chapter 3 diagonalizableming/diagonalizable.pdf · 教學卓越線性代數...

教學卓越線性代數

指導教授：陳啟銘老師

1 應用數學系

Chapter 3 Diagonalizable (對角化)

章節引導：

A：a diagonal matrix ⇒Ak =

k

nn

k

k

a

a

a

0...0

0.0.

.0.0.

.0.0.

.00

0...0

22

11

其中 A∈Mn*n(ℜ )

主對角線

A =

nna

a

a

0...0

0.0.

.0.0.

.0.0.

.00

0...0

22

11

DefinitionDefinitionDefinitionDefinition：：：：(Similar)(Similar)(Similar)(Similar)

RemarkRemarkRemarkRemark：：：：

（1）A～A ⇔ ∃ I∈Mn*n( ℜ)

∋ A = I-1AI

A、B ∈ Mn*n(ℜ)

A is similar to B

⇔ ∃ P∈Mn*n(ℜ)，invertible(可逆)

∋ B = P-1AP

∃ Q = P-1

∋ B = QAQ-1

P：invertible

⇔ ∃ P-1 ∋ PP-1 = P-1P = I

P：invertible

⇒P-1：invertible

∥

Q



2 應用數學系

（2）A～B⇒ B～A

⇕ ∃ P：invertible ∃ P-1：invertible

∋ B = P-1AP ⇔ ∋ A = (P-1)-1BP-1

B = P-1

AP

⇒BP-1 = P-1APP-1 = P-1A

⇒PBP-1 = PP-1A = A

⇒ (P-1)-1BP-1 = A

（3）If A～B，B～C

⇒A～C

ProveProveProveProve：：：：

∵ A～B ∴ ∃ P：invertible ∋ B = P-1AP ∵ B～C ∴ ∃ Q：invertible ∋ C = Q-1AQ ⇒ C = Q-1AQ

= Q-1

(P-1

AP)Q

= (Q-1

P-1

)A(PQ)

= (PQ)-1

A(PQ)

⇒ ∃ PQ：invertible

∋ C = (PQ)-1A(PQ)

⇒ A～C

（4）A～0 ⇔ A = 0 (zero matrix)

（5）A～I ⇔ A = I


∵ A～I ∴ ∃ P：invertible ∋ I = P-1AP ⇒PIP-1 = A

∥ I



3 應用數學系

ExampleExampleExampleExample：：：：

A、B、C、D ∈ Mn*n(ℜ )

If A～B、C～D

⇒ AC ≁ BD

SolutionSolutionSolutionSolution：：：：

舉反例

A =

01

10

B =

1-0

01

C =

00

01 = D

（1）C～D is ok （2）A～B

∵ ∃ P =

1-1

11

∋ B = P-1AP

∴ A～B

思考路徑：

（2）證 A～B

令 P =

dc

ba

⇒B = P-1AP

PB = AP

⇔

dc

ba

1-0

01 =

01

10

dc

ba

⇒

dc

ba

-

- =

ba

dc



4 應用數學系

TheoremTheoremTheoremTheorem：：：：

T：V→V is li8near

β 1、 β 2 ：two bases of V

⇒ [ ]1

T β ～ [ ] 2T β


[ ] 11

Tβ

β [ ] 12

Iβ

βv = [ ] 1

2I

β

βv[ ] 2

2T

β

β

[ ] 11

Tβ

β [ ] 12

Iβ

βv[ ] 2

1I

β

βv = [ ] 1

2I

β

βv[ ] 2

2T

β

β [ ] 21

Iβ

βv

∥

[ ] 11

Iβ

βv

[ ]1

T β = ( [ ] 21

Iβ

βv)-1 [ ]

2T β [ ] 2

1I

β

βv

⇒ [ ]1

T β ～ [ ] 2T β

[ ] 21

Iβ

β ≠ I

↑ ↑

identity operator identity matrix

Why？


V →I V

β 1 β 1

β 2 β 2

β 1 = {u1、u2、…、uk}

β 2 = {v1、v2、…、vk}

[ ] 11

Tβ

β [ ] 12

Iβ

βv = [ ] 1

2I

β

βv[ ] 2

2T

β

β

[ ] 12

Tβ

β [ ] 12Tβ

β

[ ] 11

Iβ

β = I

[ ] 21

Iβ

β ≠ I



5 應用數學系

[ ] 21

Iβ

βv =

.....

...

...1211 aa

≠ I

I(u1) = u1 = a11v1+a12v2+…

.

.

.

[ ] 11

Iβ

βv =

1...0

...

0...100

0...010

0...001

Iv(u1) = u1 = 1*u1 + 0*u2 + 0*u3 + … + 0*uk

Iv(u2) = u2 = 0*u1 + 1*u2 + 0*u3 + … + 0*uk

.

.

.


（1）A、B ∈ Mn*n( ℜ )、A～B ⇒ Ak～Bk ，k ∈ N


∵A～B ∴ ∃ P ∈ Mn*n( ℜ ) ，invertible ∋ B = P-1AP

Bk = (P

-1AP)

k

= (P-1

AP) (P-1

AP) (P-1

AP)*…*(P-1

AP)

k times

= P-1

APP-1

APP-1

AP*…*P-1

AP

= P-1

AkP

∴ Ak～Bk



6 應用數學系

（2） f(x) ∈ Fn[X] f(x) = a0 + a1x + a2x2 + … + anxn

A～B ⇒ f(A)～f(B)


∵ A～B ∴ ∃ P ∈ Mn*n( ℜ )，invertible ∋ B = P-1AP

f(B) = f(P-1

AP)

= a0I + a1(P-1

AP) + a2(P-1

AP)2 +…+ an(P

-1AP)

n

= a0 (P-1

AP) + a1(P-1

AP) + a2(P-1

A2P) +…+ an(P

-1A

n P)

= P-1

(a0I)P + P-1

(a1A)P + P-1

(a2A2)P + … + P

-1 (anI

n)P

= P-1

(a0I + a1A + … + anAn)P = P

-1(f(A))P

∵ f(A)～f(B)

（3） A，B ∈ Mn*n( ℜ ) A～B ⇒ AT～BT


∵ A～B ∴∃ P∈Mn*n( ℜ )

∋ B = P-1AP

Hence

BT = (P

-1AP)

T

= PTA

T(P

-1)T

= PTA

T(P

T)

-1

( = ((PT)

-1)

-1A

T(P

T)

-1)

∴ AT～BT

DefinitionDefinitionDefinitionDefinition：：：：

A～B ⇔ ∃ P ∈ Mn*n( ℜ ) ，invertible

∋ B = P-1AP

⇔ ∃ Q ∈ Mn*n(ℜ ) ，invertible

∋ B = Q-1AQ

此時 P = Q-1

（1）(AB)T = BTAT (ABC)

T = C

TB

TA

T

（2）(A-1)T = (AT)-1



7 應用數學系

（4）A～B A is invertible

⇒B is invertible


∵ A～B ∴ ∃ P ∈ Mn*n( ℜ ) ，invertible

∋ B = P-1AP

∵ A is invertible

∴ AA-1 = I = A-1A

令 C = P-1A-1P 則 BC = (P-1AP)( P-1A-1P) = P

-1AP P

-1A

-1P

= I

CB = (P-1

A-1

P)( P-1

AP)

= P-1

A-1

P P-1

AP

= I

∴ B is invertible


B ∈ Mn*n( ℜ )，B is invertible

⇔ 1. ∃ C ∈ Mn*n( ℜ )

∋ BC = I

稱 C 為右反矩陣

2. ∃ D ∈ Mn*n(ℜ )

∋ DB = I

稱 D 為左反矩陣

（5）A～B ⇒ det(A) = det(B)



8 應用數學系


∵A～B ∴∃ P ∈ Mn*n( ℜ) ，invertible

∋ B = P-1AP

Hence

det(B) = det(P-1

AP)

= det(P-1

)det(A)der(P)

= (det(P))-1

det(A)det(P)

= det(A)

（6）A～B ⇒ tr(A) = tr(B)

tr = trace ，跡數


∵ A～B ∴ ∃ P ∈ Mn*n( ℜ ) ，invertible

∋ B = P-1AP

tr(B) = tr(P-1

AP)

= tr(P-1

(AP))

= tr((AP) P-1

)

= tr(A)

A =

nnn

n

n

aa

aaa

aaa

..

..

..

...

...

1

22221

11211

tr(A) = a11 + a22 + … + ann

（7）A～B ⇒ 1. Nullity(A) = Nullity(B)

2. Rank(A) = Rank(B)

Hint：Rank(AP) = rank(A)

（1）det(AB) = det(A)det(B) （2）det(A-1) = (det(A))-1



9 應用數學系


2. ∵ A～B ∴ ∃ P ∈ Mn*n( ℜ ) ，invertible

∋ B = P-1AP

Rank(B) = rank(P-1

AP) = rank(AP)

= rank(A)

（8）A～B ⇒ 1. charA(x) = charB(x) ∈ 特徵函數

2. JA = JB ∈ Jordan form

3. λ A = λ B ∈ eigenvalue ∈特徵值

※相似性⇒ 保 det，trace，Nullity，rank，特徵函數，Jordan form，eigenvalue

DefinitionDefinitionDefinitionDefinition：：：：(invariant subspace)(invariant subspace)(invariant subspace)(invariant subspace)


T：V → V is linear

⇒ （1）{0} is a T – invariant subspace

ProvProvProvProveeee：：：：

T(0) = 0 ∈{0}

（2） Ker(T) is a T – invariant subspace

（1）V：a vector space W：a subspace of V

（2）T = V→V is linear 則 W is called a ＂T – invariant subspace＂ ⇔ T(W) ⊆ W

⇔ ∀ w∈W，T(w) ∈W

W W

T(W

T

T

V V



10 應用數學系


∀ u ∈ Ker(T)

⇒T(u) = 0 ∈ Ker(T)

（3）Im(T) is a T – invariant subspace


思考路徑： ∀ w ∈ Im(T) ⊂ V

∴∃ u ∈ V ∋ T(u) = W

Hence

T(w) ∈T(Im(T)) ⊂ T(V) = Im(T)

T：V→W Im(T) = {w ∈ W ￨∃ u ∈ V ∋ w = T(u)} Ker(T) = {u∈V￨T(u) = 0}

∀ W ∈ Im(T)

∴ T(w) ∈ T(Im(T)) ⊂ T(V) = Im(T)

（4）If W is a T – invariant subspace

Then T(W) is a T – invariant subspace


∀ u ∈ T(W)

⇒ T(u) ∈ T(T(W)) ⊂ T(W)

↑ (∵W is a T – inveaiant subspace ⇔ T(W) ⊂ W)

（5）If W1，W2 are two T – invariant subspaces

then W1 ∩ W2，W1 + W2 are also T – invariant subspaces



11 應用數學系


1. ∀ u ∈W1 ∩ W2

⇒ u∈W1 and u∈W2

⇒T(u)∈W1 and T(u)∈W2

⇒T(u) ∈W1 ∩ W2

2. ∀ u∈W1 + W2

u = w1+w2，where w1∈W1，w2∈W2 ⇒T(u) = T(w1+w2) = T(w1) + T(w2) ∈ W1+W2

↑ T is linear


V is a vector space，dim(V) = n T：V→V is linear If W is a T – invariant subspaces of V

and β 1 = {u1,u2,…,uk} be an ordered basis of W

then ∃ an ordered basis

β 2 = { u1,u2,…,uk , u+k1,uk+2,…,un }of V

∋ [ ]2

T β =

2

1

A0

CA , where A1 = [ ]

1

Tβ

w

∵ β 1 is w basis 且 w 為 V 的子空間 ∴由獨立擴張定理得知 ∃ β 2 為 V 的一組基底

T￨w(u1) = T(u1) = a11u1 + a21u2 + a31u3 + … + ak1uk T￨w(u2) = T(u1) = a12u1 + a22u2 + a32u3 + … + ak2uk

.

.

. ⇒ [ ]1

Tβ

w =

T￨w(uk) = T(u1) = a1ku1 + a2ku2 + a3ku3 + … + akkuk = T(uk+1) = a1k+1u1 + a2k+1u2 + a3k+1u3 + … + ank+1uk

.

.

= T(un) = a1nu1 + a2nu2 + a3nu3 + … + annuk

V

T：V→V

W ⊂ V

T￨W：W→V

W

W

uk+1

.

.

u

T(uk+1)

w U1 U2 . . Uk

TW

kkkk

k

k

aaa

aaa

aaa

...

..

..

...

...

21

22221

11211



12 應用數學系

⇒ [ ]2

Tβ

w =

+

+++

+

+

+

nnkn

nkkk

knkkkkkk

nkk

nkk

aa

aa

aaaaa

aaaaa

aaaaa

...0...00

.....

.....

...0...00

......

.

.

......

......

1,

,11,1

1,21

21222221

11111211

det

2

1

0 A

CA = det(A1)det(A2)


V：a vector space ，dim(V) = n If （1）T：V→V is linear

（2）W1，W2 are two T – invariant sudspaces （3）V = W1 ○+ W2 （4） β 1 = {u1、u2、…、uk} is an ordered basis of W1

β 2 = {v1、v2、…、vk} is an ordered basis of W2

⇒ 1. β 1 ∪ β 2 is an ordered basis of V

2. [ ]βT =

2

1

0

0

A

A ，where A1 = [ ] 11T βw ，A2 = [ ] 22T βw

T(u1) = a11u1 + a21u2 + … + ak1uk + 0 + … + 0

.

.

T(uk) = a1ku1 + a2ku2 + … + akkuk + 0 + … + 0

T(uk+1) = 0 + … + 0 + ak+1,k+1uk+1 + … + ank+1un

.

.

T(un) = 0 + … + 0 + ak+1,n+1uk+1 + … + annun

A1

A2

C



13 應用數學系

⇒ [ ]βT =

+

+++

nnkn

nkkk

kkkk

k

k

aa

aa

aaa

aaa

aaa

...0...00

.....

.....

...0...00

0...0...

.

.

0...0...

0...0...

1,

,11,1

21

22221

11211

A2

A1



14 應用數學系

§§§§ 3 3 3 3----2 2 2 2

eigenvalue(eigenvalue(eigenvalue(eigenvalue(固有值固有值固有值固有值、、、、特徵值特徵值特徵值特徵值))))，，，，eigenvector(eigenvector(eigenvector(eigenvector(固有向量固有向量固有向量固有向量、、、、特徵向量特徵向量特徵向量特徵向量))))



（1）u ∈ ker(T) ⇒ u is an eigenvector of T


∵u ∈ ker(T)

⇒ T(u) = 0 = 0*u

Hence ∃ λ = 0 ∋ T(u) = λ u = 0*u

（2）u is an eigenvector of T wuth respect to λ ， λ ∈ F

( ⇔ T(u) = λ u)

⇒ α u is also a eigenvector of T wuth respect to λ ，α ∈ F


T(α u) = α T(u) = α * λ u = λ (α u)

�T is linear

∴α u is also a eigenvector of T w.r.t.，λ

V：a vector

T：V→V is linear （1） λ ∈ F λ is called an eigenvalue of T

⇔ ∀ u∈ V，u ≠ 0 ∋ T(u) = λ u

（2）u∈V u is called an eigenvector of T

⇔ ∀ λ ∈ F ∋ T(u) = λ u



15 應用數學系


If T

y

x=

24

31

y

x，and x1 =

−1

1，x2 =

4

3

Then （1）x1，x2 are eigenvectors of T

（2）Find the eigenvalues of T w.r.t. x1，x2，respectively


（2）Let λ 1， λ 2 be the eigenvalue to T w.r.t. x1，x2，respectively

24

31

−1

1 =

2

2-

∥

Then T(x1) = λ 1x1 ⇒ T

1-

1 = λ 1

−1

1 =

1

1

- λ

λ⇒ λ 1 = -2

T(x2) = λ 2x2 T

4

3 = λ 2

4

3 =

2

2

4

3

λ

λ⇒ λ 2 = 5

∥

24

31

4

3 =

20

51

如何去找 eigenvalue λ ？

λ is an eigenvalue of T

⇔ ∃ u∈V，u ≠ 0， ∋ T(u) = λ u

⇔ ∃ u∈V，u ≠ 0， ∋ (T- λ I)(u) = 0

⇔ T- λ I is not invertible

⇔ det(T - λ I) = 0

F： V→V is invertible ⇔ def(F) ≠ 0

⇔ if F(u) = 0，then u = 0

⇔ ker(F) = {0}

F is not invertible

⇔ ∃u ≠ 0 ∋ F(u) = 0

A∈ Mn*n(ℜ )

⇔ A is invertible

⇔ ker(A) = {0}

⇔ det(A) ≠ 0

A is not invertible

⇔ det(A) = 0



16 應用數學系



Find the eigenvalues of A =

11

14 = ？


charA(x) = det(A-xI) = 0

det(A-xI) = x

x

−14

1-1

= (1-x)2-4

= x2 – 2x -3

∴charA(x) = x2-2x-3 = 0

(x-3)(x+1) = 0

x = -1 or 3

the eigenvalues of T are -1 or 3


Define T： [ ]x2ℜ → [ ]x2ℜ By T(f(x)) = f(x) + xf’(x)

Find the eigenvalue of T ？


β = {1,x,x2} is an standard ordered basis of [ ]x2ℜ T(1) = 1 + x*0 = 1 = 1*1 + 0*x + 0*x

2


charT(x) ≜ det(T- λ I) = 0

�the characteristic equation of T

charT(x) = det(T- λ I)

� the char polynomial of T



17 應用數學系

T(x) = x + x*1 = 2x = 0*1 + 2*x + 0*x2

T(x2) = x

2 + x*2x = 3x

2 = 0*1 + 0*x + 3*x

2

⇒ [ ]βT =

300

020

001

charT(x) = det (T-xI) = det( [ ]βT -xI) = 0

det([ ]βT -xI) = x

x

x

−

−

300

020

00-1

charT(x) = (1-x)(2-x)(3-x) = 0

∴x = 1,2,3


相似保固有值，特徵多項式

If A,B∈ Mn*n(ℜ )，A～B

Then （1） λ A = λ B

（2）charA(x) = charB(x)


∵A～B

∴∃P∈Mn*n(ℜ )，invertible

∋ B = P-1AP

（2）charB(x) = det(B-xI) = det(P-1AP)

= det(P-1

(A-xI)P)

= det(A-xI)

= charA(x)



18 應用數學系

DefinitionDefinitionDefinitionDefinition：：：：(eigenspace) (eigenspace) (eigenspace) (eigenspace) 固有子空間固有子空間固有子空間固有子空間

RRRRemarkemarkemarkemark：：：：

（1）{0} ⊂ V( λ )

0∈V( λ )

（2）V( λ ) ≜ {u∈V￨T(u) = λ u}

= {u∈V￨(T- λ I)(u) = 0}

= ker(T- λ I)

※ V( λ ) = ker(T- λ I)

V( λ ) = ker(A- λ I)

（1）V：an vector space

（2）T：V→V is linear

（3） λ is an eigenvalue of T

Define V( λ ) ≜ {u∈V￨T(u) = λ u}

We called V( λ ) is an eigenspace of T(w.r.t. λ )

= A∈Mn*n( ℜ )

λ is an eigenvalue of A

V( λ ) ≜ {u∈Mn*n( ℜ )￨Au = λ u}



19 應用數學系

LemmaLemmaLemmaLemma：：：：

V( λ ) is a T – invariant subspace of V

ProProProProveveveve：：：：

證：＂∀ u ∈V( λ )⇒T(u) ∈V( λ )＂

For any u∈V( λ )⇒ T(u) = λ u

Hence

T(T(u)) = T( λ u)

= λ T(u)

� ∵ T is linear

LemmaLemmaLemmaLemma：：：：


⇔ V( λ ) ≠ {0}


（方法一）


⇔ ∃ u∈V， ∋ T(u) = λ u

u ≠ 0

∴∃ u ≠ 0，u∈V ∋ u∈V( λ )

⇒V( λ ) ≠ {0}

（方法二）


⇔ ∃ u ≠ 0，u∈V， ∋ T(u) = λ u

⇔ ∃ u ≠ 0，u∈V， ∋ (T- λ I)(u) = 0

⇔ ker(T- λ I) ≠ {0}

⇔ V( λ ) ≠ {0}



20 應用數學系


V( λ ) = {0}

⇔ λ is not an eigenvalue of T


A =

14

11

Find the eigenvalues , eigenvectors are eigenspace？


（1）charA(x) = det(A-xI)

= x

x

−14

1-1

= x2-2x-3

x2-2x-3 = 0

(x-3)(x+1) = 0

x = 3 or -1

∴the eigenvalues of A are 1，-3

（2）

V(-1) = (A-(-1)I)

= (A+I) = {u ∈F2*1￨(A+I)u = 0}

V(-1) = {u ∈F2*1( ℜ )￨A(u) = -u}

� -1 的 eigenspace

= {u =

y

x∈F2*1( ℜ )￨

14

11

y

x = -

y

x}

= {u =

y

x∈F2*1( ℜ )￨

+

+

yx

yx

4 =

−

−

y

x }

= {u =

y

x∈F2*1( ℜ )￨y = -2x}



21 應用數學系

= {u =

− x

x

2￨x∈ ℜ }

= {t

− 2

1￨t∈ ℜ }

-1 的 eigenvector = { t

− 2

1￨t∈ ℜ 且 t ≠ 0}


Find the eigenvalues，eigenvector and eigenspaces of A =

022

1-22

1-13


charA(x) = det(A-xI)

=

x

x

x

−

−−

−

22

122

11-3

= -x3+5x

2-8x+4

= -(x-1)(x-2)2

∴1,2,2 為矩陣 A 的 eigenvalues

V(1) = ker(A-I)

= {u =

z

y

x

∈F3*1( ℜ )￨

022

1-22

1-13

z

y

x

= 1

z

y

x

}

= {

z

y

x

∈F3*1( ℜ )￨

+

−+

−+

yx

zyx

zy

22

22

3x

=

z

y

x

}

= {t

2

0

1

∈F3*1( ℜ )￨t∈ ℜ }



22 應用數學系

⇒相對 eigenvalue 1 的 eigenvectors

= {t

2

0

1

∈F3*1( ℜ )￨t∈ ℜ，t ≠ 0}

V(2) = ker(A-2I)

= {

z

y

x

∈F3*1( ℜ )￨

022

1-22

1-13

z

y

x

= 2

z

y

x

}

= {

z

y

x

∈F3*1( ℜ )￨

+

−+

−+

yx

zyx

zy

22

22

3x

=

z

y

x

2

2

2

}

= {t

2

0

1

￨t∈ ℜ }

⇒相對 eigenvalue 2 的 eigenvectors

= {t

2

1

1

￨t∈ ℜ，t ≠ 0}



23 應用數學系


（1）T：V→V is linear

（2） λ 1 , λ 2 , … , λ k are distinct eigenvalues of T

⇒（a）V( λ 1) , V( λ 2) , … , V( λ k) are independent

（b） If ui∈V( λ i) for all i = 1 , 2 , … , k

then {u1 , … ,un} is linear independent

Prove Prove Prove Prove ：：：：

（a）設 n = 2，V( λ 1)，V( λ 2)：independent

If u1 + u2 = 0，u1∈V( λ 1)，u2∈V( λ 2)

T(u1) = λ 1u1� � T(u2) = λ 2u2

then u1 = u2 = 0

T(u1 + u2) = T(0) = 0

∥

T(u1) + T(u2)

∥

λ 1u1 + λ 2u2

⇒ λ 1u1 + λ 2u2 = 0 – (1)

λ 2(u1 + u2) = λ 2*0 = 0

⇒ λ 2u1 + λ 2u2 = 0 – (2)

(2) - (1)⇒ ( λ 2- λ 1)u1 = 0

⇒ u1 = 0 代入（1）

⇒ λ 2u2 =0

∴u2 = 0

設 n = k-1 時，V( λ 1) , V( λ 2) , … , V( λ k-1) are independent

當 n = k，

if u1 + u2 + … + uk = T(0) = 0，ui∈V( λ i)，i = 1 , 2 , … , k

then T(u1 + u2 + … + uk) = T(0) = 0

⇒T(u1) + T(u2) + … + T(uk) = 0

（1）S1 , S2 , … , Sk

Si ⊂ V，subspace S1 , S2 , … , Sk independent

⇔ if u1 + u2 + … + un = 0，ui ∈Si

then u1 = u2 = … = uk = 0

（2）S is linear independent

⇔ if α1u1 + … +αnun = 0，

u1,…,un ∈S

thenα1 = … =αn = 0



24 應用數學系

⇒ λ 1u1 + … + λ kuk = 0 －（1）

λ k(u1 + … + uk) = λ k*0 = 0

⇒ λ ku1 + … + λ kuk = 0 －（2）

（2）-（1）

∴( λ k- λ 1)u1 + ( λ k- λ 2)u2 + … + ( λ k- λ k-1)uk-1 = 0

�w1 �w2 �wk-1

∵V( λ 1) , V( λ 2) , … , V( λ k-1) are independent

∴( λ k- λ 1)ui = 0，∀ i = 1 , 2 , … , k-1

∵λ k ≠ λ i ，∀ i = 1 , 2 , … , k-1

∴ui = 0，∀ i = 1 , 2 , … , k-1 (代回（1）)

λ 1*0 + λ 2*0 + … + λ k-1*0 + λ kuk = 0

⇒ λ kuk = 0

⇒ uk = 0

（b）Ifα1u1 + … +αnun = 0，where ui ∈ V( λ i)，i = 1 , 2 , … ,k

Then (1) αiui ∈ V( λ i)，∀ i = 1 , 2 , … , k

(2) αiui = 0，∀ i = 1 , 2 , … , k

(∵V( λ 1) , V( λ 2) , … , V( λ k) are independent )

⇒α1 = … =αk = 0

(∵ui ≠ 0，∀ i = 1 , 2 , … , k)



25 應用數學系


（1）若λ 1， λ 2 are independent

則 λ 1 ≠ λ 2

⇔ V( λ 1) ∩V( λ 2) = {0}

（2）1. V1 is an eigenvector of T w.r.t. λ 1

若 2. V2 is an eigenvector of T w.r.t. λ 2

3. V1 + V2 ≠ 0

則 λ 1 ≠ λ 2 ⇔ u1 + u2 is not an eigenvector of T

( λ 1 = λ 2 ⇔ u1 + u2 is an eigenvector of T)

Prove Prove Prove Prove ：：：：（2）

＂⇒＂(反證法)

Suppose u1 + u2 is an eigenvector of T w.r.t λ

Then T(u1 + u2) = λ ( u1 +u2)

∥

T(u1) + T(u2)

∥

λ 1u1 + λ 2u2

⇒ λ 1u1 + λ 2u2 = λ u1 + λ u2

⇒ ( λ 1- λ )u1 + ( λ 2- λ )u2 = 0

�V( λ 1) �V( λ 2)

⇒ ( λ 1- λ )u1 = 0

( λ 2- λ )u2 = 0

⇒ λ 1 = λ

λ 2 = λ ………..(�)

＂⇐＂(反證法)

Suppose λ 1 = λ 2 = λ

Then T(u1 +u2) = T(u1) + T(u2) = λ u1 + λ u2



26 應用數學系

= λ (u1 +u2)

⇒ λ u1 + λ u2 is an eigenvector of T ………. (�)

（3）u is an eigenvector of T w.r.t. λ

⇒αn is an eigenvector of T w.r.t. λ

（4）u1，u2 is an eigenvector of T w.r.t. λ

⇒ u1 +u2 is not also eigenvector of T w.r.t. λ

SolutionSolutionSolutionSolution：：：：((((反例反例反例反例))))

u = u1 is an eigenvector

-u = u2 is an eigenvector

⇒ u + (-u) = 0 ∴不為 eigenvector


Find the eigenvalue of A =

3-4-613-

0501-2

1-11071

0001-2-

00034


charA(x) = det(A-xI)

=

x

x

x

x

−−−−

−−

−−

−−−

34613

05012

711071

00012

0003x-4

=

x

x

x

x

x

−−−

−

−−

−−−346

050

7110

12

3-4

= [(x-4)(x+1) + 6][-(x-10)(x-5)(x+3) + 42(5-x)]

= (x2-3x+2)[-(x-5)(x

2-7x-30+42)]

= -(x-1)(x-2)(x-5)(x-3)(x-4)

A =

2

1 0

AC

A

det(A) = det(A1)det(A2)



27 應用數學系

DefinitionDefinitionDefinitionDefinition：：：：(diagonalizable) (diagonalizable) (diagonalizable) (diagonalizable) 對角化之判別與方法對角化之判別與方法對角化之判別與方法對角化之判別與方法


T：V→V is linear，dim(V) = n，A∈ Mn*n( ℜ )

If A～ [ ]βT ， β is a basis of V

Then ∃ β ` is a basis of V

∋ A = [ ] 'T β


T：V→V is linear

T is diagonalizable

⇔ ∀ β ：a basis of V

（1） T：V→V is linear

T is diagonalizanle

⇔ ∃ β is a basis of V

∋ [ ]βT is a diagonal matrix

（2）A∈Mn*n( ℜ )

A is diagonalizable

⇔ ∃ P∈ Mn*n( ℜ ) is invertible

�不唯一

∋ P-1AP is a diagonal matrix

� = D



28 應用數學系

[ ]βT ：diagonalizable


＂⇒＂

T is diagonalizable

⇔ ∃ β ` is a basis of V

∋ [ ] 'T β is a diagonalizable matrix

Hence for any basis β of V

（1）若 β = β `，則 OK

（2）若 β ≠β `，則 [ ] 'T β = P-1 [ ]βT P，其中 P = [ ]'

VIβ

β

⇒ [ ]βT is diagonalizable


是對矩陣 A =

14

11作對角化？

（1）charA(x) = det(A-zI)

= x

x

−14

1-1

= x2-2x-3

（2）令 charA(x) = 0

則 x2-2x-3 = 0

x = -1，3

（3）V(-1) = {t

2-

1￨ t ∈R}

V(3) = {t

2

1￨ t ∈R}

（4）P =

22-

11



29 應用數學系

（5）P-1AP = D

-1

22-

11

14

11

22-

11 =

30

01-


Find A =

−

−

411

121

221

（1）the eigenvalues，eigenvectors，eigenspaces

（2）對 A 作對角化，並求其一轉換矩陣


（1）

1. charA(x) = det(A-xI)

=

x

x

x

−

−−

411

121

22-1

= - x3 + 7x

2 – 15x + 9

let charA(x) = 0

⇒ x = 1,3,3

∴eigenvalue = 1,3,3

2. V(1) = {

z

y

x

∈ M3*1( ℜ ) ￨

−

−

411

121

221

z

y

x

= 1*

z

y

x

}



30 應用數學系

= {

z

y

x

∈ M3*1( ℜ ) ￨

++−

−+

++

zyx

zyx

zyx

4

2

22

=

z

y

x

}

= { t

−

1

1

2

￨ t∈ ℜ } the eigenspace of T w.r.t. 1

V(3) = {

z

y

x

∈ M3*1( ℜ ) ￨

−

−

411

121

221

z

y

x

= 3

z

y

x

}

= {

z

y

x

∈ M3*1( ℜ ) ￨

++−

−+

++

zyx

zyx

zyx

4

2

22

=

z

y

x

3

3

3

}

= { t1

0

1

1

+ t2

1

0

1

￨t1 , t2 ∈ ℜ }the eigenspace of T w.r.t. 3

3. V(1) = { t

−

1

1

2

￨ t ≠ 0 t∈ ℜ } the eigenspace of T w.r.t. 1

V(3) = { t1

0

1

1

+ t2

1

0

1

￨t1 , t2 不全為 0∈ ℜ }the eigenspace of T w.r.t. 3

（2）P =

101

011-

112

，則 P-1AP = D =

300

030

001



31 應用數學系

DefinitionDefinitionDefinitionDefinition：：：：((((代數相重數代數相重數代數相重數代數相重數，，，，幾何相重數幾何相重數幾何相重數幾何相重數))))


A =

022

1-22

1-13

，找代數相重數，幾何相重數？

（1） the eigenvalues of A：1,2,2

⇒m(1) = 1

m(2) = 2

（2）V(1) = { t

2

0

1

￨ t∈ ℜ }

V(2) = { t

2

1

1

￨ t∈ ℜ }

⇒ gm(1) = dimV(1) = 1

gm(2) = dimV(2) = 1

（1）T：V�V is linear，dim(V) < ∞


⇒ (a) charT(x) = 0 中的重根數

稱為λ 的代數相重數，記作( λ )

(b) dimV( λ ) = dimker(T- λ I)

稱為λ 的幾何相重數，記作 gm( λ )



32 應用數學系

此矩陣無法對角化

∵ 2 = gm(2) ≠ m(2) = 1


（1）T：V�V is linear，dim(V) = n


⇒ 1 ≤ gm( λ ) ≤ m( λ ) ≤ n

ProveProveProveProve：：：：※※※※幾何相重幾何相重幾何相重幾何相重數數數數≤代數相重數代數相重數代數相重數代數相重數

（1）1 ≤ gm( λ )

gm( λ ) = dimV( λ )


⇔ ∃ u ≠ 0，u∈V， ∋ T(u) = λ u

⇔ ∃ u ≠ 0，u∈V， ∋ (T- λ I)(u) = 0

⇔ ker(T- λ I) ≠ {0} �∃ u ≠ 0，u∈V， ∋ u∈ker(T- λ I)

⇔ V( λ ) ≠ {0}

⇔ dimV( λ ) ≥ 1

（2）dim(V) = n

⇒代表 charA(x)的最高次方為 n

∴m(λ ) ≤ dim(V) = n

（3）gm( λ ) ≤ m( λ )

∥

k

(a) k = gm( λ ) = dimV(λ )

(b) V( λ ) is T-invariant subspace of V

ProveProveProveProve：：：：(b)(b)(b)(b)

∀ u∈V( λ )

⇒T(u) = λ u



33 應用數學系

⇒T(T(u)) = T( λ u) = λ T(u)

⇒T(u) ∈V( λ )

choose β 1 = {u1 , u2 , … , un} is a basis of V( λ )

and β = {u1 , u2 , … , un} is a basis of V

then [ ]βT =

C0

BA =

−

−++

xC

xC

nn

kk

.C"

C'.

.

.

1,1

λ

λ

A = [T￨V( λ )]1β =

k*k

.

.

λ

λ

charT(X) = det( [ ]βT -XI)

=

−

−

−

++

xC

xC

x

x

nn

kk

.C"

C'.

.

.

-

1,1

λ

λ

= ( λ -x)kg(x)

= (-1)k(x- λ )kg(x)

m( λ ) ≥ k

0

B

B

0

0



34 應用數學系

TheoremTheoremTheoremTheorem：：：：((((可對角化的充要條件可對角化的充要條件可對角化的充要條件可對角化的充要條件))))

若 T：V�V is linear，dim(V) = n

則下列敘述是等價

(a) T is diagonalizable

⇔ (b) 1. chart(x)可在 [ ]XF 上完全分解成一次因式乘積(允許重根)

2. gm( λ i) = m( λ i)，∀ λ i：eigenvalues of T

⇔ (C) 若λ 1 , λ 2 , … , λ n 為 T 的所有 eigenvalues

則 V = V( λ 1)○+ V( λ 2) ○+ … ○+ V( λ k)


(a) ⇒ (b)

Suppose λ 1 , λ 2 , … , λ n are all disjoint eigenvalues of T

then V( λ 1) , V( λ 2) , … , V(λ k) are all eigenspaces of T

Let β 1 = {u11 , u12 , … , u1m1} is a basis of V( λ 1)

β 2 = {u21 , u22 , … , u2m2} is a basis of V(λ 2)

.

.

β 1 = {uk1 , uk2 , … , ukmk} is a basis of V(λ k)

and， β = β 1 ∪ β 2 ∪ … ∪ β k is a basis of V

such that



35 應用數學系

[ ]βT =

k

k

λ

λ

λ

λ

λ

λ

.

.

.

.

.

.

.

.

2

2

1

1

⇒○1 m1 + m2 + … + mk = n

○2 charT(x) = det([ ]βT -xI) =

−

−

−

−

−

x

x

x

x

x

x

k

k

λ

λ

λ

λ

λ

λ

.

.

.

.

.

.

.

.

-

2

2

1

1

= ( λ 1-x) 1m ( λ 2-x) 2

m … (λ k-x) km

0

0



36 應用數學系


T： 33 ℜ→ℜ by

T

++

−−

=

z

zyx

zy

z

y

x

33

32

找一個基底 ß，使得 [T] ß 為 diagonal matrix 並求此對角矩陣

想法:

（1）利用 standard basis ß 1 of 3ℜ ⇒ 求出 [T] 1ß

（2）對[T]1ß做對角化


ß 1 =

1

0

0

,

0

1

0

,

0

0

1



37 應用數學系

T

⋅+

⋅+

⋅=

=

1

0

0

0

0

1

0

1

0

0

1

0

0

1

0

0

0

1

T

⋅+

⋅+

⋅−=

−

=

1

0

0

0

0

1

0

3

0

0

1

)2(

0

3

2

0

1

0

T

⋅+

⋅+

⋅−=

−

=

1

0

0

1

0

1

0

3

0

0

1

)3(

1

3

3

1

0

0

A= [T]1ß=

−−

100

331

320

另解 [T]1ß

T

++

−−

=

z

zyx

zy

z

y

x

33

32

=

⋅

−−

z

y

x

100

331

320

\\

[T]1ß

（1）

Char A (x) = det (A-XI)



38 應用數學系

=

x

x

x

−

−

−−−

100

331

32

= -(x-1) 2 (x-2)

∴eigenvalues of T : 1，1，2

（2）

V(1) =

⋅=

−−

z

y

x

z

y

x

z

y

x

1

100

331

320

=

ℜ∈

−

+

−

2121 ,

1

0

3

0

1

2

tttt

V(2) =

ℜ∈

− tt

0

1

1

1

⇒ A 可以做對角化，且其對角方矩 D =

200

010

001

P =

−

−−

010

101

132

: 可逆 ∋ P 1− AP = D

改成

⇒取 ß = ∋

−

−

−

0

1

1

,

1

0

3

,

0

1

2

[T]1ß =

200

010

001


A ∈ M 33× (C)



39 應用數學系

A =

−

101

010

111

（1） Find the eigenvalues

（2） Find an invertible matrix P∈ M 33× (C) ∋ P 1− AP = D : a diagonal matrix


（1）

Char A (x) = det(A-XI)

=

x

x

x

−

−

−−

101

010

111

= -x 3 + 3x 2 - 4x + 2

= -(x +1)(x 2 - 2x + 2)

= -(x - 1)(x- 1 - i)(x- 1 + i)

∴ eigenvalues of T: 1，1+i，1-i

（2）

V(1) =

ℜ∈

tt

1

1

0

V(1+i) =

ℜ∈

t

i

t

1

0

V(1-i) =

ℜ∈

t

i

t 0

1



40 應用數學系

⇒ P =

i

i

11

001

10

，D =

−

+

i

i

100

010

001

DefinitionDefinitionDefinitionDefinition：：：：((((同步對角化同步對角化同步對角化同步對角化))))

Remark: Remark: Remark: Remark:

A， B∈ M nn× (F)

A，B 可對角化

⇒AB≠ BA

Lemma:Lemma:Lemma:Lemma:

A， B∈ M nn× (F)

A，B 同步對角化

⇒AB = BA


Q A，B:同步對角化

P∃ ∈ M nn× (F) ， invertible

∋ P 1− AP = D 1 ，P1− BP = D 2 ，where D 1 、D 2 : diagonal matrix

⇒ A = P D 1 P1− ，B = P D 2 P

1−

⇒ AB = (P D 1 P1− )( P D 2 P

1− ) = P D 1 D 2 P1− = P D 2 D 1 P

1−

（1）T，G : V →V are linear

T and G are simultaneously diagonalizable

ß∃⇔ : a baisi of V

∋ [T] ß ，[G] ß : diagonal matrices

（2）A，B∈M 33× (F)

A，B : simultaneously diagonalizable

P∃⇔ ∈ M 33× (F)，可逆

∋ P 1− AP = D 1

P 1− BP = D 2 diagonal matrices



41 應用數學系

= ( P D 2 P1− )(P D 1 P

1− ) = BA

Lemma:Lemma:Lemma:Lemma:

A，B ∈ Mn*n( F )

（1）A，B：diagonalizable

（2）AB = BA

⇒A，B：可同步對角化

chapter 3 diagonalizableming/diagonalizable.pdf · 教學卓越 線性代數...

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chapter 3 diagonalizableming/diagonalizable.pdf · 教學卓越線性代數...