## Characteristic Vectors of Matrix

In this page characteristic vectors of matrix we are going to see how to find characteristic equation of any matrix with detailed example.

Definition :

The eigen vector can be obtained from (A- λI)X = 0. Here A is the given matrix λ is a scalar,I is the unit matrix and X is the columns matrix formed by the variables a,b and c.

Another name of characteristic Vector:

Characteristic vector are also known as latent vectors or Eigen vectors of a matrix.

Example :

Determine the characteristic vector of the matrix

 1 1 3 1 5 1 3 1 1

Solution :

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(1-λ) - 3] + 3 [1 - 3 (5-λ)]

= (1-λ) [5-5λ-λ+λ²-1] - 1 [1- λ - 3] + 3[1-15+3λ]

= (1-λ) [λ²-6λ+4] - 1[-2 - λ] + 3[-14+3λ]

= λ² - 6λ + 4 - λ³ + 6λ² - 4λ + 2 + λ - 42 + 9λ

= - λ³ + 7 λ² - 10λ + 10λ + 6 - 42

= - 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. By using synthetic division we have found one value of λ that is λ = 3.

Now we have to solve λ² - 4 λ - 12 to get another two values.For that let us use factoring method.characteristic vectors of matrix

λ² - 4 λ - 12 = 0

λ² - 6 λ + 2 λ - 12 = 0

λ (λ-6) + 2 (λ-6) = 0

(λ+2) (λ-6) = 0

λ + 2 = 0                    λ - 6 = 0

λ = -2                             λ = 6

Therefore the characteristic roots are x = 3,-2 and 6

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  ------ (1)

1x + 2y + 1z = 0  ------ (2)

3x + 1y - 2z = 0  ------ (3)

By solving (1) and (2) we get the eigen vector   characteristic vectors of matrix  characteristic vectors of matrix The eigen vector x =

 1 -1 1

Substitute λ = -2 in the matrix A - λI characteristic vectors of matrix

=

 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  ------ (4)

1x + 7y + 1z = 0  ------ (5)

3x + 1y + 3z = 0  ------ (6)

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

 -1 0 1

Substitute λ = 6 in the matrix A - λI

=

 -5 1 3 1 -1 1 3 1 -5 The eigen vector z =

 1 2 1

Let P =

 1 -1 1 -1 0 2 1 1 1

The column of P are linearly independent eigen vectors of A . Therefore the diagonal matrix =

 3 0 0 0 -2 0 0 0 6

 Questions Solution

Question 1 :

Determine the characteristic vector of the matrix

 5 0 1 0 -2 0 1 0 5

Question 2 :

Determine the characteristic vector of the matrix

 1 1 3 1 5 1 3 1 1

Question 3 :

Determine the characteristic vector of the matrix

 -2 2 -3 2 1 -6 -1 -2 0

Question 4 :

Determine the characteristic vector of the matrix

 4 -20 -10 -2 10 4 6 -30 -13

Question 5 :

Determine the characteristic vector of the matrix

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

characteristic vectors of matrix Characteristic Vectors of Matrix to Characteristic Equation 