CHARACTERISTIC VECTORS OF MATRIX

About "Characteristic vectors of matrix"

Characteristic vectors of matrix : 

Here we are going to see how to find characteristic equation of any matrix with example problems.

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 names of characteristic Vectors of matrix :

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 



 The eigen vector x =

 
1
-1
1
 

Substitute λ = -2 in the matrix A - λI

                             
  =
 
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
 

Practice questions

(1)  Determine the characteristic vector of the matrix

 
5 0 1
0 -2 0
1 0 5
 

Solution

(2)  Determine the characteristic vector of the matrix

 
1 1 3
1 5 1
3 1 1
 

Solution

(3)  Determine the characteristic vector of the matrix

 
-2 2 -3
2 1 -6
-1 -2 0
 

Solution

(4)  Determine the characteristic vector of the matrix

 
4 -20 -10
-2 10 4
6 -30 -13
 

Solution

(5)  Determine the characteristic vector of the matrix

 
11 -4 -7
7 -2 -5
10 -4 -6
 

Solution

Related topics

After having gone through the stuff given above, we hope that the students would have understood "Characteristic vectors of matrix"

Apart from the stuff given above, if you want to know more about "Characteristic vectors of matrix", please click here.

Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.

Widget is loading comments...