## 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 -λ

=  (-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

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