Diagonalization of Matrix 5

In this page diagonalization of matrix 5 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 5 :

Diagonalize the following matrix

11 -4 -7 7 -2 -5 10 -4 -6

Now we have to multiply λ with unit matrix I.

To find roots let |A-λI| = 0

 -λ(λ²-3λ²+2) = 0

λ = 0

Now we have to solve λ²-3λ²+2 to get another two values. For that let us factorize 

   λ²-3λ²+2 = 0

λ²-1λ-2λ+2 = 0

λ(λ-1)-2(λ-1) = 0

(λ-1)(λ-2) = 0

 λ - 1 = 0

 λ = 1

 λ - 2 = 0

 λ = 2

Therefore the characteristic roots (or) Eigen values are x = 0,1,2

Substitute λ = 0 in the matrix A - λI diagonalization of matrix 5

A-λI=

11   -4   -7 7   -2   -5 10   -4   -6

 

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

11x - 4y - 7z = 0  ------ (1)

7x - 2y - 5z = 0  ------ (2)

10x - 4y - 6z = 0  ------ (3)

By solving (1) and (2) we get the eigen vector

Substitute λ = 1 in the matrix A - λI

=	10   -4   -7 7   -3   -5 10   -4   -7

 

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

10x - 4y - 7z = 0  ------ (4)

7x - 3y - 5z = 0  ------ (5)

10x - 4y - 7z = 0  ------ (6)

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

Substitute λ = 2 in the matrix A - λI       diagonalization of matrix 5

=	9   -4   -7 7   -4   -5 10   -4   -8

 

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

9x - 4y - 7z = 0  ------ (7)

7x - 4y - 5z = 0  ------ (8)

10x - 4y - 8z = 0  ------ (9)

By solving (7) and (8) we get the eigen vector

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