The Matrix of a Linear Transformation
Important result (general case):
Let L :V  W be a linear transformation of an n-dimensional vector
space V into an m-dimensional vector space W ( m  0, n  0 ) and let
S  v1 , v2 ,, vn 
and
T  w1 , w2 ,, wm 
be bases for V and
W, respectively. Then, there exists a unique m n matrix A such that
Lx T  AxS ,
where
A  Lv1 T
Lv2 T
x V ,
 Lvn T .
Note: the matrix A is referred to as the matrix of L with respect to the
bases S and T.
Note: as L :V  V and S  v1 , v2 ,, vn  is a basis of V, then
Lx S  AxS ,
where
A  Lv1 S
Lv2 S
x V ,
 Lvn S  .
the matrix A is referred to as the matrix of L with respect to the basis S.
Intuition:
1
L
Lx 
x
A
xS
Lx T
 AxS
Example:
L : R  R is defined by Lx   2 x . Let S  v, T  w. Suppose v  3w, u  5v .
Then,
Lu   2  u  2  5  v  5  2  v  5  2  3w  5  6w  30w .

Lu T
 30  6  5  Lv T u S .
( u  5v  u S  5
Lv T
 2v T  6 wT  6) .
Example:
Let L : R 3  R 2 defined by
  x1  
    1
L x   L  x2    
  x   1
  3
1
2
 x1 
1  
x2  Ax
.
3  
 x3 
Let
1 0 0 
1 0 


S  0, 1, 0 , T   ,   
0 1 
0 0 1 








2
Then, since
 x1 
1
0
0
 x1 
x   x2   x1 0  x2 1  x3 0  xS   x2   x
 x3 
0
0
1
 x3 
and
 x1 
1 1 1    x1  x2  x3 
1
0 




L x   
x


x

x

x

x

2
x

3
x
2
3  
 1 2 3 0  1
 2  
1 2 3  x   x1  2 x1  3x3 
 
1
 3
 x1 
 x1  x2  x3 
1 1 1   1 1 1
 Lx T  
  Lx   1 2 3  x2   1 2 3xS  AxS
x

2
x

3
x
2
3

x  

1
 3
then
Lx   Lx T  AxS  Ax .
Example:
Let L : R 3  R 2 defined by
  x1  

  x  x2 
L x   L  x2     1
.
2x
3

 x  
  3
Let
1 0 1  
1 0 


S  v1 , v2 , v3   1, 1 , 2 , T  w1 , w2    ,    .
 0   2  
 0   4   3  
      
Find the matrix of L with respect to the bases S and T.
[solution:]
A  Lv1 T
Lv2 T Lv3 T 
3
Thus,
 1 


Lv1   L 1  
 0  
  
1  1 2
1
0 


2

0
 2  0  0 
0 
2  2 w1  0 w2

  
 
 
 2
 Lv1 T    .
0 
 0  

 0  1 1
1
0 
Lv2   L 1    
    1   4    1w1  4 w2

0 
 2
 4   2  4  8
 
1 
 Lv2 T   
 4
 1  

 1  2 3
1
0 
Lv3   L 2   


3

3
 6 
0 
2  3w1  3w2
2

3






 
  3 
 
3
 Lv3 T    .
3
Therefore,
2
A
0
1
4
3
3

.
General Procedure for Computing A:
Let
L : Rn  Rm
be a linear transformation. Let
S  v1 , v2 ,, vn 
be bases for
Rn
and
Rm ,
and
T  w1 , w2 ,, wm 
respectively. Then, the matrix of L with
respect to the bases S and T can be obtained via the following steps:
4
1. Form the m  n  m augmented matrix
w1
w2  wm Lv1  Lv2   Lvn .
2. Transform the augmented matrix into the reduced row echelon matrix,
I nn
A.
The matrix A is the matrix of L with respect to the bases S and T.
Example (continue):
Let L : R 3  R 2 defined by
  x1  

  x  x2 
L x   L  x2     1
.
2x
3

 x  
  3
Let
1 0 1  
1 0 


S  v1 , v2 , v3   1, 1 , 2 , T  w1 , w2    ,    .
 0   2  
 0   4   3  
      
We can use the above procedure to find the matrix of L with respect to the bases S
and T.
1. Form the augmented matrix
w1
1 0 2 1 3
w2 Lv1  Lv2  Lv3   

0 2 0 8 6
2. Transform the above augmented matrix into
I 22
1 0 2 1 3
A  
.
0 1 0 4 3
Therefore, the matrix of L with respect to the bases S and T is
2
A
0
1
4
5
3
3

Example:
Let L : P1  P2 defined by
Lx  L p(t )  tp(t ) .
Let

.
S  v1 , v2   t , 1, T  w1 , w2 , w3   t 2 , t ,1
Find the matrix of L with respect to the bases S and T.
[solution:]
A  Lv1 T
Lv2 T   Lt T L1T 
Thus,
Lv1   Lt   t  t  t 2  1  t 2  0  t  0  1  1w1  0w2  0w3
 Lv1 T
1
 0
.
0
Lv2   L1  t  1  t  0  t 2  1  t  0  1  0w1  1w2  0w3
 Lv2 T
0 
 1
.
0
Therefore,
1
A
0

0
Example:
6
0
1
.
0

Let L : P1  P2 defined by
Lx  L p(t )  tp(t ) .
Let

.
S  v1 , v2   t , 1, T  w1 , w2 , w3   t 2 , t  1, t  1
Find the matrix of L with respect to the bases S and T.
[solution:]
A  Lv1 T
Lv2 T   Lt T L1T 
Thus,
Lv1   Lt   t  t  t 2  1  t 2  0  t  1  0  t  1  1w1  0 w2  0 w3
 Lv1 T
1
 0
.
0
Lv2   L1  t  1  t  0  t 2 
 Lv2 T
1
1
1
1
 t  1   t  1  0w1  w2  w3
2
2
2
2
0
 
 1 
 12 .
 2 
Therefore,
1

A  0

0


7
0 

1 
2
.
1
2

Example:
Let L : P2  P1 defined by


L at 2  bt  c  a  b t  c .
Let


S  v1 , v2 , v3   t 2 , t , 1 , T  w1 , w2   t ,1 .
Find the matrix of L with respect to the bases S and T.
[solution:]
Lv2 T Lv3 T   Lt 2 T Lt T L1T
A  Lv1 T
Thus,
 


Lv1   L t 2  L 1  t 2  0  t  0  1  1  0t  0  1  t  0  1  1w1  0w2
1
 Lv1 T    .
0 


Lv2   Lt   L 0  t 2  1  t  0  1  0  1t  0  1  t  0  1  1w1  0w2
1
 Lv2 T    .
0 


Lv3   L1  L 0  t 2  0  t  1  1  0  0t  1  0  t   1  1  0w1   1w2
0
 Lv3 T    .
 1
Therefore,
1
A
0
1
0
8
0
 1
.
NOTE: in the above example,
L
Lv   a  b t  c
 a  b w1   c w2
v  at  bt  c
 av1  bv 2  cv3
2
vS
a 
 b 
 c 
1 1 0 
A

0 0  1
Av S
a 
1 1 0   

 b 
0 0  1  c 
 
a  b 


 c 
 Lv T
9