Review 2:
91. 6. 13
1. (Section 7.1 & 7.2)
 Definition of linear transformation and linear operator
 Check if L : V  W is a linear transformation.
 Definition of kernel and range.
Example 1:
Let L : R 2  R 3 defined by
 xy
  x   

L
  y    y  .
   x 
 
Is L a linear transformation?
[solution:]
x 
x 
 1

L is not a linear transformation since for u   1   R 2 , v   2   R 2 ,
y
y
2

x  x  y  y   x y  x y  x y  x y 
  x1  x 2    1 2 1 2   1 1 2 2 1 2 2 1 
Lu  v   L 
y1  y 2
y1  y 2
   


y

y
 1 2  
 

x1  x2
x1  x2

 x1 y1  x 2 y 2   x1 y1   x 2 y 2 
 x   x 
  y1  y 2    y1    y 2   L  1    L  2    Lu   Lv 
y
y
 x1  x 2   x1   x 2    1     2  
2. (Section 7.3)
 Definition of the matrices of a linear transformation with respect to
the standard basis, with respect to the basis S and T, with respect to
the basis S and with respect to the basis T.
 Find the matrices of a linear transformation with respect to different
bases.
1
 The relationship between the matrix of L with respect to the basis S
and the one with respect to the basis T.
Example 1:
Let the linear transformation L : R 2  R 3 defined by
 x
  x 
   y
L


 y 
 .
  

0

Let
1 0 0 
1 0  


S  v1 , v2    ,   , S  e1 , e2 , e3  0, 1, 0 ,
0 1 
0 0 1 
      


1  0  0 


T  w1 , w2 , w3   0,  1 , 1 
0  1 1 
      
(a) Find the standard matrix of L (i.e., the matrix A such that
Lx  Ax ).

(b) Find the matrix of L with respect to the bases S and S .
(c) Find the matrix of L with respect to the bases S and T.
[solution:]
(a)
  1 
A  Lv1  Lv2    L   
  0 
Since
2
 0   
L    ,
 1 
1
0 
 1   
 0    
Lv1   L     0, Lv2   L     1 ,
 0  0
 1  0
 
 
thus
1
A
0

0
0
1
.
0


(b) The matrix of L with respect to the bases S and S is
Lv1 S
Lv2 S .


Since
1
1
 1   



Lv1   L     0  1e1  0e2  0e3  Lv1 S   0
 0  0
0
 
and
0
0
 0   
Lv2   L     1  0e1  1e2  0e3  Lv1 S   1
 1  0
0
 
thus
1
0


0
0
1

0


is the matrix of L with respect to the bases S and S . Note that it is
exactly the same as the standard matrix of L in (a).
Note: let L be the linear transformation
S  e1 , e2 ,, en 
and
3
L : R n  R m and let

S   e1 , e2 ,, em

Rn
be the standard (natural) bases for
and
R m , respectively,
where
1
0 
0 
0 
1
0 




e1 
, e2 
,  , en   ,



 
 
 
0 
0 
1
and
1
0 
0 
0 
1
0 







e1 
, e2 
,, em   .



 
 
 
0 
0 
1
The standard matrix of L (the matrix such that
Lx  Ax ) is

exactly the same as the matrix with respect to the bases S and S .
(c) the matrix of L with respect to the bases S and T can be obtained via
the following procedure:
Step 1: Form the augmented matrix
w1
w2
w3
Lv1 
1
Lv 2   0
0
0
0
1
1
1 1
0
0 1 .
0 0
1
Step2: Transform the augmented matrix into the reduced row echelon
matrix
I 33
1

A  0

0
0
0
1
1
0
0
0
1
0
Thus, the matrix
4
0 

1 
2 .
1
2 
1

0

0


0 

1 
2
1 
2

is the matrix of L with respect to the bases S and T.
Example 2:
Let
 3  4  
 1  2  
S  v1 , v2    ,   , T  w1 , w2    ,   ,
 2   2 
 2   2 
be bases for
2
R , and let
 2 7 
A

  3 7
be the matrix of L : R 2  R 2
with respect to the basis S. Find the matrix of L with respect to the basis
T.
[solution:]
The matrix of L with respect to the basis T is
P 1 AP  PT  S APS T ,
where
P  PS T  w1 S
w2 S .
P can be obtained via the following procedure:
Step 1: Form the augmented matrix
v1
v2
w1
 3 4  1 2 
w2   

 2  2 2  2
Step 2: Transform the augmented matrix into the reduced row echelon
matrix
I 22
1 0 3  2
P  
.
0 1 2  1
5
Therefore,
 3  2
P  PS T  
.
2  1 
Thus,
  1 2
P 1  PT  S  

  2 3
and The matrix of L with respect to the basis T is
  1 2  2 7 3  2  2 1
P 1 AP  






.
 2 3   3 7 2  1  1 3
3. (Section 8)
 Definition of the eigenvalue, eigenvector, the characteristic
polynomial, the orthogonal matrix.
 Find the eigenvalues and eigenvectors of a matrix.
 Diagonalization of a matrix and diagonalization of a symmetric
matrix.
Example 1:
Let
 2
A   2
 7
2
1
1
0 
1 
.
 3
(a) Find the eigenvalues and eigenvectors of A.
(b) Find the matrix P and the diagonal matrix D such that A can be
1
diagonalized via P AP  D .
[solution:]
6
(a)
 2
f ( )  det( I  A)   2
7
2
0
 1    1  3  4  0
 1
2
 3
   1, 3,  4 .
1. As
 1,
 1
1  I  Ax   2
 7
2
0
2
0   x1 
 1  x2   0 .
4   x3 
t
1
 x1   2   2 




t
1
 x   x2   
t
, t  R.
4   4 

 x3   t 
1 


 
Thus,
 1 
 2
t  1  , t  R, t  0 ,
 14 


are the eigenvectors associated with eigenvalue
2. As
 1.
  3,
2
 1
3  I  Ax   2
 7
2
2
0   x1 
 1  x2   0 .
6   x3 
 x1    t    1 
 x   x2    t   t  1 , t  R.
2
2
 x3   t   1 
Thus,
7
 1 
t  1 , t  R, t  0 ,
2
 1 
are the eigenvectors associated with eigenvalue
3. As
  3.
  4 ,
 6
 4  I  Ax   2
 7
2
5
2
0   x1 
 1  x2   0 .
 1  x3 
t
 1 
 x1   13 
 13 


 x   x 2    3t   t  3 , t  R.
13
 13
 x3   t 
 1 


Thus,
 1 
 13 
t   3  , t  R, t  0 ,
 13
 1 
are the eigenvectors associated with eigenvalue
  4 .
(c) The column vectors of P are the eigenvectors of A,
1 
 1
1
2
13 

1  3 
P 1
4
2
13

1
1 
 1

and the diagonal elements of D are the eigenvalues of A,
1
D  0
0
0
3
0
8
0 
0 
.
 4
Example 2:
Let
 3 1 0 
A   1 2  1 .
 0  1 3 
(a) Find the eigenvalues and eigenvectors of A.
(b) Find the orthogonal matrix P and the diagonal matrix D such that A
T
can be diagonalized via P AP  D .
(c) Find
A n , where n is a positive integer.
[solution:]
(a)
f ( )  det( I  A) 
 3
1
0
1
 2
1
0
1
 3
   1  3  4  0
   1, 3, 4 .
Thus, we obtain these eigenvectors associated with the above
eigenvalues,
1
r 2, r  R, r  0  the eigenvectors associated with
1
 1
s  0 , s  R, s  0  the eigenvectors associated with
 1 
1
t   1 , t  R, t  0  the eigenvectors associated with
 1 
9
 1.
  3.
  4.
1  1  1 
2,  0 ,  1
(b)       are 3 orthogonal eigenvectors. By normalizing the 3
1  1   1 
eigenvectors, we obtain 3 orthonormal eigenvectors
 1

 2

 1



6

,
6

6



 1
  1
3

2 


,  1
0
.
3
 1
 

1

2 

 
3


The column vectors of P are the orthonormal eigenvectors of A,
 1

6
2
P
6
 1

6


3

1
3

1
3 
1
1
2
0
1
2
and the diagonal elements of D are the eigenvalues of A,
1
D  
0

0
0
3
0
0
0
.
4

(c)
1
1
1  n

6
2
3  1
2
n
n T
 1  0
A  PD P  
0
6
3
1
1
1   0

6
2
3 

1 

6
 1
n 2
1 
6  
6
1 

6 

2
6
 1
2
0 
6
6
  1
0 
0
2
n
4  1
1

3
3
0
3n
0
 1 

2
1   3 n  0   1
0
6 
2
 1  

2 
 1 

3
 1
n  1
1
4 
3   3
3
 1 

3 

10
1

3 
1

2 

6
1 
2
1 
3 
1
1  3  3 n  2  4 n
1
  2  2  4n
6
1  3  3 n  2  4 n

1  3  3n  2  4 n 

2  2  4n

n
n
1 33  2 4 
2  2  4n
4  2  4n
2  2  4n
11