In this page diagonalization of matrix 2 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 2 :
Diagonalize the following matrix

Let A = 

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

Now we have to multiply λ with unit matrix I.
λI = 

AλI= 

 

= 

= 

AλI= 
 
= (1λ)[ (5λ)(1λ)  1 ]  1[1  λ  3] + 3 [1  3 (5λ) ] = (1λ)[ 5  5 λ  λ + λ²  1 ]  1[ λ  2] + 3 [ 1  15 +3 λ ] = (1λ)[ λ²  6 λ + 4 ] + 1 λ + 2 + 3 [  14 +3 λ ] = λ²  6 λ + 4  λ³ + 6 λ²  4 λ + λ + 2  42 + 9 λ =  λ³ + λ² + 6 λ²  6 λ  4 λ + λ + 9 λ + 4 + 2  42 =  λ³ + 7 λ²  10 λ + 10 λ + 6  42 =  λ³ + 7 λ²  36 = λ³  7 λ² + 36 
To find roots let AλI = 0
λ³  7 λ² + 36 = 0
For solving this equation first let us do synthetic division.diagonalization of matrix 2
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
λ²  9 λ + 18 = 0
λ²  3 λ  6 λ + 18 = 0
λ (λ  3)  6 (λ  3) = 0
(λ  6) (λ  3) = 0
λ  6 = 0
λ = 6
λ  3 = 0
λ = 3
Therefore the characteristic roots (or) Eigen values are x = 2,3,6
Substitute λ = 2 in the matrix A  λI
= 

From this matrix we are going to form three linear equations using variables x,y and z.
3x + 1y + 3z = 0  (1)
1x + 7y + 1z = 0  (2)
3x + 1y + 3z = 0  (3)
By solving (1) and (2) we get the eigen vector
The eigen vector x = 

Substitute λ = 3 in the matrix A  λI
= 

From this matrix we are going to form three linear equations using variables x,y and z.
2x + 1y + 3z = 0  (4)
1x + 2y + 1z = 0  (5)
3x + 1y  2z = 0  (6)
By solving (4) and (5) we get the eigen vector
The eigen vector y = 

Substitute λ = 6 in the matrix A  λI
= 

From this matrix we are going to form three linear equations using variables x,y and z.
5x + 1y + 3z = 0  (7)
1x  1y + 1z = 0  (8)
3x + 1y  5z = 0  (9)
By solving (7) and (8) we get the eigen vector diagonalization of matrix 2
The eigen vector z = 

Let P = 

The column of P are linearly independent eigen vectors of A . Therefore the diagonal matrix = diagonalization of matrix 2 

Questions 
Solution 
Question 1 : Diagonalize the following matrix

 
Question 3 : Diagonalize the following matrix

 
Question 4 : Diagonalize the following matrix diagonalization of matrix 2

 
Question 5 : Diagonalize the following matrix diagonalization of matrix 2

