## Diagonalization of Matrix 5

In this page diagonalization of matrix 5 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 5 :

Diagonalize the following matrix

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

Let A =

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

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=

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

-

 λ 0 0 0 λ 0 0 0 λ

=

 (11-λ) (-4-0) (-7-0) (7-0) (-2-λ) (-5-0) (10-0) (-4-0) (-6-λ)

=

 (11-λ) -4 -7 7 (-2-λ) -5 10 -4 (-6-λ)

A-λI=

 (11-λ) -4 -7 7 (-2-λ) -5 10 -4 (-6-λ)

=  (11-λ)[(-2-λ)(-6-λ)-20]+4[7(-6-λ)+50]-7[-28-10(-2-λ)]

=  (11-λ)[(2+λ)(6+λ)-20]+4[-42-7λ+50]-7[-28+20+10λ]

=  (11-λ)[12+2λ+6λ+λ²-20]+4[8-7λ]-7[-8+10λ]

=  (11-λ)[λ²+8λ-8]+32-28λ+56-70λ

=  11λ²+8λ-88-λ³-8λ²+88λ-98λ+88

=  - λ³+3λ²-2λ

=  -λ³+3λ²-2λ

=  -λ(λ²-3λ²+2)

To find roots let |A-λI| = 0

-λ(λ²-3λ²+2) = 0

λ = 0

Now we have to solve λ²-3λ²+2 to get another two values. For that let us factorize

λ²-3λ²+2 = 0

λ²-1λ-2λ+2 = 0

λ(λ-1)-2(λ-1) = 0

(λ-1)(λ-2) = 0

λ - 1 = 0

λ = 1

λ - 2 = 0

λ = 2

Therefore the characteristic roots (or) Eigen values are x = 0,1,2

Substitute λ = 0 in the matrix A - λI diagonalization of matrix 5

A-λI=

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

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

11x - 4y - 7z = 0  ------ (1)

7x - 2y - 5z = 0  ------ (2)

10x - 4y - 6z = 0  ------ (3)

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

The eigen vector x =

 1 1 1

Substitute λ = 1 in the matrix A - λI

=

 10 -4 -7 7 -3 -5 10 -4 -7

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

10x - 4y - 7z = 0  ------ (4)

7x - 3y - 5z = 0  ------ (5)

10x - 4y - 7z = 0  ------ (6)

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

The eigen vector y =

 -1 1 -2

Substitute λ = 2 in the matrix A - λI       diagonalization of matrix 5

=

 9 -4 -7 7 -4 -5 10 -4 -8

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

9x - 4y - 7z = 0  ------ (7)

7x - 4y - 5z = 0  ------ (8)

10x - 4y - 8z = 0  ------ (9)

By solving (7) and (8) we get the eigen vector

The eigen vector z =

 2 1 2

Let P =

 1 1 1 -1 1 -2 2 1 2

The column of P are linearly independent eigen vectors of A . Therefore the diagonal matrix = diagonalization of matrix 5

 -2 0 0 0 1 0 0 0 2

 Questions Solution

Question 1 :

Diagonalize the following matrix

 5 0 1 0 -2 0 1 0 5

Solution

Question 2 :

Diagonalize the following matrix

 1 1 3 1 5 1 3 1 1

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 5

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

Solution