.-q.F"- --- ;""8F;:;- . r+j:-ri:Y!4S;F*;;+-. ::

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

i,d

i1

'"4

..:.1

,til't,

'1

'i'tt,

4L4l42424243434445

46474849

7t79798283848585

9l

2.82.8.r2.8.22.8.3

J

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'