Question

Let A be a 4¡Á4 matrix with real entries that has all 1's on the main diagonal (i.e. a11 = a22 = a33 = a44 = 1). If A is singular and ¦Ë1 = 3 + 2i is an eigenvalue of A, then what, if anything, is it possible to conclude about the values of the remaining eigenvalues ¦Ë2, ¦Ë3, and ¦Ë4 ?

Answer: ¦Ë1 = 3 + 2i, ¦Ë2 = 3 ¨C 2i , ¦Ë3 = 0, ¦Ë4 = -2.

Could someone explain why that is the answer? I understand that given ¦Ë1, ¦Ë2 must be an eigenvalue, but how should I have known that 0 and -2 are eigenvalues?

Thanks.

Answer 1

If the matrix is singular, then the determinant must be zero.
One way to get singularity is to have two identical rows, such as:

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

Now we find the eigenvalues by equating to zero the determinant of the following matrix:

1-λ 1 0 0
1 1-λ 0 0
0 0 1-λ 0
0 0 0 1-λ

which is
((1-λ)^2-1)(1-λ)(1-λ)=0

Since two solutions of λ are known, we can substitute into the above equation as follows:

((1-λ)^2-1)(1-3+2i)(1-3-2i)=0 ...(1)

Solving equation (1) for remaining λ gives:
λ=0, -2