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
x0
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 )  SpanT ( x1 ), , T ( x n )  Span( S )  SpanT ( 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  10  11 , 2  10  21  , 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 nn (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.