How do inverse matrices affect eigenvalues?

[Old Guy]

Homework Statement

Given a matrix with eigenvalues \lambda_{i}, show that if the inverse of the matrix exists, its eigenvalues are \frac{1}{\lambda}.

Homework Equations

The Attempt at a Solution

This shouldn't be so hard. I've come up with a few trivial examples, but I would like to get a general proof in 3 (or more) dimensions. I've tried solving for the eigenvalues, getting the eigenvectors and trying a unitary transformation. I've tried substituting the product of the matrix with its inverse in the characteristic equation and playing around with that before actually calculating the determionant. I've tried other symbolic brute force attempts, but they get very complicated very quickly. I've researched a bunch of determinant identities to no avail. I feel that there must be some key relation that I'm just missing. Any help would be appreciated.

 

[borgwal]

Just write down the definition of eigenvalue and eigenvector, and then apply the inverse matrix to that equation. Nothing complicated involving determinants and/or unitary transformations are necessary.

 

[Old Guy]

Is this the idea?
$\begin{array}{l}<br /> \Lambda \left| \psi \right\rangle = \lambda \left| \psi \right\rangle \\ <br /> \Lambda ^{ - 1} \left( {\Lambda \left| \psi \right\rangle } \right) = \Lambda ^{ - 1} \left( {\lambda \left| \psi \right\rangle } \right) \\ <br /> \left( {\Lambda ^{ - 1} \Lambda } \right)\left| \psi \right\rangle = \left( {\Lambda ^{ - 1} \lambda } \right)\left| \psi \right\rangle \\ <br /> I\left| \psi \right\rangle = \lambda \Lambda ^{ - 1} \left| \psi \right\rangle \\ <br /> \left| \psi \right\rangle = \lambda \Lambda ^{ - 1} \left| \psi \right\rangle \\ <br /> I = \lambda \Lambda ^{ - 1} \\ <br /> \end{array}$

 

[Avodyne]

Fine except for the last line; you can't drop the state because it's not a generic state; it's an eigenstate of Lambda. You can, however, multiply both sides of the next-to-last line by 1/lambda.

 

[borgwal]

Except for the very last line, you're there!

 

[Old Guy]

Yes, I see. Thanks very much!