Let A and B be $n$ x $n$ matrices, let det A $\neq$ $0$, and suppose, for some v $\epsilon$ $\mathbb{R}$
A v = $\lambda_A$v and B v = $\lambda_B$v.
Show that v is an eigenvector of A$^{-1}$B.
Find $\lambda_{A^{-1}B}$, the eigenvector associated to v as an eigenvector of A$^{-1}$B.
Here was my logic for the first part: I multiplied A$^{-1}$ to both sides of B v = $\lambda_B$v
A$^{-1}$B v = A$^{-1}$$\lambda_B$v
= $\lambda_B$A$^{-1}$v
= $\lambda_B$$\lambda_C$v *
= $\lambda$v
$\therefore$ v is an eigenvector of A$^{-1}$B.
*my logic here was since det A $\neq$ $0$, v is an eigenvector of A$^{-1}$ since v is an eigenvector of A, so there is some $\lambda_C$ such that A$^{-1}$v = $\lambda_C$v
I'm not sure if this is correct and I'm not really sure how to start the second part of the question.
It's actually much easier.
If you start out with $Av = \lambda_A v$, and $A$ being invertible, then just apply $A^{-1}$ to the above equation to get: $$ A^{-1}(Av) = A^{-1} (\lambda_A) v \implies A^{-1}v = \frac{v}{\lambda_A} $$
(Note that the eigenvalue has to be non-zero, since the determinant of $A$ is the product of it's eigenvalues, and that is non-zero)
Hence, the $\lambda_C$ which you have mentioned, is now known to you. That is , $\lambda_C = \frac 1{\lambda_A}$. Now, you can figure out the answer to both the first and second question (they come in one package).