Diagonalization of Matrix 3





In this page diagonalization of matrix 3 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 3 :

Diagonalize the following matrix

 
-2 2 -3
2 1 -6
-1 -2 0
 




   Let A =

 
-2 2 -3
2 1 -6
-1 -2 0
 

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=
 
-2 2 -3
2 1 -6
-1 -2 0
 
-
 
λ 0 0
0 λ 0
0 0 λ
 
 
                      
  =
 
(-2-λ)   (2-0)   (-3-0)
(2-0)   (1-λ)   (-6-0)
(-1-0)   (-2-0)   (0-λ)
 
 
  =
 
(-2-λ)   2   -3
2   (1-λ)   -6
-1   -2  
 
 

  =  (-2-λ)[ -λ(1-λ) - 12 ] - 2[-2 λ - 6] - 3 [-4-(-1)(1-λ) ]

  =  (-2-λ)[ -λ + λ² - 12 ] + 4 λ + 12 - 3 [-4+1-λ ]

  =  (-2-λ)[ λ² -λ - 12 ] + 4 λ + 12 - 3 [-3-λ ]

  =  (-2-λ) [λ² -λ - 12 ] + 4 λ + 12 + 9 + 3 λ

  =  -2λ² + 2λ + 24 - λ³ + λ² + 12 λ  + 4 λ + 12 + 9 + 3 λ

  =  - λ³ - λ² + 2λ + 12 λ  + 4 λ + 3 λ + 24 + 12 + 9

  =  - λ³ - λ² + 21λ + 45

  =   λ³ + λ² - 21λ - 45

To find roots let |A-λI| = 0

   λ³ + λ² - 21λ - 45 = 0   diagonalization of matrix 3

For solving this equation first let us do synthetic division.

By using synthetic division we have found one value of λ that is λ = -3.

Now we have to solve λ² - 2 λ - 15  to get another two values. For that let us factorize 

   λ² - 2 λ - 15  = 0

λ² + 3 λ - 5 λ - 15 = 0

λ (λ + 3) - 5 (λ + 3) = 0

(λ - 5) (λ + 3) = 0

 λ - 5 = 0

 λ = 5

 λ + 3 = 0

 λ = - 3

Therefore the characteristic roots (or) Eigen values are x = -3,-3,5

Substitute λ = -3 in the matrix A - λI

                             
  =
 
1   2   -3
2   4   -6
-1   -2   3
 
 

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

1x + 2y - 3z = 0  ------ (1)

2x + 4y - 6z = 0  ------ (2)

-1x - 2y + 3z = 0  ------ (3)

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



 The eigen vector x =

 
0
0
0
 

Substitute λ = 5 in the matrix A - λI  Diagonalization of Matrix3

                             
  =
 
-7   2   -3
2   -4   -6
-1   -2   -5
 
 

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

-7x + 2y - 3z = 0  ------ (4)

2x - 4y - 6z = 0  ------ (5)

-1x - 2y - 5z = 0  ------ (6)

By solving (4) and (5) we get the eigen vector Diagonalization of Matrix3



The eigen vector z =

 
-24
-48
24
 

Let P =

 
0 0 -24
0 0 -48
0 0 24
 

Eigen vectors of x and y are linearly dependent. So we cannot find diagonal matrix.  diagonalization of matrix 3


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 4 :

Diagonalize the following matrix

 
4 -20 -10
-2 10 4
6 -30 -13
 




Solution

Question 5 :

Diagonalize the following matrix  diagonalization of matrix  3 diagonalization of matrix 3

 
11 -4 -7
7 -2 -5
10 -4 -6
 




Solution







Diagonalization of Matrix3 to Characteristic Equation
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