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
 



Solution

Question 2 :

Determine the characteristic vector of the matrix

 
1 1 3
1 5 1
3 1 1
 



Solution

Question 3 :

Determine the characteristic vector of the matrix

 
-2 2 -3
2 1 -6
-1 -2 0
 



Solution


Question 4 :

Determine the characteristic vector of the matrix

 
4 -20 -10
-2 10 4
6 -30 -13
 



Solution

Question 5 :

Determine the characteristic vector of the matrix

 
11 -4 -7
7 -2 -5
10 -4 -6
 



Solution

characteristic vectors of matrix







Characteristic Vectors of Matrix to Characteristic Equation
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