Eigenvalues of an Invertible Matrix

[phatalerror]

My Linear Algebra textbook omits a proof for if lambda is an eigenvalue of an invertible matrix (non-zero of course), then 1 / lambda is an eigenvalue of the inverse of said matrix. Anyone care to share?

 

[galactus]

Suppose $\displaystyle Ax={\lambda}x$ where A is invertible.

Then $\displaystyle x=A^{-1}Ax=A^{-1}{\lambda}x={\lambda}A^{-1}x$

Since A is invertible, we know that $\displaystyle {\lambda}\neq{0}$.

Therefore, $\displaystyle A^{-1}x=\frac{1}{\lambda}x$

So, $\displaystyle \frac{1}{{\lambda}}$ is an eigenvalue of $\displaystyle A^{-1}$ and x is the eigenvector.