tr r5 4 3lol=l-ro z -ol-sl L-e 6 -sJ.*ro=[-?
-14 =6
-3o
=-6
" b=Agg=
-107
-6
?l =-u-l2
tg7MATRD( AICEBRACOLLEGE LINEAR AI.GEBRA136
rI -1 O O i I o O Ollo g o o:-t I ool-lo o | -2 i 4 2l olLo o -2 I : -5 -3 o lJ
We multtply second rowtY In-roo:too"l
_lo r o o'-* + o ollo o t-2:4 2 t olL" o -z l:-s-B o r_l
We add second row with the llrst row.
f, o o o: 3 + o rl_lo r o o'_+*ool
lo o t-2i 4 2 t ol[_o o -2 1:-s-B o U
We mulflply third row by 2 and then add with the fourth row.
[, o o o, 3 ] o rl-lo I o o,_++ool
lo o r-zi 4 21olfooo-s:st21_l
We:nultiply fourth row by (-'JT, o o o, Z + o ol
_lo I o o,_++oollo o I -2 | 4 2 r oIlo o o ,,-i-+-3-il
We multiply fourth row by 2 and then add with the third ror'
t-'o o'3+o I i:,, .,'' !'Y
lo r o o'-|+ o ol \-'-lo o t o,i t-i--31 =rl+A-rl
Lo o o r:-r-*-3-r"l \
[t :i:.,i;l] [i :i:jhl]We multiply third row by 3 and then subtract from the
first row.
[roo' B+-g I-Lsi?:-i ki .1=bA-'t
f l-r O O:lOOOlsoru,on,rar.l=l-l 3 ? -3 ; 3; ? 3 I
Le t-2 l:ooolJ
fl -1 O O : I O 0 Ollo s o o:-lrool-lo -6 | -2: 6 o r olLo e-2 li-8oorl
f s-tt 9fHence A is invertibre and *, =l -i i i IL-z lo 5JExample 16. Find the lnverse of the matrix
otO IUV ustng only row transformations
-2 lto reduceA to I.rJ
rl-l olr 2 oA=l-o o lLg t-z
We subtract flrst row from sOcond row. We multiply ftrstrow by 6 and then add wtth the thiid row. Also we multiplyfirst row by 8 and then subtract from the fourth row.
We multiply second row by 2 and add with the third row.
We also multiply second row by 3 and then subtract from thefpurth row
MfiTRIX AI,GEBRA 139
T38 COLLEGE LII{EARAI'GEBRA
Hence A is invertible and
f z l o rl135 II r 1 o oll-5 T ' ,l
^_,=l z ,f_r-TlL-l-5;E-Ei
Eramp!9l1?' Solve the following linear equations with the
herp of matrice" i:;J: ll (1) i,Solutlon : Flrat Proccss t tt" "y"t"m
of linear equations
can be urritten in matrix -form as
t? lllXI=El Qt
tz' l.*= $l,"=El,*"'.t.*rcte= Lr _z
(2)wegetN{=L B)
I-€t D be the determinant of the matrlxA' then
lz r l-- 4-r=-5*o.,=1, _zl=_*-L--vrv. ^_, ^-r-+a
So the matrix A is non-slngular and hence A-l etdsts'
Now the cofactors of D are'
A.11 =-2, A12=- I
A91 =- 1, Ar2=2'
rhererore, Adr A:If ,r]';ttr-,,1
*o
6r =fi aa1 n = - | [:? -ll =Li i,]
We multiply botl: stdes of equation (S) by A-t'
A-rA)K= A-rL
or, IX = A-r L Jstnce A-lA = Ior.X=A-rL tand D(=X
rhusX=[i t] EI =[i ;]= t lt
*, [;l = Lil HenG ;= -l]Sccondurrcesg
, fA. ""!*."iia matrix of the given system of llneari;i/ations is
', i'" tALl= [? :, : [] Ir,"*hange flrst and second rouls.
t t -2 : 3I We multlply flrst row by 2 and then- lZ I : lJ subtract from the second row'
- 1|, -u
-?l ." muftiplysecond -*uY |.
-ll -i i-ilNow the system is in row canonical form'
Then formtng llnear system we have r = - I and x- 2y = 3'
.'.x=2!+3=-2+3= 1.
Thus the requirgd soluflon of the system is x =1 and y = - 1'
Erampt4^/**. the following linear equations with the
herp ormatrrce" , ?Xi 7- ,fr: :E I D.u. p. rss{2x+fu -Sr=2O )
Eolutlon I Ftrst Procerg : The given linear equatlons can
be written in matrix-forrn as
lzi -,i\[i]{;al u,,
14O COLIEGE LINEARAI,CEBRA
suppose,*,^=[! 3 _rl" = [i]*,r"=[;A ]then the equation gven by (I) reduces to A)( = L (2)
Let D be the determinant of the matrix A' then
ls 5 - 7 I =s(-s + lo8)- s(- tz +24r-z(s6-2rp= 14 t -t2l=irs_60_ 2#=tZ*o.lz 9 - 3l-ote-w--w
So the matrix A ls non-singular and hence A-l odsts-
We multply both sides of equation no (2) by A-t on the left,
ThenweBetA-t A)K=A-rLor, D( = A-l L
or,X.=frL (3) {sinceA+A = I and
Now the cofactors of D are
^,, = l3 : l'l = r*. Arz = Fr) lt -'?l = - rr,
o,,=lt li =*,
&r = Fr) ls : Il =- Qg,, r*=13 -Zl=u,...,.
ls-zl lq-qlar, = l7 - irl = - ss, As = (-r) li-- iri =s.
ls slo*=ln il=-tz.Therefore,
r lo5 -r2 34 T I lo5 -48AdiA=l- a8 5 -r7 I =l-12 5" L-ss I -r7) L 34 -t7
-53 I8'l-17 )
MATRIX AI.CEBRA
- r 1O5 4€! -53 Iand A-r =$
^o ^= # f 17 -rtr -r? lNow from equation no (3) we get
[;l=+[-tr ]i ]l[;]=+[-ffi i,B]ffi1=+[ 'tr ]ff1
"" [i]=+t{l=[i]n** r=LJ
ryiotaProccs. : The augmented matr[r of the gtven linear
eqqaUons ts
rs s -7 : IQr #:,"ffi:f*:H['ffit*-rAIr = Ll I -!r' irZ J ffi fryg",i#f* subtract
-[i -,i *'i.JAl *ruf*ifHt3',f"T**
- l-l ,r? i i :s'n-lw..aa second rowwith ttrtrd ror'"Lo rz 5 : gaj
'l -n't]*" ","tuoly oeond row bY f +)and thtrd run' bY {-11'
t4l
rI-44-lo-tz-oLo o-I
rr 4 4 . -71-Ls I t . aJ
I42 COLLEGE LINEARAI,GEBRA
Now the system is in- row canonical form, Then formingthe linear system, we have z = O,
6y+ Wz=2,x-4y-42=-7ot,z=O,y =2, x=8-7 = l.Thus x= l,y = 2, z = O ls a solutlon of the gtuen equations.
EXERCISESI - 31.(dro= [i 33i J,"ou=[?Ilnd the matrices, SA, A + B, A- B, 3A - 2E}.
Ansners,[,333131 ,IZ-L-1 3] :. :
t-l I 3 .i I*o hB 3 '3"-? I: , ,;,,,. i,:1:
(bl Sho.w tllat the six matrices
,=[3 ?],^=lA -91 "+[* f]
"=j I aj * ] satisff the reration"'.A2 =W =.C2 =I, AB ! D, AC: BA= E.
z.trA= [i -B],u= [3 -1] *0"= [-i -i] ,
i fh"ri show that [i) A (B + C) = AB + AC
-r -2 5lo -3 4I
. r. i:1.,, ,
r-l3.{alltA=l 4
Lo
(ii) (A + B) CrAC + BC. [R.U. S. I:fftl3 2] ft -2
-2 5 land e=12 3 -r -3-j Ls 2
find the matrices AB and BA. '
6l_3 I
*"u. that A3 - 4A2 -A+'41='o'
MATRIX AI,GEBRA r48
lR.u,P_. \w4l
[L U.S Lg7sl
[b) Construct the products AB and BA'
roll0-1 ro 1I Ol
where"=ll33llu=l:l 3 3 1lLorlo] Lo-t-t ol
and also show that
[33 3lAB_BA=_4c,where"=LB
B B ?lr15 15-2lr-3 8 -rll
Ansrwers: tall 25 - 4 11 I,l 4 - I 22 l,"l-z -15 2JLg t2 tz )f-2'o o'21 ': l2 oo 21
b)AB=l 3 3 S 3l*=l 3 3 3 3ll-z o o z) l'z o o -2)I I -I Il l-l 2 3l
+. Ifa=l-3 2 -rlande=12 n u_]
L-z I ol LI23then show that AB * BA.
r2-3 -51 T-l 3 5l5,rfa=l-l 4 sl,s=l I-3-slana
Lr-s-+) L-l 3 5l
, f :4 6z.tra=l r B
[-r -4
r 2 -2 -4l|C.=l -1 3 ltnen shoWthat
t tt2 -3i ; ,
AR = BA,= O, AC = A and CA =tC. i
I44 COLLEGE LINEARAI,GEBRA
r I -4 -r -41I z o s-+le.IfA=l-r t-Z elL-r 4-t 6j
Prorre that A4 - 5A3 + 9A2 *7A+ 2I= O.
9. Veriff that (AB)r =S,$ wtrere
11 I l-1 ro I -l 1a=lz 2 s lard e=l -3 2 4l^-LiZ eJ*- Li i oJ
r1 -1 1-'tro.ffa=|2 -i o ltn ttshowthatLr o oJ
A3 = A2 . A =,{,. .$ = I and hence ffnd A-l -
r0 o I 1 p. u. H. leE& R- U. S. L#ill
Ansrcr : A-r =l o -l 2 |
rt. Lr -l lJ6r.\*"O the adJotnt matrices and the lnverse matrlces of
eathpf the followlng matrices :
@ ^=[? -ll .,B=f_i i ilL o I ol
o-r?-,J
r-l O(Iil) c=l o t
Lo o
rrngrtrs : (i) AdJ A = [ -: ] I, o-' -* [j I I
'i! *"=[_i i ,i]*,+[i i i],*l Adrc=[i.' l-, -?]-'{i i +]
UATRIX ALGEBEA
12. (i) Find the lnverse of the matrlx12 -l -l -l
e=l t-z tlLl -l 2)
(il) Flnd the lnverse of the matrixr5 4 41
e=17 4 -s Ilz I 6J
t L -L;lAoswcrlB: tilrr=l i ? tl
L-6 -6 zJ !
(ii) *'=#[-?i -,?, e),r. oo =l I -?
-1 I ,,,0 , =i-i -" -u, lL-s ; -;J-'""-1 5 L 2)
Find the matrlces aB, en' A-r' B-r' (aB)-t' Check your
results by veriffing the relatlons'iagf-l lfY a-t, AA-r =I, EriB=I.
ADsrGr,,;=|-'i i iBl,*=1":3 ? ill--li -i-sl' It 4 rsj-r 7 19 I1
"-'=#f 11 '+ ,3 ]'rf-2 3 ql
B-r =#l'g -'J _; lGBrr=ig[ ,,$ -;* -i}]
rt o rl rg l ll14. rf "=Ll I ?l*'au=ll I ll
Show that A3 - N +A-28=O'
r*5
p. u. IL 1s751
p.u.P. 19791
Ic. U. P. 1fr161
!,t{F?r ahebrq-lq
MATRX ALGEBRA r47
2o.Findtheinverseofeachofthefollowin$matriceslby
""i.rg-""t-the elementary row operations (row equivalent
canonical matrix) :
COLI,EGE LINEAR .AI,GEBRA
f2 3 41 r1 2 3lU a=la 3 2larrrdl=lz s 4l*'-'-12 t 3J Ls+ IJShow thai (A + B)2 * A2 + 2AB +* '
-t ::: ;, P'U'H' 196'C'U'tI.lg77l
r-l 2 3l l-l t llre=l l3 slr"ae=l 12 9lLrs;zl Lt4 eJ
Find the values ofA-r B'and AB-r tD. U. P. 19761
[r-r-il [o roiAnscrers,e-'s=l o I ? l*-'=l'? 3lt ; I l-o-zz-lLo o 5J
146
t5.
3ltl8_l
$.nj.{i !4.
l',I:l
l- 3t rZI z -2I s 5,lr'-2l-z !
To I 0 -21T-5 4 -31 l_z I -1 1 I
(iv) I ro -z g lM [_i i _i o I :: :\ :,.-'-'La -6 EJ L-i _; i s.l:r^ 3 5 l_1l-r 4 4 ra I
l, 5 I '5 I
tu,l I -. , ,'i' -4 fB I :
It -a-4 -4a I i
lrlrlLo 4 '-4t' 16J' i ,t
I r'i s ?,1 tn.r,rinla a3 '\ ,o, - +a -'21' iru21.tr^=l? "o ?,1 #;;;ffiiar."- , lLl -r ]J.---i t, , .. ii ., i i, i I : Er.u.H.f.+p,fti.,
Answers, {*0"+,r;[ ii -i _:l ' , i
r 1'3 4'1u"''iIs -r olL-r 5 Il11 f2 5 2
?1,13 3 3rl l+ rz'o
ll'',[i io1 [1 -l 2
* j'l? : 'l
-;],r[-',; # li]*
u) l3
(i) l-3
41g ltut-rlr-1 2
(iv1 l2 I14 -2
ADswers:
16.
o orI ? I
o-o i n-' * al, ID. tt.P' ts77tr1
rz. ffa=l1Lr
Arrswer:
:t-3
18. IfA=12Lo
19. .Find
the
[_ie=l ILs
Answer:
l;r'i{,:i .ri..i.f 1}i
i
I
{
l
cI
i
i
I
;i
11 0 Ol* fa-, *ay =l o t O ,l= l.
.LO 0 rJ ,r.
-3 41-3 4 I P.orr" ttrat A3 = A-1--i il ,linverse of the matrix '
o 121t 2 tlo I 1l )
: rT 'r l::il: au, 'tr;Fti ':1,=t €e 6' 6 I
lr,rr;1 ! tl*rlr'R-r= l ,-2: .U" =?i,,',,! ;l i,:l'-; rl 6 -6
r
L I' rrQ -t o J',':,, !l
\l*"ffi
148 COLLEGE LINEAR AI,GEBRA
22. Find the inverse of the matrtx
MATRIX AT.GEBRA I49I l-r_t
{o .I5l2 r"l-G {51I ll
__l
G\i5J
rt 2 3-te=12 E s l
Lr o 8J
[*
f -4O 16Answer:A-l=l 13 -5Ls -2
23. Show that the matrlxl-cos0 sin0 Ol
e=l-sine coso ol--Lo o tJ
_31
-rl
are hrro matrices, then lind A'n "t
a e g'.p. u. H. T. lsl, rs?l
Aoswers , A'B =* [3 -3 -il l, "
E' =* [-B -Z
3 I"Lo o -zJ "L-12 r7-s)
is invertlble for all values of 0 and flnd A-l .
then prove that (AB)-r = B'. A' p. U. tI. T. 1$3, fS426. Prove that the followlng matrlces are orthogonal :
(i) Itr3 -":L% ]
oo = [3 i -il t]ren show t]rat
4:g,t and.SA are both symmetrie matrices.Show that the matrix
rI I l rtrlr -r r -rlA=rlt t _l _f li"orthogonal.Lr -l -r 'rl
Prove that the matrixr [r r+rIe= 6 li-i - -il is unitary.
show that the matrix o=l-? -l -l l" idempotent.L-q 4 -sJ- ------r
show that the matrix " = [6 i :1 I ," involutory.
r. Ls -l -sJ
x+y+z=6)(ii) x-y+z=21
2x+y -z= I )
[D. U. p. ts77l
27.
28.
t-cos0 -sln0 O-1
Ancwer: frr -l sin 0 cos e O ILo o tJr1 2 3-] rl I l'l
z+.ffe=ll 3 Slanan=l I 2 slLr s LxJ Lt 4 5J
31.
rl t 1-'t 12 5 31zs. tr
^=Ll ? B.]*"=L? L ?l
;-cos0 -sln0 OlfiflLST' T'?J
r I 2_ Z1I 3 3 -3 I
'' | 1*, ? IL-s 5 5J
r! ? 2_1l-3 3 3l
(,,) Ig-l 3llz 2 t ILs s -BI.4
:l
):1,ri
,flil
(r) iJJrrt= ,. )
150t..
'3x+2Y-z=2Ol(iii) 2x+ iY - 3z = 7 |
x-Y + 6z= 4l )
3x-Y + z=-2 I, (iv) *:rilr:'::8|
rc. u. P. Lsnl
x+2Y+z= 2 I(v) ?x-_ lr:32__: l. I
151
lD. u.P. rwsl
ID.U.P. 19851
[D.u. H. 1*]31
tD.u.P. 19861
Answers:(i)x=2iy=3,(ii)..tr='=,t=2'z=3'-(iii) x= 5,Y = $, 7=l' .0{)' x= -2:Y =O'z= 4'
(v) x=-2'Y =l'z=2'36. Solve the following systems of linear equations with
the helP of matrices :
5x-.6Y + 4z= 15 ) +(i) 7x + 4y - i; = Ls I p' u'P' LsTs'J' u' rr l98ol
A3swer:x=3,Y=4,2=6.2x-3Y+42= 1l
(ri) 3x+ 4y -t, = rc t p'u'P' 1966' J' u' rL 19781rrr' ;;-i;+22= 3 l
Ainsg3r 3X=2, Y=1, z=o,
x+!*z=91(iii) 2x+ iY + Zz = 52 I2x+Y-z=v)Aoswef i X= 1, Y =3,2=5.
x+2Y- z---l )Aivl 3x+ 8Y * 22 = 28 |
//""' ;;*si- z=14)Ansmtr. 2 x=-26,Y = 13,z= l'
x1 +3x2 +Zx, =?1{v) 2xi+ 'rq+3x3=1f\v'
5r, * 2i2 + trs = 4)
2174Aaswer: xl =O' &=T, xs =-0'
37. Solve the following syst6ms of linear equations by
using row equlvalent canonical matrix (by elementary row
transformations) :
MATRIX AI,GEBRA
x+ Y+z =1'l(vi) 2x+5Y+52 = 2l
4x+9Y + l2z=3 )
llAEsrertX= 1.y=S,z=-3.
2x- Y- z=6 )(vii) x+3Y*?"=! I3x-Y-52'iiAngr€rix=3, Y=-2,2=2'
x+ y+ z=3 )(viii) x+2Y+22=41
x+ 4Y +92=6 )
A[ss.tE: x=2,Y=l,z=O'2x+ Y+ z=3)
(ix) 3x-iY+22= 9l4x+3Y-z=-l )
Aos;wer : x= 1, Y = - I' z = 2'
x+zy+32+ + =O I(x) 2x+iY +52+ Z =O I.^' iiiiv*a'* 1o=o J
Aqgref i x=-3, Y = 1, z=- l'
2x-Y+z=1.1(i) x+4Y-sr=-?l (ii)
3x+2Y- z= O J
x+2Y+32+t=3 .)
x+ v* z-t=S t(iii) x+ y- z+t=4 Ix- Y+ z+t= 2 )
COLLEGE LINEARALGEBRA
Ic. u.,P. 19841
tD.u.H. 1s761
x+2y + z=2 )3x+ Y-22=l I4x-3y- z=3 I
2x+4Y+22=4 )
AosurcrB: (l)x=o,y= 1' 7=2' 01)x= L'y=o'z=L'(llt} x= 1, Y = - l, z =2' t= -2'
tD.u. H. 19781
t52 COLLECE LINEAR AI.CEBRA
x+ Y+ z+ t=5 I2x+ v+32- t=14 I(iv) 3x+ By - 2z + 2t= | t4lt-2Y+ z-3!=6 )
-tAnswer: A
39. Find the inverse of the matrix
by using onlY the elementaryequivalent canonical matrix)'
Angwer:2 4 5-3
-1 3-3 I-1[= 2 -3 4
l-l 0
7 -2 2-IL
p.u.P. 1S761
row operations {row
Aosrcr:.x= 1, Y=2,2=r',i;,*, 2ls o z
38. FindtheinverseofA=12 I -lLtol rl
11 2 t 2 3llz 3 l o I I
e=lz.z I o o lIt I I I I I
Lo-z o 2 -2)
ro I o -21l-z l -l 1l=l-r I -1 olLr -2 I 3l
I21
ToITIT
-2
o
40. lf A and B are two s5rmmetric matrices of the same
*a.i'ur"";i;& th;i; n"i""'"u.y and sufliclent condition for
tt. *"t i, AB to be symmetric ls that AB = BA'41. Show that eiery non-sin$ular idempotent matrix is
an identitY matrix.-' -;i: nJAis an idempotent matrix and A + B = I' then prove
that B is arl idempotent matrix q"d AB = BA = O'- 4g. If A and-B are n-rowed'unitary matrices' then prove
that BA is also unitarSr matrix'
CIIAFTTR SffiVECTOR SPACES
6.1 Blnary opcradon (or composldoa) on a sct
It ts a rule by whlch two elements can be combined
together to glve a new element. Such a rule may be addition'
sribtractton, multiplication and so on. The most fundaqeptalconcept fo; studylng algebraic structures is that of btnary
operatlon on a set.
Defrnltlon : Irt S be a non-empty set' Then
S xS = {(a, b) : a eS, b eS}. AbFary olrcratlon on a set S is a
funcfion (or mappin$ from S xS into S: i' e lf f : S xS -+S' then
f is satd to be a bhary olrcratlon on the set S'- Often we use *rJ symUots *, o, * , X, "' etc to denote the
binary operations on set. Thus 'r'wlll be a binary operation
on S if and only if a *b.e S for every a, b eS and a*b is unique'
A btnary operatlon on a set S is also sometimes called a
blnary "ompo"ltion
in the set S. If a*b e S for every a' b e,S'
then we also say that S is closed with respect to the
composition denoted bY '*'.
A set having one or more binary operations is known as
algebratc stnrcJure- :
Nowwewillde{ineSome.,algebraicstructuressuchasGroups, Rings, Fields and Vector Spaces using btnary
operations.
9.2 Detrnltion orgroup wlth examples
a group C is a non-empty set of elements for which a
binary operation * is defined. This operation satisfi5s the
following axioms :
(i) Closure, IfgL g-p implies that a * b eG;
(ll)Assoctattvity. If d, b, c e G implies that(a*b-)*e=a*[h*n]
trl.
214
i (ui) Identlty,.There odsts a unique element e e G (called the
ldentityelemcnt) suchthata t e i e * a= afora[a eG'
(iv) Inversc. For eve4r a e G there exists an element a'e G(called the inversc of a) such ft121 a,* a'= a'* a = e'
When the binary operation is addltlot G is called an
addltive group and when the binary operatlon ls
e gro.rp G is called abellan (or commutatlve) if for every
1p$a*b=b*ap:ample 1. The set {1, - l} is a group with respect to the
blnary operation of tnirltiplication.
Example 2. fhs,sef, {1' -1, i, - i} where i = a/Jis a group
with respect to the blnary operation of multiplication'
p-xagde 3- The set oJall integers, i. e' {"" -3' -2' -2' -L' A'
L, 2,'3,...1 iq-a group #ih respect to the binary operation of
addition. :
A rtng R is an addltlve abelian bo'p with the following
additional ProPerties :
. (i) The group R is closed with respect to the binary
operation multiplication. i.e a, beR + abeR
" (ii) Multiplication is assoclatlve i. e'
", - thb)g =gQ*}for all4. b. c eR
(iii) Muluplication is dlstrlbutlve with respect to addiuon
on both ttre left and the right' that is'a(b+c)=ab+acl'--^6 .
"r"J = ;;; !l| ror all a' b' c eR
Example 1. The setZ = {O, t 1, t2, .'.!is a ring under the
VECTORSPACES 2I5
Eamnlc 2. Conslder the set Z = {O. 1. 2. 3, 45}Z is ring under the binary operations of addition and
multiplicatton modulo 6.6.4 Dcf,nttlon of field with cxamptesA fieme ring with urut eiement rn which
every non-zero element has a multiplicative imrerse.
Examples of fields are the ring of rational numbers, thering of real numbers and the ring of complex numbers.
For every pair of real numbers a and b there corresponds a
real number called their sum and is denoted by a + b. Theaddition composition has the'following properties :
.SlLa + b = b + a for every a, b in R(comsnrtattve).
-S]pL{6+ b) + c = a + (b + c) for wery a, b, c in R (Assocfatke)
AJglThere odsts a real number, vD O such thata + O = O + a = a (cdstcnse of addtttv,c fdenfity)
A t4l To each real number a, there corresponds a realnumber, vlz (-al such that (-a) + a = a + (-a) = o
(edstence of addttise luverse)Also for every palr of real numbers a and b there
corresponds a real'number called their product and is denotedby ab. The multiplication composition has the followingproperties :
tr4-OJ ab = ba for every a, b in R (CoqmutattvelU fZ1(ab) c = a (bc) for anery a, b, c in R (Asgpciatfw)
IVI t3) There exists a real humber, viz 1 such that for every
ae R la = al'= a- (e.rdstence of multlplicetive identity')M (4) Io each real number a * O, there corresponds
another, lri, | ",r"r, ** (]) . =. ("f = ,
(odsteace of muttlpllcatlve lnverse)
AM (1) ah + c) = atr + ac for every a, b, c in RfDistributive law)
COLLEGE LINEARAICEBRA
!;i,it
' i,..*,}
.l
tl
tl
ll"i
:lrit,ol,'f
!,.
ilIJ!rII,f
.'iIitit*Itil.:.i:il
,,lf
,,ti
i$,
binary operations of ordinary addition and multiplication.
COLLEGE LINEAR AICEBRA
spage over an arbitrary field F is a non-empty set
rt{ti
i
tI
ii,lt
V, whose elements are called vectors for which two operauons
&fe, pr€scfibed. The first operation, called vector addltlon,assigns to each pair ofvectors u and v in V a vector denoted by
11 + v, called their sum. The second operation, called scalarmultlplication assings.to each vector v in V and each scalar a
in F a vector denoted by a v which is in V. The two operations
are required to satisff the following adoms :
4ljrAddttioa is nfrmmutstfue.
For all vectors u' v € V, u + v = v + u.
A t2)Additlon fs eggegfagggFor all vectors u. v, w eV, (u + v) + w = u + (v + w)
A (3) Ddstence of O (zeno vector).
There exists a vector O eV such that for all v eVv+O =O+v=v.,A (4) Hsiercc of neEatlv'e.
For each v e V there is a vector
- v eV forwhich v + (-v) = (- v) * v = 0
][lUForanyscalaroe Fand anY t'vectors u, v e V, cr (u + v) = c[u + 0v
M(21 For any scalars a,P e F and anY
vector v eV, (ct, + P)v = anr + Pv.
!!p[For any scalars o,P e F and anY
''"' M (E Fo, each v eV, lv = vwhere I is the unit scalar and I e F.
For some applications it is necessary to consider vector
spaces where the scalars are complex numbers rather than the
real numbers. such vector spaces are called complex vector
spaces.
Vector spaces are also sometimes called linear spaccs'
VECTOR SPACES 217
6.7 tranplcs dtcctot sPacc
nrnrrllrlc 1. IR2= (a, b) la, b e lR } is a vector space over the
Iield F = IR rvlth respect to the operations of vector addition in
IBPand scalar multiplicatio., on IR?.
dr"pr.""rrts the sei of all points in the plane'
E*unFle 2. Ep = {@' b, c) |.a, b, c e IR} is a vector space over
the field F = IR with respect to the operations of vector
addition in IR3 and scalar multiplication on IRS '
IBP asplssgnts the set of all points in space.
Example 3. kt V be any plane through the origin in R3 '
Then the points tn V form a vector space under the standard
addition and scalar multiplication operations for vectors in
IRP .
Erampte 4. For any arbitrary fie1d F and any integer n, the
set of all n-tuples (ur, u2, ..., ur,) of elements of F is a vector
space over F under vector addition and scalar multiplication
given by (u1 , u2; ..., uJ + (vl ,vz, ..', vJ = (u1 +v1 , 1r2*Y2, "', urr,+ vn )
o(u1, u2, ..., uJ = (cru1 , c[r2, ..., ou,r) where u1, v1, a eF'
Thls vector space is generally denotet by F". Th.e zeto
vector in F is tlle n-tuple of zxro i. e. O = {o, o, ..., o).
Example 6. Let M be the set of all m xn matrices with
entries from an arbitrary field F' Then M is a vector space
over F with respect to the operations of matrix addition and
scalar multiplication given bY
l-ar r e,tz aln I [b, r btz
| "r, dzz ^r! I*l o, bzzI ... I I ...
La-, a-z ... a-r, i Lb-, b",,
br" Ibz, I
I
b-r --l
218 COLLEGE LINEAR AI,GEBRAVECTOR SPACES 219
We claim that wtth these definitions of vector addition
and scalar multipltcation IR o becomes a vector space'
EsasPlG 1O' The set of all continuous functions
y = fH, - "' ' "':* oo satisffin8
frre differential equation
y" -y' - 2y =O is a vector space' tln fact any solution of this
differential equation-is a linear combination of y = e-x and
y=e\. _^_^Theorem 6'1 l,et V be a vector sPace gver an arbitrary fteld
F. Then, \ n(i) For any scalar cr eF and 0 eV' cro = O'
(ii) For o eF and anY vector v eV' o1
(iii) For creF and v eV' (-c) v = cr(-v) = - crV'
(id If arr = O' *h"" aeF and v eV' then er - o orv = O'
i*"t: (r) 0O = tt'(O + 0) = oO +cto
Adding- uO to both sldes' we get
(40) + cro = (-oo) + (c[o] Cto)^ -,1
=(-cro+aO)+600=O+fl'Oor,O=O+q0:00 "'oO=O:(u) O = ov + [ov) = (o+ o) v + (-ov)
= (ov + ov) + (-ov)
= (ov + (Ov + - ov) = ov + O = ov
oll=O'
(iii) O = sQ = cl(v + (-v) )= crrr + cr(-v)
Adding - cff to both sides' we get
- ca/ + Q = - Gv + (cn'r + cr (-v))
. = (- cnr + crv) + cr (-v)
=O+o(-v)or, - ay =ct(--v) .'' cr (-v) = - ot.
Again, Q = ov = (o+ t-cr)) v = crt/ + (- ct)v
3ml=[
Ettt +ozt +
brr dtzbzt dzz
8lta2n
*bt, I+bz, I
I
+ b-r, J
+ btz+ bzz
"" -[
+
?tr rozt
bmt amz
?ttzdzz
I [oar t cl?r2
I I o"r, clAzzl=1...I Lo"-, @n2
4lndzn
o-n
Oaln IoJa,zn I
.1
cra-r, I2ml ano
where &1, bu and a e F'
Erample 6. f€t V be the set of all continuous real valued
functions defined on the closed interval Io, u. For'imy f' g e vand cre IR, deffne f + g and odby (f + d (.ld = f (d + g (x)
(cd (x) = cd (d for every xin [o' U'
Under these operations V becomes a vector space over IR '
(sirace the sum of continuous functions and scalar multtple of
any continuous function are continuous)
Example 7.I-etV be the set of all polynomials
ao .lF + a1 ;F-r+ az /'-2 + ... + a*r x + a,, with co-eflicients ai
from ap arbitrary fteld F. Then V ls a vector space over F with
respect to the usual operations of addition of polynomials and
multiplicatton bY a constant
Example 8. L,et S denote the set of all pairs of positive real
numbers. ir = (ur, uz), v = (v1 , v2),
Deffne u + v = (ur vr, tTzvz)and am = (uto, uzo) (o is any scalar)
where u1v1 and u2v2 are the usual prodqcts of real numbers
and u1s and u2q are the cr.th powers' S together with this
prescription for addition and scalar multiplication is a vector
space which we denote bY M2'
E)rample 9. kt m', Ue the set of all positive real numbers
and define forx, Y = fr, a vector sum by x*y = xywhere the
product on the right is the usual product of numbers' If a is
any number and ,a R, define d x = f ' tllat is' the number x
raised to the a Po\Yer.
22O COLLEGE LINEARAICEBRA
Adding - crv to both sides, we get
-crv+O=-crv+(mr+(-c)v)= (- cry + crv) + (- alv
= (- cl) + a)v + (- alv
=ov+(-mr)=O+(-a)v'or, - cn/ = (- a)v .'. (- s) Y = - ov.
(iv) Suppose that crv = O and cr* o, then, there odsts a scalar
a-leFsuch thato-rs= l, hence
v = lv = (a-l o)v = a-l (crv) = a-l (O) = O.
Ilefinition of Cartesian or Euclideaa space
Let n be a positive integer. The cartesian n-space denoted
by IRn is the set of all sequences (u1 , :u2, ..., u,r) of n real
numbers together with two operations(ur, uz, ..., uo) + fu1 , v2, ..., vJ = (u1 + v1 , Ll2 *Y2, ..., u,, + v,r)
and cr (ur, uz, ..., uJ 1(aur, mrz, ..., mrJ'ln particut"r, ql = IR is the set of all real numbers with
their usual and multiplication.2 For each positive integer n, Euclidean space
space.
: We shall have to show t]:at IRn satislies all axioms
(i) Le.tu = (u1, u2, ..., u.,) and v = (v1, v2, .'., vr,) 5s in IRnthen
u + v = (ur, uz, ..., uJ + fu1 , v2, ..., vr,)
=(ut *v1 ,u2 *Y2,..., ur+vJ
= (v1 +ur, v2 +uz, ..., v, + u,r) = v + u. So axiomA (l) is true.
(ii) l€t u = (ur ,t)2, ..., uJ, v = (vr, vz, ..., vJ and
w = (wr ,w2, ...,wJ be in Rn. Then
{u +v) +w=(ur *v1 ,u2 *Y2,..., u'r+vJ + (wr,w2,..., wn)
=(ur *v1 *w1 , i.jr2+Y2*w2,..., ur+vrr+w,.,)
= (u1 , u2, ..., uJ + fu1 + wr, vz *Y,12, ..., vr, + wJ
= u * (v + w). So axiom A (2) holds.
IRN
t
I
t.
i\
ofa vector space.
VECTORSPACES 22I
(iii) t€t O = (o, o'..., o) be lrr IRn ' Then for any
u=(ur. u2, ..', uJ m ff we wlll have
u + b = (u1, u2, .... uJ + (o,o,"',o) :
= (u1 + o, u2 + o, '... ur, + o)
= (u1 . u2, ..., r-r,J = s.
Moreover, if v = (vr, vz, .'., vJ is any vector 6 IRn such
ttratu +v= u' then (ur *V1 , u2 *Y2, ""un +vJ i (ur' uz' "" uJ
Therefore, u1 * v1 = ut implies vr =^o I
l? * "' :.* 'il]"" ::.=
o t
un + vn = un impues vr, =o )
i. e. v = (o, o,...,o) = O, Thus O ts the unlt vector witli
property that u + O = u and so tlre a:dom A (3) hotds'
(iv) Ietu = (u1, tl2, ..-' u'J and set
-u = Fur, -uz, ...' -uJ, Then
u + (-ut = (ul, u2. ..', uJ + (<rr ' - uz' "" - uJ
=(u1 -u1, :Uq--u2,.-.' un:uJ(o'o'""o)=O' I '
Moreover, lf w = (w1 ,w2' ..., wrr) is any vector ln ff such
thatu+w=O, then (u1 *w1 .u2 *w2, .''r urr+wJ = (o' o' "" o) and
therefore, u1 *w1 = o implles v/l =-ulu2 * w2 = O imPlies w2 =-ttz
.
un +wn = 0 imPlieswn =-lln '
i. e. w = - u. Thus - u is the uniqu€ vector with the property
that u + (-u) = O and so axiirm A (4) holds'
(v) kt cr be a real number (scalar) and u = (ul' u2' "" u") and
v = (v1 ; vfi, ...,vr) be vectors tn IRn, Then
/ct(u
COLLEGE LINEARALGEBRA
+ v) = 61 (ul + v1 ,t)2 * Y2,..', ur, + vJ
= (cr(u1 +v1), cr(u2 +vz), ..., cr(u, +vr,))
= (du1 * crV1 , c[u2 + Gu2, "', clun + s/n)
= (cru1 , c[r2, ..., ouJ + (ov1 , ov2, "', crtd
, = cr(ur, u2, ..., uJ + o(vr , Y2, "" vJ = ou + crv'
So that a:dom M(1) holds'
. (vi) I,et cr, p be the real numbers (scatars) and
lJ=(u1,u2;.-., uJbein ni, Then
(cr+ 0 u = ((a+ 0 ur, (cr+ 0)uz' "', (s+ 0 uJ- ' = (ctu1 + Fur, clu2 * 0r:2, "" c[tln+ BuJ
= d{ul, t72r ..., uJ + P(u1, uz: "" uJ '
= cru * Bu So axiom M (2) is satisfied'
(vii) L;et a, p be real numbers (scalars)'and u = (ut' !:z' "" uJ
be in i*. T'tren (op) u 1 aP fur, u2, "', uJ = (apu1' "'' a0uJ
= a(Bur,puz, .., FuJ
= o (0(ur, uz, "','uJ) = a (0u)
,, $ adom M (3) holds(viii) Let I be the unit scalar and
' ; = (ur. uz, ..., uJ be in IRn ' Then
lu = i(rr, *, ..., uJ - (Iur, lu2, ";, luJ
- = (u1,,u2, ..,. uJ = uSo a$9rl M (4) is satisfied'
Theorem6.S.I-€tVbethdset.of.allfunctionsfromanon-emptV set S into an arbitrary lield F' For any functions f' I eV
arri any scalar oe F let f + g and cr 'f be the functions in V
defined by (f + g ) (x) = f(x) +,8(x) and (cr0 (x) = af (x) for every
xe S.
Prove that V is a vector space over F't'Ii'i
VECTORSPACES 223
Proof : Since S is non-empty, V is also non-empty. Now we
have to show that all the axioms of a vector space hold.
t0 t€t f, geV, Then(f + g 1:d = fl.{ + g($ = gH + IId
= (g + 0 [rd for anery xe S.
Thus f + I = g + f. So adomA (l) holds.(ii) t€t f, g, treV, To show that (f + g) + h = f + {g + h),.It is
required to show that function (f + g) + h and thi functionf + (g + h)) both assign the same value to each x e S. Now
(f+ g + h) 19 = (f + g B + h (, = (flr)+ g(rd) + h E.- (f + G + h)) (d = f (S + g + h) (.d = flS + €(d + h(d).for every xe S.
But f (x), g(x) and h (x) are scalars in the f.ield F whereaddition of scalars is associative.
Hence (fld + g0d) + h(x) = f(d + €H + h(rd)
Accordin$y, 6+ I + h = f + 19 + h).
So axiom A (2) is satisfied.
(iii)"I.et O denote the zero function.
O(xl = O for every xeS.
(f+ O) (d = f{d + O E = f (.rf + O = f (d
for every.xeS.
Thus f + O = f and O is the zero vector in V.
So axiom A (3) holds
tiv) For any function f e V,' let - f be the fr.Inction defirted
by (-0 x= - f (1.
Then (f+ (-0) (x) = flx) + (-0 A =f(fi -f(x) = O = O (dfor every xe S. Hence f + (-fl = I
So axiom A (4) is satisfied.
COLLEGE LINEARN'GEBRA
;.Y', I
I224
(v) lrt ce F and f' g e V'
ttren tatf + d) H = a ((f + d (d)'= s 614 + gH)
=;;;d;= (ol) H + (ad (d = (of + ugf (d forevery xe s'
lslnce cr' f(x)' 6 (x) are scalars.tl t:e field F where
multiplication is distrtbutlve over addttionl
frerrcea[f+d=qf+og'So a:dom M (1) is Pattsfied'
(vi) Let o, 0 eF and f eV'
Then (a + g) 0 (d = (cr + 0 f(d = 06(.d + pf 0d
=(od)(d+00(d= (af + F0 (d for every xeS'
Hence(cr+9)f=cf+0f'so a:dom M (2) holds
(vll) l*ts"PeF and f e V'
;;" (CIP)0 d = (aS) fl$ = a (Ffk))---=;
dlo m = (o Pol [d for wery ,as'
(vlil) I-et f eV' tf'"" Ot the unit scalar leF"=iiot*
= lf (d = f['$ for every xe s'
Hence lf = f' J 1:*- M t4} rs
' Stnce alt tne-Jims are satlsfled' so V ts a vector space
'",#. [:n T: #;; :' 1 ]1.:: :::Ti L:il:"H;
li'lJ;* :il ffiTT#;;;':n:'"::,H;ifr :$ f:::ffiffiffiffi; oJn"* *':.1'^'T':i'*Yi I ;;:Sil#tl'#H].";il':;;-e w' cr' 0 eF implies that
crwl +$r4 eW'
vECToRspicps 22:t
69 nrnrtrflcrlofrutap*CcExeoplcl.I,etVbeanyvectorspaoeoverthefieldF.Then
the set lol conststlng of the zero vector alone and also the
Example 2' consider tlte vector space EP= {(a' b) la' b € R}
ttrenWl ={(a, 0 laeIR}, Wz ={(0,b) lbeIR} and',W3 =(a,b) la=bandabe IRlaresubspacesof d'Wr andWz represent the sets of aU points on the x-a:ds and
on the y-a;ds resPectivelY
Also W3 represents tfre set of all points on the line-5r = '16'
Uamptc 3. Consider thevectorspace
mP=11a,b,c) la,b,ce IRI \
Then W1 = {@, b. 0 la, be IR}, Wz = {0, b, c) lb, c e IR}'
and w3 = {(a, o' c} la, c e IR} are subspaces of IRP ' w'' w'andW3 represent the sets of all points in the X-Y,Y-Z and'Z-X
planes respectivelY
fxanple 4. Consider tlie vector space
E9={h.b,d la,b,ce tsP1.
ThenWl ={@,O,O) IaeIR},W2 ={(O,b'q lbeIR} -
and W3 = {(0,0,c) lc e IR} are subspaces of IBP ' W1, W2 and
W3 represent the sets of all points on the x-axls, y-axis and
z -a:ds respectivelY.
fiample 5. I-et M22 be the vector sP*1jr of all squate 2 x 2
matrices. Then the set wof all2 x,2 matrices having zeroes on
the main diagonal is a subspace of the vector space M22'
Example 6. Let V be the vector space of all square n xn
matrices. Then the-set consisting of those matrices A = [aq] tbr
which au = aJr called synrmetric StricSs
is a subspace of v'
Linear,tlgebra-t6 ,#,
Y?.
r,
226 COLLEGE LINEAR AI,GEBRA
Erample 7.i-etV be the vector space of all functt1nl of a
real variable x and let W be the set of all functions f eV for
which f(5) = O. Suppose that f' geW and let h = f+ g1' Then (5) = O'.
g(O = O and so h(5) = fl5) + g[5) = O + O = O'-' -r,
follows that heW and so W is closed ""1"j, "Oli*"^Again let k = crf where cr is any scalar' k(5) = ofls) = ct'O = O
*i.rr"" ke W. Clearly, W is closed under scalar multipllcation'
Thus W is a subsPace of V'
'--- ft"-ptu A. Tht set of all continuous functions is a
subspace of the vector space of all real valued functions'
' Example 9. IJt C be the set of all complex numbers' Then C
is a vector space over the real field IR'
l€tW = fiU lu e R i = '/eflt' ThenW is a subspace of C'
tlreorcm.6.4WisasubspaceofVifandorrlyif(i) W is non-emPt5r
(ii) W is closed t"tiut vector addition i' e' v"w eW implies
thatv +we W'
(iii) W is colsed under,scalar multiplication i"":'Y t W
impliesthatove Wforevery cre F' ., ..,"Proof:Supposethattheglventlrreeconditions.(i)(ii)(iiil
' tnd closed under vectorhord in w; then * o
1o".:*::1 e
in waddition and scalar multiplication- Since the vectors
reiong to V axioms A(1)' A(2)' M(1)' M(2)' M(3) ana Uf4 hold in
W. Hence we have only to show that A(3)' A(4) hold rn W' By
condition (i) W is non-empty' say ueW' then by condition (iii)
;;= ; .; *rd v + o = v for every vew' Hence A(3) holds in w'
Again. if veW then (-1) v - - veW and v + (-v) = Q
#;"; A(4) holds in w' Thus w is a subspace of v' -
;;;;;;;;iv, if w is a subspace of V then ctearlv the given
conditions (i)', {iil, (iii) hotd in W' Hence the theorem is proved'
. VESIORSPACES 227
Tlreorqn 6.6 A non-empty subset W of a vector spg.ceV over
the fleld F is a subspace of V if and only if(l) u, v e W+u -v € W (ii) oe F, u e W+au q W.
Pr,oof : The conditlons are neeessar5r I j,''i ,,, . : , :, ;
If W is a subspace of V, then W is an abelian group.with
respect to vector addition. : , , ,,. . ,
i, e, u, veW+u, -v€W+u+(-v)eW=+U -V eW.
' Also W must be closed under scalar multiplication, i, e
creF, ue W + crueW. Thus conditions (i) and (O are necessary.
The condltlons are suffcientI€t w be a non-empty subset of V satisffing the two given
conditions, From condition (i), we have
ueW=+u-u.eW= OeW.
Ttrus zero vector of V belongs to W and it will also be the
zero vector of W.
NorOe W, ueW+ O-ueW=+-ueW.Thus additive inverse of each element of W is at6b.in W.
Again, ueW, veW=+ueW, -veW- :
+u-(-v)eW :
=u+veW. '"i : tl
Hence W is closed with respect to vector addition.Since the elements of W are atlso the elernents .of V.
therefore vector addition will be cor-nmutative as well asassociative in W.
Hence W is an abelian group under vector addition. Alsofibm condition (ii) W is closed under scalar multiplication.The remaining postulates of a vector space.will hold in Wsince they hold in V of which W is a su.bset. Thus W is a
subspace ofV.
,?, COLLEGE LINEARAIrEBRA
Theorem 6.6 A non-ernpt5r subset woof a vertoi spac€ v overthe lield F is a subspace of V if and only tf
d, 0€F and u, v e W=oru + pveW;
hoof : ltc condltlontcnecesearyIf w ls a subspace of v. then wmust be closed undervector
addltion and scalar multlpltcation, Therefore,6ge IR, and ueW=+oUeWpeFand veW=+fl,veW.
Now cueW and pveW+ou +Sv€W.Thus the condition ls necessary.trhe condltloa fsgrficacmt
. suppose that w is a non-empty subset of v satisfirlng thegiven conditlon,.i.e o, pEFand u, t eW= cru + pveW.
Irt o = 0 = I, then leFand u, veW+ lu + lveW1.e,u+veW
[since ueW=+ueVand lu = u€V.lThus W is closed un&r vector addiflon,Now takingo = - I, F = o, wesee that if ueW,then (-1) u + o ueW+ - {lu) + O eW=+ - ueW.Therefore, the additirre imrerse of each element of W isralso
Now ueWand - ue'W.
.'. u + (-u) eW=+Oe W.
Thus zero vector of V berongs to w and it udll arso.be thezero vector of W.
Since the elements of .W arE also the elements of V,therefore, vector addition will be associative as well ascornrnutative in w, Thelefore, w is an abelian group withrcspect to vector addition,
yECToRSPACES 22g.
Now taking v = O, we see that if o, peF and ueW, thencru+poewl.eorr+oew =+cruew. i
Thus w ts closed under scalar multiplication. Theremalnlng postulates of a vector space will hold jn W sirlcethey hold in v of which w is a subset. Hence w is a subspace ofv.
I
Dellnltlon : kt S and T be two sets.,By S n T uib rnean thelntereeectlon of S and T, the set of atl element"
"orr_o, to Sand T, By S U T we mean the unlon of S and T, the set of all
Theorem 6.Z.Ihe intersection of two subspaces S and T of avector space V is also a subspace ofV.
Proof : Since S and T are subspaces of V, theyempty and clearly O e S and 0eT. Therefore, OeShenceSnT*@.
Now let u, v eS O Tthen u, v e S and u, v eT. Snce S rra fare subspaces of V, u, v e S implies that ou + pv e S whereo,0 eF. Similarly, u, v eT tmplies that au + pveT wherea, p eF.
Hence u, v eS fi T implies that au + pv.e S O Tfor,o, F eF. .
Therefore, S nT is subspace of the veetor space V.It is to be.noted that the union of S and T is not a subspace
of V unless S CT orT C S,
[the sSrmbol "C" tneans contai:eed ir:lTheorem'5'g The intersection of any fam,y of subspaees
ofa vector space is a subspace.koof : kt {S, liet} be an5r family of subspaces oli vector
space V over the fleld F. Let g = O 51.
tel
are non-fl Tand
.l'
{''|
$T
I,#
,"1$l
,ro
tr,ffi
28tVECTORSPACES
230 COLLEGE LINEAR AI.CEBRA
,Since 0 eS1for every i e I, we have OeS' Hence S is a non-
empty subset of V. Now u, v eS.and o, pe F impltes that u €S1'
.Jl
v€'q for every i e I as S = nst and hence cru + pv eS1 for every
;.; ig tr
I eI, since each 51 is a subsPace ofV'.'. Gn + 0v eS.
Therefore, S is a subsPace of V'
WORTED OI,T E:TAMPLES
ffil. Show that S = {(a o, c) : a, c e IR} is a subspace
of the vector sPace IR3.
pr,oof : For 0 e R3, O = (o, o, o) eS
Since the second comPonent of O is o,
Hence S is non-empty. -
For any vectors u = (a, o, c) and v = (a" o' c') in S and any
scalars (relrl numbers) s, P, we have
,cu + $v's (a, o, c) + 0h'. o' cl
= (oa, o, uc) + (9d,o'Pc')
,., = (ca + Pa, o, ctc + Pc')
'. Since second comPonent is zero'
Thus cu + pve S and so S ls a subspace o{.'*''
. rx^r"r'fle 2. Shorythat
T= (a, t, ",
Oe IR4t 2a-3b+ 5c-d) is a subspace of IRa'
pnoof : Foroe R4,0 = (o, o, o, o)eT
Sir:ce 2. o - 3-o+ 5.o - o = o
Hence T is non-emPtY.
Suppose ttrat-u = (a, b, c' d) and v = (a" b-' c" d-) are in T
then 2a - 3b + 5c - d = o and 2a' - 3b' + Sc'd' = o'
Now for any scalars (real nurnbers) cr' B
We have au + pv = cr(a' b' c' d) + P(a-' b" i' d')
= (oa, clb, ac' qd) + 0a" 0b- Fc., Pd')
= (oa + Pa,' cr,b + Pb-' cc + Pc-'dd.^f''
Also we have 2(aa + pa1 - 3(ob + 9bl + S(cc + g"') - (qd + Pd')
= a(Ya- 3b + 5c - A)'+ FtZa -3b' + 5c- - d') = co + po = o'
Thus cru + FveT and so T is a subsPlce of d '
' ra'e'r"lilc'3' w=-;';:; i"'u'ce IR anda-'o*3=5]
is not a subsPace of d'
For O. d, o = (o' o' o) €W' Since o-2'o a 3'Q = o * 5'
Exanple 4' l-et'V be the vector space of all square n xn
matrices over tlee real fleld F = IR'^'^-
;;;;thatw ls a subsPace ofvwhere :
(i) w consi*t"-or all svmml:T t"tttces' that is' all
matrices e = {all for which at1 = a1t' ij = 1' 2' 3' """" n'
(ii) W consoo tii'-"t"""" which commute with a given
matrix T; that '"'
*': LV : AT = TA)' tD' u' H' T 19sol
Proof : (0 OEW since all entries "j ":: ::trj:T:Proof i (u ve 19 = [bi,l belong to W.
equal. Now suPPose that A = [ag] anc
that is, 4t= Etandbg = b''
. Then fo' t"y."""tt'" a' I e F' oA + pB is the matrix whose
ij ""*T, * i"JJ,:i r",, + pbli'rhus aA + pB is also a svmmetric
matrix. Therefore' aA + FB eW'
Hence W is a subsPace of V'
(ii) O eW since OT = O = TO' Now suppose that A' B e W;
that is, 61 = TA and BT = TB',il
fl
41'
.232 COLLEGE LINEARALGEBRA
For any scalars cr,peF(aA + pB) 1= (cA) T+ (pB) T
=0(AT)+BGrO=aEA)+pEB)
T(0A) +T(FB)
=T(uA+pB).Therefore, oA + pB commutes with T. Thus aA + pB eW.Hence Wis a subspace ofV.eample 5.I€tV= IRP , show thatW is not a subspace of V
where: lD. U. P. t9;n2l^
(i) W = {(a, b, c) I a2O} i.e W consists of those vectors ofIRP whose first component is non-negative.
(ii) w = {(a, b, d la2 + b2 + cz s l} i.e W consists of thosevectors of IRP whose lengfhs do not exceed l.
(iii) w = {(a, b, c) I a, b, c e O} i.e w ""i"?",1J.:;'tr:,:lof IR9 whose components are rational numbers.koof : (i) ktv = (2, g,S) eWand c = - Se IR= F.
, Then ov = -3 (2,5,5) = (-9, - g,-f E) GW, since _6 is negative.Hence W is not a subspace of V.(ii) kt u = (O, l, O) and v = (O, O, t)Then u(W since 0P + 12 +G = I < Iand veW since @ +G + 12 = l < l.Now for any two scalars 0 = p = le IR = F,
' ocu+0v= I (Ol,O)+I(O,0, t)=(O;l,O)+(O,O, t)=(O,1,l) 6y
Sirrce o2 + 12 * 12 =2 <1.' Hence W is not a subspace of V.
VECTORSPACES 233
(CI Let, =;, S. T ew and o = Ge IR= F.
rhen * = G @,s.2)= (sfi, s.15, z13 ) e,since its components are not rational.HenceW ts not a subspace ofV.Erample O. Let V be a vector space of alt 2x2 matrices over
the real field IR. Show that W is not a subspace of V where :
(i) W consists of all matrlces with zero determinant;(ii) W consists of all matrices A for which ,{2 =Anoor: (il l-.tA= [3 3J andB= [? 3 IThenA, Be[ since det (A) = O and det @) = g.
ButA.'= [3 il . [ 3l= [? 3] **sincedet(A+B)=-l*O.
Hence Wis not a subspace ofV.
(iil The unit matrix , = [; !J ororg" w, since
,,=[3 ?] [ ?] = B ?l =r
But nr=lt ? Jo*" not belong to w
since (4t)2=tfi .!] tf;. lJ=[|u,!J*nr
Hence W is not a subspa.ce of V.6.lO f,inear combiration of vectorsLet.V be a vector spaie over the field F and Iet v1,. .., vre V
then any vector v e v is cared a *near "o-uioatiootory1,y2,.'.., v,, if and only if there exist gpalars d1, d,2, ...,*.;;
such that v = Gr v.t + %,yz + ... + okrVn =l?",.
23i- COLLEGE LINEARAI-GEBRA
Examplc 7. Consider the vectors v1 = (2' L' 41'v2 = (1' -1' 3)
and v3 = (3,2,5) ; UP. sno* that v = (5' 9' 5) is a linear
combination of v1 , v2 and v3..
Proof : In order io sho- that y is a linear combination of
v1 , v2 and v3, there must be scalars o,t' ez and oP in F such that
v=0lVl +azvz +C[3v3
i, e, (5, 9, 5) = or (2' L' 4l + wi (1, * I' 3) + oe (3' 2' 5)
=.{2c,1, sr, 4or) + @2, -a'2, Wl + (3",'-2%' 1'l= (2ouy'+ a"2 +3ca, o1- g, + 2oa, 4rl,1+ 3a'2 * 1'u)
Equating corresponding components and forming linear
systemwe get
2o1 + c2 +3cr3 =51cr1 - W+2as=9 I (1)
4o1 +3a,2+fus=5 J
Reduce the system (1) to echelon form by elementary
op.r"Uorr". Interchange first and second equations Ttren we
hlve the equivalent system
o1 - o'2 + 2'c4 =9)2a1+ ca+3o.=5; Vl4o,1 +Saa+5ctr=51
We multiply first equation by 2 artdby 4 and then subtract
from the second and third equations respectively' Then we
have the equivalent sYstem
C[1 - A,2+2a3= ?^l3W - cr3 =- t3.i (9
7g'2-343=-31 )
We multiply second equation ty I ana then subtract from
the third equation. Then.we have the equivalent syptem
dr- dz+20s = 9l :
W- crs=-I3! {42 2l "_ {t=_ i )
VECf,ORSPACES 235
From the third equation' we have Gg = 1' Substituting
o3 = 1 in the second "qt'"uo" we get 3CI2 - | = - 13 or' 3c2 =- 12
Of ' d'2 =-4'
Agatn substltuttn g W = - 4' uB = I in the first equation' we
getcr, + 4+2=9 "'0'1
=tSothesolutionofthesystemisctl=3,aa=_4'o3=1.Hencev=Svt -4v2+vg'
Therefore, v is a linear combination of vl ' v2 and v3'
Exemple 8' Is tl:e vector v = {2' - 5' 3) in Ef ls a linear
combinatlon of the vectors'
' vl = (1, - 3,21,v2 = {2' 4'-1) and vs - ll' -5' n' I
Solutlsn : trt v = o1 v1 + wv? + cts v€ where o't' oQ' and o3
are unknown scalars'
i.e (2, -5, 3) = crr (1, *3, 2) + wz 12' 4'-1) + cls' {l' - 5'n
= (or, -3crr' 2sr) + (W' - 44a' - o'zl +(oe' - Sos' 7%)
Equating correspondtng components and forming the
linear sYstem we get
a1 +2a2 + cls = ?l-g", - 4az - 5cr3 = -5 |2ot- *2+7o4= 3)
Reduce the system to echelon form by the elementary
operations- We multiplv lst equation O' .'.1Y::
tn*
subtract from 2nd & 3rd equations respectlvely' Then we have
the equivalent sYstem
dt + 2W+ os = 2-12W -2a3= I f
'5a'2 + Sttg = -L )
---'*'*-r<rifa*#
236 iI
COLLEGE I,INEAR AI,GEBRA
We multiply 2nd equation by ! ana then add with the 3rd
equation. Then we have the equivalent systema1 +2a2 * 0,3=2] or +2o,2+ 0e=21
2a,2 -2qs =^l l= 2e, -2o, = 1 |o+ o ='u)- o =rul
This system has an equation of the for* O = 3whtch is not true. Hence the above system is inconsistent
i.e it has no solution. Thus the vector v is not a linearcombination of the given vectors vr, v2 and v3.
Example 9. For which value of l. will be the vectorv = (1, 1",5) tn IR9 is a linear combination of the vectorsv1 = (1, -3, 2) and v2 = (2,-I, l).
Solutlon : kt v = o(l vt + aq vq where o, 1 and a2 arr-unknourn scalars. i.e (1, l, 51 = sr (1, * B, 2l + a2 @, -1, Ll
={or, - Scl1, 2o1) + (zae, -o"2, ad= (c1 + 2*j, -3q - o,2,2a1 + o2)
Equating corresponding components and forming thelinear system, we have
cr1 +2a2=11-3ar- ez'=)\l (*)
2o4+ %=5 )
Reduce the system to echelon forrri by the elementaryoperatlons. We muldply Ist equation by -S and 2 and thensubtract from 2nd and Srd equations respectively. Then wehave the equivalent system
a1 +2fr2=l Ifue ='l'+31
-3o, =3 !
VECTORSPACES,lWe multrfly 3rd equatton bY-i. Then
equlvalent system. ct1 +2a2=l
1fu = l"+3I :
e2 =-| )
This system. has solutlon for l. = - 8 and the qolutlon isOt =3, U2=- 1
t.e th-e systern (*) has solutiori for'}, = - 8.Hence v ls a linear combinatio[ v1 and v2 tf l" =:8..ffiplc lo:writethematrix"= [?-] I*"Itnear comblnatlon of the matrlces
^, = [; -i I,
^, =,Li ,1tr *o o, =[] I I
II zsz
we have the
E U.E T. 1S4Sokrttoa : Set A as a linear combinatlon of .$1, A2 and A3
usqg the unknourns rr,r, (b and os.
A1c1A1 +'tbzAz ttbAs.rlratis, [?:i I *o, [i-\ I .* [-l"'' [?: il = [3'-;'l . [-ff 3'l . tr' ;"'l
'Ja, +oa+gu; 0,t+b-ael='l,o-CIz+o -or+o+o IEquating co,mesponding components and. rrrming the
0,t+&Z*d'3= 3lC[1 +g]2- OS =-1 [
-o"2 = t Iol-c[1 =-2 )
ttrence the ttre solution of the system is ct1 = 2, wz = -1, n, =2Therefore, A = 2Ar -A2 +2Asi .
that lis, A'is.a linear comb{nation of ,{1, A3 q$ 43.
Exemple ;ll. write the matrix " = [?'- il "" a lirrear
combination of the matrices
o,={l 3] ,"=[? ?] andAs=[3-?]
I I*". Ii ;l
iI
I,.J
238 COLLEGE LINEARALGEBRA
Solution : Set A as a linear combination of ,{1' A2 and Ag
using the unknown scalars dt, h and cr3.
A=Ctr
[?-tl=Ar + de rA,2 + caAg
*[l 3J .*t? ?l .*[3-?l
eomponents and forming the
i]
=[xl f'].[3, LI .[8-ff]=[Il .*:l r?,:']Equating corresponding
linear system, we get
01 -B01 +2A3 =C[1 + 0,2 =
W- 03=-
Hence the solution of the systern is ct'1 = 3, wz = - 2, ac = - |Therefcire-A = SrAr - 2Az - AA :
Ttr:ur4 is a linear combination of A1, ,{2 4nd A3.
flf f:near qran of a shbcet of avrcetor spacc
If S is a non-empty subset of a vector gPace V, then L (S)'
the linear span of S, is the set of all linear combinations of
finite sets of elements of S.
Example 12. The vectors e1 = (1, o, o), .e2 = (o, 1, o) and
e3 = {o, o, l) generate thevector space mt..fo, any vector
(vr, vz, v3) € mP i" a linear combination of e1, €2 8rrd €3,
specifically (vr , vz, vs) = vr (1, o, o) + v2 (o, 1, o) + ve (o, o, l)= Vl€l +vz e'z + v3 €3'
Theorem 6.9 I€t S be a non-empty subset of a vectOr space
V, ttren L (S) is a subspace of V containiqg S. Furthermore' if W
is anrother subspace of V containlng S,.then L (S) c W.
VBSTONSPACES 239
Proof : If ueS, then lu = u€Sl hence S is a subset of L (S)'
Also L (S) ls non-empty' slnce S is non'empty'
Now suPPose that u, veL (S)' then
u = ctlul + ot2v+ "' + cl"u-and
v = prvr +52v2 + ..' + Prrvr, where
ui, v1 eS and oi, ft are scalars in F' Thus for )" peF
l"u + pv = l,(cttut + %!z + "' + ot"uJ t
+ F $rvr +fuv2 + "' + P"vJ
= (l,or) u1 + (la2)u2 * "r * (l'otJ u*+
ftrpr)vr + U02) Yz + "' + 00Jv"
which is a linear combination of the elements of S and so
is again in L (S). Hence L (S) is a subspace of V'
Now suppose that W is a subspace of V containing S and.
u1 ,1r2, ..., u.eSCW' then all multipl€s cr1u1 ' azutz' "" cq"u*eW
where cqeF and hence the sum c[1u1 * buz + "' + oqnum€w; that
is; W contains all linear combinations of elements of S'
Consequently, L (S) C W' Hence the theorem is proved'
Theoreq 6.1O : If S and T are subsets of a vector space V
over the field F' then'
(i) sGT+L(S)GLrD(ii) L(SLrI)=L(S)+L[I)(iii) S is a subspace of V if and only if L (S) = S
(iv) L(L(S))=L(S)Pr,oof : (i) Lrt u = c[Iul + %rJz + "' + a"u" eL{S)
where {ur, uz, "" ur} is a finite subset of S and
c^t, se,..., c&r€F' Since S gf' therefore'
{u1, u2, ..., ur,} is also a linite subset of T'
So u = (t1 u1 * uavz + "' + crru,'el- fI)4'e,
frn,,t.
6$'.
?
,14O COLLEGE LINEARAISEBRA
thusuel.is):+u€LfD. .'. t($ ELfD
HenceS e:r+L(s) cL(I)(ii) I-et uel, (SU'I) then
g =orur + &zW+..., +CIru" + Frvr + 01,.vz + "' + F-v-
where {ur, trz, ..., lrn' vt, v2. "1' vm} ts a flnite $ibset of SUT
such that frr, rr, ..., uJ G S and tvr ' vzr "" vJ Ef' 'Now ct1u1 + o,2W + "' + qtu"eL (S)
and brvr.+ gzvz + "' + P,rrv,rreL[I]'
Therefore, uel. (S) + L (I).
Eudently USUT) EL (S) + LG) (A)'
Let w be any element of L (S) + Ifl)then w =.1t * v for ueL (S) and v eI{t)'
Now slnce u is a llnear comblnatlon of a linite ndmber of
elements of S and v is a linear combinatlon of a linite number
;;;"*s ofT. Ttrerefore, t'* l9o" a linear combrration
of a finite numbers of elements of SUT' Thus u + vel. (St].D'
Therefore, from (A) and (B)' we get
. USIjD = US) + L(D'
(iii) Suppose that S is a subspace of V' Then we have to
prrye that US) = S. Irt ueUS)'
Then g r o,1u1 + e2W + "' + onuowhere c[1 ' cr2""' on€F
and ur, u2 , ... u,'eS. But S is a sublace ofv
Therefore, it is closed with respect
multiplication' and vector addition'
Hence u = 0lul + d2W + "' + crru,,eS'
Therefore, L{S} E S' Atrso'S g -Iu,{S}i
to scalar
VECTOh.SPACESl.,-F -r-
24t' i , L --r. .. ,!: -,t
Conversely, suppose tJrat L (S ) = S. then we hayg to prove
Now we lmov that US) iq a subspac" of t., , , .- : r, , , r
Slnce S = US), Therefore, S is also a subspace of V.
(M We kr:aw that L (S) is a subspace of V.
thefiore, by part (iii), it follows th,tt L 1q511 = L (S).
gle 12 (a) Show tlrat tJ:e vectors u i (1, 2,'B), rv:s (0' 1,
2) andyT= (0, O, 1) generate d. E;U"P. 19?51
Proof : We must determine whetl:er an qrbitrary veei<ii
v'= (d, b, c) in dc"n be expressed as a linear combinati#v'= xu + yv + zvr of the veetors u, v and w. Eryressing this
equation in terms of components gives.
(ab.c)=x(1,2,31+y(O, 1,2)+z(O, O 1) ;
- k 2x 3{ + (o, y, 25r) + (O, O, z)
= {x"2x+y, 3x+ 4 + z) i
Equating corresponding components and forming the
x =a ) z+2y+3rcc1 ri : :l2x+ y=b i+ y+2seb I .
3x+2y +z=c ) ra )
The above system is in echelon form and is consistent. Infact, tlre sys@ has the sohrtion x= 4, Y =b' 2a, z = c - 2b + a
rAu6 vand w generate (spanl mF.
ple 13, Determine whettrerrreeto.rs Yr =(2,, -1" 3J
yz=(4,1.2)andve={8,-l,S)span d. I
Sokrtloa : We must determine whether an arbitrar5rvectorv = (a, b, c) in d,W be orpressed as a Unear combinati6n
v = orvr * o'2v2 * ct3vs of the vectors v1 , v2 and vg.&rprqssing ttlis equation in terms of components, we get
Ltnear Ngebra-I6
w?(ab, c) = at (2,- 1, 3) + oe 14, t,2l +% (8, - 1, 8)
r',r..: ,i j,..., I .. ':r.t , '-. .i,: i i
= (2a1, - cr,1, kr) + @o"2, aa,zoel + (t3os, - as, Scts)
= (2o4 + 4uz + S%, - or + q2 - oa, Sar + fyq.+ *6)Equati+g. cg:IfFPold[g.3opp11nent3 and formin$ the
2a1 + 4uz + 8cr3 = 41
llnear system, we havC'd I + 'W - os = b iir-i r ir."r: i' r 3O1 + Zoq + 8o3 =9.,
: ?hts pftiblem thus reou*eesito deterrrliirihg whether or not
Wrsystspl$ consistent for all values of a' b and c. No-w thisssstefn y{!l F-e iaorlsiqtg$t,fo{ all a, .b and e, if and only if thematrix qf co-efficients.ii{):l'.rr1rri',r i,:,i .. .r2 4 81'' '"' A=l'-1 ll -1 lis invertibte.
L s 2 Bl
= 2(B + ii - a1-g+ 3) + S(4 - 3)
=2O+20-40=O.
COLLECE LINEARALGEBRA
il:.
4",
l'{
I
lz 4 8hl=l-r I -l
ls2 8
Hence the co-efficie nt matrix A is'not invertible and
"orr".qycftly, vr, v2 urnd v3 do not span IR3.
u1 = (1, i,b, ri, - (-2,-4,-8) and u3 = (3,9,27) generate
(spud d., Solution : We must determine whether an arbitrary vector
u = (a, b, c) e tri3 "r, be expressed as a'linear combination
' u=xrur +huz*x3u3 of thevectorsrr,u2,zlnd u3'
Bxpiessing this equation in teil of compbnentS, we S
11!,")="('' l'') .*F2'4'-+r)+ xs(3'e'2v :
: (*''gu;' ]"1),f, tr-,zn' -4xi',8*,) + ts*' s" z z*11 '
VECTOR SPACES 2A&
Now form the system of llnear equations by equating the
eorresporldlng components.
.'. n=*,(sa-zb +2c). ' )ii
x1-2x2+ 3x3=2iixt -4h + 9x, =b Iixt-8;h,+27a5=c)
or, equivalentlyx1 -2x2 + 3x3 = a1"'"' '- ''x1- 8x2+ 18xs=2bf (1)x1 - 32x2.+ loSxs =4c J
Now we reduce tHe system to eeHelon ,,form by the
elementaty transformations. We subtract ftrst equatiori frorr$lthe second and thtrd equattons respectively,'Then we haveE
theequivalentsystem , ,, .: r r .eEI
xt-2xz + 3x3 = 3 I !)-.6:q +15x3=2b-al (21 E
-3Oxz+lo5xs;4c-a) "9We multiply second equation by 5 and then subtra"t fro*fl
the third equatlon. Thus we'get tlib' equiValent system
x1-2x2 + 3x3 =4 I-6xz+15x3=2b-a : i6t
which is in echelon formFrom the third equation, we get xs =i (2a- 5b + 2cl
Putting the value of x3 in the second equation,
we get - 6xz + 2a- 5b + 2c = 2b * aor, 6x2 = 2a - 5b + 2c - 2b +a = 3a - 7b + 2*
Again, putting the values of x2 and xs ilr.the first equationwe get r, -l ts" -zb +2c). i {2"- 5b + 2c) =-r' . ,
:
Ii
,l
.{"{I{I''l
tia?_ coLLEGE t INBanALcEBRA
"oi, iS"i - rsaigbu- loc+6a- tsb+6c= 15a
or, 15x1 =%La-2'Ob+4cxr =* {%La-zr)b+ 4cl.
Therefore. (a, b, c) = fi \Z+^-ZOA+ 4c) u1 +
* tg" - zb + ?.cl uz +* {zr - 5b + 2c) us.
Hence ur, u2, u3' sPan trf .
Yerifioatlon of tlrc result(i) *@q"-20b + 4")-i $a-7b+2nl +*tz"-5b + 2s)
=]utz+^-%)b +4c- 15a + 35b- loc + 6a- 15b + 6c)
,, .: *$ {SO"* 15a + 35b- 35b + l0c - loc = fi {tSu') = *
", (ti;, $ .6tza- lob + 2c)-i tou- 14b+4c) +i 6a- 15;b+6c)
'':'" ''ofigza- lob +?,c-Soa+7ob-20c+ t8a-45b+ 1&)
= S tsO^ * 3oa + 7Ob - 55b + ?,oc - ZAd
=ft {rsu) =u.(iii) * to"- 5b + c) -ittz"-2{3b + 8d "*
(18a-45b + 1&)I
=fi tOa- SU.+ c- 6Oa + 14Ob *4Oc + 54a- 135b + 54c)
=StOO"-6Oa+ 14Ob - 14Ob+55c-40c)
=* (15c) = c.
Hence tlee result is verifled.
6.12 Row qnae end aolumn sllEcc of e ua'rtx
Irt A be an arbitrary m xn matrix over the real field IR:
l-arr dt2 arn Ia=1"" 4zz a.zn
II ... I
L"*, an2 a-n IThe m rows of A are Rq= (a11, a,12,,"', arr,)
Rz = (azr, dzz, ..., aar). .,., Rj, = (a*rr 462,"',a6j' j
,:1,: - i. I' These m rours viewed as vectors in Rn span e subspace of Rn
called the tos sPacc of A"
.YECTORSPACES :. ?lp
Agaln, the columns of A are ; , :; ,, ,.i {
[arr I larz I [ar" Ic, =l ::' I o=l?.'.' I .,
"" =l ::" I.L;;,J-L;;,J L;;"] ;
These n columns of A viewed as vectors ln Rm' span a
subspace of Rm called tlle cohrmn slnce ofA ,
Defialtioa : For any trvb subspaces'S'"rrd T of 'a vbctor
space V, thelr sum S + T is the set given by {u+v 'l ue S and veT}'
Thdorem 6.11 For any two subspaces S and TofavectorspaceV, S +Tls a subsPace ofV.
Proof : CIearlY, 0 = 0 + OeS +T : So S +T * oi,e.S+Tisnon-emPtY.
i:, 1.,:. ,
i,
IJtrcye S+T, ot, peF.Thenx=ul +vr'Y=rrz*Y2
where ur, uz eS and v1 , v2€T.
Thirs ax+ ff = o (u1 +rlr'1 + Pfu2 + v2)
= o(ur * cnrl + fr:s + pv2
=coltl*Frz+cr,ir'+frz'(l) " : 'Now stnce S and T'are subspaces of V' ' i
crul *frrze S and cnrl + Pv2e T. ' ' : i
Consequenfly, (1) Cfives that ax+ pyeS + T.
Hence S +Tts a subsPace ofV.
6.19 Dir.ect sum of nftcPaccs :
The vector spaee V is sald to be the dlreat gufoirbf'lts
subspaces S and T, denoted by V = S @ T if every vector veV can
be written in one andonly one way as v ,.'u + w, where ueS and
weT.?heorem 5.12 The vector sPace V is the direct sum of lts
subspaces S and Tif and only if (i) V = S + T and (CI SnT = {O}:
Proof : Suppose that V = S (DT. Then any ve! 9,9 be
uniquely written in the form v = u * w, where ueS and weT.
"*fl\1l,) *:il,iii
.
VECTORSPACES 2e7246 COLLEGE LINEARET,bBARA
Thus in particular ; V = S + T.Now suppose that veS O T, Then{1) v = v + O, where veS and
'eT.(21v = O + v, urhere OeS and veT.S-ince gugh a srrrry for v mlpt be gnique..'. v = o; Accordingly SOT = {O}.
01 the other hand, suppose that V = S + T and SftT = {O}
kt ve V since V = S + T, there odst ueS and weT such thatv = u + w. Now we have to show that such a sum is unique. If ttresum ls not unique, letv = u' + rV' rvhere u'e S and r/ e T. Thenu + ur = u'+ w- 45rd so u - p'= w'- rr. qut u- u-€$ and w'-weT,hence by SnT = {O}, u-u' = O and w' - W_ = 0.
Theref,ore, u = p and w = w'. TtU6 FuFh a sum for ve V isuniqueasdV=S(ET.
ffpnte tF. Let S endTbe suFsp4cep of na 4"g4ea ryS = {a, b, o, o} la, beF} ana T = f(O. O, c, d} h, a .F}. One can
easlty verts that both fhese subgets qre subspaces of IRa.
The zpro vector of Ba !s e, o, o, e. TtpF z "
ta, b, O, S= {O, O, c, d) for sorne ai b, c. {e!lifpp[s tha]ir = b = c = O. Flenqe z.* O, &th4t S OT= pf.
Therefore, S + T = S (ET, Npry pver.y {et,b;r pr, 4r} " #"un
be expressed as {a1, br,O. o} * @..P, cr,dll urtth (ar, F,,. O, Ol e S
::' , pfpB0l6pff-ff {4,l. [,.etu = (1. 9, l, -Z),v= (3, O, 4, U a&C
'',,"* :. {6, 3, - 3. 0 be thp veciors in Uf . Compute"the foilowingretons: {02u +v-w(il) 5u-3v-w.
,ffaryrerll: {i) (-1, f,9. -3}{rt} F to, 7,-4.- r3).
2 . {0 flrrd xe d cuch that u +x=v, : l
$rherc u = (2, -1, O, 3. CI andv = (O'l' 2, -L,-2) D' U' P' 19ml(lt) Solve the vector equatlon u + x = v for the following
palrofvectorsuandvm d:u=(| s' l- o' r) :' '!'t''':
Anmr: (i) x= (-2,2,2,4,4)(ulx=(0, r'-t'-tz'o). d: , :
3. Let u1 = (-1, 3, 2, O), u2 = {2,9, 4., - 1)' u.a f F, I' l, '*l nnd
E+ = (CI, 9, l, 2). Flnd scalars crr, a4, Qs ?r1d ct*.6pch thato1u1 * &zrtz + ds us + %u* ={p' 5, 6, -3)
Ars-rEr i o1 = L,or2= 1, 06 =-1. ao = l. ,:,.:.: _
4. LetV be a vgctor sPace over the fleld F lrt u :m{ v any
Shorv that (t) (-1) v = -v(ii) s(u - v) = cu - cw-
S.Verify whether the followtng sets are subspaces of. . a','vg (lR):
r i, l
{i} t0q 2Y' 5) : x YelR }
(ii) fiiq x+y,*z),& Y, z€IR l.{nf*f* 0l t!€ eet ls rpt a subspace' '
9f EP.qv tlr,at lil = ltra, h, c, d) l[a" U, c' 4) :b - 2s- d = 0, is a strhPace of B4 .
F. U',P, l{QtfnaBn4, I ,. '.;. 1
p-, F, s lpsfl,that lV
= {(a, b,
"l e ts" h+b+c.: !1,!+,1,pghPfapE
, pp:q lF{b, c, d) eB" la +b + c + d
= 0l is athat V = fia,
Itfl
i
,,{
fI'}.
I!l
,},
of ts4
)
I
i
I
l
[{:s{*{1, , i. .
.,(r),S^;, {k y,. z) eIR3 ly - oz = ol
, Fr T= !,n_\o =IR3 l**y ::; rl
COLLEGE LINEARAICEBRA';i..i' '::j.
of the following are subspaces of IR3?, :.,';i,.. VECTORSPACES
19" .Write the vectors (1, O, O) and (O, O, I) aseombinattons of the veetors (1. 0, - l), (O, I, O), (1, O, l)
Aorrtl8 : (1, 0, q =;(1, O,- I) +0 (0, l, o) +i(t, o, U
249
linear
-4 (o, o. 1) =-)0,Q -t) + o (0, r, q +; (1, o. 1)
WWrtte (8, 6, 0) as a linear combination of (-I, 2, Ol,B,f,q.@,-t,oland(o, t,-1) [D.u.pnel: ts3]
-An-srer: (5.6,0)=2(-1,2,Ol+ t (g,1,2)+ I (4,- I,O)+2(O, l._l).@. Wnrther or not the r"rror- (1, Z,;i't"; ,rr.",
cd:dination of the vectors u1 = (2, l,O), r:2 = (I, -1, 2)and us = (0, 3, -4) , [C. U. p. Ig?gfAnawer : (1, Z, 6) can not be written as a linear
coTF\)affon of u,, r.9 ard us.
ffiWnte the vectors (2,8, -Z, B) and (- 4,6, -lB, 4) as aiirr# comblnaflon of the vectors
, v1 = (2, I, O, g), v2 - (3, -1,5,21, vs = pl, O,2, t)Ansrcr 3 ti (U3, -7, 3) ;.2v1 - vz : vs
rO0- (iil F4, 6, -13, 4) = 3vr - Bvz -vs.\fl O.turorine u,tether or not the vector (8, g, -4, -2) is a
lineAr-eombtnatlon of the vector set {(1, -2, O, A). {2,8, O, _l),(2, -t,2, Lll D. U. p. 1S4l D. U. II, T. 19Sl
Aorser: (3, 9, 4, -Zl= I (1, -2, O, g) + S(2, B, 0, -t)
24. Express the potynomial p = t2 * n - . 3J1 *;? "'L*,comtrinatlon of tlrs polynomtals .. 1 .,
pt =F -^.+5, k *h, -St and ps =tt3.ADsmr:p=-3p1 +!p +4ps:' I .
25.|n tjre veetor space tsP erpress the v = (1, - 2, S) as aIinear combinaUon of the vectols vectors v, = (1, l, 1),
v2 = (1, 2, 3) and vs = (2. -1,.1).Anccr:v={vl +S,ir2 +2vg.
t
ID. U. P. 19851
Urit W = {(a, b, "t lr;
U, ce IR and 2a - b+ c = 10} isofVs 0R ). ID. U. P. 19341
Shovs that each of tJ:e following subsets of the vectorls a subspace of EP :
{i)w=11*b,c)la+b=6; - D.U.II.T.1986l(ii) W= {(a, b, c) lb + c = 0h'(ui) W= {(a, b, c) lc + a = 0}.
13. Let W = {(a, b, t) I ", U. lR } be a subset of the vector
spacetffP . Then show tirat W is not a subspace of IRp .
p. u-H. T. resl,:
, L3',*1* = q'b' c.) la2 b) be.a subset of the vector space mP .
Then show that W is not a subspace of IRF . tD. U. IL T. fggElI5. Show that W = (a, a, a) la e IR l ts a subspace of Eg .
B. U. IL 1e851
16. show that each. of the follorrrng subsets of -thq-retorspac€ mP is a subspa.ce of IBp : .
:, r," (flWl ={(ab,O) la,belR} (DWe=(O,b,c) lb,ceIR}.l*r#"',t7.,J-e* Y be a vector space of all n-square matrices.over areal field IR. Show that W is a subspaee of V if W consiits of
B.U;II. fry&&.*-"y""inetrie matrtces (i.e At =-A).' '' i& Shornr'that w is a subspace of IRa where**iirir
=(a;lb. q o) la, b, ceIR l*r' ($W= flxr, h. xs,*] la1x1 +&z xz+ssxs +%x4=eqeIR I
q.i
25O COLLBGE UNEARAI'GEBRA
36. Determine rvhettrer (4,, 2, L, O) is a-linear combination
of ich of the foUprvi,qg sels of vectors. If oo'find s'e such
combinatio$.(il l{L,2,-1, OJ, (1,3, l' ?)' (6: 1, O. 1}
{iil 93, I, o, u, [1,2' 3. 1), to' q'_6'-61]
iliit flo, -t' ?, u, (1. v,4.-21. (8, l' o' o)' (3' 8' -2-' -1)l
emrg" i (4 (4' 2, L, ol ls not e llnpar co-mbfnat{on
iiii- ti, 2, t, ol is ngf a linear corrpfnat[on'(iii) @,2, L, O) is a linear combination and
{4, ?,1' 9: 216. -L,2, 1) +.1 (L' 7' 4' -21
-3(3,1,O,0) + O(3' 3' -2.-1)
z?.Express the matrix " = [-3 ll * a [near
combrnauon of the matrrces ^,
= [l 8l' * = [l -i I
srd AB = [-? el p.u.Er. lsEN
f*rFGr:4=!$'*SP*tt& l9 rl28. Express (if popqlplel the mqtrtx A = [-1 -eI +s a
ltnear combinatlon Pf the 4PfftcPa*;=[i-i[;:i-l ll ;;*=[l 3l
p.rr.E.t 19O$8-['s rgpq
fArppf I A parr no.t be erpreesed as a llnear conrbtrt+4qr of
Ar, Ae a4d A3.- ?i. Whlch of the following matrices are llnear
paq+HnaHons of t-he matrlcPa
^=[-l 3l'"=E 11,".[fi 3f'
',[3 3l *[-3 rlltprrp, u,[3 8f =n*n*c
u,l [-3 :il "2A-,ts+c'
VEC-TIORAPACES
, bt s =ll,2,l), (3, 5, o)l and
l), (3, 5, o), (2, 3, -1)l
251
30.In qB3
f ={1,2,e-whether< S> * <T>. p. U-P. fgFq
h Show that the veetors Er = (1, U eld u2 = (1,-!f spap
R2.
,..3),6eterrnine whpther ol not the followtng vpptors span B3" (il ,r, = (1, l, 21, rt2 = (1, -1, 2), us = (1, O, l)
-4W, = (-1, 1, O), v2 = F|, O, 1), v3 = (1, 1, l)Aemr: (i) ur, uz.ue span IR3
lfi ut,v2,vq sPgn m'3.
p9,.6noW that the space generated by the vectors
u1 I (1, 2, -1, 31, *z*{2,4, L,-9ilandtrs = {3, 6, 3, -n and tIrcspace generated by the vectors v1 Z 1L, 2, -4, I U arld
v2 = {2, 4, - 5, 14) are equal. te.u.P. tq7g|33.{il Show thet Ep = ffI,8, U, {&, l. g), (l;-1, 8}
Q.u.Er. tgp$(til FInd a condttlon on a, b, c so ttrat v r (+, b, Q
ic a llnear combinqtion of v1 = (1., -S, 8) a$
vz- lP, -1, -U, that !B , so thatvbbngB to spn [vr, vzl.
fncr: a+b+c=O.*. WU ard W be the srrbspea of IBp deflnpd by
ShorY that Ro = U@tt/' , P-.V. A. fffi35. 6hgv that eEch of the follourlng subsets S is a subspace
of the lndicated space V:{il Ye lR3. S is thp collectlsn pf all trlples {x y, al auch
r$a!x =yard z*Q,(ii) V= Ifa. S ia tlre cpllection of all rF-tuples (x, r,z,$)
wchthatx-Y=?+L
,,25i2 COLLEGE- UNEAR AI.GEBRA
36. Show tltat:e, p1aneW = (a, O, c) in R3 is generated by
(i) (1, O, l) and (2. O. -1)
. tifl) (1, O,21, (2,O, 3) ard (3, O, 1).
,37.li) ProvethatS={(xy, z,tleIRa lx +y-z +t=O..and
Zx=yl is a subspace of IRa
(ii) Pfove thatT = {(xr, xz, .,,xJeRI lx1+x, +... +;q 1O}isasubspace of IRn.
38. kt Wr and W2 be the subspaces of HF deflned by
Wr=fia,b,O) labelRl l
arldw2 = (O, O, c) b€IRI. Show that IRP =Wr@V/2. .
39. kt W1 and W2 be the subspaces of IRP aefinea tyW1 =(a,o,O) laeIRl andwz=(o,b,d lb,ceIRl.
Show ghat EP =Wr (il[2.
N.Ir:LS andTbe strbspaees of IBf dellned by
' S = (a, 0, O O)' I a e IRI and T = (O b, c, d) I u, e, d e IRl.
Show that IHf = S(ET.
4r.tew (IR) = {[: ] J , .,b, c, d . ts']. rhen strow that
V(IR)=Wr Orillz where
iwz = {[: 3 I ,". R].
:
42. I,et V be the vector space over ttre lield F.
Shour that V = Wr@ W! where Wl and Wz are the subqraces
of symmetrlc and skew-symmetrlc matrices respectively
over F.
II/ECTOR SPACES 258
Linear dcpeodoacc md llncer lndependcncc.Definitloa : Irt V be a vector space over the lteld F. The
vectors v1 . V2, ..., Vm eV are sald to be ltncarly dcgendcnt overF or slmply dcpcndcnt if there exists a non-trrvial linearcombination of them equal to the zero vector 0. \
That is, flrv1 * %vz +.., + oqnvm = O whene ar * oforht treast one i.
On the other hand, the vectors v1 , v2, ...., vm inV are said.to be llaearly independeat over F or simply in@GudGat if
the only lrnear eombination of them egual to.o (zero vector) isthe trivial one. tleat is,
e['1v1 * ezvZ +... + oqnvm = O if hnd only trdt=&Z=...=(LIr=O
N A single non-zero vector v is qecesgarily independent.
*'iffi". crv = O if and only if a = o.
X@"eoElrn 6.13 The non-zero vectors v1, v2, ..., v' h a vector(\/space v are linearly dependent if and only if one of ilre vectorsV1 is a linear combination of the preceding vectorsVr, v2 . ..., Vk-r.
RroOf : Ifvs=GtVl +%vZ +...+Gk_rvr_r.
fi, ttrencrlvl +%vz +... +Or_rvk_r + (-I)v1 *ov1a1 *...*ovr,=e.
\!. and hence the vectors yt,v2, ..., Vn are linearly dependent.Conversely, suppose that the vectors vr, v2, ..., vn'are
linearly dependent. then srvl + a.2v2 *... + c'vn = O where tllescalars q1 * o, ttren c1v1 + %vz + ... + c1v1 + 0vL*1 +'...'+ &r, = 0
Or, C,rvl + %vZ + ,.. + 01V1 =Q.
Now if k = l, this implies that c1v1 -O.
B$4 COLLEGE LINEARALGEBRA
r*rlth o1. + g" 90 that v1 = 0, Slying a contradiction' lrqequse
the v1 ls a non-zero vectof' Hence k > I and we may
wr,tevp=-(fJ,' (tr) Y2-..- (H)*-,gfvlng v1 ?s s llnear combinatioh of v1' Vz' "" vL-t'
Thus the theorem ls Proved'
L.fi6r 1. L€t V be a vector space which is spanned by a
flnlte set of vectors, saY {vr ,Y2, ..',vr, }' Let U = {ut' u2. "" uit} be
an itidelibndeflt set in V. Then k S'n.
i*Oof : Since &r, vz, ..., v,,l spans V' eactr vectoi lrt U is a
llneal combihaUon of {vr ;ur, .",v',}, i'e'
lI1 = O11V1 * d'21v2 + "' + okrvn IU2=O12V1 *d22v2 +..'+Clm2vn | (1)
{
t\ = 61r*Y, * 0.2YY2 + "' + a4vt 'l
Suppose that k>n, then tl:e system
C[11X1 * A12X2 * "' + C['1XJQ. = O
' CL2tXr + dZ2h + "' + CZtXf. * O
'.:ClntXl + A19X2 + "' + Ar1X1 - O
has at least one non-trivial solution, say fbr ' b""" h')
c[11b1 + apb2 +... + cr11b1 = Ol
02rb1 + o.22b2+...+cr2kbk =Of (3)t
,cnrbr * a",.2b2+ ... + ohkbk = 0
Thpn b1u1 +t:1.ttz +'.. +Lrr.gr.
,: . = !, (cr11v1 + ctafz.+ ... + oktv.,). +
+ bg(c1r.vr *,&2YY2 + "' + q4vrr) .
(2t
w@nspAcEs
= (orrbr + o,12b2 + ... + ag brlvl +
(aerbr + o.4bz + ... + trertt) vz +
+ (d"rbr + oab2 +... + qlbp)v,0v1 +0v2+...+Ovr,=Q. r.
Thus tf k>n, we can chooge b1, b2; ..., bk not all zero suehthat b1u1 + ... + hur = O t. e. {ur, uz, ..., ur.} ts a dependent aetwhlch ls a contradiction to the given eorldition, go weconclude that k < n. Hence tlte lerrtrna is proved.
IfAAa 2.lx:t {u1,;Ir2, ...,u.} be an tndeireUdent set lii dvector space V. Let W be the sub3pac€ spanned by {ur, uz, ..., rlr}.[.et v be a vector ruhich is in V, but not in W then
{ur, uz, ..., u, v} is an ttrdependent set.
Proof : Letclul + o4uz+ ;.. +ol rtr + cv= O (l)We will show that et = % - ... + c(r = c[ = o.
8s.6
If o * o, solve for v. This gives. iGr A.o- 0-v=-EUr-o=u2-...- Ar+,:
The right hand s!d9 of fhip expressin (2) is a linearcomblnation of u1 , t)2, ,..,, F" 1!fl |re4ee in W,,b,ut the left handsl<le of (2) is not in Wt which is q eontradiction..Bewe eonelirdelltirt a = o:,Tlri€ $ives ixiui + buz,+ ... + qk{rr,iQ.;,; i': /,.,, r, :
Since {u1, rr2,-,..,.;iu. },is arr,independent s8t, 'rilU dttrisit,trave' 't\-
Hencewe have stfown thdt Ul =&) =... =cq. ='g=6 .
'lhercfore, {ur , uz . .J:,"u;,'v} is an iirdependent set. , , '
, . .,,:- , ., i; _l- r:
So the lemma is proved.'' :.: :ii;1 .rt ', j: ,i r.' ri i r: I :t 1 . ... t'
Thcorem 6. 14 The non-zero rows of a matrix in echeloni _i ; , :
hrrrrr ilre llnearly independent.
r::)(2)
all *(X2 =..,=0(r=O.
COLLEGE LINEAR AI'GEBRA
%of. , Suppose that the set of flon-zero rows say
&o-r, ...,Rr) is linearlv dependent rhen:l:*:11''*"
say R- is a linear combination of the preceding rows i: e'
R- = ocrn+t &n+r + oLn+z &n+z + "' + *t ^ll ia *c.Brs
f'r
II
iii
tF
il;;$#;;;; rtr' "o*ponent
orR{n i8 its flrst non-
zero erltsy.Then slnce the matrix is -ln
'echelon form' the ktll
components of ";.;
"' Rn are all zero and so the kth
cornponent of (r) is ctt-l o + cr"'*2 o + ao*2o + "' + crno = 0' But
this contradicts tf'"1"1'*puon that the kth comn3ne3t of R*
is not zero. Ttrus Rr'Rz' "" Rn are linearly independent'
Hence tlre theorem is prored' . r^-^-ttcorgm 6'15;;''' t'' "" Vm be rn linearly independent
vectors and a vector u is a ltrpar combination of vi' v2' "" Vm
i.e. u = cl1v1 * hvz +"' + clmvm where €t1 dfe scalars' Ttren the
above representation of u is unique'
Proof : Giveng=cl1v1 tCL1v2 + "'+o(mvm {I}
Suppose tl.at the representation of u is not unique'
' Ttren let u = Frvr + P2v2 + "' + F-v-where Fr are scalars'
Nowby subtractin$ (2) from (1)' we get
o=u-u = (crr-Fr)*ta, - Fzlvz*-'-'*(cr',,-FJv-
But the,t"to*-', ' w"""vm are linearly independent'
So that this implies or - 0t = o' :"' oqn- F* = o
i. e, ol =Ft' "', cqrr= grn'
This proves that u = Glvl * c,2Y2 + "' + c*vt is the unique
representation of u as a lineai comblrration of v1' v2' ""' vrr'
Thcorem O'fe Supose ttrat the set {vr' vz' ':" vrn} is lireearly
dependent' Then w is a linear combinatign of the vectors
!t'Y2" "' vm
ITECT,OR, SPACES 257
Ibdof : $foicc tXre cet {vr ,Y2, ..., v*. \il} is llnearly depeqdent
there exist scalafs o1, oa' ,.'. q* F not all o (zero), such that
o1v1 f hvz +... + (&nvm + Prv = O'
If p . o thert one of ttre o{ ls not zero and
c1V1 * c.2v2 * "' + o'mvm = O' But this contradicts the
hypothesls that {vr,v2, ..., v6 } is linearly independent'
Accordingly P* o and so
o,* (-?) ",.(-?}v2rr..+( r") *That ls. w is linear combination of the vectors v1' V2, ;', V6'
fi\"or" G.1? I,etur, uz, ..., un be any n linearly,'
lndependent vectois in a vector sPace V. Then any (n + l)vectors y1,Y2,..., Vn+I, each of whiCh is a linear cornbination
of u1, u2, ..., un are llneally dependent, /Proof : We prove the resuft by inducton on n. If any of the
v,' s ls zero, then trivialty given (n + 1) vectors are linearly
dependemt. So we suppose that none of the v1' s is zero. Now as
v1 and v2 are both ltnear combinations of u1, :u2, ..', "', u' withn = I, We $€t vl = a{rr ' vz =- Garrt with a1 * o' o'2 * o'
This gives,r, =*,*, =fior, \ -Y=ur -ur= o.' d1 g-z
-lor, vr - argi v2 =OFlence v1 and v2 are linearly dependent.To apply itxduction suppose that the result holds for an5r k
linearly independent vectors rvhere k < n. We can now writeV1 =Ct11u1 +Cl.lzU'z +"'+0lnutvz = GztUt +ePP2 +."' + 02run
Vn+l = On+llUl * Ctn+tZU2 + ... + OQr+lnUn
fut sume sealars ary in F.
Llnear Algpbra-I7
,1, .
'. iri
zSE,
r$l , t,
COLLEGE LINEARAICEBRA
I[qr1,:f o forevery i=1,2,..., n + I t]ren each of vl is allnearcombination of (n - 1) vectors u1,Ir2, ..., urr-t and the inductionhypothesis will yield that v1 , Y2, ..., vn. are also linearlydependent and consequenflyVl , v2 ... , vn+l are also linearlydependent
So we suppose tl:at at least one of a6 * o.
Irt q, + o, ttren for each I = 2, 3,.., n + Ivl - 0rr, 0irr-lv1 = (air - Cftrr(l1101r-l) u1 *(0;2 - C[inC[,12gln-l) ttz * ... + (cti'rFr - 06C[1'n-1C[1rr-l) Un*l
So by induction h5rpothesis the vectors
wr = Vr - ornorn- lv1 (2<i <n + 1) are linearly dependent.
Conseguently, there gdst 92, fu, ..., Bn*r in F; not all zero, such
that fow2 + fuws + ... * p,r+twn*l = O i. e.
fu2 (v2 -rlu2nalr,-lvr ) + Fr (vg - d3ncllrr- rv1 ) +
. + &,*r (v,,+r - 0ea1'11C[1rr-lv1 ] =Q
i. e. - {fu20,.2notn-l + foo3rrcrlrr-l + ... + pnal crn'1'1o1rr-l) vr
+ Fzvz +ftva + ... + p.,*r vn+t = O.
This gives that v1 , Y2, ..., Vn.'1 ore linearly dependent'
Hence the theorem is Proved.
Theorem 6.18 Every set of vectors containing a dependent
subset is dependent and every subset of an independent set is
independent.Proof : Eirst PortionSuppose that the set S = {vr. vz, ....v-}
constains a dependent subset' say {vi, v2.. ..., vr}
Since {vr, vz, ...,vr} is linearly dependent, there exist
scalars G.t, a'J,..., oq not all o, such that
O(1V1 * UAVZ +.'. + O(V. =Q
vEcToRSpACES 25g
Hence there exlst scalars c,t, o.2,..., o. o, .., o not all o
(zero), cuchthatc,lvl * o,2Y2+...*orVr *ov111 +... + ov-=Q'Accordtngly, S ls dependent.
Eccond portlon : kt T = {v1, v2, ..., v,,} be an independent
set then Prvt + Fzvz +... + pr,v- = O implies that
0r=b=...=Br=ONow let {vr, vz, ...,vr} k < m be a subset of T, then there exist
scalars 9r, fu,..., Pk whereeach p1=o, i= T,2,..., ksuchthatprvr + hvz +... + fuv1 = Q the subset {vr, vz , ..., v1} islndependent.
OUT EXAMPI,ES
Prove that the set ofvectors
lQ, 1,4, @, t,-l), (4, 3, 3)) is linearly dependent.
Proof : First Procesa
Set a linear combination of the given vectors equal to zero
by using unknown scalars &y, z: .
x(2, 1,2) + y (O, l, - 1) + z(4,3,3) = (O, O, O) , .
or, (2x, x. %d + @, y, * y) + (42, 32,32) = (0, O, O)
or, (2x+ O + 42, x+y + 32, 2x-y + 3zl = (O, O, O)
Equating corresponding components and forming thelinear system, we get
2x+O+42=O) ",!x+y+{"=6 I trl 4
2x-y +32=O )
Reduce the system to echelon form by the elementarytransformations. Interchange first and second equations.Then we get the equivalent system
x+y+32=O)2x+O+42=O I el2x-y+32=O J
".",1
2@ coLLEGE LINEARAIcEBRA
We multiply first equation by 2 and then subtract from the
second and third equations respectively. Ttren we get the
equivalent sYstem
x+ Y+32=O I--iv -zr=9 I (3)
-3i-32=o )
Divide second and thlrd equations by-2 and-3 respecttvely'
Thenwe get the equivalent sYstem
x+Y+32=O Iy+ ,=ol (4)
Y+ z=O )
Since second and third equatlons are identlcal' we can
disergard one of tJ:em' Then we have the equivalent system
x+y+rz=3 1 (5)y+
This system ls in echelon form and has only two n.orr-zero
*u"*n" in three unknowns' hence the system has non-zero
solution. Thus the original vectors are linqarly dependent'
Sccond Proccss : Form tlte matrix whose rows are the
given vectors and reduce the mltr{ to row echelon form by
Gfttg the elementary row operations'
12 | 21lo I -1 I
L; 3 3J
. we multiply first row by 2 and then subtract from the third
we subtract second row from the third row'
12 I 21-lo I -l I-16 o ol
Thls matrlx is tn row echelon form and has a zero row:
hence the giverllectors are linearly deperrdent'
r.xatnPle 17. Show that ttre set of vectors
(3, o, 1, - 1)' Q, -l'o' 1)' (1' 1' 1' -2)) is linearly dependent'
ID. U. S. 19841
Proof : Ftrst lEocesg
Set a linear combination of the glven vectors equal to the
zero vector ,ltttg unknown scalars x: y' z t
x(3, O,'1, - f) + y (2' - 1' O' Ll + z (l' L' l' -2)= (O' O' O' O)
or, (3x o' x -d + (2y'-y' o' yl + (z'z'z'-2zl =(o' o' o' o)
or, (3x+ 2y + z,O - y + z' x + O + z' - x + y - 2zl = (O' O' O' O)
Fornt the homogeneous linear equations by equating the
corresPondin$ comPonents'
VECTORSPACES26t
Reduce the system to echelon form by elementary
transformations' Interchange first and third equations'
Then we have the equivalent sYstem
x+O+z--O)-Y+z=q I el
3x + 2Y+ z,= l) |
-x+Y.-22=O )
we multiPlY second equation bY -1' ,
3x+2Y+z=Ol-!+z=O L trtx+b+z=o I
-x+Y-22=O )
row.' 12 | 21
-lo I -r I
Lo I -1J
t.
l,
T
J,tt
262 COLLEGE LINEARAI,GEBRA
We multiply first equation by 3 and then subtract from the
third equation. We also add first equation with the fourthequation. Then we get the equivalent system.
y- z=O I
iY -22=o | (3)
y- z=O )
Again, divide third equation by 2.Then we get the
equlvalent system
' Since second, thtrd and fourth equations are identical, we
can disregard any two of them. Thus the system (4) reduces to
*. 9: z:31 (s)
This system is in echelon form and has only two non-zero
equaUons in three unknowns ; hence the system has a non-
zero solution. Thus the original vectors are linearlydependent.
Sccond Process : Form the matrix whose rows are the
glven vectors and reduce the matrix to row echelon form by
usirlg elementary row oPerations :
13 0 I -ltlz -l o rlLt I I -2)
Interchange lirst and third rows'
l-1 I I -21 --lz -1 0 1lLs o I -ll
VECTORSPACES 263
We multiply flrst row by 2 arrd 3 and then subtract frorn
the second and thtrd rows respectively'
rr I I -21-lo -3 -2 5lLo -3 -2 5J
We subtract second row from third row'
{n u..torc are linearlY dePendent'
18. Show ttrat the vectors (2' - l'4) (3' 6' 2) and
ir, *,:4) are linearlY indePendent'
,l"i,:i
,1,
x+0+ z=0 ): ;-1,=|f e,y- z=O )
Proof : First ltocesg
Set a linear combination of the glven vectors equal to the
zero vector usin$ unknown scalars )c'y' z "
x(2,-1, 4) + y (3' 6' 2l + z(2' 10' - 4) = (O' O' O)
or, @x - x. +$ +(3y' 6y' zyl + t2z' ltz' - 42) = (0' o' o)
oy, (2x+ W + Zz'- x+ 6y + LOz' 4x+ 2y -4zJ = (O' O' 0)
Form a homogeneous system of linear equations equating
tJre corresPonding comPonents :
2x+3Y+ 2z=O I '-- x+6Y+ loz=0 | (1)
4x+fu- 4z=O )
Reduce the systenr to echelon form by the elementary
transfromations. Interchange first and second equations'
Then we have the equivalent sYstem
-x+6Y+1Oz=OlZx+5v + 2z=O I' 4x+ iY - 4z=O )t
:t
{,
{21
264 COLLEGE LTNEARATIEBIA
We multiply first equation by -l and we divide thirdequation by 2. Then we have the equiValent system
x-6y- loz=O )2x+3y+22=Ol (3)2x+y - 2z=O )
We multiply first equation by 2 and then subtract from thesecond and third equations. Then the system reduces to
x- 6y-loz=Ol' t5y +222=0 | t4)- l3y+l8z=O J
We multiply second equation by |f .ra then subtract
from the third equation. Then we have tfre equivalent systemx-6y-l0z=Ol
,ur.* ?r:=
o^l (5)-16z=O)
which is in echelon form.In echelon form there are exactly three equations in flrree
unknowns; hence the system has only the zero solutJon x = O,
y = O, z = O. Accordingly, the vectors are linearty independent.Second Pnocess
Form the matrix whose rows are the given vectors andreduce the matrix to row echelon _form by elementar5r rowoperations
f2 -l 41ls 6 2lLz lo -4)
We divide third row by 2 and then interchAnge with thefirst row.
VECTORSPACES 265We multtply flrst row by 3 and 2 and then subtract fi.orn
the second and thtrd rows respectively.
rl 5 -2]'-lo -e 8lLo -lr 8_J
We multiply second row by $ and then subtract from thethird row.
"[r 5 -2 1
- | O -9 9^ lwhich is in row echeton form.Lo o -;t -l
Since the echelon matrix has no zero-row, the veetors arelinearly independent. fi
Exanple 19. Let rf v ana w are independent vectors, Showthat u + v, u - v, u - 2v + w are also independent.
p. u. s, 198(), rsglProof : Set a linear combination of the given vectors equal
to tlre zero vector using three unknown scalars x, y, z:x(u + v) +y (u -v) + z (u-2v + w) = eor,,rr.r + x/ + yu -)n/ + zt - ?,nr + zut = O
or, (x+ y + z) u + (x-y- 2z)v + ant =A (l)Sfuece u, v and w are linearly independent, the co-efflcients
in the above relation (l) are each O (znro), tlrat is,x+y+ z=O )x-.y-22=O I
z=O )The only solution to the above system is x = O, y = O, z = O.
Hence the given vectors u + v, u - v and u - 2v + w areindependent.
F=rrnplG 20. Test the dependency of the following sets :
(r) (r, 2. - 3), (2, O, - U, r, 6. - rl))(ii) {(2; 0, -1), (1, 1, o), (o, - I, l)}
1I
I)1
{
1
.!
tiii
I1l
{t
rI 5 -21-ls 6 2l
Lz -r 4)ID. U. P. 19841
266 COLLEGE LINEARAI,GEBRA
Solution : (i) Set a linear combination of the glven vectorsequal to zero vector using three unknown scalars )c y, z 1
{.L,2, -3) + y (2, O, -t) + z (7,6, -11) = (O, O, O)
or, (x 2x - 3,{ + {2y, O,-y) + {72,6z,, - llz) = 19, 6, 6;or, (x+ 2y + 72,2x+ A + 62, -3x-y - llzl = (O, O, O)
Equattng the corresponding components and forming tJ:elinear system, we get
x+Zy +72=O )2x+ O+ 6z=0 I (t)3x- y-llz=O )
Reduce the system to echelon forrn by the elementqrlroperations. We multiply llrst equaflon by 2 and then subractfrom tJ:e second equation. We also multiply tfre flrst equaflonby 3 and then add with the third equation. Then we have theequivalent system
x+2y +72=O 1
-4y- 8z=O I @)5y + LOz=O )
We divide second equation by - 4 and the ttrird equation by5. then we have tlle equivalent system
x+2y +72=O )y +22=O I (3)y+22= O J
Since second and third equations are ldentical, we candisregard one of them, Then the system (3) reduces to
x.'{iU:8 I (4)
The system, in echelon fornt, has only two non-zeroequations in the three unknowns; hence the system has anon-zero solution. Thus the original vectors are linearlydependent.
VECTORSPACES 267
(ii) Set a linear comblnatlon of the given vectors equal to
the zero vector using unknown scalars x, y, z :
/i2,O,- l) + y(1, l. 0) + zl0 - t, 1) = (O' O' 0)
or, (2x 0, -,S + (y, y, O) + (0' -2, zl = (O' O' O) ',or, (2x+Y + O, O +Y - z,-x+ O + z) = ( O' O' O)
Equattng corresponding components, and forming the
linear system we get
2x+Y+O=O IO+y- z=O I (I)
-x+O+z=O )
Reduce the system to echelon form by the elementary
operations. We multiply third equation by - 1 and then
interchange with the first equation. Then we get the
equivalent systemx+O-z=O IO+y-z=O l (21
2x+Y +O=O J
We multiply flrst equationby 2 and then subtract from tllethird equation. Then we get the equivalent system
x+O- z=O ly_ z=O | (S)
Y +22=O )
r We subtract second equation from the third equation'
Then the system (3) reduces to
x+0-z=O1,;"=31 (4)
From.tlte third equation we get z = O.
Putting z = 0 is the second and lirst equations we get y -- O.
x = 0 respectively. Thus x= O, Y = O, z=O, Hence the given
vectors are linearly independent.
268 COLI,EGE LINEARALGEBRA
rx-mple 21. I-et V be the vector space of all 2 x 3 matricesover the real lield IR. ShowthatA, B, C eV:
"= [; *n il'"= fi -l -;l
,.,a c = [! fr -T I *" rnearly a.p"r,a".,t.
[D.U.H.T. 19S61
hoo,f : IrtxA+yB + zC = Owhere xy, ze 8..
*.",[i -'^il.r[i -l -tl.,,E;3 -l]=[333]." [rx
-X'x] . [d u{ 1'rr]
.Equating corresponding components and forming thelinear system, we get
x+ y+ 3z=O l2x+ 4y + 2z=O I
2x- y- 8z=O L4x+ 5y +lbz= O I3x+ 4y + 7z=O
Ix-2y- z=O )Reduce the system to echelon form by the elementary
operations. We muiUply lst equation,by 2, - 2,4, 3 and,,l andthen subtract from 2nd, 3rd';'4th sth and 6th equationsrespectlvely. Then we have the iquivalent system.
x+ y+32=O'12y -42=O I
{:"2=3 I-{;3:=3 J
.l3z;i: -'z] = [B B BJ
",,lrXl -I I 7, -T; {r; ?& 'i: t{ :': I = IS S S I
vEcroR$PACps 269.
We rnultiply 2nd equation bV i "tr.a 6th equation by -1.
Then we have the equlvalent system
x+Y+32=O ) :
y-22=O Iy-22=O Ly-22=O ty-22=O I
Y-A=O )
Now 2nd, 3rd, 4th, 5th & 6th equations are identical, we
can disregard any four of thegr.
Thus we have the equivalent flstemx+y+32=OI
y-22=O I.This system is in echelon, fgrnr lraviag two equations in
three unknonrns. So the syslem has 3 - 2 ='I.free variable
which is z. Thus the system has non-z€rci:Bolutions.
Letz= l,theny =2artdx=-5. 'r
.'.-5A+28+C=0Henethe$lvenmatrices A" B and'C are tinearly dependent.
fampte 22. ,etP(t) be the vector "*". of all polynornials
of degree < 3 over. the realtfi.Id m.. Determine whether the
followlng polynomials in P[t) are linearly dependent or
independent:
u=f + 4t2 -2t+3, y=13 a61z 1L+'4and ''
w= 3ts +8€ * 8t+7.
Solution : Set a linear combination of the gjven
polynomials u, v and w equal to the zero polynomial using the
rrnknown scalars x y and z: thatis, xr + yv + zttr = O.
, -,1..:*:
27O COLLEGE LINEARAI-CEBRA
Thus xl1+4t2-2t+ 3) +y (ts+6t2-t + 4l +z {3t9+ 8t2-8t+Z) = O
or, xt3 + 4xt2 -2t +3x+yts + 6yP -fr + 4y + 3zts + 8zt2
-8zL+72=O
or, (x+y+32) 1s'1 (zlx+6y+ 8zl t2 + l-2x-y-&zl t + 3x + 4y +72 = O
Setting the coefllcients of the powers of t each equal to 0
(zerol,*Jg.i the following homogeneous llnear system :
x+ Y+32=O l4x+6Y +82=O L
-7.; *\;?',=3 J
loar"" this system to echelon form by the elementary
operatlons. We multipty lst equation by 4' -2 and 3 and then
subtractfrom2nd,3idand4thequationsrespectively'Thenwe have the equivalent sYstem
Interchange 2nd
equivalent sYstem
x+ Y+32=Ol2Y-42=Ol
Y -22=O I
. Y-22=O )
and 4th equatlons' Then we have the
x+ Y+32=O IY-22=O L
Y-22=O I
2Y-42=O )
We multiply 2nd equatiolt by 1 and 2 arrd then subtract
from the Srd & 4th equations respectively' Then we have the
equivalent sYstemx+Y+32=O I-i;*Bl=
".Ily=BlO+ O= 0 J
VECTORSPACES 27t
This system le ln echelon form harring two equations in
three unknowns. So the system has 3 - ) = I frree variable
whlch ls z.
Hcnce the system has non -zcro solutions' that ls'
ru + )w + zN,t =O does not imply that x= Y = z = O'
Thus the given polynomials u, v and w are llnearly
dependent.PXERCISESB(B}'
l. Show that the vectors (l' O, O)' (O, 1' 0) and (1' 1' 0) in
Vs (IR) a2\earlV dependent' ID'U'P' 19801
- Z. rttticn of the following sets of vectors in lR 3 are linearly
rnffindent?(i) (1, O, 1), (-3, 2.6),14,5' -2)l(ii)(1,4,2),(3,-5, LI,p,7,8), (-1' 1' 1)l
-/'nnsrf : (i) Independent (ii) Dependent'
./.,{Vrou. that'the followlng sets of vectors in Rs are
llnearly independent; ^4
(i) (1, O, 2). (-r,1, O), (O,2, 3D
r (ii)(l, 1, 1), (o, l, 1,) (0, o, 1)l :
4. (i) Prove tl:at the set of vectors (2' 1, 1); (3, - 4' 6)
and (4, -9, 11) is linearly dependent in m'3' tD' U' S' I#tl(ll) Test the set {1, O, 1)r (O' 2,2), (,3,7, 1)} for linear
rk'lrcnclence on IR3.
Anewer : (ii) the set is linearly independent'
!-r. Sl:ow that the set of vectors
ri = l(1, o, O), (O, 1,0), (o, o, 1)' (1' 1' 1)lr" R1 is linearlv
rlr.;rt.rrtlcnt brrl that any set of three of thern is linearly
Inr ['llcnrlt'trt.
4l* ,:,*S,;::C
272
,{COLLEGE I,INEAR AI,GEBRA
-oA**r"*er ttre foltowir4g sets are rinearry lndependent :
" -tuuo, o, 1), (1, r.-21 p' a' lD
(iil t(l, -1, s'(1' 4, r,l' @ -3'nl'(tB ttlA-1)' (1' 2' 3)' (4' 5' -3$ p'U'A' 1S1' &g' P' 1S1l
;};, (0 lrdependent (ii) Independent ({iif'dependent'
nZ/Wrrrine the llnear dependence of the fotlowing two
"udasof R3:----n
til's = (1, O; -L), ts3' 2' 61' (4' 5' -2)l'
G0 T = (1, r'rt' io o' o)' (1' r', 0)l . tD'u' P' 1s8l
els;Grs : (fl S is linearly independent'
(ii) T is linearlY dePendent'
8. (8 Decide wt*ttee* s = i(r' 4' 21' (3' -5' 1)' (-1' 1' 1)l is a
lineaely independent subset of IRa p'U'P' 19&n
(il} Conrsrder tlte subset {(l' 3' 2}' (1' :7' -81(2' 1' -1)} of IRs'
Test the &Pendu,r€e of8rc subset' tD'U' P' 19?91
Aocrctt : (il'S is lbaedt' rndepcmdent' . , '
($"The subeet ls linearly dependent'
9. $ftrow ** L veetors u'= (6' 2' g' 4l'v = (O' 5' -3' 1)
and w = (O o, 7, -?tare llnear{y independerrL ID'u' s..198o
IO. If'ffre vector set {tr, v2' v5} is indeperrdent prove that
(a) the set tvr +vl-2trs' vr -Y2 -v3' vt + v3) is independent
a$d (b) therct {v1 + v2 - 3vs' v1 + 3u2 - vg' vz + v3} is dependent'
'/ P' u' P' 1s4l
'*t/ru,t'*ine whether or not the following vectors in IR3
are hnearly dependent or hdependent :
(i) vr = $, - 2'1)' vz = (2' 1' -1)' vs = ff ' 4' Ll [C'U'P' 19731
(fl) u = $,-2'4' 1)' v ={2' l'O''3} dndw= (3' -9:t-3*Ic"U'P' 19781
Ailtr€rg: G) vectors are linearly dependent
(ii) vectors are linearly independent'
VECTORSPACES ZZs
12. Determlnc whether each of the following sets are
ilncarly depenrlent or linearly independent :
0l l(0, r, o, 1) (r,2,3, -u(8, 4,3,21. (o, 3, 2, O)l
(u)(1.3,2), (1, -7,-8), (2, 1, -l)l D. U. Pr,ql" 19831
AnasGrs: (i) vectors are linearly independent.
13. (l) Show that the set {(4, 4, O, O), (0, O, 6, 6) (-5, O, S, S)} isIlnearly independent in IRa. ';
(ii) kove that the set {(1, O,2,41. (0, l. 9, 2), (-5,2, 8, ,.1,0}is linearly dependent tn IR4.
. 14. Determine whether the following sets of vectors in IRa
are linearly independent :
(i)fi3, o,4, l), (6, 2,-r,2), (-1, s, 5, l), (-3, 7.8,3))(ii) (4 - 4,8, O), {2,2,4, O), (6, O, O, 2), (6, 3, - 3; O)l
Angwers : (i)The set ls linearly independent.(ii) The set is Unearly independent.
A5. Show that the followlng sets of vectors in II{3 are
ffiury dependent :
i (i) {(1, l, -r), {1,2. sI, {3, 5, -3)} i(ii) {(2, r, l), (3, -4, 6) (4, - e, l l)}
16.,Show that the lbllowing sets
linearly independent :
(l) (1, -r, 3), (1, 4,5), {2, --3,7)}
(ii) lu. -2. Ll.(2. t. -t),(7 4. ri)
rure linearly depenclerlt or inclcpendent :
(0 {(r, -2, 1). (2, 1, -Ll.t7, -4, 1)}
(ii) {(--1, l, l), (1, ',i,2) (3, -5, l)}Ansrpers : (i) Lineirrll' rlc:pt'rrtlcrrt
(ii) I.irrcrrrly intleperrdent.
[.lnear Algebra- lI3
17. Detennine whether the f<tlkrwing sets ol vectors in IR3
i ID.u q. rfflll' tD. u. s. re83l
:
of vectclrs irr [R3'arre
lD. v.H. Lg/,4ll
lc. u. P. 19731
rc. u. P. L97SI
lD.u. P. 1982I
27.4 . CoLLEGE UNEARAT-GEtsRA
r ',18. 'koverthatithe following set of vectors in &.3 is linearlydependent:
{(s, o,'-a),r(-r; l, D, @.2,-2),(2, t, t)}.19. Which"of"Ihe following sets of vectors in IR3 is linearly
(r l(2, -I, 4) (3, 6, 2t, Q, t0, -4)l(u) (1, 3, 3), (o, l, 4), (5, 6, 3), (7, 2, -\l
Aadwcrs : (i) Linearly independent (ii) Lirrearly dependent
2O. Determine whether the following sets of vectors in IRa
are linearly dependent or independent :
(, {{1, 2, t, -2), p, -2,-2, O) (o, 2, s, 1), (3, O, -3, 6)}
(CI {(3, O,4, l), (6, 2, -1, 21, F1,3, 5; 1), (=3, 7, 8,3)}
Ansmcrs:(i) Linearly independent (ii) Ltinearly. independent.
2I. For which real values of l. do the following vectors
;,. , Ansmur: )'=-2, ),'=1.
22. Show l-hat the set of vectors in IRa
(3, -2,4. 5), (-1, 3,2,6), t4,2,5.12)) is linearly
rd w are linearly independenl vectors inathen show that the vectors
\,.i \tr, w+u{ril I-I -1'\r, ll -'1r, ll - 2v + rv
a,c lrlr;r.r linearlv independenl .
L
YECTORSPACES
24. Show that the vectors i i'r"-' u
(1. |,2. 4). (2, -t, -5,2). (1, _1, _4, O) and (2, t, l, 6)
25. L€t V be the vector space of all 2 x 2 matrleeso$s;theroal flcrd IR. show that the n'-ratriees A, B, c e V are rtnearlyln<lependent where
o= [i i],r= [d ?l *"= [ ;l R.u.M.sc.p. reesl26.|n vector space V (IR) of all2 x 2 matftces determine
whether the following matrices are linearty,depeitdCriti:."-.-,". , i ,,
tJ ?l,tS -jJ *t' rSIAnewcr: Linearly dependent ;. _ ,,., :.
27' ,.etv(IR) be the vector space of an 2 xs matrices over .
1,,::::l] i.:1j::1:-jprhe fouowing qatrisq.s in v(rR) asellnearly
IIA=W lJ, , =l-;; -SI andc =Ii :l Zlza. tq{v, (IR) be the vector ".;; ar a, n^r.-^-r-r
275
It2 +t+
<3. Determine whether the following
th_e vector space of all polynomials of
+ t, 3t2 + 2t + 2] of polynor.ntals is ltnearly
V(IR) be the vector space of allp. u. P. rs8?l
Rolypopiah orpolynoriithls arellrrnrly dependent or Independent :
(t) t3 +Zt2 +4t:1, Zts_t2_3t+E, ts _4t2 +A+S(tt) t3 - 3t2 _ 2t + 3, 2t3 _ 5t2 _ 5t + 7, t3 _ 2t2_3t + 3Aarpcrc: (i) Linearly tndependent
C.ft Brdl d dnenen darcctorrycccDr0altloa : l,et V be a vector spaee and {v1, V2 , ..., \,,,jlnllr ect of vectors ln V.
276 COLLEGE LINEARAI.CEBRA
We call [vr, vz, .... vn] I basle for V lf and only if(i) lvr ,y2, ..., v,rf ls linearly indepe4d-ent.
(ii) {vr ,y2, ..., vrr} spansv. ,ii :1, ",..,
. De0nitlon I let e1 = (1,0, 0, ,.., O), 6, = (O; l, O, ..., O), ...,
;.rgn Fr(O, O, .,.,.O, U. Then {e1 ,r,2,..., en} ts'a ltnearly
independent set tn IRn. Since any vector v = (vr , !2, ..., vr'! ln
ffipq4,be-$fiSt€Las v. = v1Q1 + rt1az t ... + v,re,r, {e1, e2', ..., es}
sp.qs, IRn .,
Ther-e{ore, {e1 , e4, .... qJ is a basis. It is called the
stsndard basls or usual basls for g1n.
Dcflnltlon : A non-zero vector sp3ce V ls called flaltedtmcaslonal if it contains flnite set of vectors [v1, v2i...., vn]
whleh forr-ns a basls;for V. If no'such set exists, V ls called
lnflnttc dlmenclond.
D;Bnttlon g The dlrrFaclsn of a linite dirnensional vectorf:.i. 1- i
spaceris the nudrber of vebtors in any basis of it.. . ;.'Or, equivalently, the dlmcnslon of a vector space is equal'
to.,{he, maximum nurnber of linearly lndependent vectors
conhrred Dt It.
".Difinltloir :'If v1 , v2, ...,'vrare vectors of a vector sPace'r...
,,';].: :
V such that every vector v e V,ean be written in the form
v = olVl:*, Qv2 +.,... * OnVp where'ot are sealars, then vg, v2,
..., v. b called a gcneratlng system of tlte vector space V.
Theorem 6.19 If V is a vector sBace of dir,lrension I1, every
generating system:pfeV.'epr1{airas'o;.1: bt{t; r.ho$r:61616]"{-hdir "tr,
linearlyindepender(.ypctors. i :".i i::i:'i';r.1-:-- '
VECTOR SPACES 277
hod: lct v1. v2, .... v,n be any generating system of V. Let rbe the ltrclterl number of llnearly lndependent vectors thatean bc ch<rcn from v1, v2, ..., v-. Then we may assume thatvr,Vl, ..,, vr are llnearty independent but that v1 .y2, ..., v1, v11j
erc llnearly dependent forl = 1,2, 3,..., m - r, Hence we have atron-lrlvlal relatlon o1v1 + ezvz +... + okvr + o!.dvr*J = O ... (l) inwhlch cr+; # O, since othetwlse the above equaUon (l) would bea non-trlval relatlon between vt, ve, ..., vr contrary to t&ellnear lndependence of these vectors. Henee the, aboveequation (t) deftnes vsay os a llnear combination of v1 , v2, ..., v.y
forJ = l, 2, ..., m - r. Therefore, lt follows that v1., v2, ..., v1, isalso a generailng system of V.
Now any set of more than r vectors of V is ltnearp''.depen-dent. Since the dlrnenslon of V is n, V.eertalnly emrtains nllnearly tndependent vectors. We must have n Sr. But elnbe Veontains r linearly tndependent vectors ul, uz, ..., ur we alsohave r sn. Hence r = rr dnd th? theorem ts proved.
Theorem 6.20 lf V ts a vector space of dimenslcin n,.ererybasls of V eontalns exactly n Hnearly lndopendent vec.to,rsl;conversely. any n llnearly lndependent vectors of v consutute'n basis of V.
Proof : Since every basis is a generafing systern.: Itcontains n but,not more than n linearly independent vectors.Slnce the vectors of a basis are llnearly independent,llrcrefore, it contains exactly n vectors tn all.
Conversely, let v1..v2, ..., v, be any n linearly indepe.nflentver:lors of V and let v be any other vector of V. Slnce n is therllrrrcnsion of V, the (n + I) vectors vr, Vz, ..., vn, v are linearlyrlrpcrrdent and there exist a non-trival ielation of the form
278 CoLLEGE LINEAR AI,GEBRA
prvl + azYz + ... t dnvr, + cv = O. Moreover 0, * 0.. Sinceotherwise we would have a non-trival relation amongvr, vz, ..., vn. Hence we have
,, = -# ", -*y2 - ...-f *It follows that every vector veV is a linear combination of
vt, yz, ..., vn and hence these n vectors form a generatingsystem. StncC they.are linearly independent, tfrey also form abasis of V. Hence the theorem is proved.
Theorem 6.21 Let V be a vector space of dimenslon n andlet v1, v2, ...,'v1, (r <n) be any r linearly independent vectors ofV. Then there exist n - rvectors v...1, v1a2, ..., Vn of V whichtogether with v1 ,y2, ..., v, constitute a bAsts of V.' Proof : Slnce r < n. the vectors vy, v2,..., v. do not generate
the whole space V. Hence there exists a vector V111, of V that tsnot In the space genereted by v1 ,v2 , . ,vr.The vectors v1 ,V2 , .. . v1,
ve..1 zlr€ therefore linearly independent, for tf0,1V1 * bVZ +... * Ofvr * OfalVsal =O.W'e must have d,r*1 = O, Slnce otherwlse vr;1 would belong
to ttre space generated by vr, y2, ...;.., v,. Ttlen tt follorvs thatdr - &z = ... of = 0 sirrce vr, vz. ..., vr are linearly independent.
Now if r + I <n, we can repeat this argument to obtain a
vector v1.,2 such that v1 , v2, ..., v1 , v1a1. V.12 or€'ltriearlyindependent and so on until we have n linearly independentvectors vi, vz, ..., v., Vr*1 , yr+2,..., vn. These n vectors constitutea basis of V. Hence the theorem is pioved.
V, then every vector ve V can ,be expressed uniquely in theform v = CIIVI + Aevz + .,. + Otnvn,
Proof : Sihce {vr, vz; ..., vr} is a basis of V, any vectorv e V can be wrltten as a linear combination of the vectorsVt, vz, .... vn i. e, v = cllvl + oevz + ... + cr,vo where cq are. scalars.Therefore. we have only to show that the co-efficients d,1. cr,2,
..., c,n are uniquely determined by v. Suppose thatv = CllVr + %vz*. ., * OkrVn = p1v1 + fov2 + ... +- P"hThen (crr - 0r) vr + @e-fizl vz + ... + (oQ, - 0.,) v, =OSince Vr, Vz. .... , vn are linearly independent, it follows
that a1 -gr = O. w2-fu =O, ot -gn= 0, thatis, G, = Ft, az =P2, ...,
0. = F.,. Hence the theorem is proved,Theorem 6.23 l.et W be a subspace of an n-dimenslonal
'rector space V. Then dim W <" n, [n particular, if dim W = nthen W = V.
Proof : Since V is of dimension n, any n + 1 or morevectors are linearly dependent. t
Furthermore, since a basis of W eonsists of linearlyindependent vectors, it can not contain more than n elements.Accordingly dim W S n. In particular;'if {w1 ,w2, ..., w,} is abasis of W, then slnee it is an independent set with n elernents,it is also a basis of V. Thus W = V, whendim W = n. Hence thetheorenr is proved.
Theorgm 6.24 lf S and T are subspaces of a finitedimensional vector space V over the field F then
clim (S + 1) = dim S + dim T- dim (Sn11
Proof : kt dim S = s, dim T = t and dim (Sffi) = r.L,et {u1. u2, ..., ur} be a basis of SfiT. Slnce SOT is a subspace
of S, We can extend the above basis to a basis of S, say {ur, uz, ,
u", vr, ... , Vs- . }. Thls basis has s elements since dim S = sSimilarly, we can extend the basis {ur , uz .... uJ to a basis of t;say (ur, u2, ..., u,-, w1 ,wz, ..., wt *)
lrtA = {us, tr2, ..., u. v1 , ..., V"_1, w1 ; ..., wt_o}.
,, 1J \II
,..diIi4l
..tlr
,ilr1
;280 COLLEGE UNEARAI-GEBRA
Clearly, A has exactly s + t - r elements. Thus the theorem
ls proved if we can show that A is a basis of S + T. Since {ur, rt}
generates S and {ui, wkl generates T, the union A = {ur, vl, wr,}
generates S + T. Now we have only to show that A ls ltnearly
independent.
Suppose that orul + oery + ... + ohur + grul + ... + fi*r.v"-r.
+ Yrwr + ... + Yt-, Wr-u = O. (l)
where ca, F:, Tr are scalars iu F.
Letv=cl1u1 *...+o!.ur+prvr +...+p"* v"-, l2l
Then form (l) we get
I v = - ylwr - lzwz yt-r wt-r, (3)
Since {u1, r1}e S, v e S by (2) and
Since {wu} cT, veT, W (3}
Accordingly veSflT.
Now since lurl is a basts of SnT, there exlst scalirrs 6r, ..., 6,
such that v = 6r ur + &uz + ... + 6rur. Thus by (3), we have
6rur + oary+... +$u, + ylwr + ... + Tt-,. wt-, =O,
But {uy, w1} is a basis of T and so is independent: Hence the
above equation forces Yr = O. ..., Tt-r = O. Substitufing thts into(l), we get o.i ul + ... + ohur + ptvr + ... + fu,. vs-r = O.
But {u;, vr} is a basis of S and so is independent. Henc,e the
ab6ve equation forces Gr = O. ...i o(r = O, 0l = O, ..., Fr- = 0.
So the equation (l) implies that q, 01 and ft are all O (zero),
So A = {u;, vJ,wr} is linearly independent and form a basis ofS+T.
'flrus dinr (S + n = s * t - r = dirn S + dimT- dim (SnT.
IIence the theorem ls proved.
VECToRSPACES ZglCorollery : If S and T are two subspaces of a flnite
dlmenslonal vector space V such that SflT = l0l thendkt, (S + T) = dim S + dim T,
Pr:oof : Since St..lT = {O}, dim (SnT) = g
Thus dim (S + n = dfin S + dtm T-dim (SnT) implies thatdim (S + T) = dirn S + dtm T, Hence the corollary is proved.
6.16 QuofrcrtspsecDefinltlon : Lrt W be any subspace ofa vector space V over
the lield F. I.et v be any element of V. Then the set W + v = lru + v: o€rM is called a rlgfit ccet of W in V generated by v.
Obviously, W + v and v + W are both subsets of V. Slnceaddition tn V ls commutative. therefore, we have W + v = v * W.Hence we shall call lv + v as simply a coset of w in v generatedbyv. '
Pnopcrtiec : (i) aV=+W+o=W(it) oleW=+W*to=W(fll) IfW + v1 arrdW+v2 are two.cosets ofW inV. thenW+v1 =W+v2 *vr -v2eW
fhcorcm€.Z5 If W is any subspace of a vector space V overthe field F, then the set V/W of all cosets W + v1 where v1 ls anyarbitrary element of V, is a vector space over F for the vectoraddition and scalar mulilplication composttions deflned as
(W +vr) + [W+vz) =W+ (v1 +v2) for every vr. v2 eV.and q fW + vr) - W + c.u1 where ae. F and v, eV.Prloof : Let v1 . v2 eV then v1 + v2e V and alsone F, vt e V =+ cnzl eV. Therefore, W + (v1 +.v2) e V/W and also
W + crvl eVlW.Thus V/W is closed with respect to addjlion of cosets and
scalar lnultiplt"cation as defined above.
:282 CoLLEGE LINEARAIEEBRA
, ktW+v1 ='W+vt'wherevl ,v1 '€V
and W *Y2 =W +v2'where v2,v2'€Y'
NowwehaveW+vl =W*v1 ':1v1 -v1
'€W
and W * Y2 =W+v2' I Y2 -v2'€W.
Since W ts a subspace of V, we have
vr -Vr'eW,v2-v2'eW 9vr - V1tY2-v2'€W
==+ (vl + vz) - fur' + v2]ew+W + (vr + vz) = W + (vr' + vzJ
= W +vr) + (W I v2) =(W+ vr') + (W + v2')
Therefore, addltlon of coeets in V/W is well-defined.
A.gain, PeF, v1 - v1'eW =+ pfur - v1')eW
+W+gvr =W+pvr'- ;
Therefore,'scatar multlpltcatlon in V/W.'ie, also-.well
defined.
(0 Addftfon b conmutatbeI.et W + v1, W + v2 be any two ebments of V/W.
Then W+vr)+[W+rr) =Y_* fur 1v1]=W+'tvj
*vr) '
=(W+v2 )+(W+vr)
. (tttlAddtttoo ts Gsoclatlve ,
l,et W + v1, W + v2 and W + v3 fe any'"tnree elements of V/W.
Then [W + v1) + [(W + vz] + (W + vs]l
=(W+vr)+[W+(v2 +v3]l
=W + [v1 + fu2 +v3]l 1
=W+[(vr +v2)+v3l
= [W + (v1 + v2)] + (!V +,v3)
= fiW + v,) + (W+ v2)l + (W + v3)
VECT0RqPACES i, Zgg(lll) Hstcnce of additlve ldgqtty- : j :{',r.j: :. r,,If O is the zero vector of V, then ,:
W+ O = WeV/W IfW + vr is aqy element ofv /W,
... W + O = W is the additive identity;(iv) Hbtence of addidve lnvcree.If W + v1 is any element of V/W, thenW + (-v1) - W - v1 dVlw. Also we have[W + vr) + [W -vr) = W + (vr-vr ) =W+ O =W..'. W-v1 is the addiflve inverse of W + v1.Thus V/W is an abellan group wtth respect to addition
composition.Further for scalar multiplication. we observe that if
a, pe FandlV+v1,W+v2eVl[ then(v) o ltw+ v1) + (W+vz)l = s [W+ (Vr + v2]l
= W+ c (v1 + v2) Sir:ce crW =W=W+(mr, +av2)
=(W+ffir)+(W+avz)=a(W+v1)+o(lV+v2).
(vi) (q + $ (w + vr) = w+ (ct + Flvr
=W+(crv1 +6r,;, = (W+anrr)+ 0M+ pvr)
=0{W+v1)+9(W+v,;(vii) a0 [W + v,; = W + (oB)v, = W + o(pv1)
= o, (W + pvr) = c{p(W + v,)l{viil) 1(W+vr)=W+ lv1 =1ry+v1 where I e F.Thus v/w is a vector space over the lield F fcrr the addition
'f cosets and scalar murtiprication. The vector space V/lV isctrlled the $uotieat space of V relative to W.
'ftre coset W + O = W is tlre zero veetor of this vector space.
284 COLLEGE LINEARAIiCEBRA ,.:
6.17 Dtmendm of gsoticnt $Pacc.
Theorem6^26lfWisasubspaceofafinitedimensionalvector space V over the lleld F' then drm V/W = dim V - drn W'
Proof:lrtmbethedimensionofthesubspaceWofthevector space V. l€t S = {wr ,}r2, ...,wJ be a basis of W' Slnce S is
linearly lndependent subset of V, therefore. tt can be extended
to a basis of V.
IJt S' = fi r, w2, ..., wn1 . v1 'Y2. ...,vrl be a basis of V'
ThendtmV=fll*r..'. dimV - dhn W = (m * rt - m = r.
So we have to Pr6/e thatdim VII/= r.
Now we claim that the set of r cosets
Sr = {W + v1, W tY2, .-,W + v.} is a basls of V/W'frrst we have to shorr that Slls llnearly tndependent''Ihe r.ero vector of V/W ts W- '
kt cr1 (W + vr) + ob (W + v2) + ... + o.r (W + vr) =\fi=* (W + crlv1) + (W + nzvz) + ... + (W + orvr)=W
=+ W + (cr1vl + %vz +..- + ot'vr) =\[:*(}1V1 +%vz+"'+OtvreW:* o1v1 + (hvz + ... + ckvr = ptwt +fuwl + "' + $rtl1'tn
Since {wr , wz, ..., w-} ts a basls of W'
:+ (11v1 t oavz + ... + oq-vr - Flwl- Frwz - "' - Bnlw'n= O
:* (I1 = O, 02 = O, ..., U, =O
Since the vectors vr, V2, ..., v, wt, wz, "', wm are linearly
independetrt. 'lJrrrs the set Sl = [W + v1, W * vz' "" W + v'] islinearly indePendent.
Now we have to shorv thal l, (Sr) =VIw'i'e 51 spansvftar'
Ir1 W + v be at1' clelrtent of v/w''Ihen veV catr be expressed as
y = 111V1 * 12w2* .:. + Y,nWrr * 61v1 +lbvz + "' +0'v,
=w *6rvl +&vz + ... +6,-v,
YECTOR SPACES
whert w - ylwr + yzwz + ... + y,nw.eW.
*lW + v.W + (w+61v1 +6,vz +... + 6.v.|
= (W + w,| + 61v1 + q2"vz + ..., + 6.v.
=W+61v1 + 62v2 +... + 6.v,
SinceweW ...W+w=W
= (lV + 61v1) + (W + &vz) + .:. + (W + ev.)
= (6, (tV+ vr) + & (W+vz) + ... + 6" (W+v,)
fhus any element W + v ofv /w can be expressed as linearcombination of 31 = {W+v1, W+ v2, ...,W +vr}
: v/w=L(Sr)
So Sl is a basis ofv /wThus dlm Y /W =r = dtm V- dim 1ry'
Hence the theorem is proved.
6.18 Coodlnatcs of rvectsrelative to ebasls ,
Let [v1 , y2, ..., vr,] be a basis of an n dimensional vectorspace V over the field F. Then any vector veV can Ue expressea
where xr €F' (fSi <n) and (-r1 , x2, ...,xn) are called theco-oldiaateg of v relative to the given basis. There is clearly
ve| apd orleredrr-tuples (xt, x2,..., ,6) with elements in F'.
Oqe ,cjt! easily vertfy thart if v, w h4ve(rr, xz. .,.d,Ur,,yZ, ..., y,,) rcspr:c:tively and oeIras co-ordinateg (xr + yr ,x2 * yr^ ..., x,, + yr)<'o-ordinates {ux1 , u.rc2, ..., a.4j.
285
co ordinates
I,',thenv+wand sv has
186 COLLEGE LINEAR ALGEBRA
6.19 solutiotr specc of a homogcncous system of llaear
rquatlonsSttxt
?:'"'
4mlXt
+ atzXz + ... + 8tr,&, = O
+ a{zi + ... + tznxn= o
* A1y,X2 + "'+ a-rr:6 =O
(1)
is a hotnogcneous systcm of m linear equations in n
unknowns over the'ieat iieta IR ' Ttie solution set W of (1)
constitutes a subspace of IBf and this subsPace is called the
solutlon sPacc of the system of linear equatlons'
As for examPles
(i) The solution set W = l(Za' a) : ae IR) of x- ? ='
is a solution space and W is a subspace of IRz :
(ii) The solution set W = (2a' -a' 4)' 'a q IRI of the linear
Isystem of equafiion , * -tr i l:3}.'" "
solution 'space ard w is
e: subsPace of lR3 '
Rcmart : The solution set W of the non-homogeneous
nlinear system il.uu n = bi (i = 1' 2' "" m) Q)'
o'er,n" ,"'i, o"ld lRdoes not constitute a sttbspace of EP '
: t6orcm O.ZZ (Imthout Fod)The follorvingithree statements are equivalent :
{i) The system of linear equations AX = B has a solutlon
' (it) B'ls a linear combination of the columns of A
(iii) The'coefflcient matrix A and thc augmented matrix
(A, B) have the same rank
VECTOR SPACES 287
ud dlmcnrloa for thG Spnctal solutloa of al5a} rrtt.ltL
ht W dcnote the fletternl solutlon of a homogeneous linear
nyalcm, 'ltte ttrtn zertt solutlon vcctors ut, u2, ..', us'are sald to
lbrm r |rrlr of W lf every solution vector oe W can be
rxprrrncd rrttlrluely as a linear combination of u1, u2', ..., u".
'llrc rrrrnrber s of such basls vectors is called the dlmenslon ofW. wrlttcn as dim W = s. UfW - {O}, we define dim W = Ol
6.21 Procedure of findtag the basls a1{ dirnelslon of therolutlm Blrac.e of e homogpnm llncar EIEtcm.
Let W be the general solution of a homogeneous linear
nynlcm and suppose an echelon form of the system has s free
vrrrlables. kt u1, u2, ..., u" be the solutions obtained by setting
rnrr <rf the free variables equal to one (or any non-zerolrrrrst:rnt) and the remalning free variables equal to zero.
't'hen dim W = s and u1, uz, ..., u" form a basis of W.
Theorem 6.26 The dimension of the solution space W ofllrc homogeneous'systeql of linear equations AX = O.ds, p-rwlrrrc n is the number cif unknowns and r is the rank of the
loelllclr:nt matrix A:
Proof : Supposeul, u2, ..., u, lbrm a basis for the eolumn
rlnr('(' of A (There are r such vectors since rank of A is r). By
lltrrrrtrrr 6.27, each system AX = ui has a solulJon say v,. Hence
A\rr =tli, Av2 =t)2, ...,Av, =u, (l)
:,l rl)l)osc dim W = $ zDd w1 . w2. .... w. fotttl a l;;rsis of W.
lll ll - {tr1 . V2. ..., Vy, W1 , W2, ..., Ur"}
\\ r claim that B is a trasis of lR" .
llrrrs nc lrr:ed to prove that B spans JR" arid that B is[ntru lt' lrrrle;rndent.
288 COLLBGE UNEARAIJGEBRA
(0 Proof of 'B.strnns- El:'Suppose ve IRr and Av = u. Then u = Av belongs, to the
column space of A and hence Av ls a linear cornbination of the
ur, sayAv = clrur + o"zq+.,. +okur {2)
l-etf =v - o1v1 - %vz - ... -: ohvr. Then using (l) artd P)
we have A(r/) = A (v - GrVr - hvz -... - okur)
=il-;;';*yti ;*,fr.*'=Av-Av=O
Thus y' belongs to the solutiqn spaee W and hence'f'ls 4,n
linear combination of w,, saYs
y'= 0rwr +\zwz + ... + F=w, =X Ftwt.
. Grvr = i-Ji.;.-u,*,i=[ i=l J=lThus v ls a linear combtnation of the elements ln B and
hence B spans IF .
(ll) Proof of 'B le llnearly lndelrndeaf,'Suppose Yrvr + ^lzvz +." * Trv. + 61w1+ Qw2+ ... + 6"w" = $
since w; e w' each Aw1 = o' usin8lthis fact and
(l) and (3), we get
1lsO=A(O)=Al I y,v,+ )
\i=l J=lo,*,)= i
,v, er, * i, un*,
rs= I yr u1 + I E1O = y, u1 -t- |2u2 + ... + Yrur.i=I j=l
Since ur, uz, .... ur are linearly independent. each Yr = O
Subst.itutingi this in (3), we get 6l wr * ... * 6uw" = O'
' :
VECTORSPACES 289However, wr. wt, ..., wr are llnearly lnclependent. Thus
each 61 ' 0. Therefore, B rc rrnearly tndependent. Accordi,gly,B lsa borlrof lR' , strree [t has r + g elements. we have r + s = n.Conroquenlly dlrn W - g - n - r. Hence the theorem is proved.
Dnnpio 23. prove that the vectors (1, 2, O), (0, b, Z) ana(* I, l, 3) form a basls fo.IR,
Proof : The glven vectors will be a basis of IRs if and onlylf they are linearly lndependent and every vector tn IRs can be .urrltten as a llnear combtnation of (1, 2, O) (0. E, 7) ana (_1, l,gl.I{rst wc ctrail pove that thc rtctors arc lincerly indepcndcrt.
For arbitrar5l scalars x, y, z.let ,,,1
d.1,2, O) +y (O, S,V +Z {-1, t,S) = (0, O, O) ., .:or. {x 2x o) + (O, Sy, Zyl + (-2, z,3z) = 19. g, 6,or, (x- z,2x + 5y + z, Zy + 3z) = (O, O, O).
Equating corresponding components and formihg linear.system, we have
. x _z=o12x*_Sy*^r=91 (t)
7y+32=O )
Reduce the system to echelon form by the elernentarytransformations. we murtipry ,rst equation by 2 and thensubtract from the second equation.
Thus we get the equtvalent systemx-z=O I
5y+32=o I el7y +32=O lor, we can write
x-z=Ot132+_5y=O | {Bl3z+7y=g )
lJnear Algebra-lg
2gA' '' ; , '; CoLLEGE LINqAR ALGEBRA
We subtract second equation from the third equation'
Then we get the equivalent sYstem ' :
' ',..1 'iiil;'.';16- Z =0.1,::' 3z*r?==oo i @)
This systqm is in echelon form and has exactly three-l
eouations in the three unkno#ns; hehce the system has only
the zero solution i. e. x = O, y = .O, z = O' So the given veotors are
linearlv independent.
tBy the application of rnalnces we can also show that the
*given vectors are linearly independentl
To shqqr that the given vectors slna R9' we must show that
an arbitrary vector v'= (h; b, c) can expressed as a. linear
combinatiorr, = (]: b, c-) = x{1,2, O) + y (O, 5, n + z (-f ' 1' 3)
Then formingf liney sYstert'r, "" g"i
';. il;,i::\ (s)
' Reduce this system to echelon folm by this elementary
operations
.Wemultiplyfirstequationby2andthensubtractfrom.the: second equation. Thus we have the equivalent system
x-z=ali.: 5y +32=b-2a | (6)
7y +32=c , )'
We subtract second equation liom the third equation'
Then wc get the equivalenl sYstem
x - z=a l5y +32=b-2a I A
VECTORSPACES 29II.i'orl the third equation, we have y = | {" -b + 2a)
Sr-rbstituting the value of y in the second equationweget r=f,va-5c- t4a)
Again substitutin gz = L (7b - 5c - L4a.) in the first equation,
wr.gr:l "=
-] 1zu - 5c - Ba)
't jrcrefore, v = (a, b, r:) = L W* ft:-Ba) (1,2,O)
* j rc-u +2al(o,b,v+it a-sc- r4a) (-"I, r,3)
Thus every vector in IR3 can be expressed as a linearcombination of the vectors (1,2, Ot, (O, 5, 7) and (-1, 1, 3). Hence
lhe vectors (1, 2, O), (O,5,7) and (-1, l, 3) form a basis of IR3 .
i) Extend ,lQ,
o.l). (1, I, 1)l to t T":_ "jT1^^.lD. u. P;19(**l
xtend the set 11'3., Z, I), (O, l, l)l to a basis of IR3 .
lD.u.s. 19ffiSolution : (i) First we have to show that the given:jI of hro
vcctors is linearly independent. Set a linear combinatiorf.oftlre two given vectors equa!,to zero by using unknown scalars
,rimd y: I .,,
,42: O,r) + y (1, I, t) = (0, O; O)
or, (2x O,-rd + $,y, y) = (0, O. O} ..:
or, (2x+ y, y. x+ y) = (O, O. 0)
Iiquating corresponding components and forming thellrrrlrr system, we get
2x+Y=O1 -,r*' *
x.*I:3 i **.rvc have x'= o'Y: o'
llcrrt't: the given two vectors i3. ffP' are linearlyIttrlt'3rcrrrlcrrl. So {(2, O, l), (1, I, l)},is a part of the basis of
lH'rrrrrl lrt'lrcc we can extend them to a basis of lR3 . Now we
292 COIJLEGE I.:INEAR AI,GEBRA/ 293
seek tJxree:independent vectors in'Ih3 which include the given
twovrctors. Thus we can easily veri$r that (2, O' l!, (1' I' l)'tO. I, O) ar,e linearly independent. So they form a basis of
,'fu.3 which ts an extenslon of the given set of vectors to a basls
of mf
(ii) First we have to show that the given set of the vectors is
linearly tndependent, Set a linear combinatlon of the two
given vectorsequal to zeto by.using unknown sca-lars xand y t
x(3,2, 1) +Y (O' I' l) ={O' O' 0)
or,, .(3x. bq $.+ (O, y' y) = (0' 0' 0)
or. {3rq Lu+ Y,x* } = (0, O' O}
';;l&quntfixg corresponding components and forming the
Iinear systGm,.we get
.- 3x =O'l2x+y = O I Thuswehavex=O, Y=0.x+y=0 J
Hence the given two vectors in IR3 are linearly
independent. so the gtven set of vectors ls a part of the basts of
.; i IR' and hence we can extend them to a basls of R3' Now we
,. seek three independent vectors ln IRs which lnclude the given
ii,'. vgctors. Thus we can easlly veri& that {3, 2, l, (o, I, I), (1, o, o}
,l_ are hgarty independent. So they form a basts of IRs which is
arf extenslon of the glven set of vectors to a basts of IR3 '
dhd** 28. Determtne a basis and the dtmension for the
i t dlution space of the followlng homogerrcous system:I
I ' x-SY+ z=O I
VECTORSPACES
Solutlon : The given linear system is
x-ry+ z=0'l2x-6iy +22=O I (1)3x-9Y +32=O I
Reduce the system to echelon form by the elementary
transformations- We multiply lirst equatton by 2 andby 3 and
then gubtract from the second and the third equations
respectlvely.Thgn we have the equlvalent system
x-3Y *z=O l' O=O | +x-$l+z=O i '
O=o J
Thls system ts in eehelon form and has o:tly one non-zem
equatton in three unknourns' So the systern has 3 - I = 2 free
variables which are y and z, Henee the dtmenslon of the
solutlon sPace is 2 [two)'set (t) y = I, z = o llllY = 0, z = I toobtain !h9 respective
solutions v1 = (3, I' O). v2 = Fl' 0' l)'Hence the set l(3' l' O)' (-1, 0, 1)I is a basts'of thdSohrttsn
, \ soaca,ltffinmplc 26. Ftnd the sotution space W of the fogov{ptg/ homogeneous system of linear equatlons :
x+zy- z+ 4t-O ) I '
2x- y+Bz+qt=9 [ p.u.H.T.tg8q4x+ Y+32+S=9 f iJ-u.H. 19881
v- z+t=O I -2x+t$- z+7L=O )
Bolutlon : Reduee the given system to echeld'iiffi'tiy tt.cletttcntary operaUons. Wemulttply lst eqtiatlcr WZ'4 andzand thcn subtract from 2nd, Snd and 5t}t' eguatlqus
rerpecllvely. Then we have the equtvalent system
x+2Y- z+4t=O I- 5y + 5z-51=O I-7y+72-7t=O t
Y- z+ t=O t- y+ 2- t=O J
I : ' x-Sy+ z=O1, 2.tc- 6y + ?.2=O l {2)i 3x-9Y+9l2=O )
i.Lt,,.,
294 COLLEGELINEARALGEBRAll
We multiply 2nd, 3rd and 5th equations by - i,- + and {-l)
respectively. Then we have the equivalent system
. x+2Y -z+ 4t=A )y-z+ t=O I
Y-z+ t=O I 'Y-z+ t=O I
. Y-z+t=O )
Since 2nd, Srd, 4th & 5th equations are-identical, we can
disregard any three of them. Then we have the equivalent
system
" * '{ -'r:-t:8 |
This system is in echelon forrn having two equaUons in 4
qnknowns. So the system has 4 - 2 = 2 free variables which
are z and t and hence it has non-zero solutions' l*t z = a and
t ='b where a and b are arbitrary reaf numbers' Puttingz -- a
an{ t = b tn the 2nd equaHon we get Y = a- b' A$airt putting the
values of y, z and t'in the lst equati<in' we $et x ='- a - 2b'
ry the.r.equi@ solutlon sPaee is
W = (-a- 2b, a -b,a, U : 4 b e IRl.
@"-rr. 27. Iet s and r be ttre following subspaces of IRa :
s - lbcY'z'g lY- 2z+t=al
T= {(x V, z, tl I x-t = O, }, -22 = Oll
find abmis and tlre dircnstron of (il S (iil T (iiil SnT'
tBof,utrm ; {i} we seek a basis of the set of respective
$rrfi*, {x,y. z, t)of ttre'equafony -22 + t = O'
T 295
f]',./i: 'r'i
t-t.r, tft
1'il
VEgfOR SPACES
The free variables are x z andt,.Pfi!,,, i..-.,.',,...,..,;(i-,,r :
i.iulc.l'luir'r 'i'\'
ilI,..il r,;rrto*rtrIa) x=l,z=Q,{=O ';11''i i i[ti'- '1 "''"t :':li'
(b) x= O,z= l't=O'(c) x = O, z = O,t = I' to obtain the respective solutions
u1 =(1,O,O'O)ru2 =(O'2' l'O)'us =(O' t'0' t)
The set {ur . trz, u3} is a basis of S and dim S = 3'
(li) We seek a basis of the set of solutions (x' y' z't) of the
equatlons
; -2i:3 )'llre freevarlables atezandt' Set (a)z= 1' t= O' (b) z=O'
t = I to obtain the respeCtive solutions u1 = (tO' 2' 1' O) and
u2 = (I, O, O' 1). The set {ur' uz} is a basis of T and dim T = 2'
(iii) SnT consists of those vectors lx' y' z' t) which qtisff
all conditions given in S and in T' i' e'
Y-22+t=9'lx_t=o Iy-22 =O J
t- 9 I srutt""t second equation
't, * ,==uo Irrom the third equation'
x-t=bly|2z = o I wtrictr is in echelon form'
t=OJ
is z. Set 7 = | to obtain the solution
Thus (ul ls a basis of SnT and dim (SO'IJ = 1'
erampte za. (i) I€t U be the subspace of IR3 teanryl
(generated) by the vectors (I' 2' I)' (O' - 1' O) and Q' A' 21' nnd a
basis and the dimension of U'
xor, Y-
.v-I
Then we have
The free variable
u = (O,2, I' O).
a
296 COLLEGELINEARALGEBRA;:. ,, (lil l,et w be tfre qub"f.". of tRs spanned u)r'tn. ,g"ior"
|t, - 2, O, O, 3), 12, - 5, * 3. -2,6), (o, 5, 15, lO. O) and (2. 6' 18, 8,6).
l'ind a basis and the dirnension of W.
Solutioa : (i) Form the matrix whose rows are given
vectors and reduce the matrix to row-echelon form by the
elementary row operations.
fl 'z, I I *. multiply first row b1r 2 and then
L; -6 i J ""utt'ct
from the third row'
fl ? L l *. multiply second row by 4 and- L; J 5 J ""utt""t
from the third rour'
fl ? I I *. multiply second row by 2 and- L;
-6 6 Jtt"" add with the rlrst row'
rl o ll': -lO -l O lrvemultiptysecondrowby-lLo o oJ -------'r-J
rl o I1-lo r o I
Lo o oJ
This matrlx ls ln fow-ecttelon form and the non-zero
rows in the matrix are (1, O. l) and (O, I, Ol. These non-zero
rows form a basis of the row space and consequently a basts
of U; that ls, Basis of U = l(1, O, U, (O, I, 0)l and dim U = 2.t.:
'' (ii) Form the matrlx whose rows are the given vectors and
reduce the matrix to row-echelon form by the elementary rorv
o 3l-26110 0lI 6Jli
-20-5 -3515618
vEc'l.oR SPACES 297
Redr,rce thts matrix to row echelon form by the elementary
row operations.
we multiply Ilrst row by 2 and then subtract from the
second and fourth nows respectlvely.
rl -2 0 0 3llo -l -B -z ol-lo 5 15 lo olLoloIs 8oJ
We multiply second rour by 5 and then add \lrith the thtrd
row.
rt -2 0 0 3'llo -l -3 -z ol-lo o o o ol.iL Loro l88oJ
Interchange third and fourth rowsrr -2 0 0 3llo -r -s -2 o I-lo ro ls 8 olLo o o o oJ
We rnultply second row by lO and then add with the thtrd
row.
We multiply second iow by -I and divide thlrd row by -12,rl -2 0 0 31lo r s 2 ol-lo o l r olLo o o o oJ
rr -2 0 0 3lIo -r -3 -2 o I-lo o -tz -t2 o ILo o o o oJ
298 clol'l-ltcll t',lNltAlt Al-cDI']ltA
..'llhe'arbovt'trri'rlrixisitrrtlrvcc;ht:lotrlbrttl"lhctror*atlifor't::ir'rtIr:ot\rsitgte{ots}:in tllt'trbovc tttittrix'ittt'tr'{1;;+21''O.:iOl;3}:
(O: '}'l}i2r':O}3 '
arrrcl (0. 0' l' l ' O)''l-tr-sc llol]-zcl'o rorvs lirrtlr lr basis lbr thc row
spilce trtrd cotrseqtrt'nth' -ar birsis of W' 'lhtrs belsis tt
\= (1. -2. 0.0'3). tO' f ' S' 2' O)' (O' O' l' i' O)) uncl climW = 3'
. Example 29' t't't W bc tltc sttbspircc llcncrirtcd by the
polyrronrials p1(t) = trt +212 - 2t + l'^
pz (t) =t3 +3t2 -t +4turdprr tt) =2t1t +t2 -71-7'
tiinci a basis aucl the climc.irsion ol W'
Solution : Clcarlr'' W is tl subspacc of the 1'1'6tor space V (F)
ol polylrotnitlls itl t olclcgrct: 5 3' Tlrus the set 51 = {l' 1' 1z' 1s} is
. basis of V (Ir)' I ncctors pr (t)' Pz (t) a.d
Now the coorclitratcs ol the givett
p3 (t) relatirre to the basis 51 arc (l '2' -2' I)' (l' 3' -I' 4) and
12. l. -7 . -7) resPectivelY'
F'orurin-g the rrratrlr rvltose rows eu'e the irbove coordinate
vcctors. rvc $et
Recluce this nrirtrix to rorv echelcltr lbnn by the elernerrtary
,or,.Up"rotioirs' We rrrultiply lst rorv by I ancl 2 irrrd tlien
sttbtrilct ltonr 2rrcl & 3rcl rows respectivel]''
fr 2 -2 ll-lo I I 3l
Lo -3 -3 -$l
rr 2 -2 Illr 3 -I 4 I
lz | -7 -7)
Wc nurltiplv 2trcl torv br' 3 arrcl t.hcn irdcl rvitlr tlre llrcl raw
tl 2 -2 ll-lo I I 3l
Lo o o oJ
VECTOR SPACES 299
This matrix is in row echelon;forrn having two nonzets .'
rows (coordinate vectors) (1, 2, -2, t151i6'1191:;1''tll"3)it'"hitfiuiill'I: ir
f,orm a basis of the vector space generated by the coordinate
vectors and so the set of corresponding polynomials is
1ts + 2t2 * 2t + t, t2 + t + 3l whtch will forrn the basis of W'
ThusdimW=2.*lr-rn$le 30. L€t U and w be the subspaces of ts.4 generated
by the set of rrectors
(1, I, O, -1), (1,2, 3, O), (2. 3, 3, -l)l and
1O,2,2. -21, Q.3,2, -3)' (1' 3' 4' -3)l respectively'
Find (il dim (U+W) and {ii} dirn runW. D. U.H. rgffil
Sof,rrtloa : (0 U + W is subspace spanned (or generated) by
all given six vectors. Hence form the matrix whose rows are
the given six vectors and then reduce this matrix to row
echelon form by the elementarSr row operations :
r1 I O -1 -1
lr 2 B ollz s s -I Ilr 2 2-2 IIe B z -s ILi s 4 -slReduce tjhis naablx to rsqr-echelon forrtr by the elementary
row op€sdions.We subtract lst row from z!rd, 4th and 6th rows' Also we
multiply Lc* rorr by 2 and ttsr subtract from 3rd and Sth
rows.1 0 -l-rl3 tl13 ll'.{ 2 -llt2-rl2 4 *zJ
3OO COLLEGE LINEAR AI,GEBRA
We subtract Znd row from 3rd' 4th & sth ro\f,rs' Also we
multiplY 2nd roulrllo
-lrloLo
by 2 and then subtract from 6th rour'
r o -l-1l B tlo o olo -l -zlo -l -2 Io -z -+J
We multlply 4th row by I and Z and then subtract frtrn sth
and 6th rows nesPecfivelY'
rl I O -l-'llo I s t Il"oo o ol-lo o -l -2 Ilo o o o ILo o o oJWe interchange Srd and 4th ron's'
rl r o -1lo I B Ilo o -l -2-lo o o olo o o oLoo o o
This matrix is ln row-echelon forrr havtn$ thee nolr-zeno
(1, l, O. -l), (O,l' 3, I) and (O' O''l' -2) whtchwill form a basls
ofU+W.ThusdimU+W)=3'(til l,et us flrst ftnd the dirn U a4d the dtm W' Forrn the
inatrixwhose ro\ils are the generators of u andtlr@redtreetbe
matrix to row-echelon form by the elementary rtrnr
operationsrl I o -l 1lr 2 3 olLz 3 s -lJ
VECTCIRSPACES 301
We mrdttply l"st row by 1 and 2 and then subtract from 2nd
and 3rd rows respectivetY.rt 1 0 -I1
-lo l s t I
Lo I 3 tJWe subtract 2nd row frorn third row. i
11 I O -1 -'l
-lo r s r I
Lo o o oJThis matri:r is ln row-echelon form having two non-zero
rows tl; l, 0, -1) ard (O. 1, 3. l) which will forrn a basis of U.
Thusdlm U =2.Again form the matrix whose rows are the generators of W
and then reduce the matrix to row-echelon form by the
elementary row operations.11 22-21lz s 2 -s I
Lt s 4 -3JWe rnultiply lst row by 2 and I and then subtract from 2nd
and 3rd lows nespectivelY.
We adtl 2nd rowwith 3td tow.
t-l 2 2 -21-lo -l -2 I ILo o o oJ
This matrlx is in row-echelon form havlng two non-lrero
rows which will form a basis of W' Thus dtm W = 2.
Now by theorem we have
rltm (U + W) = dtrn U + dlm w - dtm runw)rrr, dlm runW = dlm U + dim W- dtm (U +ril =2 + 2- 3 + t'
.'. dim {UnW1o I (one).
r1 2 2 -21-lo -r -z I I
Lo t z -rJ
3O2 COLLEGE LINEARALGEBRA
Exanple 31. Ift V be the vector space of 2 x 2 rnatrices over
.the real field lR Find a basis and the dimension of the
subspace W ofV sPanned bY
o= [-i'r],"= [? -i I, "= F'? tr *u"=[-, gl
Solution : The coordinate vectors of the given matrices
relative to the usual basis of V are as follows :
lAl = (1, 2,-r,}l,tBl = (2, 5' 1,-1). [C] = (5, 12, 1, I]and
ID I = @,4'-2,5).
Form a matrix whose rows are the coordinate vectors and
then reduce this matrix to row-echelon form by the
elementary row operations and join successive matrices by
the equivalence sign - :
T1 2 -r 3llz b r -rIls 12 l llLs 4 -2 sJ
' We multiply 1st row by 2, 5 & 3 and then subtract from
2nd, 3rd and 4th rows respectivety.
T1 2 -1 3llo I 3 -7 1 ;
, -lo 2 .6 -I4 I
Lo -2 | -4J
Wemultiply2ndrowby2anrl_2and'then*strbtractfrom..rs8rd and 4th rorvs resPectivetY
tl 2.. -T 3ll'o I 3 -7 1 's-loo o olLo o z -rBl
VECTORSPACES 303Interchange 3rd and 4th rows.
[1 2 -] 3tlo I 3 -rl. -lo o z -18 I
Lo o o olThis matrix is in row-echelon form having three non-zero
rows (1, 2, -1,3). (O, L.3, -T) and (0, O. 7, -lB) which arelinearly independent.
Hence the corresponcling matrices [_] 3 I tS _] I *o
E -rB I form a basis ofw and dim w = 3.
Example 3.2. Let V be the vector space of 2 x2 matrices overthe real field lR- Find a basis and the,dimeflsion of .thestrbspace W of V spartned by the matrices
" = U tl "= [-i iJ," = [-; fl.*o, = [_u, -il
solution : The coorcrirate vectors of the given matricesn'lirlive to the usual birsis of V are as lbllorvs :
[A]=(l.,;5,1.2).tB]=(1,1.-1.5).tC]=t2'4._5,.7):urct [DJ= (1. -7. -5. 1)
.
Iiornr a lnirtrix.rr4rose rot\l$ arcl the goorclinatq riectors,4ndllrt'rr rt-dtrce .tltis. rltatri-r to r.or,l,-echqlorr ..form.,.l:y , ther'l('ulcltti'r11: l'()\\' opcrations irrrrl -ioin succqssjt,e t4a{ficgs b1,
llrt't'rlrrir-irlt-I-tt.c siett - : :, .. . ,.,,..j ,ii it I
rl -5 -l 2rl ',lr r _.t s.l12 + -'5 T i' I
i r -r -r--, I _l
304 COLLEGE LINEAR ALGEBRA
We multiply lst row by l' 2' and I and then subtract from
2nd. 3rd and 4th rows respectively'
TI -5 4 21lo 6 3 3l-lo 6 3 3lLo -2 -L -lJ
i.
We multiply 2rrd row by I and - | and then subtract from
3rd & 4th rows resPectivelY
rl -5 4 21lo 6 3 3l-lo o o olLo o o oJ
Thts matrlx is in row-echelon form having two non-zero
rows (1, -5, - 4. 2l and (O' 6' 3' 3) whleh are ltnearly
independent. Hence the corresponding matr{ces
l-i f I *o tS S I to'- a basis or\M and dim lsr = 2'
i u1 = (1' O' 2)'ErmPle 3& Pr<rre thatthevectorl
u2 = (-l' 1, O) and u3 = (O' 2' 3) fornr a basls of lR3and find
the co-ordtnates of the vectors v = (l -1. l) and w = {-1'8' 'tU
relative to this basis'
Proof : Flrst Portlon
The given vectors will be a basis of IR3if and only if they
are linearly independent and every vector in lR3can be
wrltten as a linear combination of u' ' u2 "rnd
u3' Flrst we shall
prove that the veetors ut' uz and u'' ar '' linearly independent'
For arbitrary scal rrs X1 ' x3 tmd x3
letx1 u1 +t81u2lJQU3=:$
or, x1 {1. o' 2l f 4 (-l l' Ol + re {O' 2' 3} = (o' O' ol
of, {xr -lQ'x2+''1x3"'2xr +Sxs) *{O'OQ
VECToRSPACDS " 3O5
Equattng correspondlng components and forming the
llnear system, we get
xl - rcz =Olx2 + 2x, =O | (1)
2x1+3x3=91
Reduce the system to echelon form by elementary transfor
matlons. We mulUply ftrst equation by 2 and then subtract
from the thlrd equation.'Thus the above system reduces to
xr-xz =Olxz +2x. =g; l2l
2x2+3x"=61
Again we multiply second equation by 2 and ttren subtract
from the thtrd equation. Then we get the equivalent system'
xr-x2 -Olx2+2xs=O1 (3) :
- xs =O J
This system is in echelon form and has exactly three
equations in thrie unknowns, hence the system has only the
zero solution i. e. x = O, y = O, z = O. According[y' the vectors are
linearly independent.
To show that u1, u2 and u3 span IR', *. must show that an
arbitrary vector v = (a, b, c) can be expressed as a linear
combination v = xlur +l0r& +x.gus
or, v: (a, b, c) = xr (1, 0, 2) + & (-1, 1, 0) + xs (0' 2' 3)
Forming the linear sYstem. lve get
xt'xz -a I)02 + 2'k =b I
2x'1 + ?*. =o j
Linear Algebra-20
COLLEGE LINEAR ALGEBRA3oo
iystem to echelon form by the elementaryReduce this s
transforrnations'We multiply first eqrration by 2 and then subtract from the
third equation, Thus we have the equivalent system
xr-xz =a )^I ii,+zxs=;^ | (5)
2;2 + 3xs = c-2a )
We multiply second equation by 2 and then subtract from
*" *t i ;t*n-on' Then we get the equivalent system
xr-X2 '-a I^r ii*zxs=\ ^ ^Li tot
- xs =c-2a-2b ) I
From ttre third equation' we have xs =?a+2b-e'
Substituung the value o{x3 in the second equatlon' we get
x2 =-4a-3b+2c'Again, substituting the value of x2 in the first equation' we
$et x1 = - 3a - 3b + 2c' Therefore'
v = (- 3a- 3b + 2c) ur + F 4a- 3b+ 2*) uz + (% +2b-c) u3
or, (a, b' c) = F3a - 3b + 2c) (1'O' 2l + (4a-3b + 2c) Fl' f ' O)
+ (%.+ 2b - c) (O' 2' 3)
Thus every vector in x-3can be expressed as a linear
'u'combinatlon of the vectors (1' O' 2)' Fl' 1' 0) and (0' 2' 3)'
HenceU"""ttoolft'u2andu3formabasisof=x-r'
Second pordon : Irtv = (1. -1, 1) = xtut + )c2W + x3u3
or, (1, -1, 1) = xl (1'' 0' 2l + n(-1' 1' 0) + xs (0' 2' 3)'
Forming linear sYstem' we have
1l*t - X] * 2*"-= _'l I
2xr +3x3= l,lt7)
VECTORSPACES .3O7
Solvlng the system, we get xt =2, h = l. xs= - l.'lhus v = (1, -1, l) = 2ur + lu2 + (-1) us.
So the vector v has co-ordinates (2, 1, -l).Similarly, let w = (-1, 8, 11) = 1l1u1 +Y1uz + Ysue
or, (-1, 8, 11) = yr (1, O,2l +Yz FI, I, O) + Ys (O, 2, 3).
Forming linear system, we have
Yr-Yz --t IYz+2Ys= 8| (8)
2Yr +35rs =lIJ
SoMng the system, we get Yr = 1, Yz = Z,ys =3
Thus w = (-1, 8, 11) - lur + 2u2 + 3u3
So the vector w has co-ordinates (1, 2. 3).
Bsarnple 34. Given the vectors (2, l, l), (1, 3, 2), (1, 3, -l)and (1, -2, 3). Test whether they are lineaily independent by
Srreep out method.
Solutlon : [,et o1 (2, l, 1) + o2 (1, 3,2) + os (1, 3, -1)
* d+ (1, -2, 3) = O = (0, 0, O),
where clr, da, cL3 and o4 are scalars. then
(2o.1 +de +os +oQ, or +3a2+3o.3-2a+,u1 +2ru2 - crg + 3&)
= (0, O, 0). Equating corresponding components from both
sldes and forming linear system, we get
2a1 + o"2
0,1 + 302a1 + 2a'2
+ 0,e+ oa=Ol+3a3-2o"a'Ol- 0s *3uo =6J
308 COLLEGE UNEARAItrEBRActg
O-+RrO-+RzO-+Rs
O-+R+= ^f tst pivotal ro\r,
O+Rs= Rz-RlO-+Ro= Rs- R*
o-+Rz=$z"a pivotal row2
O-+Rs = Re-8 Rz
o _!3 o+Rs= S sta Pivotal row
From the pivotal rows, we havelllar +icLz+;s3+; a4 =O (i)
o,z+%- qa=O
cr. -i an =O
This system is Ir echelon form and has three equations in4 unknowns and hence 4 - 3 = I free variable which is aa.
Thus the system has an inftnite number of non-zero
solutions.kt o4 = t, where t is a scalar.
Then cr3 =tt,or=-it.and0,1 =-1Slnce all o's are not zero, so tJre given vectors are linearly
dependent. l
rr'*.lrr[fu 8li. ArE these vectors (2, 1, U, @, 4, n, (4, -9, l1)
dependent ?. Test by Swecp out method.
Solution:Irta1 (2, t, 1) +oq(i,4,V+43(4,-9, 11)=(O,0,O)
Or, (2q + 2a,2+ 4a4, a1 + Aoe- 9cr3, a1+ 7a2+ llo3) = (O, O, O)
ll3-2-1 3
2113t2
o
(lr)(rii)
VECTORSPACES 309
Equating corresponding components from both sides andformtng the linear system, we get
2a1 +2o"2 + tlas =O)a1 +4or2- 9as =O.la1 +7a,2 + llq =QJ
Ol O"2
o-rRrO-+RzO-+Ra
O-+R+= lst pivotal row
O+Rs= RzO-+Re= Re
O-)Rz=f Z"O ptvotal row
O+Ri = Re -6Rz
o O-+Rg= fr sra pivotal row
From the pivotal rows, we havecr +(& + ?aa=Q (i)
o, -* as =O (ii)
0g =O (iii)Therefore, the svstem has zero solution i. e o,1 =az =o,a = O.
Hence the given vectors are linearly independent.F-orlrlllc 36. FIxd the rank and the basis of a given set of
raectors {2, -1,5, 4, (O, t, Z, gl, (4, O, 6, U, (O, - 2, 4, nby usingSwccp out mcthod.Solutlon : Rank of a glven set of vectors is the number of
linearly independent vectors of that set and these linearindependent -vectors form a basis of thqt set.
224t4-gI 7 ll
3to COLLEGE LINEARALGEBRA
2-rb4orz3406lo-247
ooo
_Rr_rRz_eRs-+ Ra
--+Rs = ! t", pivotal row
-eRo = &- ORu*Rz = Rs- 1Rs-eRe=Ra-ORs
-*n Y 2nd pivotal row
-eRro= R7 -2Re+Rrr = Rs +2Rs
+Rrz= -$e
sta pivotal row
-+ &g= Rrr- SRrz
Now the rank of the given set of vectors is equal to the. number of pivotal rows = 3. A basis of the given set of vectorsis {(1, - i,lorl,@, r,z,s), (0, o, t,#)}
Since cr, (t, -i, i,Z) * oa(o,lp,B) + os (O, O, t,i3)= (O, O, O, O)
implies dt = &z = ct€ = O, (7aro solution).
ETRCISES - 6 (C)
l. Prove that ttre vectors (1, 1) and (t, O) form a basis of IR2.
2. (i) prove that {(2, - i. tl, E, 2, l), (O, t, l)} is a basis of IR3.
(ii)Provethat{(1, I, l, l), (0, l, t, I), (O, O, I, t).(O, O, O, r}}isa basis of lRr.
3. (i) Extend lQ, O,O, -l), (1, 3, -I, O)l to a basis for IRa.
ID.U.S. 1e84t
o1232-4-7*247
-8 -13813
VEC'TOR SPACES 3Il(ii) Extend {(1, 2, 0, 3), (2, *I, O, O) to a basis of IRa.Answers: (, (1, S, _t, O), (2, O, 0, _t), (0, O, l, O), (0, O, O, l)l
(ii) {(1, 2, o. 3), (2. _1.0, o), (o, o, I, O), (0, o, o, t)}4. Decide whelher S = (2, t, l), (t, 0, O), (S, I, t)I is a linearty
dependent subset of IR3. what is the dimension of thesubspace spanned by S? ID. U. S. r9g2l
Answer : s'is linearry dependent and aimension of thesubspace ls 2.
5. The subspace U of IR{is spanned by the r,,ectors (I, O, 2, S)*d (O, l,:1, 2) and the subspace Vof IRais spanned by(l,2,3,4), (-1, *1, 5, o)md (0, o, o, r).Find the dimension of U, V, UOV and U + V.Ansnret$ : dim U = 2. dim V= 3.
dim (UnU = I, dim (U + V; =4.6' Find a basis for the subspace of lRaspanned by the givenvectors: '
(l) (1, l, -4, -3), (2, O, 2. -2), (2.-1, 3, 2)(ri) (t, 1, o, o), (0" o, l, l), {1, 0,2,21 (o,_3, O, 3),*1**: (r) (1, r, -4, -3), (o, l,-5, -2t, @,0, t, -))
(U) (1' 1,0, O), (0, I, I, U, (O, O, l, l), (O, O, O,t)}.7. Pind the dtmenston of the subspace generated by the set
{(1, 2, l), (3, l, 2) {1, -S, a)} of Vs GR). tD. U. p. r9z9lAnsrer: The dimension of the subSpace is S.8. Let W be the subspace generated by the polynomials
1" -[s -2P + 4t+ l.vz =2t3 -3t2 +ft_ I,vs = f3 + 6t- 5, va = 2ts - St2 + Tt+ S.
Find the basis and dimension of W.
IGU. P. 19ZI),,86;.L U. H. 19g6lAnswer: Basis: {ts *!12 +4t+ l, t2 + t_S}and dimW= 2. i
312 COLLEGE LINEAR ALGEBRA
9. F-ind a basis and the dimension of the subspace W of P(t)
spanned by the PolYnomials(i) Pr (t) = t3 + 2t2 -3t+2,pt (t) = ts +2P -2t + 3, and
Ps (t) = 2F + 3t2 -5t - 5'
{ii) -Pr (t) = ts + t2 - 3t + 2, P2 (t) = 2F + t2 + t - 4, and
Pg (t)=4ts+3t2-5tt2.Ansnrers: (0 BasisofW= {ts+ lfz -3t+2;t2 -t+9, t+ U'
dimW = 3
(ti) Basis ofW = {t3 + f2 -3t + 2' P -7t+ 8, 2}
dlmW = 3
1O. Determine whether the given set of vectors is a basls
for gl3over IR:(0(f. I'O)'(l'O, 1),(o, 1. 1)l
(li) (-1, L,21,(2' -3' U!.(10' -14'O)lAnsweis : (r) Set of vectors ie abaQis for m.3 .
(ii) Set of vectors ls not'h b$is for m,'
11. I.et {vr, va . vs} be Uasts foi a vector space V' Show that
{ur , uz. u2} is also a basis, tilt = vl, ', = '1 + v2 and
us = vt * v2 * v3' :tors12. LetW be the subspace of g1o generated by the ve<
lL, -2,5, -3), (2. 3, l, -4) and (3; 8, -8, -5), Itnd a basls and the
dimenslon of W. lD.u.s. rpsslAnswer: Basls lll,-2,5, -3J. P. 7, *S' 2)), dimW = 2'
13. Consider the following subspaces of IRs:
U = span {(1, 3, -2, 2, 3}, U, 4, -3,4,21' (2,3, -l' -2' 9)}
W = span (1, 3, O, 2,.U' (1, 5, 4, 6, 3)' (2' 5. 3' 2' 1)l
Find a basis and the dtmeusion of
(i)U+W (iD Unw.eniwers : (i) Basis of U + W
lD.u.H. 19S71
= {(1, 3, -2,2.31. (0, I, -1, 2, -1)' (O' O. 2' O, -2}}
dlm(U+W)=3.
IfECTOR SPACES 3T3
' (ii) P."i" qf UnW = {(1, 4, -3,4,211dim flJnw) = 1'
14. I.et S and T be the folloulng subspaces of gla :
S=[(xY,z,t)lY+z+t=o]T={(x y,z,t) lx+y= O,z=2t}
Fired the basts and the dimension of' '{r} s.,liil T (tu} snt
{al, o, o, o), (o, -t, I. o), (o, -1, o. I)}, dim s = 3.
(li) Basis : {F1, t, O, O)' (O' O' 2, l}, dimT= 2:: (lti) gasls : {(3,'3, 2, lU, aliri tSnT) = l.
15. LetW = (a, b, d la, b, ceRand 2a +b + 2* = Ol
Flnd a basis and.dlrnension of W. ID. U. Prcl" rgEglAnsrer : {{I, l, -U; (-1, 2. O)} is a basis of W and dim W.= 2;
16. Frovethatin Uf tfr.vectors (1, -1, O)' (O. l, -1) forma-: , ,
basis for the subspace U = {(x Y, zl ets3 l x+ y + z = Ol-
17, Ffnd " Uasis of eat!of.thefollorrrtngsubsPaces.of nR3 :
(t) U = {(x Yr z) 11' z=Ol ",' , i- '
(tt)V= llXy,zl l'x=O,andy+z=Ol ':: 'i ' '
Ariwerlg : (i) (2, l, 9r, (i, O, f)l it a basis of U. :' .
(ii) {(O, -1. 1), (0, 1, -1)} is a basis of V.
18. I€t V be the vector space of all 2 x2 matrices orrcrlhe
real field F. Prove that V has dimension 4 by exhiblting abasis for V which has four elements.
19. I.et V be the vector space of 2 x2 matrices over the real
fteld IR; Find a basls and the dirmenslon of the subspace W of V' rices"=[i 31,"= E i J* "; [3 ? Ispanned by the matrices ":.lrgl;": qr3 I and u = [S
AEEscfs: Basisoflf,r= tll 31, I-? l]}dimW=2.
314 COLLEGE LINEAR ALGEBRA
20. lr:tV be the vector space of 2 x 2 matrlces over the real
field IR. Find a basis and the dimension of the subspace w of V
spanned by the matrices
o= [-l 3l,r= [?-? I*o .=l-3 tAnswers : Basis "* = t[-l 3l tB
2x1 + 2xz - X3 +.,cs = O'l
-X1 -x2+2xs-3xa+rE=O Ixr * *, -Tr* * ;,t;3,[
Ansrrcr : Basis : {(-f ,1,0,0'0), (*l' O' -1' O'
dimension = 2.z
I
-il [? i,l]dimW=3.
2L,l*tV be the vector space of 2 x 2 matiices over the real
field IR. Detennine r.vhether
o=[lll,u=H l],"=[? ?l*r=[3 ?lform a basis for V.
Answer : They form a basisforV.tt
.6\d- 22. Find the dimension and a basis of the solution space Wt
of the following sYstem
ID.U.S. lSOl
: dimW= 3"
ID.U.P. 19841
1)l
for the following homogeneous linear equations :
315
space
ID.U. P. 1S4l
x1 +2x2 -xs *4rq =0')2x1-x2+3x3+3x+=014x1+x2+Sxs+9xn=O7
x2-x3+xa1Ol2x1+ 3x2 - xs * 7x4 =O ) '
Ansnrer: Basis : {(*1, I, t, O), {-2, -1, O, f}}
'L dimension = 2.
WU. orri_a basis and the dimension of the solution space
for the following homogeneous linear.systemx+2y- z+3s-4t=O I2x+4y-22- s+5t=O I2x+4y-22+4s-2t=O )
Basis: l@,-t, O,0, O), (1, O, I, O, O))
'l' ai*.rrsion = 2.p'26. Find a basis and the dimension
system of linear equations :
x1 +2x2-2xs+2xn-.6=O1x1 +2x2 - xa + 34 -2x" =Ql
2x1+4x2-7xs+ &+&=OJAnsrers : Basis : {(-2, I, O, 0, O); (-4, O, -f , 1, 0), (3, O, f , O, l)}
dimension = 3.
27. Prove that the vectors u1 = (1, 2, | -2),
u2 = (0, -2, -2, O), u3 = (o, 2, 3, 1) and ue = (3, O, -3, 6)
form a basis of tRo and find the coordinates of the vectors
v = (5, O, - 8, -1) and w = (-9, 20, g4, - 25) relative to thislrirsls.
Answer : v and w have coordinates 12, - l,- 3. 1l andI
13, -2,5. -41 respectively. \
lRU.r{. r9B5l
of the following
I D. U. H. 19881lR u. H. 1988 |
3T6 COLLEGE LINEARAI'GEBRA
28. Usin$ Swccp out mcthod prove that the following set of
vectors is linearlY indePendent :
{1, g,21, (1,-7, -8), (-3' -l' -4}}'
zg.F]ndtherarrkarrdthebasisofagivensetofvectors
AnsEGr: Rank is 3 -, - q
Basrs = (1, 1,5.P, l, -r\(0, o' 1)l'
3O. Shour ttrat the set S = {(l' O' O}' (1' I' O)' (l' I' l}} is a basis
of IRs and hence find the coordinates of the vector (a' b' c) with
respect to the above b:asls'
.AoErtr : (a-b'b - e' c)'
31. Shonr that if {u' v, w} is a basts of IRs ' t}ren
32. IfWl ls a subspace of IRa generated by a set of veltors
' Sr = (1, 1' O, -l)' (1, 2' 3' O)' A' g'g'-lD andWz !s a subsPace
of IRa generated by the set of vectors
Sz = ((1,2, 2,-21.@'g'2'-3)' (l' 3' 4' -3))'
flnd (i) dtrn (Wr +Wd (it) dim (Wr nW2)'
Anglrlre : (f) dim (Wr +Wd = 3' (ii) dtm (Wr oW2) = 1'
33. Prove that the vectors v1 = (1' 2' Ol' vz = (O' 5' 7)' and
v3 = (-1, 1,3) form abasis of IR3and find the coordinates of
the vector v = (2' 3, 1)' ['' U'Ia 1990' 9U
Aosrtr: M = [O. l,-21. !94. LetS be the following basis of the vector syf W of
2x2ralsYmmetric matrlces : -ili';l Ii ll l-; -?D
Find the coordinate vector of ttre matrix AeW relative to
the above o""o *r'.rJr;='1 ; -: I '"0 (b)A = lL '^l
Ansrlr: (a) [2. -r, U O) [3' r' -21'