Finding Eigenvalues and Eigenvectors What is really important?

Finding Eigenvalues and Eigenvectors

What is really important?

04/19/23DRAFT Copyright, Gene A

Tagliarini, PhD2

Approaches

Find the characteristic polynomial– Leverrier’s Method

Find the largest or smallest eigenvalue– Power Method– Inverse Power Method

Find all the eigenvalues – Jacobi’s Method– Householder’s Method– QR Method– Danislevsky’s Method


Tagliarini, PhD3

Finding the Characteristic Polynomial

Reduces to finding the coefficients of the polynomial for the matrix A

Recall |I-A| = n+ann-1+an-1n-2+…+a21+a1

Leverrier’s Method– Set Bn = A and an = -trace(Bn)

– For k = (n-1) down to 1 computeBk = A (Bk+1 + ak+1I)

ak = - trace(Bk)/(n – k + 1)


Tagliarini, PhD4

Vectors that Span a Space and Linear Combinations of Vectors

Given a set of vectors v1, v2,…, vn The vectors are said span a space V, if given

any vector x ε V, there exist constants c1, c2,…, cn so that c1v1 + c2v2 +…+ cnvn = x and x is called a linear combination of the vi


Tagliarini, PhD5

Linear Independence and a Basis

Given a set of vectors v1, v2,…, vn and constants c1, c2,…, cn

The vectors are linearly independent if the only solution to c1v1 + c2v2 +…+ cnvn = 0 (the zero vector) is c1= c2=…=cn = 0

A linearly independent, spanning set is called a basis


Tagliarini, PhD6

Example 1: The Standard Basis

Consider the vectors v1 = <1, 0, 0>, v2 = <0, 1, 0>, and v3 = <0, 0, 1>

Clearly, c1v1 + c2v2 + c3v3 = 0 c1= c2= c3 = 0 Any vector <x, y, z> can be written as a linear

combination of v1, v2, and v3 as<x, y, z> = x v1 + y v2 + z v3

The collection {v1, v2, v3} is a basis for R3; indeed, it is the standard basis and is usually denoted with vector names i, j, and k, respectively.


Tagliarini, PhD7

Another Definition and Some Notation

Assume that the eigenvalues for an n x n matrix A can be ordered such that|1| > |2| ≥ |3| ≥ … ≥ |n-2| ≥ |n-1| > |n|

Then 1 is the dominant eigenvalue and |1| is the spectral radius of A, denoted (A)

The ith eigenvector will be denoted using superscripts as xi, subscripts being reserved for the components of x


Tagliarini, PhD8

Power Methods: The Direct Method

Assume an n x n matrix A has n linearly independent eigenvectors e1, e2,…, en ordered by decreasing eigenvalues|1| > |2| ≥ |3| ≥ … ≥ |n-2| ≥ |n-1| > |n|

Given any vector y0 ≠ 0, there exist constants ci, i = 1,…,n, such that y0 = c1e1 + c2e2 +…+ cnen


Tagliarini, PhD9

The Direct Method (continued)

If y0 is not orthogonal to e1, i.e., (y0)Te1≠ 0, y1 = Ay0 = A(c1e1 + c2e2 +…+ cnen)

= Ac1e1 + Ac2e2 +…+ Acnen

= c1Ae1 + c2Ae2 +…+ cnAen

Can you simplify the previous line?


Tagliarini, PhD10


If y0 is not orthogonal to e1, i.e., (y0)Te1≠ 0, y1 = Ay0 = A(c1e1 + c2e2 +…+ cnen)

= Ac1e1 + Ac2e2 +…+ Acnen

= c1Ae1 + c2Ae2 +…+ cnAen

y1 = c11e1 + c22e2 +…+ cnnen

What is y2 = Ay1?


Tagliarini, PhD11


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Tagliarini, PhD12


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Tagliarini, PhD13


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Tagliarini, PhD14


Note: any nonzero multiple of an eigenvector is also an eigenvector

Why? Suppose e is an eigenvector of A, i.e., Ae=e

and c0 is a scalar such that x = ce Ax = A(ce) = c (Ae) = c (e) = (ce) = x


Tagliarini, PhD15


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Tagliarini, PhD16

Direct Method (continued)

Given an eigenvector e for the matrix AWe have Ae = e and e0, so eTe 0 (a

scalar) Thus, eTAe = eTe = eTe 0 So = (eTAe) / (eTe)


Tagliarini, PhD17

Direct Method (completed)

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Tagliarini, PhD18

Direct Method Algorithm

m)(i and )r( While3.

1i i

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yy

xy

Ay x

x

xy

Do 2.

. and compute and 0 iSet

. and 0,m 0, Choose 1.

T

T

Ayx

0y


Tagliarini, PhD19

Jacobi’s Method

Requires a symmetric matrixMay take numerous iterations to converge Also requires repeated evaluation of the

arctan function

Isn’t there a better way? Yes, but we need to build some tools.


Tagliarini, PhD20

What Householder’s Method Does

Preprocesses a matrix A to produce an upper-Hessenberg form B

The eigenvalues of B are related to the eigenvalues of A by a linear transformation

Typically, the eigenvalues of B are easier to obtain because the transformation simplifies computation


Tagliarini, PhD21

Definition: Upper-Hessenberg Form

A matrix B is said to be in upper-Hessenberg form if it has the following structure:

nn,1-nn,

n1,-n1-n1,-n

4,3

n3,1-n3,3,33,2

n2,1-n2,2,32,22,1

n1,1-n1,1,31,21,1

bb000

bb

b00

bbbb0

bbbbb

bbbbb

B


Tagliarini, PhD22

A Useful Matrix Construction

Assume an n x 1 vector u 0 Consider the matrix P(u) defined by

P(u) = I – 2(uuT)/(uTu)Where

– I is the n x n identity matrix– (uuT) is an n x n matrix, the outer product of u

with its transpose– (uTu) here denotes the trace of a 1 x 1 matrix and

is the inner or dot product


Tagliarini, PhD23

Properties of P(u)

P2(u) = I – The notation here P2(u) = P(u) * P(u)– Can you show that P2(u) = I?

P-1(u) = P(u)– P(u) is its own inverse

PT(u) = P(u) – P(u) is its own transpose– Why?

P(u) is an orthogonal matrix


Tagliarini, PhD24

Householder’s Algorithm

Set Q = I, where I is an n x n identity matrix For k = 1 to n-2

– = sgn(Ak+1,k)sqrt((Ak+1,k)2+ (Ak+2,k)2+…+ (An,k)2)– uT = [0, 0, …, Ak+1,k+ , Ak+2,k,…, An,k]– P = I – 2(uuT)/(uTu)– Q = QP– A = PAP

Set B = A


Tagliarini, PhD25

Example

55143)sgn(3)sgn(aThen

1.k valueonly the k takes1, 2-n since and 3n Clearly,

.

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321

Let

22231

22121

33

aa

x

A


Tagliarini, PhD26

Example

?

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aaaaa


Tagliarini, PhD27

Example

? Find .

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001

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Tagliarini, PhD28

Example

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001

5

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001

100

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so , Initially,

QPQ

IQ


Tagliarini, PhD29

Example

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001

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5

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so , Next,

A

PAPA


Tagliarini, PhD30

Example

25

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Hence,

A


Tagliarini, PhD31

Example

what?So

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once executesonly loop thesince Finally,

AB


Tagliarini, PhD32

How Does It Work?

Householder’s algorithm uses a sequence of similarity transformationsB = P(uk) A P(uk)to create zeros below the first sub-diagonal

uk=[0, 0, …, Ak+1,k+ , Ak+2,k,…, An,k]T

= sgn(Ak+1,k)sqrt((Ak+1,k)2+ (Ak+2,k)2+…+ (An,k)2) By definition,

– sgn(x) = 1, if x≥0 and– sgn(x) = -1, if x<0


Tagliarini, PhD33

How Does It Work? (continued)

The matrix Q is orthogonal – the matrices P are orthogonal– Q is a product of the matrices P– The product of orthogonal matrices is an

orthogonal matrix

B = QT A Q hence Q B = Q QT A Q = A Q– Q QT = I (by the orthogonality of Q)


Tagliarini, PhD34

How Does It Work? (continued)

If ek is an eigenvector of B with eigenvalue k, then B ek = k

ek

Since Q B = A Q,A (Q ek) = Q (B ek) = Q (k ek) = k (Q ek)

Note from this:– k is an eigenvalue of A

– Q ek is the corresponding eigenvector of A


Tagliarini, PhD35

The QR Method: Start-up

Given a matrix A Apply Householder’s Algorithm to obtain a

matrix B in upper-Hessenberg form

Select >0 and m>0– is a acceptable proximity to zero for sub-

diagonal elements– m is an iteration limit


Tagliarini, PhD36

The QR Method: Main Loop

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Tagliarini, PhD37

The QR Method: Finding The ’s

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Tagliarini, PhD38

Details Of The Eigenvalue Formulae

?

. Suppose

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Tagliarini, PhD39

Details Of The Eigenvalue Formulae

)det()(

)(

))((

Given

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Tagliarini, PhD40

Finding Roots of Polynomials

Every n x n matrix has a characteristic polynomial

Every polynomial has a corresponding n x n matrix for which it is the characteristic polynomial

Thus, polynomial root finding is equivalent to finding eigenvalues


Tagliarini, PhD41

Example Please!?!?!?

Consider the monic polynomial of degree nf(x) = a1 +a2x+a3x2+…+anxn-1 +xn

and the companion matrix

01000

0000

0010

0001121

aaaa nn

A


Tagliarini, PhD42

Find The Eigenvalues of the Companion Matrix

?

01000

0000

0010

0001121

aaaa nn

IAI


Tagliarini, PhD43

Find The Eigenvalues of the Companion Matrix

100

00

001)1()(

1000

000

0010

001

121

1

121

aaa

a

aaaa

n

nn

nn

AI

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Finding Eigenvalues and Eigenvectors What is really important?

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