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

Now we have to multiply λ with unit matrix I.diagonalization of matrix4

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

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

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

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