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






HTML Comment Box is loading comments...
Enjoy this page? Please pay it forward. Here's how...

Would you prefer to share this page with others by linking to it?

  1. Click on the HTML link code below.
  2. Copy and paste it, adding a note of your own, into your blog, a Web page, forums, a blog comment, your Facebook account, or anywhere that someone would find this page valuable.












Featured Categories 

Math Word Problems

SAT Math Worksheet

P-SAT Preparation

Math Calculators

Quantitative Aptitude

Transformations

Algebraic Identities

Trig. Identities

SOHCAHTOA

Multiplication Tricks

PEMDAS Rule

Types of Angles

Aptitude Test