## Diagonalization of Matrix 2

In this page diagonalization of matrix 2 we are going to see how to diagonalize a matrix.

Definition :

A square matrix of order n is diagonalizable if it is having linearly independent eigen values.

We can say that the given matrix is diagonalizable if it is alike to the diagonal matrix. Then there exists a non singular matrix P such that P⁻¹ AP = D where D is a diagonal matrix.

Question 2 :

Diagonalize the following matrix

 1 1 3 1 5 1 3 1 1

Let A =

 1 1 3 1 5 1 3 1 1

The order of A is 3 x 3. So the unit matrix I =

 1 0 0 0 1 0 0 0 1

Now we have to multiply λ with unit matrix I.

λI =

 λ 0 0 0 λ 0 0 0 λ

A-λI=

 1 1 3 1 5 1 3 1 1

-

 λ 0 0 0 λ 0 0 0 λ

=

 (1-λ) (1-0) (3-0) (1-0) (5-λ) (1-0) (3-0) (1-0) (1-λ)

=

 (1-λ) 1 3 1 (5-λ) 1 3 1 (1-λ)

A-λI=

 (1-λ) 1 3 1 (5-λ) 1 3 1 (1-λ)

=  (1-λ)[ (5-λ)(1-λ) - 1 ] - 1[1 - λ - 3] + 3 [1 - 3 (5-λ) ]

=  (1-λ)[ 5 - 5 λ - λ + λ² - 1 ] - 1[ -λ - 2] + 3 [ 1 - 15 +3 λ ]

=  (1-λ)[ λ² - 6 λ +  4 ] + 1 λ + 2 + 3 [ - 14 +3 λ ]

=  λ² - 6 λ + 4 - λ³  + 6 λ² - 4 λ  + λ + 2 - 42 + 9 λ

=  - λ³ + λ² + 6 λ² - 6 λ - 4 λ  + λ + 9 λ + 4 + 2 - 42

=  - λ³ + 7 λ² - 10 λ + 10 λ + 6 - 42

=  - λ³ + 7 λ² - 36

=  λ³ - 7 λ² + 36

To find roots let |A-λI| = 0

λ³ - 7 λ² + 36 = 0

For solving this equation first let us do synthetic division.diagonalization of matrix 2 By using synthetic division we have found one value of λ that is λ = -2.

Now we have to solve λ² - 10 λ + 24 to get another two values. For that let us factorize

λ² - 9 λ + 18 = 0

λ² - 3 λ - 6 λ + 18 = 0

λ (λ - 3) - 6 (λ - 3) = 0

(λ - 6) (λ - 3) = 0

λ - 6 = 0

λ = 6

λ - 3 = 0

λ = 3

Therefore the characteristic roots (or) Eigen values are x = -2,3,6

Substitute λ = -2 in the matrix A - λI

=

 3 1 3 1 7 1 3 1 3

From this matrix we are going to form three linear equations using variables x,y and z.

3x + 1y + 3z = 0  ------ (1)

1x + 7y + 1z = 0  ------ (2)

3x + 1y + 3z = 0  ------ (3)

By solving (1) and (2) we get the eigen vector The eigen vector x =

 -1 1 0

Substitute λ = 3 in the matrix A - λI

=

 -2 1 3 1 2 1 3 1 -2

From this matrix we are going to form three linear equations using variables x,y and z.

-2x + 1y + 3z = 0  ------ (4)

1x + 2y + 1z = 0  ------ (5)

3x + 1y - 2z = 0  ------ (6)

By solving (4) and (5) we get the eigen vector The eigen vector y =

 -1 1 -1

Substitute λ = 6 in the matrix A - λI

=

 -5 1 3 1 -1 1 3 1 -5

From this matrix we are going to form three linear equations using variables x,y and z.

-5x + 1y + 3z = 0  ------ (7)

1x - 1y + 1z = 0  ------ (8)

3x + 1y - 5z = 0  ------ (9)

By solving (7) and (8) we get the eigen vector diagonalization of matrix 2 The eigen vector z =

 1 2 1

Let P =

 -1 0 1 -1 1 -1 1 2 1

The column of P are linearly independent eigen vectors of A . Therefore the diagonal matrix = diagonalization of matrix 2

 -2 0 0 0 3 0 0 0 6

 Questions Solution

Question 1 :

Diagonalize the following matrix

 5 0 1 0 -2 0 1 0 5

Solution

Question 3 :

Diagonalize the following matrix

 -2 2 -3 2 1 -6 -1 -2 0

Solution

Question 4 :

Diagonalize the following matrix  diagonalization of matrix 2

 4 -20 -10 -2 10 4 6 -30 -13

Solution

Question 5 :

Diagonalize the following matrix diagonalization of matrix 2

 11 -4 -7 7 -2 -5 10 -4 -6

Solution  