Diagonalization of matrix1 :
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 1 :
Diagonalize the following matrix

Solution :
Before finding diagonalize matrix first we have to check whether the eigen vectors of the given matrix are linearly independent.
To find eigen vector first let us find characteristic roots of the given matrix. Diagonalization of matrix1 Diagonalization of matrix1
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= 
 
= (5λ)[(2λ) (5λ)  0]  0 [(0  0)] + 1 [0 (2 λ)] = (5λ)[ 10 + 2 λ  5 λ + λ²]  0 + 2 + λ = (5λ)[ 10  3 λ + λ²]  0 + 2 + λ = (5λ)[λ² 3 λ10] + 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 AλI = 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
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 (or) λ  4 = 0
λ = 6 (or) λ = 4
Therefore the characteristic roots (or) Eigen values are x = 2,4,6
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.
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
The eigen vector x = 

Substitute λ = 4 in the matrix A  λI
= 

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 = 

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.
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 = 

Let P = 

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

(2) Diagonalize the following matrix

(3) Diagonalize the following matrix

(4) Diagonalize the following matrix

(5) Diagonalize the following matrix

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