## Characteristic Vectors of Matrix 1

In this page characteristic vectors of matrix1 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.

Question 1 :

Determine the characteristic vector of the matrix

 5 0 1 0 -2 0 1 0 5

To find eigen vector first let us find characteristic roots of the given matrix.

Let A =

 5 0 1 0 -2 0 1 0 5

The order of A is 3 x 3. So the unit matrix I =

 5 0 1 0 -2 0 1 0 5

Now we have to multiply λ with unit matrix I.

λI =

 λ 0 0 0 λ 0 0 0 λ

A-λI=

 5 0 1 0 -2 0 1 0 5

-

 λ 0 0 0 λ 0 0 0 λ

=

 (5-λ) (0-0) (1-0) (0-0) (-2-λ) (0-0) (1-0) (0-0) (5-λ)

=

 (5-λ) 0 1 0 (-2-λ) 0 1 0 (5-λ)

|A-λI|=

 (5-λ) 0 1 0 (-2-λ) 0 1 0 (5-λ)

=  (5-λ)[(-2-λ) (5-λ) - 0] - 0 [(0 - 0)] + 1 [0- (-2 -λ)]

=  (5-λ)[ -10 + 2 λ - 5 λ + λ²] - 0 + 2 + λ

=  (5-λ)[ -10 - 3 λ + λ²] - 0 + 2 + λ

=  (5-λ)[λ² -3 λ-10] + 2 + λ

=  5 λ² - 15 λ - 50 - λ³ + 3 λ² + 10 λ + 2 + λ

= - λ³ + 5 λ² + 3 λ² - 15 λ + 10 λ + λ - 50 + 2

= - λ³ + 8 λ² - 4 λ - 48

=  λ³ - 8 λ² + 4 λ + 48

To find roots let |A-λI| = 0

λ³ - 8 λ² + 4 λ + 48 = 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 = -2.

Now we have to solve λ² - 10 λ + 24 to get another two values. For that let us factorize  diagonalization of matrix1

λ² - 10 λ + 24 = 0

λ² - 6 λ - 4 λ + 24 = 0

λ (λ - 6) - 4 (λ - 6) = 0

(λ - 6) (λ - 4) = 0

λ - 6 = 0

λ = 6

λ - 4 = 0

λ = 4

Therefore the characteristic roots (or) Eigen values are x = -2,4,6

Substitute λ = -2 in the matrix A - λI  characteristic vectors of matrix1

=

 7 0 1 0 0 0 1 0 7

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

7x + 0y + 1z = 0  ------ (1)

0x + 0y + 0z = 0  ------ (2)

1x + 0y + 7z = 0  ------ (3)

By solving (1) and (3) we get the eigen vector   characteristic vectors of matrix1

The eigen vector x =

 0 -1 0

Substitute λ = 4 in the matrix A - λI

=

 1 0 1 0 -6 0 1 0 4

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

1x + 0y + 1z = 0  ------ (4)

0x - 6y + 0z = 0  ------ (5)

1x + 0y + 4z = 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  characteristic vectors of matrix1

=

 -1 0 1 0 -8 0 1 0 -1

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

-1x + 0y + 1z = 0  ------ (7)

0x + 8y + 0z = 0  ------ (8)

1x + 0y - 1z = 0  ------ (9)

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

The eigen vector z =

 1 0 1

 Questions Solution

characteristic vectors of matrix 1  characteristic vectors of matrix 1  characteristic vectors of matrix 1 characteristic vectors of matrix 1

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 matrix1characteristic vectors of matrix 1
characteristic vectors of matrix 1

Characteristic Vectors of Matrix1 to Characteristic Equation
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