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
 




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 5

 
4 -20 -10
-2 10 4
6 -30 -13
 




Solution






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