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
 




Solution

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






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