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

Now we have to multiply λ with unit matrix I.

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

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

Eigen vectors of x and y are linearly dependent. So we cannot find diagonal matrix.  diagonalization of matrix 3

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Diagonalization of Matrix3 to Characteristic Equation