linear t r a n s f o m t i o n s

8/12/2019 Linear t r a n s f o m t i o n s

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UNIT 15 LINEAR T R N S F O m T I O N SND LINEAR EQU TIONSStructure

15.1 IntroductionObjectives15.2 Linear Transformations

15.3 Rank and Nullity .15.4 Special Linear Transformations15.5 Linear Transformations and Linear Equations15.6 The Complex Case15.7 Summary15.8 SolutionsIAnswers

15 1 INTRODUCTIONIn Unit 14,we introduced some aspects of Euclidean spaces, with a view to study asystem of linear equations. n his unit, linear transformations from one Euclidean spaceto another Euclidean space are introduced. Such transformations help us to understandthe structure of the solution set of a system of linear quations. The notions of rank andnullity of such transformations are first introduced. These notions are then used tounfold the structure of the solution set of a linear system. Important properties ofone-one, onto and invertible linear transformations are also discussed. Finally, theextension, involving complex numbers,isbriefly pointed out.

The main objectives of this unitareintroduce linear transformations from one Euclidean space to another,

* understand the structure of the solution set of a system of linear equations,point out some important properties of special linear transfonnations, andinclude complex case and related aspects.

15 2 LINEAR TRANSFORMATIONSConsider the Euclidean spacesR and R . Let

T : R n - . R mdenote a function mapping).Thus T is a rul which associates a uniqu vector ofRwith every vector of R .As usual, let T X ) R denote the unique vector associatedby the rule T with the vectorX .A mapping T s called a linearh.ansfonnaCwn iffor all Y nR and all a R the following conditions are satisfied:a) T C X + Y ) = T Q + T Y )b ) T a X ) = a T X ) .

Let us understand the implications of the conditions a) and b). The left hand side of a)requires the operations:AddX and Y using the arithmetic ofR n and then findT X+ Y n R according to the rule T. Similarly, the right hand side of a) requires


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the operations :Find T X) and T Y) in R according to the rule T and then addT X ) and T Y) using the arithm etic of R . Condition a) requires that the twovectors T X Y nd T X) T Y ) n R , obtained by different sequences ofoperations, be equa l for allX and Y in R .A similar interpretation can be given forcuiidition b) also.Note that T :R - R s a linear transformation if and only ifT a X + p Y ) - a T X ) + f 3 T Y ) f o r a l l X , Y + E R n a n d a l l a , P - + E R .Using condition b) of any linear transformation T, we observe that

T 0 ) = T QX)= OTIX) = 0.Thus the vector 0 of R s mapped to the vector 0 of R y every linear transfornlationT R - R n other words, if T ( 0 ) * 0 for any mapping T, then T can not be alinear transformation.xample

The mapping T R .R defined byT xi,xz) = xi x2 1 , b l -x2, xi + 3x2) is not linear because

xample 2 :Consider the mapping T :R - R defined by

Then,T 1,O) = 1, 1) and T 0 , l ) = 1 ,-1)

and T [ 1,O) 0, 1)] = T 1, 1) = 2,O).Also,

T 1,O) T 0, l) = 1 , l ) 1 ,-lj = 2 , O ) .This verification does not mean that condition a) is satisfied. For some oth erchoice o f X and Y this condition may not be satisfied. For example,

T [ 1 , l )+ L O ) - T 5 1) = 5, 1)and T 1 , l ) T 1,O) = 5 0 ) 1 , l) = 3, l ) .Hence, this mapp ing is not linear.

Example 3 :Consider the mapping T R . defined by the followin g rule

T X I , x2) = x?+ x2,*1- X 2 , X I 3x2).Let X X I , X Z ) and Y = y l , 2) be any arbitrary vectors of R 2. ThenX Y = X I yl, x2 y2) and using the rule defining T,

X I +y1) + x2+y2),2 X I +yl)- x2 + ~ 2 , ~ 1 ~ l 3 x2+ ~ 2= X I X Z , 2 x 1 - X Z, X I 3x2 @I +y2,2yl -Y2, l +3 ~ 2 )CY+ Y)= I

arithmetic in R 3,T X ) T Y) .

Similarly,T aX ) T a I, a x2)


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= ( a x 1 + a x z , 2 c r x l - 2 a x z , a x l + 3 a x 2 ( r u l e 0= (XI+ ~ 2 ~ 2 x 1 -2 , xi + 3x2) (arithmetic in R )= T (x).

Hence, the transformation T, defined in example 3, is a linear transformation.It may b e noted that in the linear transformation, given in example 3, each of the threecomponents of T (xi, x2) is a linear function o fx l and x2 such that T (0) = 0. We nextconsider a linear transformation from R to R of this type.Let T :R .R e a linear transformation and let {A l,Az, ...An ) be any b asis of R .bs in g the rule T, w e can find the vectors T (Al), T (A2), ...,T (An) in R '. We, now,sho w that the transform T 3 f anyXER can be obtained by using only thearithmetic in R and R ' and the conditions (a) and (b) of a linear transformation.Choose any X .Since {A l, A2, ...A, ) is a basis of R ,we can find uniquely thescalars a l , a2, ..., a,, (i.e., th e coordinates of Xwith reference to the given basis) suchthatThen, we must have

T ( X ) = T [ a l A ~ + ( a z A z +..+a n A n ) ]= T (a lA1) T(azA2 + + a n & ) (use (a)= T (a lA1 ) T (a2Az) + + T (a nA n) (use (a) repeatedly)= a T (AI) +a2 T (A2)+ ...+ nT(An) (use (b) a

Thus, w e have s how n that if T (Al), T (A2), ...,T (An) are fixed , then the re quirem ents(a) and @ of a linear transformation determine T X )uniquely for everyX .Auniqu e line ar transform ation can, thus, be defined by prescribing the transforms of thebasis vectors of the domain space.xample :

Let us determine the linear transformation T :R .R such thatT (1, 0) = (2, 3, 0) and T ( l , 1 ) = (1,2, 4). We first note that ((1, O), (1, 1) ) is abasis of R 2 Therefore, a unique h e a r ransformation satisfying the givencondition s must exist.Choose any X = (XI,xz) in R 2. A quick calculation shows that

Therefore,

Wh ich of the following transformations are linear :(a) T (x, Y, z) = Cv z)(b) T(x ,y ,z ) = ( x , & - 1 , 3 x + y , z + 2 , 3 x )(c) = ( l x I + l y I ,x -y ,. +y ).(d) T (1, y, z) = (0, a )


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i.e. T (S) s the set of all images of the vectors in s as determined by the transformationT. Clearly T (S) R . We use the linearity of T and the fact thatS is a subspace ofRto show that T (S) s a subspace of R .Choose any Yl and Y2 in T (S).Then we can findXi andX2 in S such that

T (XI)= Y1 and T x ~ )Y2

= T (Xi+X2) ( use linearity of T )SinceS is a subspace we must haveXI +X2 S. Hence Yl + Y2 E T (S) .Similarly

a Y I = a T (Xi) = T (aX I )E T (S).Hence we have proved that T (S) s subspace of R whenever T :R R s a lineartransformation and S is a subspace of R . n particular T (R ) s a subspace of RThe subspace T (R ) is called the r nge sp ce ofT. The dimeilsion of T ( R ) is calledthe r nk ofT and s denoted by T). Clearly

T ) = dimension of L { T (Ai),T (Az),...,T (An))where {Al,Az, ...,An) is any basisof R .Again let T R R be a linear transformation and choose any Y ER . Let usdefine

i.e. T-' Y) is the set of all those vectors in R which are mapped to Y by T. ClearlyT-' ( Y ) for every Y E R . It is possible that T-' Y) = for some Y ER .Clearly T Y) if Y 4 T (R ).However T-' (0) * because T (0) = 0. Wefirst show that T - (o) is subspace ofR for every linear transformation T defined onR .Chose any Xi and X2 in T-' (0).Then we must have

T (Xi) 0 and T (X2) = 0.Since T is linear

T (Xi +X2) = T (Xi)+ T (X2)

HenceXi +X2 T-' (0).Similarlyi.e. Xi T- (0) or all a E R.Therefore T- (0 ) s a subspace of R .We note that if anyX * 0 is in T-' (0), henT (0 ) ontains an infinite number of non-zero vectors of R becausea X E T -' (0 )for all a E R. The subspace T -' (0 ) f R is called the null sp ce of the transformationT. The dimension of T (0) s called the nullity of T and is denoted by T).Interestingly the subspaceT (0 ) lays a vital role in the determinationof T - I Y) forany Y E R . Suppose that T-' (Y) * for some Y E R . Then we can choose atleastoneXI E T- (Y) . Thus we haveXI E R and T (Xi) = Y.Now choose any X E T-' (0), .e. T X ) = 0. Using the linearity of T, we observe that


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Thus, we have shown that for any XI E T- Y) and X E T- (O), the vectorXi +X E T- (Y). In other words, T- (Y) contains atleast as many vectors as doesT- (0) provided there is atleast one vector in T- (Y). We now show that every vectorin T- (Y) is indeed of the fo rmxi+ X wherexi E T- (Y) is fixed and X E T- (0).Suppose that Xi E T- Y) and X2 E T- (Y) so that T (Xi) = Y and T (X2) = Y. Then

T (Xi X2) = T (Xi) T (X2)= Y-Y = 0 ,

i.e.,Xi -X2 E T- (0). Thus, the difference of any two vectors in T- (Y) is in T- (0).Now choose any Z E T- (Y). Then

Z = Xi + (2-Xi)= xi + x ,

where X = Z -Xi is in T- (0). Thus, every vector in T- (Y) is of the fonnXi +X,where Xi E T- (Y) is fixed and X E T- (0).The foregoing discussion shows that the determination of T- (Y) depends upon ourability to find someXl such that T (Xi) = Y and the null space of T, i.e., T- (0). ThenT- (Y) can e viewed as thatparallel translate of the subspace T- (0) which containsXi.It may be noted that since T- 0) is a subspace of R , it can be completely describedby any one of its basis. Hence, T- (Y) can be described in terms of any basis of T- (0)and any one Xi stisfying T (Xi) = Y.If T- (Y) * and a non-zero X is in T- (0),then T- (Y) contains an infinite number of vectors of R .There is an interesting relationship between the rank, p (T), and nullity, v (T), of alinear transformationT. Let T :R .R be a linear transformation. Then v (T) is thediinension of T- (0). Let v (T) = k and choose any basis {Ai,A2,...,Ak) of T-I (0).Then, we can find vectorsAk, 1 , .. ,A, such that {Al, A2, ...,Ak, Ak + 1 , ..,An) is a basisof R .We shall prove that {T (Ak+ 1) , ..,T A,)) is a basis of T (R ) , .e., the rangespace of T.If possible, let {T (Ak, 1) , ..,T (A,)) be linearly dependent. Then, we can choosescalars a k + 1 , ...,a, such that

and atleast one of the a s is not zero. Using linearity of T we get

i.e,, a k + Ak+1+ ...+ anAnET- (0). This is not possible in view of our choice of A s.Hence {T (Ak i ) , ..,T A,) ) is linearly independent.Further, choose any Y E T (R ) so that T X ) = Y for someX E R .Since{Al, A2, ...,An) is a basis of R ,we can determine p s uniquely so that


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because T A;) = 0 for i = 1 to k. Thus any YE T (R ) can be expressed as a linearcombination of { T (Ak 1 , .. T (An) ).Thus, we have proved that { T (Ak ...,T A,) } is a basis of T (R ). Since this basiscontains ti-k vectors, we have p 0 = n k = n T). Hence, we have therelationship

p T ) Y T ) = n = dimension ofR .The rank, p (T), of a linear transformation T can be easily determined. The aboverelation can be used to determine the nullity,Y T). Hence, the size of T-' Y), which isidentical to Y T) is some sense, can be determined for any Y E R .

xample :Let us use the linear transformation T :R r R defined by

to illustrate the various concepts discussed in this section.First, consider the following subspace of R

Check that [ ( 0,1,- 1 ) , ( 1 -1,-1 ) is a basis of S.Since,

T(0,1,-1) = (0, -3) and T(1,-1,-1) = (-1,0),therefore, T ( ) = L [ ( 0,-3 ), ( -1,O ) ] = R2.We also know that [ ( 1,0,0),( 0,1,0),( 0,0,1 ) ] s a basis ofR 3. Since,T(1,0,0) = (1 ,1) ,T(0,1,0) = (1,-l),andT(O,O,l) = (1,2)therefore

T ( R3 ) = L {( 1 l ),( 1,-1 ),( 1,2) = R2.Hence,p(T) = 2 an d v ( T ) = 3 - 2 = 1. Further,

Since Y ( T ) = 1,any non-trivial solution of the linear system

can be chosen as a basis of T ( 0 ). Check that (XI,x2, x3 ) = ( 3, 1 , 2 is oneof the solutions. Thus, { ( -3, 1, 2 ) is a basis of T-' ( 0 ) , .e.,

T - ' ( 0 ) = L { ( - 3 a , a , * r ) l a R } .Now, let us choose Y = ( 1,l ) in T (R3) i.e., R2.. ThenT- '(Y) = ~ - l ( l , l ) . = ( x I , x ~ , x ~ ) ~ x I + x ~ + x ~~ , x L - X Z + ~ X ~I } .Since ( -2, 1 , 2 ) satisfiesxl +x2 +x3 = 1, andxl -x2 k 1, we haveidentified one member of T-' ( Y).Hence,

It may be noted that T-' ( 0 ) is a stmight line in assin^ in^ through the points( 0, 0, 0 ) and ( -3, 1,2 ).Further, T-' ( 1, 1 ) is also a straight line in R3 passing

inear forn nut ions andinear quations


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Determine a linear txansfom ation T d R~ w k e xange space isL { ( 1, z 3 ), 1, 1, 1 )].

E 7Give any exam ple of a linear transformation such that p ( T ) = v ( T ).

E 8Let T be the linear transformation

T ( x , y , z ) = ( x + Y - ~ z , Y - z )Detennine T- ( 0 , O ) and T- ( 5 3 ).

through t eparticular. olution ( -2 , 1 ,2 ) and parallel to the straight line T- ). ThusT- ( 1, ) is a parallel translate of T - 0 )

EFind the rank and nullity of the f ollo win g linear txansformationsa) T ( x , y , z ) ( x - y + z , y - z , x ) .b) T ( x , y , z , t ) = x + y - z + t , 2 ~ + 2 y , 3 ~ + 2 y - z + 2 t ) .


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15 4 SPECIAL LINEAR TRANSFORMATIONSS o far we have studied some aspects of a line transformation T :Rn -.R '. We, now,impose some additional conditions on a linea transformation Tand examine thereslllting consequences.A rransfonnation T :Rn m is called a one-one l i n a trdns o n io I if T is linear and

T ( X 1 ) = T ( X ~ ) + X I - X2.W e know that a linear transf ~rm ations uniquely determined when w know thetransform of a basis set of the domain of the transformation. Let { A I,42, ... ,A n be abasis of R and T :R -+R ' be a one-on e linear transfonnation . W e nc w show th at theadditional condition of one- one o n T implies thaq T (A1 ), T (A2, ..., i (An ) islinearly independent.

* T a i A l + a z A z + . . . + a , A , ) = O = T O ) (sinceTisli?ear)- A1 + a z A 2 + ...+a n A , = 0 ( since T is 01 e--one )- a 0 for = 1 to n ( since {A I, ~ ,..,An is linearly indep ~ de nd en t.The above sequence of arguments shows that { T (A1 ), T ( A 2 ), ...,T A, ) is linearlyindependent. Hence p ( T ) = n and consquently, ( T ) = 0, Thus T-' 0 = 1 0 }and, for any Y E R m T- Y ) is either empty or else contains just on e vector

I Since { T (A1 ), T (A2 ), ...,T (A,, ) has to be linearly independent in R ', theretore, fora one-one linear transformation T :R -.R ' ,we must haven m.A transformation T :R - R s called onto linear transfonn ation if T is linear andevery vector in R rn is the T-image of some vector in R . Thus, if T is od o, we musthave T ( R ) = R ' . t is possible that T (S ) = R or som e subspaceS of R . Seeexample 5). Clearly, we must have m s n for the existence of a onto lineartransformation T :R R ' .A linear transformation T : - R which is both one-one and onto is called aninver tib le linear hs f o n n a tw n . For any invertible linear transformation, we must havei) . m =

(ii) ~ - ' ( 0 ) [ O) , a nd(iii) T-' ( Y ) contains a single vector for every Y E R .Let T :R -.R be an inve ~tib leinear transfonnation and { AI, A2, ...,A, } be any basisof Rn. Let T (A; ) - Bi for i = 1 o n. The d iscussion, above, shows that{BI,B2, ...,B n } is also a basis of Rn. We know that we can define a unique lineartransformationH : -.R such that H ( Bi ) - A;. Thus, the transformationHsatisifes

Th e transformation H associated w ith a n invertible transfonnation T is called theinverse of T and is usually denoted by T-'.rom he abov e discussion, w have the following very important result.

Theorem :Let T :R - be a linear transfonnation. Then the following statements areequivalent :(i) T is invertible(ii) T is non singular(iii) T is onto


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Proof : We know that rank (T ) nullity ( T ) = n .Now T is non-singular if and only if'nullity T) = 0 , and the range of T is R if andonly if rank T)= n. Since rank plus the nullity is n, the nullity is zero preciselywhen the rank is n. Therefore, T is non-singular if T (R,) =R,Thus if either condition (ii) or (iii) holds, then the other is satisfied as well and Tis invertible.

xample :Let {AI,AZ,...,An be any arbitrary basis of R . For any XER ,we can uniquelydetermine a ' s such that X = a A1+ a 2Az + ...+a, An. We show that the tram-formation T :R .R defined by

T t X ) = ( a l , az , ...,a n )is an invertible linear transformation.Let X = a A1 +a2Az + ...+ anAnand Y $IAI + $2A2 ... fin Anso tha tT(X) = ( a 1 ...,a ...,a n ) a n d T ( Y ) = ($l,BzAz . . . , f i nSince

X + Y = ( a 1 + $ 1 ) A l + ( a 2 + 8 2 ) A 2 + ..+( a n + B n ) A ntherefore,

T ( X + Y ) = ( a l + B i , a z + p t , ...,a n + $ , )= ( a ~ , a z ,.-,an + ( 1, Pt ...,Bn)= T ( X ) + T ( Y ) .

Sin~ilarly,how tllai T ( a X ) = a X ) for all a ER. Hence, the transforrna-tioil T, as defined above, is a linear transformation.Further,

T X ) = T Y ) => ( a ~ a z...,a n ) = (Pl ,Pz ...,p n )=> a; = p for = 1 to n= > X = Y .

Thus, T is a one-one linear transformation. Moreover, if we choose any( y l y2, ...,y ) E R , then we can obtain .Z = y lA l+ yzA2 + a ynAnClearly,T (Z ) = ( y1,ya . n ).Thus, 'is a onto linear transformation. Hence, T is aninverlible linear tra~lsformatioi~.This example shows that a particular basis can be used to define an invertiblelillear transformation. In fact, every invertible line transformation is of this typeand corresponds to the choice of some basis.

xample :Let us compute the inverse of the trailsfonnation T :R . R defined by

T ( . Y ~ , . Y ~ , X ~ )( x ~ + x z + T ~ , x z + x ~ , x ~ ) .We know that { (1,0,0), (O,l,O). (0 ,0,1 ) ) s a basisof R~ andT ( l , O , O ) = ( l , O , O ) , T ( Q , l , O ) = ( l , l , O ) , a n ~ ( 0, 0 ,1 ) = ( l , l , l ) .Since ( 1,0,O ), ( 1,1,O ), ( 1, 1, 1 is also a basis of R we first conclude thatT is an invertible linear transformation. Therefore, the inverse transformation T-has to satisfy~ - ~ ( i , o , O )( l ,O,O) ,T- ' (1 , lO) = (O,l,O),andT-I( l ,l .l)=(O,O, 1


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( x I , x ~ , x ~ )( X I- X Z ) ( 1 , 0 , 0 ) + ( ~ 2 - ~ 3 )1 . 1 , o )therefore, the transfonnation T- is given y

T ( x I , x ~ , x ~ ) ( X I - X ~ , X ~ - - X ~ , X ~ ) .Thus, T- s the transfonnation discussed in example 6when the ba[ A I A ~ A ~ ]s { , O , O ) , ( 1, 1, O ) , ( 1, 1, 1 11

Which of the following transformationsare invertible :( a ) T (x , y , z) ( x + y + z , y + z , 2 ) .b) T ( x , Y , z ) = ( 2 ~ + ~ - ~ , 2 r - y - ~ , 2 r + y - z ) .c ) T ( X I Y , Z ) ( X , Y , ~ Z ) .

In E9, determ ine the inverse transformations wherever possible.

A linear transformation is su ch thatT ( l , 2 3 , 4 ) = ( 1 , 1 , 1 , 1 ) , T ( 1 , 2 , 3 , 0 ) = ( O , l , l , l ) , andT ( l , 2 , 0 , 0 ) = ( 1 , 3 , 3 , 3 ) .Is it possible to define T 1, 0, 0, 0 )su ch that T is invertible?


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15 5 LINEAR TRANSFORMATIONS AND LINEAREQUATIONSYou must have observed by now that linear transformations are very intimatelyc o ~ e c t e d ith linear equations. For the sake of clarity, we examine this connectionmore precisely now.

Let us consider a linear transformation T :R . R .Recall that (El, E2, ...,En ) is abasis ofR where Ei R is a vector whose ith component is and all the othercomponents are 0. Let T Ei ) ai for 1 to n.. Then, clearly

where X ( ~ 1 ~ x 2...,x , , = xlEl+xzEz+ ...+ nEn.Clearly, the set of solutions of the single linear equation

Let usnow consider m linear transformations Ti :R . R where Ti for = o m isdefined by

where Ti(Ej) = ajifori = l t o man d j = l t o n .Then, it is easy to show that T : . R defined by

is a lineartransfomation.Now amsiderth following system of m linearequatiom inn variables

We associate the linear transfomtion T, defined above, with this system of linearequations.We observe that the solution set of the above system is T- ( b ) whereb = ( bl, b~ ...,b ). Recalling the information we have about T- b ) , we make thefollowing observations about the solutions of a system of linear equations :(a) Since T- ( b ) g if b T R ), therefore, the system of linear e quations hasno solution if the vector b does not belong to the range space of the linear

transformation.(b) If b T R ), then the system of linear equations has atleast one solution.

Further.(i) If T ( { O } hen th system of linear equations haa unique solution.(ii) If a non-zem X belongs to T-' , then he system of linear equationshasan infinite number of solutions. he olutions, in this case, can be viewed as aparallel translate of T- 0 )

Thus, we obserhre that a system of linear equations can have a) no solutions, or @)unique solution, or (c) an infinite number of solutions. In other words, a linear system


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with a finite numb er (other than or 1) of solutioils does not exist.W e also know that the Gau ss elimination method can be used effectively to determine apa rtic ula r solution of a system of linear equations. The sam e method can also be used todetermine a basis of T- ).Thus,we can obtain all the solutior~s f a system of linearequations.Th e importance of linear transformations is to unfold the structure of the solution set ofa syste m of linear equa tions. It is only after this structure is understood that a meth od,like the Gauss elimina tion m ethod, helps us to actually co mpu te all the solutions.

2De term ine the structure of solutions of the following systems of linear equations:

(a) x + y + z = 1 b) x + y + z = 1 (c) x + y + z = 2

15 6 THE COMPLEX CASE

LinearTruwfomutioosmdLhear Equations

So far we hav e intentionally confined our attention to som e aspects of the Euclideanspace R . W e assum e that you are familiar with the set C of all complex numbers andthe op eratio ns along with their properties of add ition, subtraction, inultiplication,division, conjug ation and absolute value involving coinplex numbers.Most of the con cepts that we have discussed abo ut R can easily be extende d if wesimply perm it the comp lex numbers to replace the real numbers. However, there aresom e important differences also. W e now point out on e of the most importantdifferences which a ccou nts for all the other differences also.Let

C n = { ( c I , c ~..., c n ) I c ; E C for = 1 to rr],i.e., C is the set of all rz-tuples of complex numb ers. We can define addition of twocom plex vectors an d scalar multiplication of a com plex vector by a comp lex numb er,componentwise, as in the case of R . It is easy to verify that all the properties ofaddition a nd sc alar multiplication, that we observed in the case of R , continue to holdgood . All the other concepts, that we have studied in R , can now be extended to Ceasily. T he on ly .difference, now, is in the concept of inner product.Let

and Y ( ~ 1 , ~ 2 ,.., ,, E C


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b h i c e ~ W e define the inner product (X, Y) in C as follows(X, Y) - ~171x 2 7 2 + .. . + ~ n &

Thus, w e first take the con jugate vector F each of whose com ponent is the conjugate ofthe corresponding component of Y, and then take the inner product as in the case of R .Note that the complex inner product is the same as the real inner product whenever bothX and Y are in R .The defin ition of com plex inner product, as defined abov e, satisfiesthe fo llowing properties( i ) ( X , Y ) = ( Y , X )( i i ) ( a X + $ Y , Z ) = a ( X , Z ) + p ( Y , Z ) f orall a , p E C-(iii) ( X , a Y + $ Z ) ~ ( x , Y ) + B ( x , z ) forall a , p E C(iv) IIXll > 0 i f X * 0 and IO I I 0

IIax+ yI12 I a 1 2 1 1 X I 1 2 + a P ( x , ~ ) + ~ ( ~ , ~ ) + l ~ 1 2 1 1 ~ IHere I I X = (X , X ) for X E C .W e invite the reader to prove these propertiesand note the differences in the corresponding propertics of the real inner product. It isthese differences in the properties of inner product that account for various differencesin some results concerning R and C For example, the following analogue of thefamiliar Pythagoras theorem

is valid in R Howe ver, this result is not valid in C Try to prove this. Ou r purpose isprimarily to point towards the possible extens ion of R to C and to the very largenumber of identical results. The differences in the results, which a re there, do notcoilcern us at this stage. As such, we do not pursue these differences..15.7 SUMMARYIn this un it, following points are highlighted:1. mapping T: R R is called a line ar tran sfo rm atio n if for all X,Y inR anda l l a E R , ( i ) T ( X + Y ) = T ( X ) T ( Y ) a n d ( i i ) T ( a X ) - a T ( X ) .

Further T (R ) is called the ra n k spac e of T. The dimension of T (R ) is calledthe ra n k of T. T-' ( Y ) is the set of all those vectors in R which are mapped to Yby .The subspace T- ( 0 ) of R s called the null spac e of T and demension ofT ( 0 ) is called nullity of T. Also, ( rank of T ) ( nullity of T ) = (dimension ofR ).

2 (a)A transformationT:Rn Rm is called one one linear transformationif T islin ea ra nd T( Xl) T ( X 2 ) e X X 2 - H e r e n s m .@ A transformation T: Rn+Rm is onto if every vectoy in R is the T - image ofsome vector in R . Here m s n.(c)A linear transformation n Rm s invertible if T is both one -one and oilto.

3. A system of equations A X b or linear transformation T R 4 as(a) No solution if b$ i(Rn )@ A unique solution if T-l ( = { 0 } and(c) An infinite number of solutions if a non -zero X T -' 0 ) .

E 1. The transformations in (a), (d) and (g) are linear.


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E , 2 T (2,3)= (1 ,Z 3).E . N0,bccrust( 4 , 5 , 3 ) 3 ( 1 , 1 , 1 ) + 1 ( 1 , 5 0 ) md 3 ( 2 , 3 ) + 1 ( 1 . 2 ) 2 (5,101E4. T(x ,y ) (X-y ,x+2~) .E5. ( r ) p 5 v 1 ;

( b ) p 3,v 1.E6 T(x,y,z)=(x+y+22,2r+y+3~,3x+y+4z).Manysnchlinur

~ f ~ r m r t i o n sxist.E7. T(x,y,z,t) z , z , t ) .138 T-' (o ,o) ( (x ,x ,x)~xER];

T - ' ( 5 , 3 ) ( ( 3 + ~ , 4 + x , l + x , ) I x R } .E9 he trrnsfonnations in (a) and (c)are invertible.E10. T - ( x , y , z ) = ( x - y , y - z , z ) f o r T i n ( a ) ;

~ ( x , y , z ) ( x , Y , : ) ~ ~ Tn c>El l . No,because{ 1,2,3,4), 1,2,3,0), 1,2,0,O)l islinesrlyindependent

(1 ,3 ,3 ,3) ~ ~ , 1 , 1 , 1 ~ + 2 ~ ~ , 1 , 1 , 1 ~ .E12. a) Unique solution, x, y,z -9,6,4 ).

@ No solutionc ) { (-2+~,4-2.z ,z)IzR] istbesdutionset


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UNIT 16 MATRICES ND DETERMIN NTSStructure

16.1 IntroductionObjectives

16.2 Matrices16.2.1 Equality of Matrices16.2.2 Sum of Matrices16.2.3 Scalar Multiplication16.2.4 Transpose and Conjugate of Matrix16.2.5 Special Matrices16.2.6 Partitioningof Matrices16.2.7 Submatrices

16.3 Matrix Multiplication16.4 Matrices and Linear Trai~ form ation s16.5 Rank of Matrix16.6 In verse of a ~ a t r i x16.7 Determinants

16.7.1 Properties of Determinants16.7.2 Laplace Expansion of Determinants

16.8 Determ inants and Matrices16.8.1 Crarner s Rule

16.9 Summary16.10 Solutions/Answers

16.1 INTRODUCTIONIn U nit 14 you have see11 the role that linea r trailsfon natio ils play in uilderstan diilg thestructure of the so luti ol~ s et of a systeill of linear equations. Th e comp utational aspectsof linear transformations can be understood by using matrices. W e introduce matricesand various operations il~vo lvingmatrices in this unit. Th e notions of rank a nd inverseof a matrix are also introduced. A fter iiltroducing the notion of determin ant of a squa rematrix some im por tai~t roperties of determinants are observed. Th ese properties areuseful in comp uting the detennin ants inore efficiently. Fillally determin ants are used tounderstand certain aspects of matrices.

bjectivesTh e inain objectives of this unit are

T o introduce the notioil of matrices and operations inv olving matrices.T o understand the relation between inatrices and linear trailsfonnation s a nd thusintroduce the conce pts of rank and inverse of a matrixTo introduce determinants of order n and give some useful properties ofdeterminants andT o understan d certai11.aspects of ina trices in term s of detenn inan ts.

16.2 M TRICES

linear t r a n s f o m t i o n s

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