The matrix A 2 1 1 1 2 1 3 3 2 has two real eigenvalues one

The matrix A = [-2 -1 1 -1 -2 1 -3 -3 2] has two real eigenvalues, one of multiplicity 1 and one of multiplicity 2. Find the eigenvalues and a basis of each eigenspace.

Solution

The characteristic equation of A is det (A- I₃) = 0 or, ³ +2²+ = 0 or, (²+2 +1) = 0 or, (+1)²= 0. Thus, A has 2 distinct eigenvalues, ₁ = ₂= -1 (of multiplicity 2) and ₃ = 0( of multiplicity 1 ).

The eigenvectors of A corresponding to the eigenvalue are solutions to the equation (A- I₃)X = 0. Thus, the eigenvectors of A corresponding to the eigenvalue -1 are solutions to the equation (A+I₃)X = 0. We will reduce A+I₃ to its RREF as under:

Multiply the 1st row by -1

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

Add -1 times the 1st row to the 3rd row

Then the RREF of A+I₃ is

1

1

3

0

0

0

0

0

0

If X=(x,y,z)^T, then the above equation is equivalent to x +y+3z=0. Hence x=-y-3z so that X=(-y-3z,y,z)^T = y(-1,1,0)^T+z(-3,0,1)^T.Thus,the eigenbasis of A corresponding to the eigenvalue -1 is {(-1,1,0)^T, (-3,0,1)^T}.

The eigenvector of A corresponding to the eigenvalue 0 is solution to the equation AX = 0. The RREF of A is

1

0

1

0

1

1

0

0

0

If X = (x,y,z)^T , then the above equation is equivalent to x+z = 0 andy+z = 0 so that x =-z and y =-z. Also, X = (-z,-z,z)^T = z(-1,-1,1)^T

Thus, the eigenbasis of A corresponding to the eigenvalue 0 is {(-1,-1,1)^T }.

1	1	3
0	0	0
0	0	0