diagonalization of matrix 2

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

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

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 =

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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