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







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