Suppose that B is the inverse of A. Show that if |psi> is an eigenvector of A with eigenvalue a not equal to 0, then |psi> is an eigenvector of B with eigenvalue 1/a. So I know that A|psi> = a|psi>, and I'm trying to prove that A^(-1)|psi> = 1/a|psi>. I tried simplifying A as a 2x2 matrix and then doing the inverse of that. And then I assumed that the inverse of A has an eigenvalue b. So then I did the determinant of A^(-1)-b = 0 in the hopes to find b and see that it's equal to 1/a. But that became really messy. Any suggestions on how to solve this problem? Thank you so much!