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Lecture 18 Eigenvalue Problems II Shang-Hua Teng

Lecture 18 Eigenvalue Problems II

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Lecture 18 Eigenvalue Problems II. Shang-Hua Teng. Diagonalizing A Matrix. Suppose the n by n matrix A has n linearly independent eigenvectors x 1 , x 2 ,…, x n . Eigenvector matrix S: x 1 , x 2 ,…, x n are columns of S. Then. L is the eigenvalue matrix. Matrix Power A k. - PowerPoint PPT Presentation

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Page 1: Lecture 18 Eigenvalue Problems II

Lecture 18Eigenvalue Problems II

Shang-Hua Teng

Page 2: Lecture 18 Eigenvalue Problems II

Diagonalizing A Matrix• Suppose the n by n matrix A has n linearly independent

eigenvectors x1, x2,…, xn.

• Eigenvector matrix S: x1, x2,…, xn are columns of S.• Then

n

ASS

11

is the eigenvalue matrix

Page 3: Lecture 18 Eigenvalue Problems II

Matrix Power Ak

• S-1AS = implies A = S S-1

• implies A2 = S S-1 S S-1 = S S-1

• implies Ak = S kS-1

Page 4: Lecture 18 Eigenvalue Problems II

Random walks

How long does it take to get completely lost?

000001

Page 5: Lecture 18 Eigenvalue Problems II

Random walks Transition Matrix1

2

345

6

000001

021

4100

21

310

41000

31

210

21

310

00410

310

0041

210

21

31000

310

100

P

Page 6: Lecture 18 Eigenvalue Problems II

Matrix Powers

• If A is diagonalizable as A = S S-1 then for any vector u, we can compute Aku efficiently

– Solve S c = u– Aku = S kS-1 S c = S k c

• As if A is a diagonal matrix!!!!

Page 7: Lecture 18 Eigenvalue Problems II

Independent Eigenvectors from Different Eigenvalues

• Eigenvectors x1, x2,…, xk that correspond to distinct (all different) eigenvalues are linear independent.

• An n by n matrix that has n different eigenvalues (no repeated ’s) must be diagonalizable

Proof: Show that

implies all ci = 0

011 kk xcxc

Page 8: Lecture 18 Eigenvalue Problems II

Addition, Multiplication, and Eigenvalues

• If is an eigenvalue of A and is an eigenvalue of B, then in general is not an eigenvalue of AB

• If is an eigenvalue of A and is an eigenvalue of B, then in general is not an eigenvalue of A+B

Page 9: Lecture 18 Eigenvalue Problems II

Example

0110

0001

0100

,0010

BA

AB

BA

Page 10: Lecture 18 Eigenvalue Problems II

Spectral Analysis of Symmetric Matrices A = AT (what are special about them?)

Spectral Theorem: Every symmetric matrix has the factorization

A = QQT

with real eigenvalues in and orthonormal eigenvectors in :

A =QQ-1 = QQT with Q-1= QT.

Page 11: Lecture 18 Eigenvalue Problems II

Simply in English

• Symmetric matrix can always be diagonalized

• Their eigenvalues are always real• One can choose n eigenvectors so that they

are orthonormal.

• “Principal axis theorem” in geometry and physics

Page 12: Lecture 18 Eigenvalue Problems II

2 by 2 Case

Real Eigenvalues

222det baccabcacb

ba

cbba

A

Page 13: Lecture 18 Eigenvalue Problems II

2 by 2 Case

so

bc

xxcb

baxIA

ab

xxcb

baxIA

cbba

A

222

2

222

111

1

111

21 xx

Page 14: Lecture 18 Eigenvalue Problems II

The eigenvalues of a real symmetric matrix are real

• Complex conjugate of a + i b is a - i b• Law of complex conjugate :

(a-i b) (c-i d) = (ac-bd) – i(bc+ad)• which is the complex conjugate of

(a+i b) (c+i d) = (ac-bd) + i(bc+ad)

• Claim:

• What can be?

TTTTT xAxxAxxxAxAxAx

AxxT

Page 15: Lecture 18 Eigenvalue Problems II

Eigenvectors of a real symmetric matrix when they correspond to different ’s are

always perpendicular

AyxyAx AA yAyxAx TTTT and and 21

What can the quantity be?

Page 16: Lecture 18 Eigenvalue Problems II

In general, so eigenvalues might be repeated

• Choose an orthogonal basis for each eigenvalue

• Normalize these vector to unit length

Page 17: Lecture 18 Eigenvalue Problems II

Every symmetric matrix has the factorization A = QQT

with real eigenvalues in and orthonormal eigenvectors in :

A =QQ-1 = QQT with Q-1= QT.

Spectral Theorem

Page 18: Lecture 18 Eigenvalue Problems II

Every symmetric matrix has the factorization A = QQT

with real eigenvalues in and orthonormal eigenvectors in :

A =QQ-1 = QQT with Q-1= QT.

Spectral Theorem and Spectral Decomposition

Tnnn

T

Tn

T

n

n xxxxx

xxxA

111

11

1

xi xiT is the projection matrix on to xi !!!!!