EigenValuesEigenvectors

8/4/2019 EigenValuesEigenvectors

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Primer on Eigenvalues andEigenvectors

Copyright Brian G. Higgins (2003)

Introduction

In this notebook we discuss the solution of the following matrix problem

(1)Ax = l x

where A is a nn matrix, x is a 1n column vector, and l is a scalar constant. Matrix problems of this form occur

frequently in the solution of differential equations, and have important physical significance in applications in the

physical sciences and engineering.

We explore various analytically methods commonly used to solve the algebraic eigenvalue problem when n is small.

We also discuss important properties of the eigenvalues and eigenvectors for these systems, and show howMathematica

can be used to perform these calculations.

The contents of this notebook are the essential building blocks for studying more advanced topics in linear algebra,

numerical methods, statistics and differential equations. Thus all chemical engineering students should master this

material early in their academic careers.

Eigenvalue/Eigenvector Calculations

In this section we illustrate how to compute eigenvalues and eigenvectors for a nn matrix A. We also discuss various

properties of the eigenvalues and eigenvectors. We start with a general definition

Definition 1

Let A be a nn matrix. A scalar l is called an eigenvalue ofA if there exists a nonzero 1n column vector x that is a

solution to the following system of linear equations:

(2)A x= l x

The solution vector x is called an eigenvector, and the scalar l is called its associated eigenvalue.

EigenValuesEigenvectors.nb 1


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Example 1: Eigenvectors of a Matrix A

Suppose we have the matrix Agiven by

A=

i

k

jjjjjjj

1 -1 0

0 1 1

0 0 -2

y

{

zzzzzzz

then

x=

i

k

jjjjjjj1

3

-9

y

{

zzzzzzzis an eigenvector since

A x=

i

k

jjjjjjj1 -1 0

0 1 1

0 0 -2

y

{

zzzzzzz i

k

jjjjjjj1

3

-9

y

{

zzzzzzz =i

k

jjjjjjj-2

-6

18

y

{

zzzzzzz = H-2Li

k

jjjjjjj1

3

-9

y

{

zzzzzzz

with l = -2. Thus the associated eigenvalue ofx is l = -2

Property 1

The eigenvalues and eigenvectors of a matrix A are determined by solving the following homogeneous system of linear

equations:

HA11 - lL x1 +A12 x2 + + A1n xn = 0 A21 x1 + HA22 - lL x2 + + A2n xn = 0

An1 x1 +An2 x2 + + HAnn - lL xn = 0

where x= 8x1 , x2 , , xn


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Example 2: Calculation of eigenvalues and eigenvectors of a matrix A: Distinct Eigenvalues

Suppose we are given the following matrix:

A=

i

k

jjjjjjj

1 2 1

6 -1 0

-1 -2 -1

y

{

zzzzzzz

Determine the eigenvalues li and the associated eigenvectors xi ofA.

First we determine the characteristic equation for the eigenvalues

DetAi

k

jjjjjjj1 - l 2 1

6 -1 - l 0

-1 -2 -1 - l

y

{

zzzzzzzE = 0

Expanding the determinant gives

PHlL = H1 - lL8H-1 - lLH-1 - lL< - 286H-1 - lL< + 18-12 + H-1 - lL< = 0Multiplying out gives

PHlL = 12 l - l2 - l3 = 0

Thus the roots ofPHlL = 0 are then

PHlL = l Hl + 4LHl - 3L = 0

Hence the eigenvalues are:

l1 = 0, l2 = -4, l3 = 3

To find the eigenvectors xi we need to solve the following system of homogeneous equations for each li :

Axi = li xi

Consider first l1 = 0. The equations we must solve are

i

k

jjjjjjj1 2 1

6 -1 0

-1 -2 -1

y

{

zzzzzzzi

k

jjjjjjjx11

x12

x13

y

{

zzzzzzz = 0

Thus we obtain

x11 + 2 x12 + x13 = 0, 6 x11 - x12 = 0, -x11 - 2 x12 - x13 = 0

We can solve this system by Gauss elimination to obtain

x11 = -1

13x13 , x12 = -

613

x13

Hence the eigenvector is



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x1 =

i

k

jjjjjjjjj

- 113

a

- 613

a

a

y

{

zzzzzzzzz, where a is arbitrary

A suitable eigenvector is

x1 =i

k

jjjjjjj1

6

-13

y

{

zzzzzzz

Consider next l2 = -4. The equations we must solve are

i

k

jjjjjjj5 2 1

6 3 0

-1 -2 3

y

{

zzzzzzzi

k

jjjjjjjx21

x22

x23

y

{

zzzzzzz = 0

We obtain the following system of equations

5 x21 + 2 x22 + x23 = 0, 6 x21 + 3 x22 = 0, -x21 - 2 x22 + 3 x23 - 0

Solving this set gives

x22 = 2 x23 , x21 = -x23

Thus the eigenvector is

x2 =

i

k

jjjjjjj-a

2 a

a

y

{

zzzzzzz where a is arbitrary.

A suitable vector is then

x2 =

i

k

jjjjjjj-1

2

1

y

{

zzzzzzz

A similar calculation for l3 = 3 gives

i

k

jjjjjjj-2 2 1

6 -4 0

-1 -2 -4

y

{

zzzzzzzi

k

jjjjjjjx31

x32

x33

y

{

zzzzzzz = 0

The eigenvector from this calculation is

x3 =

i

k

jjjjjjj2

3

-2

y

{

zzzzzzz

To summarize: In this problem we found 3 distinct eigenvalues, and 3 associated eigenvectors



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Example 2: Mathematica Solution

Mathematica has several built-in functions for determining the eigenvalues and eigenvectors of a square matrix. To

calculate the eigenvalues the function is Eigenvalues; to find the eigenvectors, the function is Eigenvectors; and to

solve for both the eigenvalues and eigenvectors one can use Eigensystem. We illustrate this function using the matrix

A.

A=

i

k

jjjjjjj1 2 1

6 -1 0

-1 -2 -1

y

{

zzzzzzz;

8Eigenvalues@AD, Eigenvectors@AD

EigenValuesEigenvectors

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