Lecture 22

Last time we discussed the notion of a steady statevector of a stochastic matrix P

,which is a probability vector

I such that

PI = I.

(( ) )

This is an example of an eigenvector of a square matrix

with "

eigenvalue" I

.

Eigenvectors and eigenvalues Let A = nxn square matrix .

A NONZERO I EIR"

is an eigenvector of A if there exists a

scalar X such that

AI = XI.

We say X is an eigenvalue of A and that I is an

eigenvector with eigenvalue 2 .

Remark : X can be 0,but I cannot .

Examples : A = nxn matrix.

① Any NONZERO I C- Nut A is an eigenvector with eigenvalue O .

② A=

I 6; I =

6.

Then

5 2- 5

AT=

6 ( ; ) - 5 I =

- 24=- Liu .

20

ooo it is an eigenvector of A with eigenvalue- 4

.

Suppose a nonzero IE IR"

is an eigenvector of A with

eigenvalue X,ie,

AI = XI.

Note XI = XIN I.

So

AI = (X In) I ⇐ AE - @ In) I = 8

⇐ (A - XIN) I = 8 ⇐ I is NONZERO vector

in Nul ( A - XIN) .

Theorem : A = nxn matrix .A nonzero IE IR

"

is an

eigenvector of A with eigenvalue X if and only ifI is a nonzero vector in Nul ( A - XIN) .

Eigenspace : The subspace Nut ( A - XIN) is called the

eigenspace of A corresponding to 7 .

Exercise : Let A=

4 2 3 x =3 .

]

- l l - 3

2 4 9

Find a basis of the eigenspace of A corresponding to the

eigenvalue 3.

Solution : we have to find a basis of

Nut ( A - 3 I, ) .

A - 3 Is =4 2 3 3 O O

- l l - 3-

O 3 O

2 4 9 O O 3

=I 2 3

- I -z - 3

2 4 6

I 2 3 I 2 3R2→ RztR ,

- I - z - 3 > O O O

Rg -7123-212 ,

2 4 6 O O o

EF of the coefficient matrix

of ft - 3 Is)I=8 .

Free variables : Xa,X,

Basic variables : X, .

X,t 2×2 t 3×3 = 0

⇒ X,= - 2×2 - 3×3 .

8. General Solution in parametric form is

-2×2-3×3 -2 -3.

Xz= ×2 I t X, o

Xz O l

so -2,

-3are a basis of the eigen space .

I O

O l

How to find the eigenvalues X of A A = nxn matrix .

X is an eigenvalue of A if and only if

(A - XIN) I = 8 has a non - trivial solution if andonly if

A - XIN is NOT invertible if and only if

det CA - Xin ) = 0 .

Observation : det (A - Xin ) will be a polynomial in X

of degree n .

Example : A=

2 4.

- I 8

A - X Iz =2 - X 4

.

- I 8 -X

fo det (A - I Iz) = (2-1) (8-1)+4 = 16-101+22+4

= Xt - lot t 20-

polynomial in X of deg 2

Characteristic Equation of A o

.This is the equation

det ( A - Xin) = O .

Example : The characteristic equation of ⇐ 4

,

is 12-10×+20=0 .

Theorem : The eigenvalues of A are the roots of the

characteristic equationdet ( A - XIN) = O .

Example : Suppose A is an upper or lower triangularmatrix .

For example , say A - I O l - Iy

'

O 2 2 I

O O O - 5

O O O 7

Then A - X In is also upper or lower triangular matrixwhose diagonal entries are

-X t diagonal entries of A .

For example , for A=

I O l - l, we get

O 2 2 I

O O O - 5

O O O 7

A - X In =I -X O l - I

0 2-X 2 I

O O -X - 5

O O O 7-X

Then det ( A - XIN) = product of diagonal entries ofA -XIN .

O

o o the roots of det ( A - X In) = 0 are precisely the

diagonal entries of A .

For example , for A=

I 0 I - I

0 2 2 I]

O O O - 5

O O O 7

det ft - XIU ) = ( t -X) ( 2 -H C -X) (7 -X) .

So the roots of det ( A - XIU ) = 0 are

X = I,2,O

,7

.

These are the diagonal entries of A .

Theorem : The eigenvalues of a triangular matrix A are the

entries on the main diagonal .

Eigenvectors with distinct eigenvalues :

Theorem : Let I, ,

. . .

, X-p E IR" be eigenvectors of a

matrix A with distinct eigenvalues . That is, if

X ; is the eigenvalue of Xi , then

Xi FXj for i * j .

Then XT , . ..

, X-p are linearly independent .

Upshot °

. If A = nxn matrix has a distinct eigenvalues=

Xp ,- . .

,Xu

and if Xi E IR" is an eigenvector with eigenvalue Xi ,then

XT ,. . .

,I

are linearly independent . ooo By the Basis Theorem ,

XT ,. . .

,In

are a basis of IRN .