Anyway, you do so in the same way you always find the eigenvectors. Nothing changes. First, find the eigenvalues as you've done. Choose one of the eigenvalues, say, , and subtract it from the diagonal entries:
Solving for in will give us the eigenvectors just like it does when we have real eigenvalues. [Added: This is from the equation , which can be rewritten as ].
We know that this matrix is singular because we made it singular. We did that by subtracting just the right eigenvalues that would make the determinant zero! So Just set and solve for . In other words, we want to make sure this is true:
Super. That means we just need to solve for in
It turns out that so we can write and know it's true. Do the matrix multiplication to convince yourself.
So you've found one basis eigenvector, which is . There's one more basis for the other eigenvalue, but I think you can work that out yourself now