Chapter 2 Linear Transformations 2-1 Linear Transformations Linear transformation T: V→W, where V and W are vector spaces over F : T(ax+by)=aT(x)+bT(y). Note : T(0)=0. b Eg. T[f(t)]= f (t )dt is a linear transformation of f(t). a (Proof) b b a a [af (t ) bg (t )]dt a b f (t )dt b g (t )dt a Null space of T, N(T): N(T)={x V: T(x)=0} V. Range of T, R(T): R(T)={T(x): x V } W. Nullity(T)=dim(N(T)), and Rank(T)=dim(R(T)) Eg. Linear operator T: R3→R2 is defined by T(a1,a2,a3)=(a1-a2,2a3). Find N(T) and R(T). (Sol.) N(T): T(a1,a2,a3)=(a1-a2,2a3)=(0,0) a1=a2=a, a3=0 N(T)={(a,a,0): a R}: A straight line in R3. R(T): a1,a2,a3 R a1-a2,2a3 R, ∴ R(T) =R2 Eg. Let T: P3(x)→P3(x) be defined by T(f)=[f(x)+f(-x)]/2. Find a basis for N(T). Find a basis for R(T). [2001台大電研] (Sol.) N(T): T(ax3+bx2+cx+d)=[(ax3+bx2+cx+d)+(-ax3+bx2-cx+d)]/2=bx2+d=0 b=d=0 N(T)={ax3+cx: a, c R}: An odd function in P3(x). A basis for N(T) is {x3,x} R(T): T(ax3+bx2+cx+d)=[(ax3+bx2+cx+d)+(-ax3+bx2-cx+d)]/2=bx2+d: An even function in P3(x). A basis for R(T) is {x2,1}. Theorem Linear operator T: V→W, and T is one-to-one N(T)={0}. (Proof) “ ”: T(x)=0=T(0). ∵ T is one-to-one, ∴ x=0 N(T)={0} “ ”: Suppose T(x)=T(y), then 0=T(x)-T(y)=T(x-y), Hence x-y N(T) x=y Onto: One-to-one linear operator T: V→W and Rank(T)=dim(V)=dim(W). Eg. (a) T: F2→F2 by T(a1,a2)=(a1+a2,a1) is onto. x (b) T: P2(x)→P3(x) by T(f)(x)=2f’(x)+ 3 f (t )dt is one-to-one but not onto. 0 Theorem Linear transform T: V→W, where V and W are vector spaces. Then N(T) and R(T) are subspaces of V and W, respectively. Theorem Linear transform T: V→W, where V and W are vector spaces. If the basis of V is β={x1,x2,…, xn}, then R(T)=Span{T(x1),T(x2),…, T(xn)}. Eg. Linear operator L: R2→R3 is defined by L(x,y)=(x+y,x-y,x+2y) (a) Find a basis for the range of L. (b) Is L onto? [1990 交大工業工程所] 1 1 x y x x (Sol.) L( ) 1 1 x y y 1 2 y x 2 y 1 1 1 0 0 (a) ∵ Rank ( 1 1) 3 , ∴ Choose the basis of R(L)= ( 0 ' 1 ' 0 ) . (b) R2→R3, ∴ No. 1 2 0 0 1 Eg. Linear operator T: R2→R2 is defined by T(1,0)=(1,4), T(1,1)=(2,5). (a) T(3,5)= ? (b) What is the null space of T? (c) What is R(T)? (d) Is T one-to-one? (e) Is T onto? [1990 交大電子所] a b x (Sol.) T(x,y)=(ax+by,cx+dy)= c d y T(1,0)=(1,4) a=1, c=4, T(1,1)=(2,5) a+b=2, c+d=5 b=1, d=1 1 1 3 8 (a) T (3,5) , or T(3,5)=(8,17) 4 1 5 17 1 1 (b) For the standard basis of R2, T . 4 1 x y 0 x0 1 1 x x y 0 0 4 1 y 4 x y 0 4 x y 0 y 0 N(T)= {0} (c) ∵ x+y, 4x+y R, ∴ R(T)=R2. (d) ∵ N(T)={0}, ∴ Yes. (e) ∵ one-to-one and R2→R2, ∴ Yes. Theorem Linear operator T: V→W, if V is finite-dimensional, then Nullity(T)+Rank(T)=dim(V). [1999 台大電研] (Pr oof ) Suppose dim( V ) n and x1 ,, x k : basis of N (T ) extend x1 ,, x k to x1 , , x k , x k 1 , , x n : basis of V It is desired to show that S T ( x k 1 ), , T ( x n ) is a basis for R(T ). 1. For 1 i k , T ( xi ) 0 2. R(T ) SpanT ( x1 ), , T ( x n ) Span( S ) SpanT ( x k 1 ), , T ( x n ) 3. If T ( xi ) is linearly independent , k 1 i n, then the proof is complete. n biT ( xi ) 0 i k 1 n n i k 1 i k 1 T (bi xi ) T ( bi xi ), k 1 i n ci F i k s.t. n k i k 1 i 1 bi xi ci xi or k ( c i ) x i i 1 n b x i k 1 i n b x i k 1 i i i N (T ) Span( x1 ,, x k ) 0 x1 , , x n is a basis (linear independent ), bi 0 i T ( xi ) is linearly independent Eg. Let T: V→W be a one-to-one linear transformation and V be a finite-dimensional vector space. Show that Rank(T)=dim(V). [台大電研類似題] (Proof) T: V→W is one-to-one N(T)={0}, ∴ Nullity(T)=0 Nullity(T)+Rank(T)=Rank(T)=dim(V). 2-2 Matrix Representations and Coordinate (Basis) Transformations In Matlab language, we can use the following instructions to find out the ijth-entry, the mth-row, and the nth-column of a matrix: >>A=[0,1,2;3,4,5] A= 0 1 3 4 >>A(2,1) ans = 3 >>A(2,3) ans = 2 5 5 >>ROW1=A(1,:) ROW1 = 0 1 >>COL1=A(:,2) COL1 = 1 4 2 Matrix representation of a linear transformation: Linear transformation T: V→W, β={x1,x2,…,xn} is a basis for V, γ={y1,y2,…,ym} is a basis for W, then ! aij F for i=1, 2,…, m and j=1, 2, …, n such m that T(xj)= aij y i i 1 a11 a A= 21 a31 a12 a 22 a32 T ( x1 ) a11 y1 a21 y2 a31 y3 for j=1, 2, …, n. That is, T ( x2 ) a12 y1 a22 y2 a32 y3 , and then T ( x ) a y a y a y , etc. 3 13 1 23 2 33 3 a13 a 23 . a33 Eg. Linear operator T: P3(R)→P2(R) by T(f)=f’, find T , where β, γ are the standard ordered bases for P3(R) and P2(R), respectively. (Sol.) 1, x, x 2 , x 3 , 1, x, x 2 T (1) 0 0 1 0 x 0 x 2 , T ( x) 1 1 1 0 x 0 x 2 , T ( x 2 ) 2 x 0 1 2 x 0 x 2 , and T ( x ) 3x 0 1 0 x 3 x T 3 2 2 0 1 0 0 0 0 2 0 0 0 0 3 Eg. Let W be the subspace sp(sin2x,cos2x) of the vector space of all real-valued functions with domain R, and let B=(sin2x,cos2x). Find the matrix representation A relative to B for the linear transformation T: W→W defined by T(f)=D2(f)+3D(f)+2f, where D presents the derivative operator. (where sp(X) denotes the set of all linear combinations of vectors in X) [2005 成大電腦 通訊所] (Sol.) T[sin(2x)]=-4sin(2x)+6cos(2x)+2sin(2x)=-2sin(2x)+6cos(2x), 2 6 T[cos(2x)]=-4cos(2x)-6sin(2x)+2cos(2x)=-6sin(2x)-2cos(2x), ∴ A= 6 2 Eg. Find the 4 by 4 matrix A represents a cyclic permutation: each vector (x1,x2,x3,x4) is transformed to (x2,x3,x4,x1). What is the effect of A2? Show that A3=A-1. [2005 北科大電腦通訊所] (Sol.) β={(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)} A(1,0,0,0)=(0,0,0,1)=0(1,0,0,0)+0(0,1,0,0)+0(0,0,1,0)+1(0,0,0,1) A(0,1,0,0)=(1,0,0,0)=1(1,0,0,0)+0(0,1,0,0)+0(0,0,1,0)+0(0,0,0,1) A(0,0,1,0)=(0,1,0,0)=0(1,0,0,0)+1(0,1,0,0)+0(0,0,1,0)+0(0,0,0,1) A(0,0,0,1)=(0,0,1,0)=0(1,0,0,0)+0(0,1,0,0)+1(0,0,1,0)+0(0,0,0,1) 0 1 0 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 1 1 0 0 0 2 3 =A-1 ∴ A= ,A = ,A = 0 0 0 1 1 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 0 0 1 0 3 AT A Eg. V=R2×2, T: V→V is a linear transformation given by T(A)= . Find [T]β, where β is a 2 1 0 0 2 0 1 0 0 basis of V: β= { , , , } . [2006 台科大電研] 0 0 1 0 2 0 0 2 (Sol.) 1 0 1 T ( ) (3 0 0 0 0 2 0 T ( ) (3 1 0 2 0 1 0 1 0 1 0 0 2 0 1 0 0 )/2 1 0 0 0 0 0 0 0 0 0 0 1 0 2 0 0 2 1 0 2 0 2.5 1 0 1 0 2 3 0 1 0 0 )/2 0 ( ) 0 0 1 0 0 2 1 0 2 2 0 0 2 3.5 0 0 3.5 0 1 0 2 0 1 0 1 0 3 0 2 1 0 1 0 0 T ( ) (3 )/2 0 ( ) 0 2 2 0 0 2 2 0 1 0 2 0 2.5 0 0 0 2 1 0 0 0 0 0 0 0 0 0 1 0 0 2 0 1 0 0 T ( ) (3 )/2 0 0 0 1 0 2 0 2 0 2 0 2 0 0 1 0 2 0 0 2 0 0 1 0 1 3 0 2 2 0 ∴ [T]β= 3 1 0 0 2 2 0 1 0 0 1 1 0 0 Eg. Let T: M2×2(F)→M2×2(F) be the linear operator defined by T ( x) x x . Find 0 0 1 1 the matrix representation for T. [交大電信所] 1 0 0 1 0 0 0 0 (Sol.) Choose the standard basis β for M2×2(F) as { , , , } 0 0 0 0 1 0 0 1 1 T ( 0 0 T ( 0 0 1 1 1 ) 0 0 0 0 1 1 1 0 ) 0 0 0 0 0 1 0 0 0 0 0 1 1 0 1 0 0 0 0 1 0 1 0 1 1 1 0 0 0 0 1 2 1 1 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 2 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 T ( 1 0 T ( 0 0 1 1 0 ) 0 0 0 1 0 1 1 0 ) 1 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 1 0 1 1 0 1 0 1 1 1 0 0 0 0 0 1 1 0 1 1 1 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 1 1 0 0 0 1 0 0 0 0 0 0 1 0 0 0 1 0 0 1 1 1 0 2 ∴ T 0 0 0 0 Eg. T: R2→R3. 1 0 0 1 . Rank(T)=3. 0 1 0 1 T(a1,a2)=(a1-a2,a1,2a1+a2). Let α={(1,2),(2,3)}, β={(1,0),(0,1)} for R2 and γ={(1,1,0),(0,1,1),(2,2,3)} for R3, then find T and T . (Sol.) 1. T (1,2) (1,1,4) a (1,1,0) b(0,1,1) c(2,2,3), T (2,3) (1,2,7) d (1,1,0) e(0,1,1) f (2,2,3) 11 7 3 3 7 2 11 4 a , b 2 , c , d , e 3, f , T 2 3 3 3 3 3 4 2 3 3 2. T (1,0) (1,1,2) a (1,1,0) b(0,1,1) c(2,2,3), T (0,1) (1,0,1) d (1,1,0) e(0,1,1) f (2,2,3) 1 2 a , b 0, c , d 1, e 1, f 0, T 3 3 1 3 0 2 3 1 1 0 Eg. Let β’={2x2-x,3x2+1,x2} and β={1,x,x2} be both the ordered bases for P2(x). And let T: P2(x)→P2(x) be defined by T(ax2+bx+c)=cx2+bx+a, then find [ T ] , [T ] '' , [T ] ' , [T ] ' . [文化電 機轉學考] (Ans.) T 0 0 0 0 1 1 0 1 0 2 3 1 ' ' 0 1 0 , T ' 2 3 1 , T 0 0 1 , T ' 1 0 0 . 1 0 0 8 8 3 1 2 3 0 1 0 Annihilator of S, S0: S0={T(x): T(x)=0 for x S}, where V and W are subspaces, S is a subspace of V. Theorem (a) S1 S2, then S 20 S10 . (b) V1 and V2 are subspaces of V, then (V1+V2)0= V10 V20 . (Proof) (a) S1 S 2 x S1 , then x S 2 If T ( x) 0 x S 2 , then T(x)=0 for all x S1 S 2 S 20 S10 (V1 V2 ) o V10 V1 V1 V2 ( b) (V1 V2 ) 0 V10 V20 o 0 (V1 V2 ) V2 V2 V1 V2 For T V10 V20 T V10 and T V20 , a V1 V2 a a1 a 2 for a1 V1 and a2 V2 T (a) T (a1 ) T (a2 ) 0 , T (V1 V2 ) 0 V10 V20 (V1 V2 ) 0 , ∴ (V1 V2 ) 0 V10 V20 m Multiplication of matrices: If C=AB, then Cij= Aik Bkj , where A is a m×n matrix, B is an n×p matrix, k 1 and C is an m×p matrix. In Matlab language, we can use the following instructions to obtain the product of two matrices: >>A=[2,5,1;0,3,-1]; >>B=[1,0,2;-1,4,-2;5,2,1]; >>C=A*B A= 2 5 1 0 3 -1 B= 1 0 2 -1 4 -2 5 2 1 C= 2 -8 22 10 -5 -7 e Eg. A 3.7 10 5 n 2 i 1 3 7 sin( 2) 0 3 3 2 a b c 1 2.3 1 3 0 . 2 0 . 3 0 . 1 1 0 b c a 5 1 2 2 1 1 2 1 1 c a b 3 2 1 1 1 2 2 3 6 , find the third column of A. [1993 中山應數所] a11 a12 (Sol.) A is a (3×4).(4×3).(3×3).(3×4)=(3×4) matrix. A a 21 a 22 a31 a32 a13 a 23 a33 0 a14 0 a13 a 24 , A a 23 , 1 a34 a33 0 0 0 0 ' ' ' ' ' ' ' ' ' 1 2.3 1 3 0 ' ' ' ' ' ' a b c 1 1 A ' ' ' ' ' ' ' ' ' 5 1 2 ' ' ' ' ' ' b c a 1 1 2 1 ' ' ' ' ' ' c a b 1 ' ' ' ' ' ' ' ' ' 3 2 1 2 0 0 3 3 0 3 0 3 ' ' ' 0.2 0.3 0.1 a b c ' ' ' 0.2 0.3 0.1 1 ' ' ' a b c (a b c) ' ' ' 1 2 1 1 2 1 1 ' ' ' 1 1 1 a b c 1 1 1 ' ' ' 1 2 3 2 3 6 6 e (a b c) 3.7 10 5 n2 i 1 3 7 sin( 2) 2 0 2 0 0 (a b c) 0 . ∴ The third column is 0 1 1 1 2 a b c 0 . (a b c) Theorem T: V→W, let β, β’ be the ordered bases of V, and γ, γ’ be the ordered bases of W. If [β]=Q[β’] and [γ]=P[γ’], we have T ' P 1 T Q . ' (Proof) T , P ' Q ' T Q ' ' P 1 T Q ' T '' ' T '' P 1 T Q x y Eg. Assume T ( ) in the standard basis. Find the matrix representation for T with y x y respect to the new basis {(1,1),(1,2)}. [1993 中山應數研] 1 0 1 0 0 1 1 0 x 0 1 x 0 1 T (Sol.) T ( ) 0 1 , T ( ) 1 1 T ( ) 0 1 0 1 1 1 0 1 y 1 1 y 1 1 1 1 0 1 1 0 1 1 2 1 1 1 10 11 , 2 10 21 , P 1 2 , P 1 1 , 2 1 0 1 1 1 2 3 T ' 1 1 1 1 1 2 1 1 Eg. A linear transform T(a1,a2,a3)=(2a1+a2,a1+a2-a3): R3→R2 and β={(1,0,0),(0,1,0),(0,0,1)}, β’={(2,0,0),(0,-1,0),(0,0,-2)}, γ={(1,0),(0,1)}, γ’={(1,1),(1,-1)}, then find T and T ' . ' (Sol.) T(1,0,0)=(2,1)=2(1,0)+1(0,1), T(0,1,0)=(1,1)=1(1,0)+1(0,1), T(0,0,1)=(0,-1)=0(1,0)-1(0,1) T 1 2 1 0 1 1 1 . (1,1)=1(1,0)+1(0,1) , (1,-1)=1(1,0)-1(0,1), P , P 2 1 1 1 1 1 1 2 1 2 1 2 (2,0,0)=2(1,0,0)+0(0,1,0)+0(0,0,1), (0,-1,0)=0(1,0,0)-1(0,1,0)+0(0,0,1) 0 2 0 3 1 1 ' 1 (0,0,-2)=0(1,0,0)+0(0,1,0)-2(0,0,1) Q 0 1 0 T ' P T Q 1 0 1 0 0 2 Another method: T (2,0,0) (4,2) 3(1,1) 1(1,1), T (0,1,0) (1,1) (1,1) 0(1,1) 3 1 1 T (0,0,2) (0,2) 1(1,1) 1(1,1) Q 1 0 1 n Trace of an n×n matrix, tr(M): tr(M)= M ii tr(aA+bB)=atr(A)+btr(B) i 1 1 2 Eg. A , tr(A)=1+(-4)=-3. 3 4 Eg. Show that tr(AB)=tr(BA) and tr(A)=tr(At). [2005 台大電研] (Proof) tr( AB) ( AB) ii i i A ik k Bki Bki Aik Bki Aik BA kk tr( BA) i k k i k Eg. Show that no matrices A and B Mn×n(F) such that AB-BA=I, where I is an n×n unit matrix. [台大電研] (Proof) A and B such that AB-BA=I, then tr(AB-BA)=tr(AB)-tr(BA)=0 n=tr(I). It is contradictory to the assumption; hence no such matrices A and B exist. A Similar matrices: M nn (F ) , B is similar to A if invertible Q Mn×n(F) fulfills B=Q-1AQ. B Eg. If A and B are similar to each other, then tr(A)=tr(B). [台大電研] (Proof) B Q 1 AQ tr( B) tr(Q 1 AQ) tr( AQQ 1 ) tr( A) 1 4 2 9 Eg. Show that A= and B= are not similar to each other. 2 3 3 7 (Proof) tr(A)=1+3=4 -5=2+(-7)=tr(B), ∴ A and B are not similar to each other.