Chapter 7
Eigenvalues and Eigenvectors
7.1 Eigenvalues and Eigenvectors
7.2 Diagonalization
7.3 Symmetric Matrices and Orthogonal Diagonalization
Elementary Linear Algebra
R. Larsen et al. (6 Edition)
投影片設計編製者
淡江大學 電機系 翁慶昌 教授
7.1 Eigenvalues and Eigenvectors

Eigenvalue problem:
If A is an nn matrix, do there exist nonzero vectors x in Rn
such that Ax is a scalar multiple of x？

Eigenvalue and eigenvector:

Geometrical Interpretation
A：an nn matrix
：a scalar
x： a nonzero vector in Rn
Eigenvalue
Ax  x
Eigenvector
Elementary Linear Algebra: Section 7.1, pp.421-422
1/53

Ex 1: (Verifying eigenvalues and eigenvectors)
2 0 
A

0

1


1 
0 
x1    x2   
0 
1
Eigenvalue
2 0  1 2
1
Ax1  
    2   2 x1



0  1 0 0
0 
Eigenvector
Eigenvalue
 2 0  0   0 
0 
Ax2  
    1   (1) x2



0

1
1

    1
1
Eigenvector
Elementary Linear Algebra: Section 7.1, p.423
2/53

Thm 7.1: (The eigenspace of A corresponding to )
If A is an nn matrix with an eigenvalue , then the set of all
eigenvectors of  together with the zero vector is a subspace of
Rn. This subspace is called the eigenspace of  .
Pf:
x1 and x2 are eigenvectors corresponding to 
(i.e. Ax1  x1, Ax2  x2 )
(1) A( x1  x2 )  Ax1  Ax2  x1  x2   ( x1  x2 )
(i.e. x1  x2 is an eigenvecto r correspond ing to λ)
(2) A(cx1 )  c( Ax1 )  c(x1 )   (cx1 )
(i.e. cx1 is an eigenvecto r correspond ing to  )
Elementary Linear Algebra: Section 7.1, p.424
3/53

Ex 3: (An example of eigenspaces in the plane)
Find the eigenvalues and corresponding eigenspaces of
  1 0
A

0
1


Sol:
If
v  ( x, y )
  1 0  x    x 
Av  
 



 0 1  y   y 
For a vector on the x-axis
Eigenvalue
1  1
  1 0  x    x 
 x
 0 1 0   0   10

   
 
Elementary Linear Algebra: Section 7.1, p.425
4/53
For a vector on the y-axis
Eigenvalue
2  1
  1 0  0   0   0 
 0 1  y    y   1 y 

     
Geometrically, multiplying a vector (x, y)
in R2 by the matrix A corresponds to a
reflection in the y-axis.
The eigenspace corresponding to 1  1 is the x-axis.
The eigenspace corresponding to 2  1 is the y-axis.
Elementary Linear Algebra: Section 7.1, p.425
5/53

Thm 7.2: (Finding eigenvalues and eigenvectors of a matrix AMnn )
Let A is an nn matrix.
(1) An eigenvalue of A is a scalar  such that det( I  A)  0 .
(2) The eigenvectors of A corresponding to  are the nonzero
solutions of (I  A) x  0 .


Note:
Ax  x  (I  A) x  0
(homogeneous system)
If (I  A) x  0 has nonzero solutions iff det( I  A)  0.
Characteristic polynomial of AMnn:
det(I  A)  (I  A)  n  cn 1n 1    c1  c0

Characteristic equation of A:
det( I  A)  0
Elementary Linear Algebra: Section 7.1, p.426
6/53

Ex 4: (Finding eigenvalues and eigenvectors)
2  12
A

1

5


Sol: Characteristic equation:
det( I  A) 
 2
12
1
 5
 2  3  2  (  1)(  2)  0
   1,  2
Eigenvalue s :  1  1, 2  2
Elementary Linear Algebra: Section 7.1, p.426
7/53
 3 12  x1  0
(1)1  1  (1I  A) x  
 



  1 4   x2  0 
 x1  4t  4
 3 12 1  4

~
       t  , t  0


  1 4  0 0 
 x2   t  1 
 4 12  x1  0
(2)2  2  (2 I  A) x  
 



  1 3   x2  0 
 x1  3t  3
 4 12 1  3

~
       t  , t  0


  1 3  0 0 
 x2   t  1
Check : Ax   x
i
Elementary Linear Algebra: Section 7.1, p.426
8/53

Ex 5: (Finding eigenvalues and eigenvectors)
Find the eigenvalues and corresponding eigenvectors for
the matrix A. What is the dimension of the eigenspace of
each eigenvalue?
2 1 0
A  0 2 0 


0
0
2


Sol: Characteristic equation:
  2 1
0
I  A  0
 2
0  (  2)3  0
0
0
 2
Eigenvalue:   2
Elementary Linear Algebra: Section 7.1, p.428
9/53
The eigenspace of A corresponding to   2:
0  1 0  x1  0
(I  A) x  0 0 0  x2   0
0 0 0  x3  0
0  1 0 0 1 0
 x1   s  1 0
 0 0 0 ~ 0 0 0   x2   0  s 0  t 0, s, t  0
0 0 0 0 0 0
 x3   t  0 1
 1 0

    

s 0  t 0 s, t  R  : the eigenspace of A correspond ing to   2
 0 1

    

Thus, the dimension of its eigenspace is 2.
Elementary Linear Algebra: Section 7.1, p.428
10/53

Notes:
(1) If an eigenvalue 1 occurs as a multiple root (k times) for
the characteristic polynominal, then 1 has multiplicity k.
(2) The multiplicity of an eigenvalue is greater than or equal
to the dimension of its eigenspace.
Elementary Linear Algebra: Section 7.1, p.428
11/53

Ex 6：Find the eigenvalues of the matrix A and find a basis
for each of the corresponding eigenspaces.
1 0 0 0 
0 1 5  10
A

1
0
2
0


3 
1 0 0
Sol: Characteristic equation:
 1 0
0
0
0
 1  5
10
I  A 
1
0
 2
0
1
0
0
 3
 (  1) 2 (  2)(  3)  0
Eigenvalue s : 1  1, 2  2, 3  3
Elementary Linear Algebra: Section 7.1, p.429
12/53
(1)1  1
0
0

 1

 1
0
0
 (1I  A) x  
 1

 1
0  1
0  5 10  0
~
0  1 0  0
 
0 0  2  0
0
0
0   x1  0
0  5 10   x2  0

0  1 0   x3  0
   
0 0  2   x4   0 
0
0
2   x1   2t  0  2
0 1  2  x2   s  1  0 


s
t
, s, t  0
0 0 0   x3   2t  0  2 
   
    
0 0 0   x4   t  0   1 
0 0
 0    2  
    
1  0  
 ,
 is a basis for the eigenspace
0  2   of A corresponding to   1
0  1  
Elementary Linear Algebra: Section 7.1, p.429
13/53
1
0
 (2 I  A) x  
 1

 1
0  1 0
1 0 0
 0 1  5 10  0 1

~
 1 0 0
0  0 0

 
 1 0 0  1 0 0
(2)2  2
0   x1  0
1  5 10   x2  0

0 0
0   x3  0
   
0 0  1  x4  0
0 0  x1   0  0
 5 0  x2  5t  5


t , t 0
0 1  x3   t  1
      
0 0   x4   0   0 
0
0
 0  
  
 5  
   is a basis for the eigenspace
1  of A corresponding to   2
0 
Elementary Linear Algebra: Section 7.1, p.429
14/53
(3)3  3
2
0

 1

 1
0
2
0
0
2
0
 (3I  A) x  
 1

 1
0 0  1 0
 5 10 0 1
~
1 0  0 0
 
0 0  0 0
0
2
0
0
0
0
1
0
0   x1  0
 5 10  x2  0

1 0   x3  0
   
0 0   x4  0
0  x1   0   0 
5  x2   5t   5


t
, t0
0  x3   0   0 
   
  
0   x4   t   1 
0
 0  
  
 5 

 is a basis for the eigenspace
 0   of A corresponding to   3
 1  
Elementary Linear Algebra: Section 7.1, p.429
15/53


Thm 7.3: (Eigenvalues of triangular matrices)
If A is an nn triangular matrix, then its eigenvalues are
the entries on its main diagonal.
Ex 7: (Finding eigenvalues for diagonal and triangular matrices)
 1 0 0 0 0

2 0 0
 0 2 0 0 0
(a) A   1 1 0  (b) A   0 0 0 0 0
 0 0 0  4 0
 5 3  3
 0 0 0 0 3
 2
0
0
Sol:
( a ) I  A 
1
  1 0  (  2)(  1)(  3)
5
3  3
1  2, 2  1, 3  3
(b) 1  1, 2  2, 3  0, 4  4, 5  3
Elementary Linear Algebra: Section 7.1, p.430
16/53

Eigenvalues and eigenvectors of linear transformations:
A number  is called an eigenvalue of a linear tra nsformatio n
T : V  V if there is a nonzero vector x such that T (x)  x.
The vector x is called an eigenvecto r of T correspond ing to  ,
and the setof all eigenvecto rs of  (with the zero vector) is
called the eigenspace of .
Elementary Linear Algebra: Section 7.1, p.431
17/53

Ex 8: (Finding eigenvalues and eigenspaces)
Find the eigenvalue s and correspond ing eigenspace s
1 3 0 
A  3 1 0 .


0 0  2
Sol:
 1
I  A   3
0
3
 1
0
0
0  (  2) 2 (  4)
2
eigenvalue s : 1  4, 2  2
The eigenspace s for these two eigenvalue s are as follows.
B1  {(1, 1, 0)}
Basis for 1  4
B2  {(1,  1, 0), (0, 0, 1)}
Elementary Linear Algebra: Section 7.1, p.431
Basis for 2  2
18/53

Notes:
(1) Let T:R 3  R 3 be the linear transformatio n whose standard matrix
is A in Ex. 8, and let B' be the basis of R 3 made up of three linear
independen t eigenvecto rs found in Ex. 8. Then A' , the matrix of T
relative to the basis B' , is diagonal.
0
4 0
A'  0  2 0 
B '  {(1, 1, 0), (1,  1, 0), (0, 0, 1)}


0
0

2


Eigenvecto rs of A
Eigenvalue s of A
(2) The main diagonal entries of the matrix A' are the eigenvalue s of A.
Elementary Linear Algebra: Section 7.1, p.431
19/53
Keywords in Section 7.1:

eigenvalue problem: 特徵值問題

eigenvalue: 特徵值

eigenvector: 特徵向量

characteristic polynomial: 特徵多項式

characteristic equation: 特徵方程式

eigenspace: 特徵空間

multiplicity: 重數
20/53
7.2 Diagonalization

Diagonalization problem:
For a square matrix A, does there exist an invertible matrix P
such that P-1AP is diagonal?

Diagonalizable matrix:
A square matrix A is called diagonalizable if there exists an
invertible matrix P such that P-1AP is a diagonal matrix.
(P diagonalizes A)

Notes:
(1) If there exists an invertible matrix P such that B  P 1 AP,
then two square matrices A and B are called similar.
(2) The eigenvalue problem is related closely to the
diagonalization problem.
Elementary Linear Algebra: Section 7.2, p.435
21/53

Thm 7.4: (Similar matrices have the same eigenvalues)
If A and B are similar nn matrices, then they have the
same eigenvalues.
Pf:
A and B are similar  B  P 1 AP
I  B  I  P 1 AP  P 1IP  P 1 AP  P 1 (I  A) P
 P 1 I  A P  P 1 P I  A  P 1 P I  A
 I  A
A and B have the same characteristic polynomial.
Thus A and B have the same eigenvalues.
Elementary Linear Algebra: Section 7.2, p.436
22/53

Ex 1: (A diagonalizable matrix)
1 3 0 
A  3 1 0 


0
0

2


Sol: Characteristic equation:
 1  3
0
I  A   3   1 0  (  4)(  2) 2  0
0
0
2
Eigenvalue s : 1  4, 2  2, 3  2
1
(1)  4  Eigenvecto r : p1  1 (See p.403 Ex.5)
0
Elementary Linear Algebra: Section 7.2, p.436
23/53
(2)  2  Eigenvector : p2
P  [ p1

p2
Notes:
(1) P  [ p2
p1
(2) P  [ p2
p3
1
p3 ]  1
0
1
p3 ]   1
 0
1
p1 ]   1
 0
Elementary Linear Algebra: Section 7.2, p.436
1
0 
  1, p3  0 (See p.403 Ex.5)
 0 
1
1 0
0
4 0
 1 0  P 1 AP  0  2 0 
0 0  2
0 1
1 0
 2 0 0 
1 0  P 1 AP   0 4 0 
 0 0  2
0 1
0 1
 2 0 0
0 1  P 1 AP   0  2 0
 0
1 0
0 4
24/53

Thm 7.5: (Condition for diagonalization)
An nn matrix A is diagonalizable if and only if
it has n linearly independent eigenvectors.
Pf:
() A is diagonaliz able
there exists an invertible P s.t. D  P 1 AP is diagonal
Let P  [ p1 p2  pn ] and D  diag (1 , 2 ,, n )
PD  [ p1
p2 
1 0
 0 2
pn ]
 
 0 0
 0
 0

 
 n 
 [1 p1 2 p2  n pn ]
Elementary Linear Algebra: Section 7.2, p.437
25/53
AP  A[ p1
p2  pn ]  [ Ap1
Ap2  Apn ]
 AP  PD
 Api  i pi , i  1, 2,, n
(i.e. the column vec tor pi of P are eigenvecto rs of A)
 P is invertible  p1 , p2 ,, pn are linearly independen t.
 A has n linearly independen t eigenvecto rs.
() A has n linearly independen t eigenvecto rs p1 , p2 , pn
with correspond ing eigenvalue s 1 , 2 , n
i.e. Api  i pi , i  1, 2,, n
Let P  [ p1
p2  pn ]
Elementary Linear Algebra: Section 7.2, p.438
26/53
AP  A[ p1
 [ Ap1
p2  pn ]
Ap2  Apn ]
 [1 p1 2 p2  n pn ]
1 0
0 
2
 pn ]
 

0 0
 0
 0 
 [ p1 p2
 PD
 

 n 
 p1 , p1 ,, pn are linearly independen t  P is invertible
 P 1 AP  D
 A is diagonaliz able
Note: If n linearly independent vectors do not exist,
then an nn matrix A is not diagonalizable.
Elementary Linear Algebra: Section 7.2, p.438
27/53

Ex 4: (A matrix that is not diagonalizable)
Show that the following matrix is not diagonaliz able.
1 2
A

0
1


Sol: Characteristic equation:
 1  2
I  A 
 (  1) 2  0
0
 1
Eigenvalue : 1  1
0  2 0 1
1
I  A  I  A  
~
 Eigenvecto r : p1   


0 0  0 0
0 
A does not have two (n=2) linearly independent eigenvectors,
so A is not diagonalizable.
Elementary Linear Algebra: Section 7.2, p.439
28/53

Steps for diagonalizing an nn square matrix:
Step 1: Find n linearly independent eigenvectors p1 , p2 ,, pn
for A with corresponding eigenvalues 1 , 2 ,, n
Step 2: Let P  [ p1
p2  pn ]
Step 3:
1 0
0 
2
P 1 AP  D  
 

0 0
0
0 
, where Api  i pi , i  1, 2,, n
 

 n 


Note:
The order of the eigenvalues used to form P will determine
the order in which the eigenvalues appear on the main diagonal of D.
Elementary Linear Algebra: Section 7.2, p.439
29/53

Ex 5: (Diagonalizing a matrix)
 1  1  1
A 1
3 1


 3 1  1
Find a matrix P such that P 1 AP is diagonal.
Sol: Characteristic equation:
 1
I  A   1
3
1
1
  3  1  (  2)(  2)(  3)  0
1   1
Eigenvalue s : 1  2, 2  2, 3  3
Elementary Linear Algebra: Section 7.2, p.440
30/53
1  2
2  2
 1 1 1  1 0 1
 1I  A   1  1  1 ~ 0 1 0

 

3

1
3
0
0
0

 

 x1   t   1
 1
  x2    0   t  0   Eigenvecto r : p1   0 
 x3   t   1 
 1 
1  1 0  14 
 3 1
 2 I  A    1  5  1 ~ 0 1 14 


 
 3  1  1 0 0 0 
 x1   14 t 
1
1
  x2    14 t   14 t  1  Eigenvecto r : p2   1
 x3   t 
 4 
 4 
Elementary Linear Algebra: Section 7.2, p.440
31/53
3  3
 2 1 1  1 0 1 
 3I  A   1 0  1 ~ 0 1  1

 

3

1
4
0
0
0

 

 x1   t   1
 1
  x2    t   t  1   Eigenvecto r : p3   1 
 x3   t   1 
 1 
 1 1  1
Let P  [ p1 p2 p3 ]   0  1 1 
 1 4 1 
 2 0 0
 P 1 AP  0  2 0
0 0 3
Elementary Linear Algebra: Section 7.2, p.440
32/53

Notes: k is a positive integer
d1
0
(1) D  

 0
0
d2

0
d1k
 0
0
 0
k
  D 
 

 d n 
 0
0
d 2k

0
 0
 0

 
 d nk 
(2) D  P 1 AP
 D k  ( P 1 AP) k
 ( P 1 AP)( P 1 AP)    ( P 1 AP)
 P 1 A( PP 1 ) A( PP 1 )    ( PP 1 ) AP
 P 1 AA    AP
 P 1 Ak P
 Ak  PD k P 1
Elementary Linear Algebra: Section 7.2, p.445
33/53

Thm 7.6: (Sufficient conditions for diagonalization)
If an nn matrix A has n distinct eigenvalues, then the
corresponding eigenvectors are linearly independent and
A is diagonalizable.
Elementary Linear Algebra: Section 7.2, p.442
34/53

Ex 7: (Determining whether a matrix is diagonalizable)
1  2 1 
A  0 0
1


0 0  3
Sol: Because A is a triangular matrix,
its eigenvalues are the main diagonal entries.
1  1, 2  0, 3  3
These three values are distinct, so A is diagonalizable. (Thm.7.6)
Elementary Linear Algebra: Section 7.2, p.443
35/53

Ex 8: (Finding a diagonalizing matrix for a linear transformation)
Let T:R 3  R 3 be the linear transformatio n given by
T ( x1,x2 ,x3 )  ( x1  x2  x3 , x1  3 x2  x3 ,  3 x1  x2  x3 )
Find a basis B for R 3 such that the matrix for T
relative to B is diagonal.
Sol: The standard matrix for T is given by
 1  1  1
A  T (e1 ) T (e2 ) T (e3 )   1
3 1 
 3 1  1
From Ex. 5, there are three distinct eigenvalues
1  2, 2   2, 3  3
so A is diagonalizable. (Thm. 7.6)
Elementary Linear Algebra: Section 7.2, p.443
36/53
Thus, the three linearly independent eigenvectors found in Ex. 5
p1  (1, 0, 1), p2  (1,  1, 4), p3  (1, 1, 1)
can be used to form the basis B. That is
B  { p1 , p2 , p3}  {( 1, 0, 1), (1,  1, 4), (1, 1, 1)}
The matrix for T relative to this basis is
D  [T ( p1 )]B [T ( p2 )] B [T ( p3 )]B 
 [ Ap1 ]B [ Ap2 ]B [ Ap3 ]B 
 [1 p1 ]B [2 p2 ]B [3 p3 ]B 
 2 0 0
 0  2 0
0 0 3
Elementary Linear Algebra: Section 7.2, p.443
37/53
Keywords in Section 7.2:

diagonalization problem: 對角化問題

diagonalization: 對角化

diagonalizable matrix: 可對角化矩陣
38/53
7.3 Symmetric Matrices and Orthogonal Diagonalization

Symmetric matrix:
A square matrix A is symmetric if it is equal to its transpose:
A  AT

Ex 1: (Symmetric matrices and nonsymetric matrices)
 0 1  2
A 1 3 0 
(symmetric)


 2 0 5 
4 3
B
(symmetric)

3
1


3 2 1
C  1  4 0
(nonsymmetric)


1 0 5
Elementary Linear Algebra: Section 7.3, p.447
39/53

Thm 7.7: (Eigenvalues of symmetric matrices)
If A is an nn symmetric matrix, then the following properties
are true.
(1) A is diagonalizable.
(2) All eigenvalues of A are real.
(3) If  is an eigenvalue of A with multiplicity k, then  has k
linearly independent eigenvectors. That is, the eigenspace
of  has dimension k.
Elementary Linear Algebra: Section 7.3, p.447
40/53

Ex 2:
Prove that a symmetric matrix is diagonalizable.
a c 
A
 c b 
Pf: Characteristic equation:
 a c
I  A 
 2  (a  b)  ab  c 2  0
c  b
As a quadratic in , this polynomial has a discriminant of
(a  b) 2  4(ab  c 2 )  a 2  2ab  b 2  4ab  4c 2
 a 2  2ab  b 2  4c 2
 ( a  b ) 2  4c 2  0
Elementary Linear Algebra: Section 7.3, p.448
41/53
(1) (a  b) 2  4c 2  0
 a  b, c  0
a 0 
A
is a matrix of diagonal.

0 a 
(2) (a  b) 2  4c 2  0
The characteristic polynomial of A has two distinct real roots,
which implies that A has two distinct real eigenvalues. Thus,
A is diagonalizable.
Elementary Linear Algebra: Section 7.3, p.448
42/53

Orthogonal matrix:
A square matrix P is called orthogonal if it is invertible and
P 1  PT

Ex 4: (Orthogonal matrices)
 0 1
0  1
1
T
(a) P  
is orthogonal because P  P  
.


  1 0
1 0 
3
5 0
(b) P   0 1
4

0
 5
 4
 3
0


5
5
1
T


0 is orthogonal because P  P  0 1
 4
3 


0
5 
 5
Elementary Linear Algebra: Section 7.3, p.449
4
5
0 .
3

5 
43/53

Thm 7.8: (Properties of orthogonal matrices)
An nn matrix P is orthogonal if and only if
its column vectors form an orthogonal set.
Elementary Linear Algebra: Section 7.3, p.450
44/53

Ex 5: (An orthogonal matrix)
 13
 2
P 5
 2
3 5


0
5 
3 5
2
3
1
5
4
3 5
2
3
Sol: If P is a orthogonal matrix, then P 1  PT  PPT  I
 1
 32
T
PP   5
 3 25

2
3
1
5
4
3 5
  13
 2
0  3
5  2
3 5  3
2
3
2
5
1
5
0
2
3 5
4
3 5
5
3 5
 1 0 0
 
I

0
1
0
 

 0 0 1 

 13 
 23 
 23 
Let p1   25 , p2   15 , p3   0 
 325 
 345 
 3 5 5 
Elementary Linear Algebra: Section 7.3, p.451
45/53
produces
p1  p2  p1  p3  p2  p3  0
p1  p2  p3  1
{ p1 , p2 , p3} is an orthonorma l set.
Elementary Linear Algebra: Section 7.3, p.451
46/53

Thm 7.9: (Properties of symmetric matrices)
Let A be an nn symmetric matrix. If 1 and 2 are distinct
eigenvalues of A, then their corresponding eigenvectors x1
and x2 are orthogonal.
Elementary Linear Algebra: Section 7.3, p.452
47/53


Ex 6: (Eigenvectors of a symmetric matrix)
 0 1
Show that any two eigenvecto rs of A  


1
0


correspond ing to distinct eigenvalue s are orthogonal .
Sol: Characteristic function
  3 1
I  A 
 2  6  8  (  2)(  4)  0
1   3
 Eigenvalue s : 1  2, 2  4
 1  1 1 1
 1
(1) 1  2  1I  A  
~
 x 1  s  , s  0


 1  1 0 0
1
 1  1 1  1
1
(2) 2  4  2 I  A  
~
 x 2  t  , t  0


 1 1  0 0 
1
 s  t 
x 1  x 2        st  st  0  x 1 and x 2 are orthogonal .
 s  t 
Elementary Linear Algebra: Section 7.3, p.452
48/53

Thm 7.10: (Fundamental theorem of symmetric matrices)
Let A be an nn matrix. Then A is orthogonally diagonalizable
and has real eigenvalue if and only if A is symmetric.

Orthogonal diagonalization of a symmetric matrix:
Let A be an nn symmetric matrix.
(1) Find all eigenvalues of A and determine the multiplicity of each.
(2) For each eigenvalue of multiplicity 1, choose a unit eigenvector.
(3) For each eigenvalue of multiplicity k2, find a set of k linearly
independent eigenvectors. If this set is not orthonormal, apply GramSchmidt orthonormalization process.
(4) The composite of steps 2 and 3 produces an orthonormal set of n
eigenvectors. Use these eigenvectors to form the columns of P. The
matrix P 1 AP  PT AP  D will be diagonal.
Elementary Linear Algebra: Section 7.3, p.453
49/53

Ex 7: (Determining whether a matrix is orthogonally diagonalizable)
1 1 1
A1  1 0 1


1
1
1


 5 2 1
A2   2 1 8



1
8
0


Symmetric
matrix
Orthogonally
diagonalizable
 3 2 0
A3  
2 0 1
0 0 
A4  
0  2
Elementary Linear Algebra: Section 7.3, p.454
50/53

Ex 9: (Orthogonal diagonalization)
Find an orthogonal matrix P that diagonaliz es A.
2  2
2
A   2 1 4 


 2 4  1 
Sol:
(1) I  A  (  3)2 (  6)  0
1  6, 2  3 (has a multiplici ty of 2)
v1
(2) 1  6, v1  (1,  2, 2)  u1 
 ( 13 , 32 , 23 )
v1
(3) 2  3, v2  (2, 1, 0), v3  (2, 0, 1)
Linear Independent
Elementary Linear Algebra: Section 7.3, p.455
51/53
Gram-Schmidt Process:
v3  w2
w2  v2  (2, 1, 0), w3  v3 
w2  ( 52 , 54 , 1)
w2  w2
w3
w2
2
1
u2 
 ( 5 , 5 , 0), u3 
 ( 325 , 3 4 5 , 3 5 5 )
w2
w3
 13
(4) P  [ p1 p2 p3 ]   32
 23
2
5
1
5
0

4 
3 5

5

3 5
2
3 5
  6 0 0
 P 1 AP  P T AP   0 3 0
 0 0 3
Elementary Linear Algebra: Section 7.3, p.456
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Keywords in Section 7.3:

symmetric matrix: 對稱矩陣

orthogonal matrix: 正交矩陣

orthonormal set: 單範正交集

orthogonal diagonalization: 正交對角化
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