Diagonalization of Matrix 1





In this page diagonalization of matrix1 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 1 :

Diagonalize the following matrix

 
5 0 1
0 -2 0
1 0 5
 

Solution:

Before finding diagonalize matrix first we have to check whether the eigen vectors of the given matrix are linearly independent.

To find eigen vector first let us find characteristic roots of the given matrix.



   Let A =

 
5 0 1
0 -2 0
1 0 5
 


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

 
5 0 1
0 -2 0
1 0 5
 

Now we have to multiply λ with unit matrix I.diagonalization of matrix 1

  λI =

 
λ 0 0
0 λ 0
0 0 λ
 
A-λI=
 
5 0 1
0 -2 0
1 0 5
 
-
 
λ 0 0
0 λ 0
0 0 λ
 
 
                   
  =
 
(5-λ)   (0-0)   (1-0)
(0-0)   (-2-λ)   (0-0)
(1-0)   (0-0)   (5-λ)
 
 
                             
  =
 
(5-λ)   0   1
0   (-2-λ)   0
1   0   (5-λ)
 
 
|A-λI|=
 
(5-λ)   0   1
0   (-2-λ)   0
1   0   (5-λ)
 

  =  (5-λ)[(-2-λ) (5-λ) - 0] - 0 [(0 - 0)] + 1 [0- (-2 -λ)]

  =  (5-λ)[ -10 + 2 λ - 5 λ + λ²] - 0 + 2 + λ

  =  (5-λ)[ -10 - 3 λ + λ²] - 0 + 2 + λ

  =  (5-λ)[λ² -3 λ-10] + 2 + λ

  =  5 λ² - 15 λ - 50 - λ³ + 3 λ² + 10 λ + 2 + λ

  = - λ³ + 5 λ² + 3 λ² - 15 λ + 10 λ + λ - 50 + 2

  = - λ³ + 8 λ² - 4 λ - 48

  =  λ³ - 8 λ² + 4 λ + 48

To find roots let |A-λI| = 0

   λ³ - 8 λ² + 4 λ + 48 = 0

For solving this equation first let us do synthetic division.diagonalization of matrix 1

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  diagonalization of matrix 1

   λ² - 10 λ + 24 = 0

λ² - 6 λ - 4 λ + 24 = 0

λ (λ - 6) - 4 (λ - 6) = 0

(λ - 6) (λ - 4) = 0

 λ - 6 = 0

 λ = 6

 λ - 4 = 0

 λ = 4

Therefore the characteristic roots (or) Eigen values are x = -2,4,6

Substitute λ = -2 in the matrix A - λI  diagonalization of matrix 1

                             
  =
 
7   0   1
0   0   0
1   0   7
 
 

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

7x + 0y + 1z = 0  ------ (1)

0x + 0y + 0z = 0  ------ (2)

1x + 0y + 7z = 0  ------ (3)

By solving (1) and (3) we get the eigen vector



 The eigen vector x =

 
0
-1
0
 

Substitute λ = 4 in the matrix A - λI

                             
  =
 
1   0   1
0   -6   0
1   0   4
 
 

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

1x + 0y + 1z = 0  ------ (4)

0x - 6y + 0z = 0  ------ (5)

1x + 0y + 4z = 0  ------ (6)

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



 The eigen vector y = diagonalization of matrix 1

 
1
0
-1
 

Substitute λ = 6 in the matrix A - λI

                             
  =
 
-1   0   1
0   -8   0
1   0   -1
 
 

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

-1x + 0y + 1z = 0  ------ (7)

0x + 8y + 0z = 0  ------ (8)

1x + 0y - 1z = 0  ------ (9)

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



The eigen vector z =

 
1
0
1
 

Let P =

 
0 1 0
1 0 -1
1 0 1
 

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

 
-2 0 0
0 4 0
0 0 6
 

Questions

Solution


Question 2 :

Diagonalize the following matrix

 
1 1 3
1 5 1
3 1 1
 




Solution

Question 3 :

Diagonalize the following matrix

 
-2 2 -3
2 1 -6
-1 -2 0
 




Solution

Question 4 :

Diagonalize the following matrix

 
4 -20 -10
-2 10 4
6 -30 -13
 




Solution

Question 5 :

Diagonalize the following matrix

 
11 -4 -7
7 -2 -5
10 -4 -6
 




Solution

diagonalization of matrix 1







Diagonalization of Matrix1 to Characteristic Equation
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