# Eigenvalues and Eigenvectors

1. Dec 14, 2006

### azone

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!

2. Dec 14, 2006

### marcusl

Start with A|psi> = a|psi> and solve for (isolate) |psi> in terms of inv(A). You should be able to take it from there.

3. Dec 18, 2006

### Marco_84

well it is not too difficult:

you supposed A invertible and B=A^(-1).

since AB=BA=1---->[A,B]=0.

(1)|psi>=BA|psi>=Ba|psi>=AB|psi>=ab|psi>.

but ab=1 so the eigenvalue b must be b=1/a