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## 特徵值與特徵向量

### 7.3 對稱矩陣與正交對角化 7.4 特徵值與特徵向量的應用

Elementary Linear Algebra 投影片設計製作者

R. Larsen et al. (6 Edition) 淡江大學 電機系 翁慶昌 教授

(2)



×

n



×

×

x： Rn

x Ax =



(3)





 

= −

1 0

0 A 2





= 0 1 x1

1

1 2

0 2 1 0

2 0

1 1 0

0

2 x

Ax  =



 =

 

= 









= −

### 特徵值





= 1 0 x2

1 1 0 −10 0 0

2

2 ( 1)

1 1 0 1

0 1

0 1 0

0

2 x

Ax  = −



 =

 

= −









= −

(4)



×

n

1

2

)

,

. .

(

1 =

1

2 =

2

)

. . (

) (

) (

) 1 (

2 1

2 1

2 1

2 1

2 1

+

+

= +

= +

=

+

) (

) (

) (

) (

) 2

(

1 =

1 =

1 =

1

(5)



= 0 1 0

1

v = (x ,y)

=

=

1 0

0 v 1

v = (x ,y)

=

=

1 0 0

0 1

0

0

1

1 = −1

(6)

=

=

1 0 0

0 1

0

0 1

2 =1

2

1 = −1

2 =1

(7)



n×n

0 )

I

det(λ −

= (1) A的特徵值為一數值

(2) A相對應於

det(

IA) = 0



 AMn×n

0 1

1

) 1

I ( )

I

det(

A =

A =

n + cn

n +L+ c

### λ

+ c

 A的特徵方程式 (characteristic equation) 0

) I

det(

A =

(λI −

)

= 0

det(

I A) = 0

0 )

I (

− =

=

λ λ (齊次系統)

(8)



= −

5 1

12

2

12 ) 2

I

( −

=

A

0 )

2 )(

1 (

2 3

5 1

12 ) 2

I (

2 + + = + + =

=

+

= −

A

1 = −1,

2 = −2 2

, 1 −

=

(9)

0 12

4

1

) 1

(

1 = −

0

1 , 4 4

0 0

4

~ 1 4 1

12 3

0 0 4

1

12 ) 3

I (

2 1 2 1 1

 ≠

 

= 



 

= 



 

⇒ 



 

 −



 



 

= 



 



 

= −

t t t

t x

x x

x x A

Q

2 )

2

(

2 = −

0

1 , 3 3

0 0

3

~ 1 3 1

12 4

0 0 3

1

12 ) 4

I (

2 1 2 1 2

=

=

=

= −

i

(10)

=

2 0

0

0 2

0

0 1

2



0 )

2 (

2 0

0

0 2

0

0 1

2

I = − 3 =

− −

= −

A

= 2

(11)

= 2





 =







 −

=

0 0 0 0

0 0

0 0

0

0 1 0

) I

(

3 2 1

x x x x

A

0 ,

, 0 0 0

1

2 0

1

+

=

=

0 ,

, 1 0 0

0 0

3

2

+

=

=

2

, 1 0 0 0

0 1

=

+

(12)



(1) 若特徵值

1

1

### 的重數(multiplicity)為k

(2) 特徵值的重數往往會大於或等於其特徵空間的維度

(13)



= −

3 0

0 1

0 2

0 1

10 5

1 0

0 0

0 1

0 )

3 )(

2 (

) 1 (

3 0

0 1

0 2

0 1

10 5

1 0

0 0

0 1

I

2 − − =

=

− −

−− − −

=

1 =1,

2 = 2,

3 = 3

(14)

1 )

1

(

1 =









=

















= −

0 0 0 0

2 0

0 1

0 1

0 1

10 5

0 0

0 0

0 0

) I

(

4 3 2 1

1

x x x x x

A

0 2 1

0 2

2 1

0 0

2 0

0 1 10

5 0

0

0 0

0

0 1





−









−









 −





s

t x

x

2 0

2 0 ,

1 0













−









=1

0 ,

, 1 2 0

0 0 1 2

0 0

0 0

0 0

0 0

2 1

0

~ 0

2 0

0 1

0 1

0 1

10 5

0 0

4 3

2





 



 +





 



=





 



=





 







 







 



− −

s t s t

t t s

x x x

(15)

2 )

2

(

2 =









=

















= −

0 0 0 0

1 0

0 1

0 0

0 1

10 5

1 0

0 0

0 1

) I

(

4 3 2 1

2

x x x x x

A

5 0 5

0 0

5 1

0

0 0

0 1

10 5

1 0

0 0

0

1 1

0 1 5 0

















= 2

0

, 0 1 5 0

5 0

0 0

0

1 0

0 0

0 5 1

~ 0 1 0

0 1

0 0

0 1

10 5

1 0

4 3

2

=

=

− −

(16)

3 )

3

(

3 =

=

= −

0 0 0 0

0 0

0 1

0 1

0 1

10 5

2 0

0 0

0 2

) I

(

4 3 2 1

3

5 0 5

0 5

0 1

0

0 0

0 1

10 5

2 0

0 0

0

2 1

0 5 0

= 3 λ

0

, 1 0

5 0

5 0

0 0

0

0 1

0 0

5 0

1

~ 0 0 0

0 1

0 1

0 1

10 5

2 0

4 3

2

= −

= −

(17)



×







= 1 1 0 0 0

2 )

(a A

= 0 0 0 0 0 0 0

0 2

0

0 0

0 0

1 )

(



 



 −

=

3 3

5

0 1

1 )

(a A

= −

3 0

0 0

0

0 4

0 0

0

0 0

0 0

0 )

(

) 3 )(

1 )(

2 (

3 3

5

0 1

1

0 0

2 I

)

( = − − +

+

=

A

a

3 ,

1 ,

2 2 3

1 =

=

= −

3 ,

4 ,

0 ,

2 ,

1 )

(

1 = −

2 =

3 =

4 = −

5 =

(18)



) (

:

) (

x

x x

x

=

(19)



=

2 0

0

0 1

3

0 3

1

) 4 (

) 2 (

2 0

0

0 1

3

0 3

1

2

+

=

+

=

− λ λ

λ λ

λ λ

2 ,

4 2

1 =

= −

### λ

2

)}

1 , 0 , 0 ( ), 0 , 1 , 1 {(

4

)}

0 , 1 , 1 {(

2 2

1 1

=

=

=

=

B

B

(20)



### 注意：

)}

1 , 0 , 0 ( ), 0 , 1 , 1 ( ), 0 , 1 , 1 {(

'

(1) 3 3

=

B

A A'

B' T

A' R

T:R B'

− −

=

2 0

0

0 2

0

0 0

4 '

0 0 − 2

(2)

(21)

### 摘要與復習 (7.1節之關鍵詞)

 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

 characteristic equation: 特徵方程式

 eigenspace: 特徵空間

 multiplicity: 重數

(22)



-1





(1)若存在一可逆矩P使得

### B

= 1

(2)特徵值問題與對角化問題兩者關係密切

-1

(23)



×

= 1

P A P

AP P

P P

AP P

B = − = − = −

− I I ( I )

I

1 1

1 1

A

A P

P A

P P

P A P

P A P

AP P

P P

AP P

B

=

=

=

=

=

=

=

I

I I

I

) I

( I

I I

1 1

1

1 1

1 1

(24)







= −

2 0

0

0 1

3

0 3

1 A

0 3

1 −

0 )

2 )(

4 (

2 0

0

0 1

3

0 3

1

I = − + 2 =

− +

−− −

=

2 ,

2 ,

4 2 3

1 =

= −

= −

= 4

)

1

( λ

= 1 1

1 (見p.501 範例5)

(25)

− ⇒

= 2 )

2

(



 =



=

1 0 0

, 0

1 1

3

2 p

p

=

=

=

2 0

0

0 2

0

0 0

4 1

0 0

0 1 1

0 1

1 ]

[

1

2

3

1

(見p.501 範例5)





=





=

=





=





=

=

4 0

0

0 2

0

0 0

2 0

1 0

1 0

1

1 0

1 ]

[

2 0

0

0 4

0

0 0

2 1

0 0

0 1

1

0 1

1 ]

[

1 1

3 2

1 3

1 2

AP P

p p

p P

AP P

p p

p P



(1)

(2)

(26)



×

) (⇒

1

=

) , , , ( ]

[ 1 2 1 2

1

n

n

=

=

=

0 0

0 0

0 0

]

[ 2

1

2 1

n

pn

p p

PD

L

M O

M M

L L L









=

(27)

PD AP

Ap Ap

Ap

AP n

=

=[ 1 2 L ]

Q

) .

. (

, , 2 , 1

,

i i

i

i =

=

n

1, 1,

,

n

n

,

,

, ,

) (

2 1

2 1

i =

i i, =1, 2,

, ] [

1

2

n

=

(28)

n

n n

n n

=

=

=

=

=

0 0

0 0

0 0

] [

] [

] [

] [

2 1

2 1

2 2 1

1

2 1

2 1

n

0 0

n

=

−1 1

1, ,

,



(29)





= −

2 0

0

0 1

3

0 3

1 A

0 )

1 2 (

I − =

−1 − =

2 =

A ( 1) 0

1 0

2

I − =

−1

−− =

2 =

A

1 =1

=

=

=

− 0

1 0

0

1

~ 0 0

0

2

I

0

1

(30)

 n×n方陣對角化步驟

=[

1

2

n]

1,

2,

n

=

=

0 0

0 0

2 1

1

=

=

n

0

0

0

0 2

1

i =

i i, =1, 2,

,



(31)



1

1 1

3

1 3

1

1 1

1





= −

0 )

3 )(

2 )(

2 (

1 1

3

1 3

1

1 1

1

I = − + − =

+

−− −

−−

=

A

3 ,

2 ,

2 2 3

1 =

= −

=

(32)

1 = 2

− −

=

0 0

0

0 1

0

1 0 1

~ 3

1 3

1 1

1

1 1

1

1I

=

=

=

1 0

1

1 0

1

0 1

3 2 1

3

1 1

2 = −2





 −





− −

−−

=

0 0

0

1 0

0 1

~ 1 1

3

1 5

1

1 1

3

I 41

4 1

2 A

=

=

=

1

1

1 1

t 2

4 1 4

1 4 1

2 1

(33)

3 = 3

− −

=

0 0

0

1 1

0

1 0

1

~ 4

1 3

1 0

1

1 1

2

3I

=

=

=

1 1

1

1 1

1

t 3

3 2 1

3

1 1

=

=

=

3 0

0

0 2

0

0 0

2

1 4

1

1 1

0

1 1

1 ]

[

1

3 2

1

(34)



=

=

k n k

k

k

n

0 0

0 0

0 0

0 0

0 0

0 0

) 1

( 2

1 2

1

1 )

2

( D = P AP

1 1

1 1

1 1

1 1

1 1

) (

) (

) (

) (

) )(

(

) (

)

2 (

=

⋅⋅

=

⋅⋅

=

⋅⋅

=

=

=

P A P

AP AA

P

AP PP

PP A

PP A

P

AP P

AP P

AP P

AP P

D

AP P

D

k k k

(35)



×

(36)



=

3 0

0

1 0

0

1 2

1 A

3

, 2 ,

2 = − =

=

1 = 2,

2 = −2,

3 = 3

(定理7.6)

(37)



B T

B R

x x

x x

x x

x x

x x

x x T

R R

T

3

3 2

1 3

2 1

3 2

1 3

2 1

3 3

) 3

, 3

, (

) , , (

:

− +

− +

+

=

T的標準矩陣為

## [ ]

=

=

1 1

3

1 3

1

1 1

1 )

( )

( )

(e1 T e2 T e3 T

A

T的標準矩陣為

) 7.6 (

3 ,

2 ,

2 5

3 2

1

A

=

=

=

### λ

(38)

)}

1 , 1 , 1 ( ), 4 , 1 , 1 ( ), 1 , 0 , 1

{(− − −

= B

) 1 , 1 , 1 ( ),

4 , 1 , 1 ( ),

1 , 0 , 1

( 2 3

1 = − p = − p = −

p

=

=

=

=

0 2

0

0 0

2

] [

] [

] [

] [

] [

] [

] ) ( [ )]

( [ ] ) ( [

3 3 2

2 1

1

3 2

1

3 2

1

B B

B

B B

B

B B

B

(39)

### 摘要與復習 (7.2節之關鍵詞)

 diagonalization problem: 對角化問題

 diagonalization: 對角化

 diagonalizable matrix: 可對角化矩陣

(40)



AT

A = 即



= 1 3 0 2 1

0

=

5 0

2

0 3

1





= 3 1 3 B 4

= 1 4 0 1 2

3

(41)



×

(42)







= c b c A a

0 )

( 2

2 − + + − =

− =

−− −

=

a c a b ab c

A

I

### λ

− = − − = −(a +b) + abc = 0 b

A c

I

2 2

2 2

2

2 2

2 2

2

4 )

(

4 2

4 4

2 )

( 4 )

(

c b

a

c b

ab a

c ab

b ab a

c ab b

a

+

=

+ +

=

+

− +

+

=

− +

≥ 0

(43)

0 4

) (

) 1

(

2 +

2 = 0

,

= =

a b c

0



= a

A a

0 4

) (

) 2

(2) (

)2 + 4

2 > 0 (

2 +

2 >

2 −(

+

)

+

2 = 0

(44)



−1 =

T

=

=

=

0 1 0 1

(a) P

P 1 PT



=

=

= −

0 1

1 0

0 1

1

(a) P 0

P 1 PT

= −

=

=

5 0 3

5 4

0 1

0 5

0 4 5

3

5 0 3

5 4

0 1

0 5

0 4 5

3

(b)

1

T

(45)



×

(46)







=

5 3

5 5

3 4 5

3 2

5 1 5

2

3 2 3

2 3

1

0 P

P1 = PTPPT = I

I 1

0 0

0 1

0

0 0

1 0

0

5 3

5 3

2

5 3

4 5

1 3

2

5 3

2 5

2 3

1

5 3

5 5

3 4 5

3 2

5 1 5

2

3 2 3

2 3

1

 =



=









=

PPT

I

⇒ =

= P PP P

=

=

=

3 2

5 3 1 3 2

5 2 2 3 1

1 ,

,

0

(47)

1

0

3 2

1

3 2

3 1

2 1

=

=

=

=

=

=

}

, ,

{

1

2

3

(48)



×

1

2

1

2

### 為正交

（二）經向金融聯合徵信中心查詢或徵授信過程中知悉其有債務本

▸ 學校在收集學生的個人資料前，必須徵得學生的同意，並向所

（二）經向金融聯合徵信中心查詢或徵授信過程中知悉其有債務本

（二）經向金融聯合徵信中心查詢或徵授信過程中知悉其有債務本

• 有一個可以耐重 W 的背包，及 N 種物品，每種物品有各自的重量 w[i] 和價值 v[i] ，且數量為 k[i] 個，求在不超過重量限制的情 況下往背包塞盡量多的東西，總價值最大為多少？.