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.