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大學線性代數初步

大學數學

前言

數學大學線性代數 , 大數 .

大學線性代數學 ,學線性代數 . 大學線性代數. 學大大 ( ) 線性

代數 , . 大學線性代數 , 大學數學 . 大大數學

, 學 , 線性代數. , 線性代數

數學 . , ., .

學 , 數學學 , 線性代數 . 線性代數 .

, . 學. , (Question).

, 大. , 線性代數 . ,

學線性代數學線性代數 , .

, ,代. , .

, . , 性, . , .

v

Chapter 9

Eigenvalues andEigenvectors

, n×n matrices eigenvalue eigenvector.eigenvalue eigenvector 性 . n×n

matrix .

9.1. Characteristic Polynomial

n× n matrix A, v ∈ Rn, 數學 Akv k 數. Fibonacci sequence F1,F2, . . . . Fk+1 = Fk +Fk−1 數 .

A =[

1 11 0

]k ≥ 2 vk =

[Fk

Fk−1

],

Avk =[

1 11 0

][Fk

Fk−1

]=

[Fk +Fk−1

Fk

]=

[Fk+1Fk

]= vk+1.

v3 = Av2, v4 = Av3 = A(Av2) = A2v2, . . . , vk+1 = Ak−1v2.k ≥ 2, Ak−1v2, Fk+1 .

k 大 , Akv . , λ ∈ RAv = λv , A2v = A(Av) = A(λv) = λ (Av) = λ 2v. A3v = λ 3v, . . . ,

Akv = λ kv. , Akv. vλ ∈ R Av = λv . .

Definition 9.1.1. A n× n matrix. v ∈ Rn, λ ∈ RAv = λv, v A eigenvector, λ A eigenvalue.

, A eigenvector . v A eigenvector,λ ,λ ′ ∈ R Av = λv = λ ′v, (λ −λ ′)v = O v ̸= O, λ = λ ′. A

eigenvector v 數 λ Av = λv. eigenvector veigenvalue λ .

189

190 9. Eigenvalues and Eigenvectors

Question 9.1. A n× n matrix, v ∈ Rn Av = O. veigenvector? eigenvalue ?

n× n matrix eigenvector eigenvalue ?eigenvalue, eigenvector. λ ∈ R A

eigenvalue, v ∈Rn Av = λv. Inv = v,λv = (λ In)v. Av = λv (A−λ In)v = O. 言 , λ A eigenvalue

n×n matrix A−λ In linear system (A−λ In)x = O nontrivial solution x = v.Theorem 3.5.9, A−λ In invertible, Theorem 8.2.6(1)det(A−λ In) = 0. 言 , A eigenvalue λ λ det(A−λ In) = 0.

λ det(A− λ In) = 0 ? A = [ai j], t 數,det(A− tIn).

A− tIn =

a11 − t a12 · · · a1n

a21 a22 − t · · · a2n...

... . . ....

an1 an2 · · · ann − t

數學 , det(A− tIn) t 數 n 數 .

t = λ 數 , λ det(A−λ In) = 0, λ Aeigenvalue. , λ A eigenvalue, t = λ det(A− tIn)

. det(A− tIn) A eigenvalue,.

Definition 9.1.2. A n× n matrix, t 數 pA(t) = det(A− tIn).pA(t) A characteristic polynomial ( )..

t = λ characteristic polynomial pA(t)λ A eigenvalue. A Rn eigenvectors,

t = τ pA(t) ( 數 ), v ∈ Rn Av = τv.Av ∈ Rn, τv ̸∈ Rn, Av = τv . , eigenvalue

characteristic polynomial .

Example 9.1.3. A =

2 0 11 3 11 1 2

, A characteristic polynomial pA(t) =det(A− tI3) = det

2− t 0 11 3− t 11 1 2− t

. rowpA(t) = (2− t)det

[3− t 1

1 2− t

]+det

[1 3− t1 1

]= (2− t)(t2 −5t +6−1)+(1− (3− t)).

pA(t) = (2− t)(t2 −5t +4) = (2− t)(t −1)(t −4). t = 1,2,4 A charac-teristic polynomial , A eigenvalues 1,2,4.

9.1. Characteristic Polynomial 191

A n×n matrix , characteristic polynomial det(A− tIn)t . determinant , . 數學

columnrow. 2×2 matrix

[a bc d

]characteristic polynomial det

[a− t b

c d − t

]t (a− t)(d − t) bc

數 , t2 t 數 (a− t)(d − t) t2 t 數

at2 − (a+ d)t . 3× 3 matrix A =

a b cd e fg h i

characteristic polynomial.row

det

a− t b cd e− t fg h i− t

= (a− t)det[ e− t fh i− t

]−bdet

[d fg i− t

]+ cdet

[d e− tg h

].

前 2× 2 det[

e− t fh i− t

]t2 t 數

(e− t)(i− t) t2 t 數 , (a− t)(e− t)(i− t) t3 t2 數.

det[

d fg i− t

]det[

d e− tg h

]t , det(A− tI3) t3 t2

數 (a− t)(e− t)(i− t) . A chacteristic polynomial pA(t)3 (−1)3t3 +(−1)2(a+ e+ i). a,e, i A diagonalentries, a+ e+ i A trace, tr(A) . 數學 ,

A = [ai j] n× n matrix , A characteristic polynomial pA(t) = det(A− tIn)t n 數 , (a11 − t)(a22 − t) · · ·(ann − t)(−1)ntn +(−1)n−1(a11 + · · ·+ann)tn−1. A diagonal entries a11 + · · ·+ann

tr(A), .

Proposition 9.1.4. A n×n matrix. A characteristic polynomial t n數 . tn 數 (−1)n, tn−1 數 (−1)n−1tr(A) 數數 det(A).

Proof. pA(t) = det(A− tIn), 前 pA(t) 數 . pA(t)數 pA(0) = det(A−0In) = det(A). �

Question 9.2. A n×n matrix. A eigenvalues?

eigenvalue . λ ∈R A eigenvalue. t = λA characteristic polynomial pA(t) = det(A− tIn) . t −λ

pA(t). (t −λ )m pA(t), (t −λ )m+1 pA(t), eigenvalue λalgebraic multiplicity (代數數) m. t = λ pA(t) , λ

algebraic multiplicity 1.

Question 9.3. Identity matrix In eigenvalue ? algebraic multiplicity ?

characteristic polynomial 性 . n × nmatrices characteristic polynomial . ,


characteristic polynomial . A,B n× n matrices, n× n invertiblematrix U , B = U−1AU , A,B similar (

). A B characteristic polynomial .

Proposition 9.1.5. A,B n×n matrices n×n invertible matrix UB =U−1AU . A B characteristic polynomial.

Proof. B characteristic polynomial det(B− tIn) = det(U−1AU − tIn).性

U−1(A− tIn)U =U−1AU −U−1(tIn)U =U−1AU − tU−1InU =U−1AU − tIn.

determinant 性 (Theorem 8.2.6)

det(B− tIn) = det(U−1(A− tIn)U) = det(U−1)det(A− tIn)det(U) = det(A− tIn).

A B characteristic polynomial. �

characteristic polynomial A AT characteris-tic polynomial.

Proposition 9.1.6. A n×n matrix, A AT characteristic polynomial

Proof. trnaspose 性 (A− tIn)T = AT − tITn = AT − tIn (Proposition 3.2.4),Theorem 8.2.6 (3),

PAT(t) = det(AT − tIn) = det((A− tIn)T) = det(A− tIn) = PA(t).

�

Question 9.4. A AT eigenvalues eigenvalue A AT

algebraic multiplicity .

9.2. Eigenspace

n× n matrix eigenvalue ,eigenvalue eigenvectors.

A n× n matrix λ ∈ R A eigenvalue. det(A−λ In) = 0,(A−λ In)x = O nontrivial solution. v ∈ Rn

x = v (A−λ In)x = O . v (A−λ In)v = O, Av = λv.v A λ eigenvalue eigenvector. , v A λ eigenvalue

eigenvector, x = v (A−λ In)x = O nontrivial solution.n×n matrix A−λ In nullspace ( {v ∈ Rn | (A−λ In)v = O}) A

λ eigenvector. nullspace vector space, .

Definition 9.2.1. A n×n matrix λ ∈ R A eigenvalue. A−λ Innullspace A eigenvalue λ eigenspace. EA(λ ) .

9.2. Eigenspace 193

λ eigenspace λ eigenvalue eigenvectors .O eigenvector, vector space O. λ eigenspace

λ eigenvalue eigenvectors O . vector space? vector space 性, vector space dimension

大 . EA(λ ) dimension eigenvalue λ geometric multiplicity( 數). eigenvalue λ algebraic multiplicity λeigenvectors , λ geometric multiplicity .

Example 9.2.2. A =

−1 4 2−1 3 1−1 2 2

, B = 0 3 1−1 3 1

0 1 1

. A Bcharacteristic polynomial −(t −1)2(t −2). 1 2 A B eigenvalues.

A B, eigenvalue 1 algebraic multiplicity 2, eigenvalue 2 algebraicmultiplicity 1. A B eigenspace.

A eigenvalue 1 eigenspace, A − I3 =

−2 4 2−1 2 1−1 2 1

null space. elementary row operations, echelon form

1 −2 −10 0 00 0 0

.EA(1) = Span(

210

,10

1

). A eigenvalue 1 eigenvector21

0

10

1

linear combination nonzero vector. v =41

2

=21

0

+210

1

Av =

−1 4 2−1 3 1−1 2 2

412

=41

2

= v.

dim(EA(1)) = 2, A eigenvalue 1 geometric multiplicity 2.

A eigenvalue 2 eigenspace, A− 2I3 =

−3 4 2−1 1 1−1 2 0

null space.elementary row operations, echelon form

1 −2 00 1 −10 0 0

. EA(2) =Span(

211

). A eigenvalue 2 eigenvector21

1

nonzero vector, A eigenvalue 2 geometric multiplicity 1.


B eigenvalue 1 eigenspace, B− I3 =

−1 3 1−1 2 10 1 0

null space.elementary row operations, echelon form

1 0 −10 1 00 0 0

. EB(1) = Span(10

1

).B eigenvalue 1 eigenvector

101

nonzero vector,B eigenvalue 1 geometric multiplicity 1. B eigenvalue

2 eigenspace, B− 2I3 =

−2 3 1−1 1 10 1 −1

null space. elementary rowoperations, echelon form

1 0 −20 1 −10 0 0

. EA(2) = Span(21

1

). Aeigenvalue 2 eigenvector

211

nonzero vector, Aeigenvalue 2 geometric multiplicity 1.

Example 9.2.2 characteristic polynomial,eigenvectors 大 , eigenspace dimension .

eigenvalues algebraic multiplicity , geometricmultiplicity .

Question 9.5. Identity matrix In eigenvalue geometric multiplicity ?

Proposition 9.1.6 A AT characteristic polynomialeigenvalue eigenvalue A AT algebraic multiplicity .

geometric multiplicity , .

Proposition 9.2.3. A n×n matrix λ ∈ R A eigenvalue. λA geometric multiplicity λ AT geometric multiplicity .

Proof. dim(EA(λ ))= dim(EAT(λ )), dim(N(A−λ In))= dim(N(AT−λ In)).Theorem 4.4.13 dim(N(A−λ In)) = null(A−λ In) = n− rank(A−λ In), A∈Mn×n

dim(N(AT−λ In)) = n− rank(AT−λ In). AT−λ In = (A−λ In)T rank((A−λ In)T) =rank(A−λ In) (Proposition 4.4.14), dim(N(A−λ In)) = dim(N(AT −λ In)). �

v A eigenvector, eigenvalue λ , v nonzero vectoreigenvalue λ eigenvector. v,w A eigenvectors

eigenvalue , v,w . v,w linearly independent..

9.3. Diagonalization 195

Proposition 9.2.4. A n×n matrix v1, . . . ,vk A eigenvectors. v1, . . . ,vkeigenvalues , v1, . . . ,vk linearly independent.

Proof. 數學 . 前 k = 2 , k − 1eigenvectors . k eigenvectors . v1, . . . ,vk A

eigenvectors eigenvalue λ1, . . . ,λk ( Avi = λivi, for i = 1, . . . ,n).v1, . . . ,vk−1 linearly independent. , v1, . . . ,vk−1,vk linearly

dependent. Lemma 4.2.4, vk ∈ Span(v1, . . . ,vk−1). c1, . . . ,ck−1 ∈ R

vk = c1v1 + · · ·+ ck−1vk−1 (9.1)

eigenvector

λkvk = Avk = A(c1v1 + · · ·+ ck−1vk−1) = c1Av1 + · · ·+ ck−1Avk−1 = c1λ1v1 + · · ·ck−1λk−1vk−1.(9.2)

(9.1) λk (9.2)

c1(λk −λ1)v1 + · · ·+ ck−1(λk −λk−1)vk−1 = O. (9.3)

vk ̸= O, c1, . . . ,ck−1 0. eigenvalue , i =1, . . . ,k−1, λk −λi ̸= 0. c1(λk −λ1), . . . ,ck−1(λk −λk−1) 0 數.

, (9.3) v1, . . . ,vk−1 linearly dependent, ,. �

9.3. Diagonalization

A n×n matrix, v A eigenvector eigenvalueλ , Akv = λ kv. v A eigenvector ?

, Akv . , v1, . . . ,v ∈ Rn Rn

basis A eigenvectors , Akv.v ∈ Rn, basis c1, . . . ,cn ∈ R v = c1v1 + · · ·+ cnvn. v1, . . . ,vn

eigenvalues λ1, . . . ,λn,

Av = A(c1v1 + · · ·+ cnvn) = c1Av1 + · · ·+ cnAvn = c1λ1v1 + · · ·+ cnλnvn.

A

A2v = A(Av) = A(c1λ1v1 + · · ·+ cnλnvn) = c1λ1Av1 + · · ·+ cnλnAvn = c1λ 21 v1 + · · ·+ cnλ 2n vn.

數學

Akv = c1λ k1 v1 + · · ·+ cnλ kn vn.

matrix .

Definition 9.3.1. A n×n matrix. Rn basis v1, . . . ,vn viA eigenvectors, A diagonalizable ( ).


diagonalizable ? v1, . . . ,v∈Rn Rn basis Aeigenvectors, eigenvalues λ1, . . . ,λn. Av1 = λ1v1, . . . ,Avn =λnvn.

A

∣∣ ∣∣ ∣∣

v1 v2 · · · vn∣∣ ∣∣ ∣∣=

∣∣ ∣∣ ∣∣

Av1 Av2 · · · Avn∣∣ ∣∣ ∣∣=

∣∣ ∣∣ ∣∣

λ1v1 λ2v2 · · · λnvn∣∣ ∣∣ ∣∣ .

(i, i)-th entry λi n×n diagonal matrix ( 線 i λi線 0),

∣∣ ∣∣ ∣∣

v1 v2 · · · vn∣∣ ∣∣ ∣∣D =

∣∣ ∣∣ ∣∣

v1 v2 · · · vn∣∣ ∣∣ ∣∣

λ1 0 · · · 00 λ2 · · · 0...

... . . ....

0 0 0 λn

=∣∣ ∣∣ ∣∣

λ1v1 λ2v2 · · · λ3vn∣∣ ∣∣ ∣∣ .

C =

∣∣ ∣∣ ∣∣

v1 v2 · · · vn∣∣ ∣∣ ∣∣, AC =CD. C column linearly

independent n column, C rank n, C n×n matrix Cinvertible ( Theorem 3.5.2). AC =CD D =C−1AC. ,

n× n invertible matrix C−1AC diagonal matrix D, C n× n invertiblematrix, C n column vectors Rn basis. AC =CD,

性 C i-th column A D (i, i)-th entry eigenvalue eigenvector.C column vectors Rn basis A eigenvectors, A

diagonalizable. 前 , U−1AU ( U n× n invertible matrix)matrix A similar matrix. , A diagonalizable

A diagonal matrix similar. diagonalizable .

n×n matrix A diagonalizable ? ,eigenvectors. A eigenvectors

A diagonalizable. eigenvectors eigenvalues,A characteristic polynomial R . λ1, . . . ,λk ∈R

pA(t) = (−1)n(t −λ1)m1 · · ·(t −λk)mk . i = 1, . . . ,k, mi λialgebraic multiplicity pA(t) 數 n, m1+ · · ·+mk = n.

eigenvalue, geometric multiplicity algebraic multiplicity.A eigenvectors , eigenvalue geometric multiplicity

algebraic multiplicity. i = 1, . . . ,k, λi geometric multiplicityalgebraic multiplicity, dim(EA(λi)) = mi. vi,1, . . . ,vi,mi EA(λi)

basis. k vectors , v1,1, . . . ,v1,m1 , . . . ,vk,1, . . . ,vk,mklinearly independent. Rn m1 + · · ·+mk = n ,

Corollary 4.3.5, Rn basis. A eigenvectors,A diagonalizable.

9.3. Diagonalization 197

v1,1, . . . ,v1,m1 , . . . ,vk,1, . . . ,vk,mk linearly dependent. 0 數c1,1, . . . ,c1,m1 , . . . , ck,1, . . . ,ck,mk

c1,1v1,1 + · · ·+ c1,m1v1,m1 + · · ·+ ck,1vk,1 + · · ·+ ck,mk vk,mk = O.

i ∈ {1, . . . ,k}, wi = ci,1vi,1 + · · ·+ ci,mivi,mi . vi,1, . . . ,vi,milinearly independent, ci,1, . . . ,ci,mi 0, wi ̸= O. wi ∈ EA(λi),

wi eigenvalue λi eigenvector. , ci, j ̸= 0,i, wi eigenvalue λi eigenvectors w1 + · · ·+wk = O. Proposition

9.2.4 , eigenvalue eigenvectors linearly independent ,v1,1, . . . ,v1,m1 , . . . ,vk,1, . . . ,vk,mk linearly independent. A

characteristic polynomial R A eigenvalue geometricmultiplicity algebraic multiplicity, A diagonalizable.

, A diagonalizable, A characteristic polynomialR A eigenvalue geometric multiplicity algebraic

multiplicity. 前前 eigenvalue geometricmultiplicity algebraic multiplicity.

Proposition 9.3.2. A n×n matrix. λ A eigenvalue geometricmultiplicity d algebraic multiplicity m, d ≤ m.

Proof. dim(EA(λ )) = d, v1, . . . ,vd EA(λ ) basis. v1, . . . ,vdlinearly independent, Rn basis v1, . . . ,vd ,vd+1, . . . ,vn. C

i-th column vi n× n invertible matrix. AC = CEE =

[λ Id M1O M2

]. 1-st column , 數學 det(E − tIn) =

(λ − t)d det(M2 − tIn−d). 言 , E characteristic polynomial (t −λ )d .A E similar ( E =C−1AC), characteristic polynomial (

Proposition 9.1.5), (t −λ )d pA(t). λ algebraic multiplicity m,m t −λ pA(t) 數, d ≤ m. �

n×n matrix A diagonalizable. v1,1, . . . ,v1,d1 , . . . ,vk,1, . . . ,vk,dk Rn

basis, i ∈ {1, . . . ,k}, vi,1, . . . ,vi,di A λi eigenvalue eigenvector,λ1, . . . ,λk . vi,1, . . . ,vi,di ∈ EA(λi) linearly independent, λi

geometric multiplicity dim(EA(λi))≥ di. λi algebraic multiplicity mi,Proposition 9.3.2

mi ≥ dim(EA(λi))≥ di,∀ i = 1, . . . ,k. (9.4)

m1 + · · ·+mk A characteristic polynomial pA(t) 數 ( ),pA(t) 數 n. m1 + · · ·+mk Rn dimension, n. (9.4)

i = 1, . . . ,k

n ≥ m1 + · · ·+mk ≥ dim(EA(λi))+ · · ·+dim(EA(λk))≥ d1 + · · ·+dk = n.


“≥” “=” ( n > n ).n = m1 + · · ·+mk ( pA(t) 數 ) mi = dim(EA(λi)),∀ i = 1, . . . ,k( eigenvalue geometric multiplicity algebraic multiplicity).

.

Theorem 9.3.3. A n×n matrix. .

(1) Rn basis A eigenvectors .

(2) n×n invertible matrix C C−1AC diagonal matrix.

(3) A characteristic polynomial 數 A eigenvaluegeometric multiplicity algebraic multiplicity.

diagonalizable matrix , Theorem 9.3.3diagonalizable , (3) .

Question 9.6. A n× n matrix. A diagonalizable AT

diagonalizable.

Example 9.3.4. Example 9.2.2 A=

−1 4 2−1 3 1−1 2 2

, B= 0 3 1−1 3 1

0 1 1

.前 characteristic polynomial −(t −1)2(t −2). A,B eigenvalue1 algebraic multiplicity 2, eigenvalue 2 algebraic multiplicity 1.Example 9.2.2 B eigenvalue 1 geometric multiplicity 1, Bdiagonalizable matrix. , A eigenvalue 1 eigenvalue 2 geometric multiplicity

algebraic multiplicity, A diagonalizable matrix. A .

A eigenvalue 1 2 eigenspace EA(1) = Span(

210

,10

1

)EA(2) = Span(

211

),21

0

,10

1

,21

1

A eigenvectors R3basis. C =

2 1 21 0 10 1 1

D = 1 0 00 1 0

0 0 2

,AC =

−1 4 2−1 3 1−1 2 2

2 1 21 0 10 1 1

= 2 1 41 0 2

0 1 2

= 2 1 21 0 1

0 1 1

1 0 00 1 00 0 2

=CD.C invertible, C−1AC = D.

, A eigenvalue λ geometric multiplicity 大 0 (λ eigenvector ) algebraic multiplicity (Proposition 9.3.2).

λ A characteristic polynomial ( λ algebraic multiplicity 1),geometric multiplicity algebraic multiplicity ( 1).

9.4. The Spectral Theorem 199

diagonalizable , algebraic multiplicity 1 eigenvaluegeometric multiplicity . , A characteristic polynomial R

( ), A diagonalizable.diagonalizable, symmetric matrix. symmetric

matrix diagonalizable.

9.4. The Spectral Theorem

symmetric matrix. symmetric matrix diago-nalizable, orthogonal diagonalizable. 數學

, , symmetricmatrix .

2×2 symmetric matrix . A =[

a bb c

], b ̸= 0 (

b = 0, A diagonal matrix ). A characteristic polynomialPA(t) = t2− (a+c)t +(ac−b2). pA(t) (a+c)2−4(ac−b2) = (a−c)2+4b2 > 0,

PA(t) = 0 λ1,λ2. λ1,λ2 A eigenvealue, A

diagonalizable. v1 =[

bλ1 −a

],

Av1 =[

a bb c

][b

λ1 −a

]=

[λ1b

b2 +λ1c−ac

]= λ1

[b

λ1 −a

]= λ1v1.

λ 21 − (a+c)λ1+(ac−b2) = 0. b ̸= 0, v1 ̸= O, v1 A

eigenvector eigenvalue λ1. v2 =[

bλ2 −a

], v2 A eigenvector

eigenvalue λ2. , ⟨v1,v2⟩= b2+λ1λ2−a(λ1+λ2)+a2. 數 ,λ1λ2 = ac−b2 λ1 +λ2 = a+ c, ⟨v1,v2⟩= 0. v1,v2 R2 basis

A eigenvectors , . diagonalizableorthogonal diagonalizable. .

Definition 9.4.1. A ∈ Mn×n, Rn orthogonal basis v1, . . . ,vnvi A eigenvectors, A orthogonal diagonalizable.

, Definition 9.4.1 ui = 1∥vi∥vi u1, . . . ,un Rn orthonormal

basis A eigenvectors. A orthogonal diagonalizable Rn

orthonormal basis A eigenvector . ui eigenvalue λi

Q =

∣∣ ∣∣ ∣∣

u1 u2 · · · un∣∣ ∣∣ ∣∣ AQ = QD D (i, i)-th entry λi diagonal matrix,A Q−1AQ = D. eigenvectors basis, u1, . . . ,un orthonormal basis ?

u1, . . . ,un Rn orthonormal basis , QTQ = In, inverse matrix性, QT = Q−1. Q column vectors Rn orthonormal basis


, Q−1 = QT. 性, n×n matrix column vectorsRn orthonormal basis , orthogonal matrix (

orthonormal matrix). A QTAQ = D, A orthogonaldiagonalizable.

Question 9.7. Q ∈ Mn×n, Q−1 = QT Q orthogonal matrix?

, Q n×n orthogonal matrix D =

λ1 . . . n×n diagonal

matrix QTAQ = D. AQ = QD, Q i-th column A eigenvalue λieigenvector, Q column vectors Rn orthonormal basis,

.

Proposition 9.4.2. A ∈Mn×n. A orthogonal diagonalizable n×northogonal matrix Q QTAQ diagonal matrix.

Proposition 9.4.2, A orthogonal diagonalizable Q,D ∈ Mn×nQ orthogonal matrix, D diagonal matrix A = QDQT. AT = (QDQT)T =

(QT)TDTQT. (QT)T =Q DT =D ( D diagonal matrix), AT =QDQT =A,A symmetric. .

Corollary 9.4.3. A ∈ Mn×n orthogonal diagonalizable, A symmetric matrix.

Spectral Theorem Corollary 9.4.3 .A symmetric , A orthogonal diagonalizable. symmetric

matrix .

Lemma 9.4.4. A ∈Mn×n symmetric, v,w ∈Rn ⟨Av,w⟩= ⟨v,Aw⟩.

Proof. , , v,w ∈ Rn ⟨v,w⟩ = vTw( v,w n×1 matrix).

⟨Av,w⟩= (Av)Tw = (vTAT)w = vT(ATw) = ⟨v,ATw⟩.

AT = A ⟨Av,w⟩= ⟨v,Aw⟩. �

n× n matrix diagonalizable characteristicpolynomial 數 . symmetric matrixcharacteristic polynomial 數 .

Lemma 9.4.5. A ∈ Mn×n symmetric, A characteristic polynomial pA(t).


Proof. λ = a+ bı ( ı 數 ı2 = −1) pA(t) , a,b ∈ Rb ̸= 0. 數 , entry 數 .

數 . a+bı pA(t) , A− (a+bı)In0. A− (a+bı)In A− (a−bı)In

(A− (a+bı)In)(A− (a−bı)In) = A2 −2aA+(a2 +b2)In.

a,b ∈ R A 數 , A2 −2aA+(a2 +b2)In 數 .det(A− (a+bı)In) = 0,

det(A2 −2aA+(a2 +b2)In) = det(A− (a+bı)In)det(A− (a−bı)In) = 0.

A2 −2aA+(a2 +b2)In singular, v ∈ Rn v ̸= 0

(A2 −2aA+(a2 +b2)In)v = A2v−2aAv+(a2 +b2)v = O.

⟨A2v−2aAv+(a2 +b2)v,v⟩= ⟨A2v,v⟩−2a⟨Av,v⟩+a2⟨v,v⟩+b2⟨v,v⟩.

A symmetric, Lemma 9.4.4 ⟨A2v,v⟩= ⟨A(Av),v⟩= ⟨Av,Av⟩,

⟨Av−av,Av−av⟩+b2⟨v,v⟩= ⟨A2v,v⟩−2a⟨Av,v⟩+a2⟨v,v⟩+b2⟨v,v⟩,

∥Av−av∥2 +b2∥v∥2 = ⟨A2v−2aAv+(a2 +b2)v,v⟩= ⟨O,v⟩= 0.

∥Av−av∥ ≥ 0, ∥v∥ > 0, b = 0. 初 b ̸= 0 , pA(t) = 0, . �

symmetric matrix characteristic polynomial ,symmetric matrix orthogonal diagonalizable. 數學 ,

2× 2 symmetric matrix orthogonal diagonalizable. (n− 1)×(n−1) symmetric matrix orthogonal diagonalizable. A n×nsymmetric matrix orthogonal diagonalizable. Lemma 9.4.5 數 λ

A eigenvalue. u1 A λ eigenvector ∥u1∥= 1. Gram-Schmidtprocess, u1 Rn orthonormal basis u1, . . . ,un. orthogonal

matrix Q=

∣∣ ∣∣ ∣∣

u1 u2 · · · un∣∣ ∣∣ ∣∣, j = 1, . . . ,n Au j = c1 ju1+ · · ·+cn jun,

AQ = QC, C = [ci j]. Q orthogonal matrix, C = Q−1AQ = QTAQ.A symmetric CT = QTAQ = C, C symmetric.

Au1 = λu1, C 1-st column

λ0...0

, C symmetric C 1-st row


[λ 0 · · · 0]. C

C =

λ 0 · · · 00... B0

.C symmetric, B (n− 1)× (n− 1) symmetric matrix. ,

B orthogonal diagonalizable, w1, . . . ,wn−1 Rn−1 orthonomal basis

B eigenvectors. R =

∣∣ ∣∣ ∣∣

w1 w2 · · · wn−1∣∣ ∣∣ ∣∣, R (n− 1)× (n− 1)

orthogonal matrix (n − 1)× (n − 1) digonal matrix D RTBR = D.

P =

1 0 · · · 00... R0

. ,

PTCP =

λ 0 · · · 00... RTBR0

=

λ 0 · · · 00... D0

.PTCP diagonal matrix, (QP)TA(QP)=PT(QTAQ)P=PTCP diagonal

matrix. Q,P orthogonal matrix, (QP)T(QP) = PT(QTQ)P = PTP = In,QP orthogonal matrix. Proposition 9.4.2, A orthogonal diagonalizable,

Spectral Theorem.

Theorem 9.4.6 (Spectral Theorem). A n×n symmetric matrix, A orthogonaldiagonalizable.

, n×n symmetric matrix A, orthogonal matrix QQTAQ diagonal matrix. , Theorem 9.4.6 , 數學步步 Q . Gram-Schmidt process, .

Proposition, 步 .

Proposition 9.4.7. A n×n symmetric matrix. v,w ∈ Rn A eigenvectorseigenvalue 數, ⟨v,w⟩= 0.

Proof. v,w eigenvalue λ ,λ ′. Av = λv,Aw = λ ′w.⟨Av,w⟩ = ⟨λv,w⟩ = λ ⟨v,w⟩. ⟨v,Aw⟩ = λ ′⟨v,w⟩. Lemma 9.4.4⟨Av,w⟩= ⟨v,Aw⟩, (λ −λ ′)⟨v,w⟩= 0. λ ̸= λ ′ ⟨v,w⟩= 0. �

A n×n symmetric matrix, A eigenvectorsRn orthonormal basis. A eigenvalues λ1, . . . ,λk,

eigenspace EA(λ1), . . . ,EA(λk). EA(λi) basis ,


A diagonalizable Rn basis. Proposition 9.4.7 , λi ̸= λ j, EA(λi) EA(λ j) . EA(λi) basis

. Gram-Schmidt process A eigenspaceEA(λi) orthonormal basis. eigenspace basis

, A eigenvectors Rn orthonormal basis..

Example 9.4.8. (1) symmetric matrix A =

0 1 11 1 01 0 1

. A characteristicpolynomial pA(t) =−(t +1)(t −1)(t −2). A eigenvalues, −1,1,2.

A , Proposition 9.4.7 eigenvector .−1,1,2 eigenvector

v1 =

−211

, v2 = 0−1

1

, v3 =11

1

.. i = 1,2,3 ui = 1∥vi∥vi,

u1 =1√6

−211

, u2 = 1√2

0−11

, u3 = 1√3

111

R3 orthonromal basis. A −

2√6

1√6

1√6

0 − 1√2

1√2

1√3

1√3

1√3

0 1 11 1 0

1 0 1

−

2√6

0 1√3

1√6

− 1√2

1√3

1√6

1√2

1√3

= −1 0 00 1 0

0 0 2

.

(2) symmetric matrix B =

5 −4 −2−4 5 −2−2 −2 8

. B characteristic poly-nomial pB(t) = −t(t − 9)2. B eigenvalues 0,9. B ,

dim(EB(0)) = 1, dim(EB(9)) = 9. v1 =

221

EB(0) = N(B) basis, v2 =−110

,v3 =−10

2

EB(9) basis. Proposition 9.4.7 ⟨v1,v2⟩= ⟨v1,v3⟩= 0,. ⟨v2,v3⟩= 1 ̸= 0, Gram-Schmidt process

v2,v3 EB(9) orthogonal basis. w2 = v2

w3 = v3 −Projw2v3 =

−102

− 12

−110

= 12

−1−14

.


u1 = 1∥v1∥v1, u2 =1

∥w2∥w2, u3 =1

∥w3∥w3

u1 =1√3

221

, u2 = 1√2

−110

, u3 = 13√

2

−1−14

R3 orthonromal basis. B

23

23

13

− 1√2

1√2

0− 1

3√

2− 1

3√

24

3√

2

5 −4 −2−4 5 −2

−2 −2 8

23 −

1√2

− 13√

223

1√2

− 13√

213 0

43√

2

= 0 0 00 9 0

0 0 9

.9.5. Application: Conics and Quadric Surfaces

symmetric matrix orthogonal diagonalizable 性線 ,

.

線 ., quadratic form . n

數 quadratic formn

∑i, j=1

ai jxix j

. x2 + 3xy− y2, 3x2 + y2 − z2 + 5xy+ xz+ 3yz 數

數 quadratic form. x =

x1...xn

, n 數 quadratic formxTAx , A n× n symmetric matrix. 數 quadratic form

ax21 +bx1x2 + cx22

ax21 +bx1x2 + cx22 =

[x1 x2

][ a b/2b/2 c

][x1x2

].

數 quadratic form ax21 +bx22 + cx23 + rx1x2 + sx1x3 + tx2x3

ax21 +bx22 + cx

23 + rx1x2 + sx1x3 + tx2x3 =

[x1 x2 x3

] a r/2 s/2r/2 b t/2s/2 t/2 c

x1x2x3

.quadratic form A symmetric, orthogonal

matrix Q QTAQ diagonal matrix

λ1 . . .λn

. 數 x =x1...

xn

t =

t1...tn

t = QTx ( QT = Q−1, x = Q t),

xTAx = (Q t)TA(Q t) = tT(QTAQ) t =[

t1 · · · tn] λ1 . . .

λn

t1...

tn

= λ1t21 + · · ·+λnt2n .

9.5. Application: Conics and Quadric Surfaces 205

, 數 quadratic form ..

Example 9.5.1. quadratic form x21 +4x1x2 −2x22.

x21 +4x1x2 −2x22 =[

x1 x2][ 1 2

2 −2

][x1x2

].[

1 22 −2

]symmetric matrix, orthogonal diagonalizable,[

2/√

5 1/√

5−1/

√5 2/

√5

][1 22 −2

][2/

√5 −1/

√5

1/√

5 2/√

5

]=

[2 00 −3

].[

t1t2

]=

[2/

√5 1/

√5

−1/√

5 2/√

5

][x1x2

][

x1 x2][ 1 2

2 −2

][x1x2

]=[

t1 t2][ 2 0

0 −3

][t1t2

]= 2t21 −3t22 .

quadratic form x22 + x23 +2x1x2 +2x1x3,

x22 + x23 +2x1x2 +2x1x3 =

[x1 x2 x3

] 0 1 11 1 01 0 1

x1x2x3

.Example 9.4.8 QT

0 1 11 1 01 0 1

Q = −1 0 00 1 0

0 0 2

Q orthogonalmatrix

−2√6

0 1√3

1√6

− 1√2

1√3

1√6

1√2

1√3

.t1t2

t3

= −

2√6

1√6

1√6

0 − 1√2

1√2

1√3

1√3

1√3

x1x2

x3

[

x1 x2 x3] 0 1 11 1 0

1 0 1

x1x2x3

= [ t1 t2 t3 ] −1 0 00 1 0

0 0 2

t1t2t3

=−t21 + t22 +2t23 .線 . 線 ax2 +bxy+

cy2 +dx+ ey+ f = 0. ,[x y

][ a b/2b/2 c

][xy

]+[

d e][x

y

]+ f = 0. (9.5)

symmetric matrix A =[

a b/2b/2 c

]QTAQ =

[λ1 00 λ2

].

數

[xy

]= QT

[xy

](

[xy

]= Q

[xy

]), (9.5)

[x y

][ λ1 00 λ2

][xy

]+[

d e]

Q[

xy

]+ f = 0.

λ1x2 +λ2y2 +d′x+ e′y+ f = 0, (9.6)[d′ e′

]=[

d e]

Q.


λ1,λ2 0 , (9.6)

λ1(x−h)2 +λ2(y− k)2 = f ′.

.

(A) λ1,λ2 :

(1) f ′ λ1,λ2 : ellipse ( ). λ1 = λ2 ,.

(2) f ′ = 0: (x,y) = (h,k) .

(3) f ′ λ1,λ2 : .

(B) λ1,λ2 :

(1) f ′ ̸= 0: hyperbola ( 線).

(2) f ′ = 0: 線.

Example 9.5.2. 線 2xy+√

2x+√

2y = 1.[x y

][ 0 11 0

][xy

]+[ √

2√

2][x

y

]= 1.

[1/

√2 1/

√2

−1/√

2 1/√

2

][0 11 0

][1/

√2 −1/

√2

1/√

2 1/√

2

]=

[1 00 −1

]數

[xy

]=

[1/

√2 −1/

√2

1/√

2 1/√

2

][xy

],

[x y

][ 1 00 −1

][xy

]+[ √

2√

2][ 1/√2 −1/√2

1/√

2 1/√

2

][xy

]= 1.

線數 x2 − y2 +2x = 1. (x+1)2 − y2 = 2,線.

2xy+√

2x+√

2y =−1, 數 (x+1)2 − y2 = 0線 x+ y+1 = 0 x− y+1 = 0.

λ1,λ2 0. λ1,λ2 0,. 性, λ1 ̸= 0,λ2 = 0 . (9.6)

λ1(x−h)2 + e′y = f ′.

.

(C) λ1,λ2 0 ( 性 λ1 ̸= 0,λ2 = 0):

(1) e′ ̸= 0: parabola ( 線).

(2) e′ = 0 λ1, f ′ : 線 ( 線 x = 0 ).

(3) e′ = 0 f ′ = 0: 線 x = h.

(4) e′ = 0 λ1, f ′ : .

9.5. Application: Conics and Quadric Surfaces 207

Example 9.5.3. 線 x2 −2xy+ y2 +4√

2x = 4.[x y

][ 1 −1−1 1

][xy

]+[

4√

2 0][x

y

]= 4.

[1/

√2 −1/

√2

1/√

2 1/√

2

][1 −1−1 1

][1/

√2 1/

√2

−1/√

2 1/√

2

]=

[2 00 0

]數

[xy

]=

[1/

√2 1/

√2

−1/√

2 1/√

2

][xy

],

[x y

][ 2 00 0

][xy

]+[

4√

2 0][ 1/√2 1/√2

−1/√

2 1/√

2

][xy

]= 4.

線數 2x2 +4x+4y = 4. 2(x+1)2 +4y = 6,線.

言 , 線 quadratic form eigenvalue λ1,λ2,λ1,λ2 線線. λ1,λ2 , ; λ1,λ2

, 線 ; λ1,λ2 0, 線 .數 , 線 degenerated ( ) ( 線,

).

Question 9.8. 線 quadratic form[

x y]

A[

xy

], A

2×2 symmetric matrix. det(A) 線 , 線線 ( )?

.

[x y z

]A

xyz

+ [ c d e ]xy

z

+ f = 0,A 3×3 symmetric matrix. A 數

λ1x2 +λ2y2 +λ3z2 + c′x+d′y+d′z+ f = 0. (9.7)

, 大 . 性,學 . , 學 .

λ1,λ2,λ3 0 , (9.7)

λ1(x−h)2 +λ2(y− k)2 +λ3(z− l)2 = f ′.

.

(A) λ1,λ2,λ3 :

(1) f ′ λ1,λ2,λ3 : , x = h, y = k z = l. , ellipsoid.

λ1 = λ2 = λ3 , ellipsoid .


(2) f ′ = 0: (x,y,z) = (h,k, l) .

(3) f ′ λ1,λ2,λ3 : .

(B) λ1,λ2,λ3 ( 性 λ1,λ2 ):

(1) f ′ λ1,λ2 : z = l , x = h, y = k線. , hyperboloid of one sheet.

(2) f ′ λ1,λ2 : z = l ,. x = h, y = k 線.

, hyperboloid of two sheets.

(3) f ′ = 0: z = l ,. x = h, y = k 線. ,

elliptic cone.

λ1,λ2,λ3 0. λ1,λ2,λ3 0,. 性, λ1 ̸= 0. .

(C) λ2,λ3 0 ( 性 λ2 ̸= 0): (9.6)

λ1(x−h)2 +λ2(y− k)2 + e′z = f ′.

(1) e′ ̸= 0 λ1,λ2 : e′z = f ′ ,. x = h, y = k

線. elliptic paraboloid.

(2) e′ ̸= 0 λ1,λ2 : e′z = f ′ 線,線. x = h, y = k

線. elliptic paraboloid. (x,y,z) = (h,k, f ′/e′)saddle point ( ).

(3) e′ = 0 λ1,λ2, f ′ : z = s ., elliptic cylinder.

(4) e′ = 0 λ1,λ2 f ′ ̸= 0: z = s線. , hyperbolic cylinder.

(5) e′ = 0 λ1,λ2 f ′ : .

(6) e′ = 0 λ1,λ2 f ′ = 0: z = s .線.

(7) e′ = 0 λ1,λ2 f ′ = 0: z = s 線..

(D) λ2,λ3 0: (9.6)

λ1(x−h)2 +d′y+ e′z = f ′.

9.6. Application: Markov Processes 209

(1) d′,e′ 0: r =√

(d′)2 +(e′)2 數xyz

= 1 0 00 d′/r −e′/r

0 e′/r d′/r

t1t2t3

λ1(t1 −h)2 + rt2 = f ′.

t3 = s 線. ,parabolic cylinder.

(2) d′ = e′ = 0 f ′ λ1 : ( x = 0 ).

(3) d′ = e′ = 0 f ′ = 0: x = h.

(4) d′ = e′ = 0 f ′ λ1 : .

Example 9.5.4. 5x2 +5x2 +8z2 −8xy−4xz−4yz+2x+2y+ z = 9.

[x y z

] 5 −4 −2−4 5 −2−2 −2 8

xyz

+ [ 2 2 1 ]xy

z

= 9.

−1√2

1√2

0−1

3√

2−1

3√

24

3√

223

23

13

5 −4 −2−4 5 −2

−2 −2 8

−1√2

−13√

223

1√2

−13√

223

0 43√

213

= 9 0 00 9 0

0 0 0

( Example 9.4.8 (2)). 數

xyz

=

−1√2

−13√

223

1√2

−13√

223

0 43√

213

xy

z

,[

x y z] 9 0 00 9 0

0 0 0

xyz

+ [ 2 2 1 ]

−1√2

−13√

223

1√2

−13√

223

0 43√

213

xy

z

= 9.數 9x2 +9y2 +3z = 9, 前 (C)(1) ,

elliptic paraboloid.

Question 9.9. 5x2 +5x2 +8z2 −8xy−4xz−4yz+2x+2y+ z = 0?

9.6. Application: Markov Processes

A ∈Mn×n diagonalizable , k ∈N Ak.v ∈ Rn, Akv. ( diagonalizable), k 大

“ ” Akv 大 . .


A ∈ Mn×n diagonalizable . diagonal matrix D =λ1 . . .λn

invertible matrix P ∈ Mn×n D = P−1AP, A = PDP−1.

A2 = (PDP−1)(PDP−1) = PD(P−1P)DP−1 = PD2P−1 = P

λ21

. . .λ 2n

P−1,數學

Ak = P

λk1

. . .λ kn

P−1.Example 9.6.1. Fibonacci sequence 0,1,1,2,3,5,8,13, . . . ,

. Fibonacci sequence Fk+1 = Fk +Fk−1 數 , F0 = 0,F1 = 1.

A =[

1 11 0

]k ≥ 1 vk =

[Fk

Fk−1

],

Avk =[

1 11 0

][Fk

Fk−1

]=

[Fk +Fk−1

Fk

]=

[Fk+1Fk

]= vk+1.

vk+1 = Akv1. k ≥ 1, Akv1, Fk+1. A characteristic polynomial PA(t) = t2 − t − 1, A eigenvalues

λ1 = (1−√

5)/2, λ2 = (1+√

5)/2. A 2×2 matrix, A eigenvalues

A diagonalizable. A λ1,λ2 eigenvector v1 =[

λ11

], λ2 =

[λ21

].

P =[

λ1 λ21 1

],

P =

[1−

√5

21+

√5

21 1

], P−1 =

1√5

[−1 1+

√5

21 −1+

√5

2

].

A A = P[

λ1 00 λ2

]P−1, k ∈ N,

Ak = P[

λ k1 00 λ k2

]P−1 =

1√5

[λ1 λ21 1

][λ k1 00 λ k2

][−1 λ21 −λ1

]=

15

[λ k+12 −λ

k+11 λ

k2 −λ k1

λ k2 −λ k1 λk−12 −λ

k−11

].

v1 =[

F1F2

]=

[10

],

vk+1 =[

Fk+1Fk

]= Akv1 = Ak

[10

]=

1√5

[λ k+12 −λ

k+11

λ k2 −λ k1

],

Fk+1 =15(λ k+12 −λ

k+11 ) =

15

((1+

√5

2)k+1 − (1−

√5

2)k+1

).


, A diagonalizable, Akv.Markov Peocesses, . 線性代數

, , .

Definition 9.6.2. Rn vector v =

c1...cn

, c1 + · · ·+ cn = 1 i =1, . . . ,n, ci ≥ 0, v probability vector. A ∈ Mn×n column vector

probability vector, A stochastic matrix. stochastic matrix Ar ∈ N Ar entry 數, A regular.

Example 9.6.3. A =[

1/2 11/2 0

]I2 =

[1 00 1

]stochastic matrix. A regular,

A2 =[

3/4 1/21/4 1/2

], entry . I2 regular, r ∈ N

In2 = I2 ( 線, entry 0).

stochastic matrix 性 .

Lemma 9.6.4. A ∈ Mn×n stochastic matrix v ∈ Rn probability vector. Avprobability vector. A entry 數, Av entry數.

Proof. A =

∣∣ ∣∣

a1 · · · an∣∣ ∣∣, v =

c1...cn

, Av = c1a1 + · · ·+ cnan. Av entriesc1a1 + · · ·+ cnan entries . ciai entries

. ai probability vector, ciai entries ci,c1a1 + · · ·+ cnan entries c1 + · · ·+ cn = 1. c1, . . . ,cn v1, . . . ,vn

entry 數, c1a1 + · · ·+ cnan entry 數. Avprobability vector.

A entry 數, a1, . . . ,an entry 數,c1, . . . ,cn 數, Av = c1a1 + · · ·+ cnan entry 大 ciaientry. c1, . . . ,cn 0, ci > 0, ciai entry 數, Av

entry 數. �

A =

∣∣ ∣∣

a1 · · · an∣∣ ∣∣ stochastic matrix, A2 i-th column

Aai, Lemma 9.6.4 , A2 column probability vector, A2

stochastic matrix. k ≥ 2, Ak i-th column Ak−1ai, 數學Lemma 9.6.4, Ak stochastic matrix. 數學 Lemma

9.6.4, Ar entry 數, k ∈N, Ar+k = Ar+k−1Aentry 數. ( ).


Proposition 9.6.5. A∈Mn×n stochastic matrix, k ∈N, Ak stochasticmatrix. A regular Ar entry 數, k ∈ N, Ar+k

entry 數.

stochastic matrix eigenvalues eigenvectors. 前, stochastic matrix ,

characteristic polynomial . , .

Lemma 9.6.6. A ∈ Mn×n stochastic matrix. 1 AT eigenvalue

v =

1...1

eigenvector. A entry 數, AT, eigenvalue 1geometric multiplicity 1.

Proof. A stochastic matrix, A column vector ai probability vector,

⟨ai,v⟩ = 1. ATv =

⟨a1,v⟩...⟨an,v⟩

=1...

1

= v. v AT eigenvectoreigenvalue 1.

A entry 數 w =

c1...cn

eigenvalue 1 eigenvector.w ̸= O, 性, c1, . . . ,cn 大 0 ( 大 0,

ci ≤ 0, −w, AT eigenvalue 1 eigenvector −wentry 大數). c j c1, . . . ,cn 大 . Aw j-th entry,

⟨a j,w⟩ = a j 1c1 + · · ·+a j ici + · · ·+a j ncn. a j 1, . . . ,a j n 數 c j > 0c1, . . . ,cn 大 ,

a j 1c1+ · · ·+a j ici+ · · ·+a j ncn ≤ a j 1c j + · · ·+a j ic j + · · ·+a j nc j = (a j 1+ · · ·+a j n)c j = c j. (9.8)

ATw = w, AT j-th entry c j, (9.8), c1 = · · ·= c j = · · ·= cn = r. w = rv, AT

eigenvalue 1 eigenvector Span(v) . AT eigenvalue 1 geometricmultiplicity 1 �

Proposition 9.1.6 Proposition 9.2.3 A AT eigenvalueseigenvalue A AT geometric multiplicity . .

Proposition 9.6.7. A ∈ Mn×n stochastic matrix. 1 A eigenvalue.A regular, A, eigenvalue 1 geometric multiplicity 1.

Proof. A stochastic matrix, Lemma 9.6.6 1 AT eigenvalue.Proposition 9.1.6 1 A eigenvalue. , A regular r ∈ N

Ar entry 數, Lemma 9.6.6 (Ar)T eigenvalue 1 geometric


multiplicity 1. Proposition 9.2.3 Ar eigenvalue 1, geometric multiplicity1, dim(EAr(1)) = 1. v ∈ EA(1), Av = v, Arv = v,

v ∈ EAr(1). EA(1)⊆ EAr(1), dim(EA(1))≤ dim(EAr(1)) = 1. 前 1 Aeigenvalue, dim(EA(1))> 0. dim(EA(1)) = 1, 1 A geometric

multiplicity 1. �

stochastic matrix A ∈ Mn×n eigenvector 性.

x ∈Rn x =

x1...xn

Ax =y1...

yn

, i = 1, . . . ,n yi = ∑nj=1 ai jx ji, j ∈ {1, . . . ,n} ai j ≥ 0,

|y1|= |a11x1 +a12x2 + · · ·+a1nxn| ≤ a11 |x1|+a12 |x2|+ · · ·+a1n |xn|...

|yi| = |a i1x1 +a i2x2 + · · ·+a inxn| ≤ ai1 |x1|+ai2 |x2|+ · · ·+ain |xn| (9.9)...

|yn|= |an1x1 +an2x2 + · · ·+annxn| ≤ an1 |x1|+an2 |x2|+ · · ·+ann |xn|

(9.9)∣∣x j∣∣ , A stochastic matrix,

j = 1, . . . ,n, a1 j + · · ·+ai j + · · ·+an j = 1,

|y1|+ |y2|+ · · ·+ |yn| ≤ |x1|+ |x2|+ · · ·+ |xn| . (9.10)

x A eigenvalue λ eigenvector, Ax = λx, yi = λxi,∀ i = 1, . . . ,n. 代 (9.10) |λ |(|x1|+ |x2|+ · · ·+ |xn|) ≤ |x1|+ |x2|+ · · ·+ |xn|.x1, . . . ,xn 0 |λ | ≤ 1. A stochastic matrix, eigenvalue λ

|λ | ≤ 1.

λ ̸= 1 . i = 1, . . . ,n ∑nj=1 ai jx j = yi = λxi.n xi , A stochastic matrixx1 + · · ·+ xn = λ (x1 + · · ·+ xn). λ ̸= 1 x1 + · · ·+ xn = 0.

A regular . eigenvalue −1 .w ∈Rn A eigenvalue −1 eigenvector. Aw =−w, Akw = (−1)kw.

k > 0 數 w Ak eigenvalue 1 eigenvector. Aeigenvalue 1 eigenvector v, A regular k > 0 數Ak entry 數 (Proposition 9.6.5), Proposition 9.6.7

EA(1) = EAk(1) = Span(v). w ∈ EAk(1) = Span(v), v,w linearly independent( v,w A 1,−1 eigenvector, Proposition 9.2.4 linearlyindependent) . A regular stochastic matrix −1 Aeigenvalue.


A regular eigenvalue 1 . Proposition 9.6.7, Ar entry 數, eigenspace EA(1) = EAr(1).

A stochastic matrix A = (ai j) entry ai j 數 . xA eigenvalue 1 eigenvector. Ax = x, yi = xi, ∀ i = 1, . . . ,n,|y1|+ |y2|+ · · ·+ |yn|= |x1|+ |x2|+ · · ·+ |xn|. (9.10) .(9.10) (9.9) . i, j ∈ {1, . . . ,n}

ai j > 0 (9.9) x1, . . . ,xn 大 00. x1, . . . ,xn 0, x1 + · · ·+ xn ̸= 0, v = 1x1+···+xn x, v

probability vector A eigenvalue 1 eigenvector. dim(EA(1)) = 1,v Rn vector. , .

Proposition 9.6.8. A ∈ Mn×n stochastic matrix, A eigenvalue λ

|λ | ≤ 1. w =

c1...cn

∈ Rn A eigenvector eigenvalue 1,c1 + · · ·+ cn = 0.

A regular, −1 A eigenvalue Rn

probability vector A eigenvalue 1 eigenvector.

Question 9.10. A regular stochastic matrix v A eigenvalue 1eigenvector. v entry ( entry 0).

A ∈ Mn×n regular stochastic matrix diagonalizable. v1, . . . ,vnRn basis A eigenvectors. vi eigenvalue λi, λ1 = 1.Proposition 9.6.8 v1 probability vector, i = 2, . . . ,n, |λi|< 1 vi =c1...

cn

c1 + · · ·+ cn = 0. P =∣∣ ∣∣

v1 · · · vn∣∣ ∣∣, A = P

1

λ2. . .

λn

P−1.entry ,

.

limk→∞

Ak = limk→∞

P

1

λ2. . .

λn

k

P−1 = limk→∞

P

1

λ k2. . .

λ kn

P−1 =P

10

. . .0

P−1.v1 entry 1, i = 2, . . . ,n, vi entry 0,

P−1P = In, P−1 1-st row entry 1.

P

1

0. . .

0

P−1 =∣∣ ∣∣ ∣∣

v1 v2 · · · vn∣∣ ∣∣ ∣∣

1 · · · 1— O —

...— O —

=∣∣ ∣∣ ∣∣

v1 v1 · · · v1∣∣ ∣∣ ∣∣ .


A regular stochastic matrix diagonalizable, k 大 Ak

column v1 , v1 Rn probability vectorAv1 = v1. regular stochastic matrix ( diagonalizable )

. 線性代數 , 數 .性, . 學 , 大

(Corollary 9.6.10).

Theorem 9.6.9. A∈Mn×n regular stochastic matrix v∈Rn probabilityvector Av = v.

limk→∞

Ak =

∣∣ ∣∣ ∣∣v v · · · v∣∣ ∣∣ ∣∣

.Proof. n = 1 A = [1] stochastic matrix, ,

n ≥ 2 . limk→∞ Ak column vector.

w ∈ Rn limk→∞ Ak =

∣∣ ∣∣ ∣∣w w · · · w∣∣ ∣∣ ∣∣

. , Av = v,Akv = v, ∀k ∈N, limk→∞(Akv) = v. v =

c1...cn

, v probability vector,c1 + · · ·+ cn = 1. limk→∞ Ak =

∣∣ ∣∣ ∣∣w w · · · w∣∣ ∣∣ ∣∣

.

limk→∞

(Akv) = ( limk→∞

Ak)v =

∣∣ ∣∣ ∣∣w w · · · w∣∣ ∣∣ ∣∣

c1c2...

cn

= c1w+ c2w+ · · ·+ cnw = w.w = v. limk→∞ Ak column vector ? i ∈ {1, . . . ,n},

Ak i-th row. Mk, mk Ak i-th row entry 大, limk→∞(Mk −mk) = 0, limk→∞ Ak i-th row entry數. i = 1, . . . ,n , limk→∞ Ak column

vector.

ai j A (i, j)-th entry k > 1 a(k)i j Ak (i, j)-th entry.Ak+1 = AkA, a(k+1)i j = ∑nq=1 a

(k)iq aq j. Ak i-th row 大

Ak (i,s)-th entry (i, t)-th entry ( Mk = a(k)i s , mk = a(k)i t ),

Ak i-th row 大 Ak+1 (i, j)-th entry (i, l)-th entry (Mk+1 = a

(k+1)i j , mk+1 = a

(k+1)i l ).

Mk+1 = a(k+1)i j = a

(k)i t at j +∑

q̸=ta(k)iq aq j ≤ mkat j +Mk ∑

q̸=taq j = mkat j +Mk(1−at j). (9.11)

mk+1 = a(k+1)i l = a

(k)i s as l + ∑

q ̸=sa(k)iq aql ≥ Mkas l +mk ∑

q̸=saql = Mkas l +mk(1−as l). (9.12)


(9.11), (9.12) A stochastic matrix , ∑nq=1 aq j = ∑nq=1 aql = 1.(9.11) (9.12)

Mk+1 −mk+1 ≤ Mk(1−at j −as l)+mk(at j − (1−as l)) = (Mk −mk)(1−at j −as l). (9.13)

at j ≥ 0,as l ≥ 0, 1−at j −as l ≤ 1, (9.13) Mk+1 −mk+1 ≤ Mk −mk.數 (Mk −mk)∞k=1 數 .

(Mk −mk)∞k=1 數 0 ?subsequence 0 . A regular . r ∈NAr entry 數. , B = Ar bi j B (i, j)-th entry

k > 1 b(k)i j Bk = Ark (i, j)-th entry. bi j > 0 b(k)i j > 0.

i ∈ {1, . . . ,n}, Mrk, mrk Bk = Ark i-th row entry 大. , (9.11), (9.12) Bk Bk+1 , (9.13)Bk Bk+1

Mr(k+1)−mr(k+1) ≤ Mrk(1−bt j −bs l)+mrk(bt j − (1−bs l)) = (Mrk −mrk)(1−bt j −bs l). (9.14)

(9.14) (9.13) 大 bt j > 0 bs l > 0. β Bentry . β , bt j ≥ β ,bs l ≥ β , 1−bt j −bs l ≤ 1−2β .

(9.13)

Mr(k+1)−mr(k+1) ≤ (Mrk −mrk)(1−2β ). (9.15)

β column entry 1 大 nβ , 0 < β ≤ 1n ≤12 (

n ≥ 2 ). 0 ≤ 1−2β < 1, (9.15) 數學

Mr(k+1)−mr(k+1) ≤ (Mr −mr)(1−2β )k,

(Mrk −mrk)∞k=1 subsequence limk→∞(Mrk −mrk) = 0, sequence(Mk −mk)∞k=1 limk→∞(Mk −mk) = 0. . �

Theorem 9.6.9 stochastic matrix , regular. Markov process 初 probability

vector , regular stochastic matrix ,. .

Corollary 9.6.10. A ∈ Mn×n regular stochastic matrix v ∈ Rn proba-bility vector Av = v. Rn probability vector w

limk→∞

(Akw) = v.

9.7. 217

Proof. w =

c1...cn

. w probability vector , c1 + · · ·+ cn = 1.Theorem 9.6.9

limk→∞

(Akw) = ( limk→∞

Ak)w =

∣∣ ∣∣ ∣∣v v · · · v∣∣ ∣∣ ∣∣

c1c2...

cn

= c1v+ c2v+ · · ·+ cnv = v.�

Question 9.11. Corollary 9.6.10 w probability vector ?

9.7.

linear operator eigenvector linear operator. basis eigenvectors , linear operator ordered

basis matrix representation diagonal matrix.linear operator ( ). square

matrix , . eigenvalue algebraic multiplicitygeometric multiplicity, square matrix diagonalizable.

, symmetric matrix.symmetric matrix diagonalizable, orthogonal diagonalizable.symmetric matrix eigenvectors orthogonal basis.Spectral Theorem, 數學 . ,

, ( ) ,.

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