## Diagonalization of Matrix 4

In this page diagonalization of matrix 4 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 4 :

Diagonalize the following matrix

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

Let A =

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

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.diagonalization of matrix4

λI =

 λ 0 0 0 λ 0 0 0 λ

A-λI=

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

-

 λ 0 0 0 λ 0 0 0 λ

=

 (-4-λ) (-20-0) (-10-0) (-2-0) (10-λ) (4-0) (6-0) (-30-0) (-13-λ)

=

 (-4-λ) -20 -10 -2 (10-λ) 4 6 -30 (-13-λ)

= (4-λ)[(10-λ)(-13- λ)+120]+

20[-2(-13-λ)-24]-10[60-6(10-λ)]

= (4-λ)[-130-10 λ+13λ+λ²+120]+20[26+2λ-24]-10[60-60+6λ]

= (4-λ)[-10+3λ+λ²]+20[2+2λ]-10[6λ]

= (4-λ)[λ²+3λ-10]+20[2+2λ]-10[6λ]

= 4λ²+12λ-40-λ³-3λ²+10λ+40λ+40-60λ

= -λ³ + 1λ² + 2λ

To find roots let |A-λI| = 0

-λ³ + 1λ² + 2λ = 0

For solving this equation -λ from all the terms

-λ (λ² - 1λ - 2) = 0

-λ = 0 (or) λ² - 1 λ - 2 = 0

λ = 0       (λ+1) (λ-2) = 0

λ + 1 = 0       λ - 2 = 0

λ = - 1            λ = 2

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

Substitute λ = 0 in the matrix A - λI

=

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

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

4x - 20y - 10z = 0  ------ (1)

-2x + 10y + 4z = 0  ------ (2)

6x - 30y - 13z = 0  ------ (3)

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

 5 1 0

Substitute λ = -1 in the matrix A - λI

=

 5 -20 -10 -2 10 4 6 -30 -12

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

5x - 20y - 10z = 0  ------ (4)

-2x + 10y + 4z = 0  ------ (5)

6x - 30y - 12z = 0  ------ (6)

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

 2 0 1

Substitute λ = 2 in the matrix A - λI   diagonalization of matrix  4

=

 2 -20 -10 -2 8 4 6 -30 -15

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

2x - 20y - 10z = 0  ------ (7)

-2x + 8y + 4z = 0  ------ (8)

6x - 30y - 15z = 0  ------ (9)

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

 0 1 -2

Let P =

 5 1 0 2 0 1 0 1 2

The column of P are linearly independent eigen vectors of A . Therefore the diagonal matrix = diagonalization of matrix 4  diagonalization of matrix 4

 0 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 5 :

Diagonalize the following matrix

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

Solution 1. Click on the HTML link code below.

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