Diagonalize the matrix 2 2 3 5 SolutionLet the givenmatrix b

Diagonalize the matrix (2 2 -3 -5)

Solution

Let the givenmatrix be denoted by A. The characteristic equation of A is det(A- I₂)= 0 or, ²+3-4 = 0 or, (+4)( -1) = 0. Hence, the eigenvalues of A are ₁ =-4 and ₂ =1. Further, the eigenvector of A corresponding to the eigenvalue -4 is solution to the equation (A+4I₂)X= 0. To solve this equation, we will reduce A+4I₂ to its RREF as under:

Multiply the 1st row by 1/6

Add -2 times the 1st row to the 2nd row

Then, the RREF of A+4I₂ is

1

-1/2

0

0

Now, if X = (x,y)^T, then the equation (A+4I₂)X= 0 is equivalent to x-y/2 = 0 or, x = y/2. Then X = (y/2, y)^T = y/2(1,2)^T. Hence the eigenvector of A corresponding to the eigenvalue -4 is (1,2)^T. Similarly, the eigenvector of A corresponding to the eigenvalue 1 is solution to the equation (A-I₂)X= 0. The RREF of   A-I₂ is

1

-3

0

0

Thus, if X = (x,y)^T, then the equation (A-I₂)X= 0 is equivalent to x-3y = 0 or, x = 3y. Then X = (3y, y)^T = y(3,1)^T. Hence the eigenvector of A corresponding to the eigenvalue 1 is (3,1)^T.

Since A has 2 distinct and linearly independent eigenvectors, therefore A can be diagonalized i.e. A = SDS^-1 , where D is a diagonal matrix with the eigenvalues of A on its leading diagonal and S has columns which are eigenvectors of A, in the same order. Then S =

1

3

2

1

Also, D =

-4

0

0

1

1	-1/2
0	0