Diagonalization of a Matrix

Diagonalization of a Matrix:

Definition of diagonalization of a matrix: A matrix A is diagonalizable if there exists a nonsingular matrix P and a diagonal matrix D such that D  P1 AP .

Example:

Let  4  6 A    .  3 5  Then,

1 2 0  1  2  4  61  2 1 D           P AP, 0 1  1 1   3 5  1 1  where 2 0  1  2 D   , P    . 0 1  1 1 

Theorem: An n  n matrix A is diagonalizable if and only if it has n linearly independent eigenvector.

[proof:]

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

1 1 0 ⋯ 0   0  ⋯ 0  D   2  ,  ⋮ ⋮ ⋱ ⋮    0 0 ⋯ n  such that

D  P 1 AP  AP  PD

1 0 ⋯ 0   0  ⋯ 0  .  Acol (P) col (P) ⋯ col (P) col (P) col (P) ⋯ col (P) 2  1 2 n 1 2 n  ⋮ ⋮ ⋱ ⋮    0 0 ⋯ n  Then,

Acol j (P)   j col j (P), j  1,2,, n. That is,

col1 (P),col2 (P),,coln (P) are eigenvectors associated with the eigenvalues 1 ,2 ,,n .

Since P is nonsingular, thus col1 (P),col2 (P),,coln (P) are linearly independent.

Let x1 , x2 ,, xn be n linearly independent eigenvectors of A associated with the eigenvalues 1 ,2 ,,n . That is,

Ax j   j x j , j 1,2,, n. Thus, let

P  x1 x2 ⋯ xn  i.e., col j (P)  x j  and

2 1 0 ⋯ 0   0  ⋯ 0  D   2  .  ⋮ ⋮ ⋱ ⋮    0 0 ⋯ n 

Since Ax j   j x j ,

1 0 ⋯ 0   0  ⋯ 0  AP  Ax x ⋯ x   x x ⋯ x  2   PD . 1 2 n 1 2 n  ⋮ ⋮ ⋱ ⋮    0 0 ⋯ n  Thus, P 1 AP  P 1PD  D ,

1 P exists because x1 , x2 ,, xn are linearly independent and thus P is nonsingular.

Important result: An n  n matrix A is diagonalizable if all the roots of its characteristic equation are real and distinct.

Example:

Let  4  6 A    .  3 5  Find the nonsingular matrix P and the diagonal matrix D such that D  P1 AP and find An , n is any positive integer.

[solution:]

3 We need to find the eigenvalues and eigenvectors of A first. The characteristic equation of A is   4 6 det(I  A)    1  2  0 .  3   5    1 or 2 . By the above important result, A is diagonalizable. Then,

1. As   2 , 1 Ax  2x  2I  Ax  0  x  r , r  R.  1  2. As   1,  2 Ax  x   I  Ax  0  x  t , t  R.  1  Thus, 1  2   and    1   1  are two linearly independent eigenvectors of A.

Let 1  2 2 0  P    and D    .  1 1  0 1 Then, by the above theorem, D  P1 AP .

To find An ,

n n 2 0  1 1 1 1 n D   n   P APP AP⋯P AP P A P  0 1  n times Multiplied by P and P 1 on the both sides,

4 n 1 n 1 1 n 1 n 1  22 0 1  2 PD P  PP A PP  A    n    1 1  0 1  1 1   2 n  2  1n1  2 n1  2  1n1        n n1 n1 n1   2  1 2  1 

Note (very important): If A is an n  n diagonalizable matrix, then there exists an nonsingular matrix P such that D  P1 AP , where col1 (P),col2 (P),,coln (P) 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 n  n diagonalizable matrix A, D  P1 AP, then Ak  PDk P 1, k 1,2, where

k 1 0 ⋯ 0    0 k ⋯ 0 Dk   2  .  ⋮ ⋮ ⋱ ⋮  k   0 0 ⋯ n 

Example:

5  3 Is A    diagonalizable? 3 1

[solution:]

5   5 3 2 det(I  A)     2  0 .  3  1

Then,   2, 2 .

As   2,

1 2I  Ax  0  x  t , t  R. 1 1 Therefore, all the eigenvectors are spanned by   . There does not exist two linearly 1 independent eigenvectors. By the previous theorem, A is not diagonalizable.

Note: An n  n matrix may fail to be diagonalizable since  Not all roots of its characteristic equation are real numbers.  It does not have n linearly independent eigenvectors.

The set S j consisting of both all eigenvectors of an n  n matrix A

n associated with eigenvalue  j and zero vector 0 is a subspace of R . S j is

called the eigenspace associated with  j .