Diagonal Ization

7/28/2019 Diagonal Ization

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Diagonalization of a Matrix:

Definition of diagonalization of a matrix:A matrixA is diagonalizable if there exists a nonsingular matrixPand a diagonal

matrixD such that

APPD 1= .

Example:

Let

= 53

64

A .

Then,

,11

21

53

64

11

21

10

021

1

APPD

=

=

=

where

=

=

11

21,

10

02PD .

Theorem:An nn matrix A is diagonalizable if and only if it has n linearlyindependent eigenvector.

[proof:]

:

A is diagonalizable. Then, there exists a nonsingular matrix P and adiagonal matrix

=

n

D

00

00

00

2

1

,

such that

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[ ] [ ]

=

==

n

nn PcolPcolPcolPcolPcolPcolA

PDAPAPPD

00

00

00

)()()()()()(2

1

2121

1

.

Then,.,,2,1),()( njPcolPAcol jjj ==

That is,)(,),(),( 21 PcolPcolPcol n

are eigenvectors associated with the eigenvalues n ,,, 21 .

Since P is nonsingular, thus )(,),(),( 21 PcolPcolPcol n are linearly

independent.

:

Let nxxx ,,, 21 be n linearly independent eigenvectors ofA associated

with the eigenvalues n ,,, 21 . That is,.,,2,1, njxAx jjj ==

Thus, let

[ ]jjn

xPcolxxxP == )(i.e.,21

and

=

n

D

00

00

00

2

1

.

Since jjj xAx = ,

[ ] [ ] PDxxxxxxAAP

n

nn =

==

00

00

00

2

1

2121 .

Thus,

DPDPAPP == 11 ,1P exists because nxxx ,,, 21 are linearly independent and thusPis

nonsingular.

Important result:An nn matrix A is diagonalizable if all the roots of its characteristic

2


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equation are realanddistinct.

Example:

Let

=

53

64A .

Find the nonsingular matrixPand the diagonal matrixD such that

APPD 1=

and find nAn , is any positive integer.

[solution:]

We need to find the eigenvalues and eigenvectors ofA first. The characteristic

equation ofA is

( )( ) 02153

64)det( =+=

+=

AI .

2or1 = .

By the above important result,A is diagonalizable. Then,

1. As 2= ,

( ) .,1

1022 RrrxxAIxAx

===

2. As 1= ,

( ) .,1

20 RttxxAIxAx

===

Thus,

1

1and

1

2

are two linearly independent eigenvectors ofA.

Let

=

11

21P and

=

10

02D .

Then, by the above theorem,

APPD 1= .

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To find nA ,

( )( )( ) ( ) PAPAPPAPPAPPD n

n

n

n 1111

10

02 ==

=

n times

Multiplied by P and 1P on the both sides,

( )

( )[ ] ( )[ ]( ) ( )

++

++=

===

+++

+++

111

111

1

111

1212

122122

11

21

10

02

11

21

nnnn

nnnn

n

n

nnnAPPAPPPPD

Note (very important):If A is an nn diagonalizable matrix, then there exists an nonsingularmatrix P such that

APPD 1= ,

where )(,),(),( 21 PcolPcolPcol n are n linearly independent

eigenvectors of A and the diagonal elements of the diagonal matrix D

are the eigenvalues of A associated with these eigenvectors.

Note:For any nn diagonalizable matrix A, ,1APPD = then

,2,1,1 == kPPDA kk

where

=

k

n

k

k

kD

00

00

00

2

1

.

Example:

Is

=

13

35A diagonalizable?

[solution:]

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( ) 0213

35)det(

2==

+

=

AI .

Then, 2,2= .

As ,2=

( ) .,1

102 RttxxAI

==

Therefore, all the eigenvectors are spanned by

1

1. There does not existtwo linearly

independent eigenvectors. By the previous theorem,A is not diagonalizable.

Note:An nn matrix may fail to be diagonalizable since Notall roots of its characteristic equation are real numbers.

It does not have n linearly independent eigenvectors.

Note:The set jS consisting of both all eigenvectors of an nn matrix Aassociated with eigenvalue j and zero vector 0 is a subspace of nR . jS

is called the eigenspace associated with j .

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Diagonal Ization