What is $C^{-1}$v and how do I prove that $C^{-1}$v is an eigenvector of $C^{-1}AC$
Solution 1:
Note that $C^{-1}$v is actually a column vector. Now just compute the required multiplication:
$C^{-1}AC(C^{-1})v=C^{-1}Av=C^{-1}\lambda v=\lambda C^{-1}v$.
For the second part, since $C$ is invertible, notice that any vector $u$ can be written as $u=C^{-1}v$ for some $v$ a vector. Thus we have, in this form, assuming $u$ is a e.v.,
$C^{-1}ACu=C^{-1}Av=\mu u$ where $\mu$ is a eigenvalue this time. Some simple rearrangement should give the answer.