The matrix A 0 0 1 1 4 4 13 13 0 0 4 8 0 0 4 8 has two dist

The matrix A = [0 0 -1 -1 -4 -4 13 13 0 0 4 8 0 0 -4 -8] has two distinct eigenvalues lambda_1 lambda_2. Find the eigenvalues and a basis of each eigenspace. Lambda_1 = -4, Basis: [_____ ______ _______ _____], [_____ ______ _______ _____].

Solution

The eigenvalues of A are solutions to the equation det(A- I₄)= 0 or, ⁴+8³+16² = 0 or, ² (²+8+16)=0 or, ²(+4)²= 0. Hence, the eigenvalues of A are ₁ =-4 and ₂ =0 ( of multiplicity 2 each). Further, the eigenvectors of A corresponding to the eigenvalue -4 are solutions to the equation (A+4I₄)X= 0. To solve this equation, we will reduce A+4I₄ to its RREF as under:

Multiply the 1st row by ¼

Add 4 times the 1st row to the 2nd row

Multiply the 2nd row by 1/12

Add -8 times the 2nd row to the 3rd row

Add 4 times the 2nd row to the 4th row

Add 1/4 times the 2nd row to the 1st row

Then the RREF of the matrix A+4I₄ is

1

0

0

0

0

0

1

1

0

0

0

0

0

0

0

0

Now, if X = (x,y,z,w)^T, then the equation (A+4I₄)X= 0 is equivalent to x = 0 and z+w = 0 or, z=-w so that X = (0,y,-w,w)^T = y(0,1,0,0)^T+w(0,0,-1,1)^T. Hence the eigenvectors of A corresponding to the eigenvalue -4 are (0,1,0,0)^Tand (0,0,-1,1)^T. Hence, an eigenbasis of A corresponding to the eigenvalue -4 is {(0,1,0,0)^T, (0,0,-1,1)^T}.

Similarly, the eigenvectors of A corresponding to the eigenvalue 0 are solutions to the equation AX= 0. The RREF of A is

1

1

0

0

0

0

1

0

0

0

0

1

0

0

0

0

Now, if X = (x,y,z,w)^T, then the equation AX= 0 is equivalent to x +y= 0 or, x = -y, z=0 and w = 0 so that X = (-y,y,0,0)^T = y(-1,1,0,0)^T. Hence the eigenvector of A corresponding to the eigenvalue 0 is (-1,1,0,0). Hence, an eigenbasis of A corresponding to the eigenvalue 0 is {(-1,1,0,0)^T}.

1	0	0	0
0	0	1	1
0	0	0	0
0	0	0	0