How to show if $ \lambda$ is an eigenvalue of $AB^{-1}$, then $ \lambda$ is an eigenvalue of $ B^{-1}A$? Statement: If $ \lambda$ is an eigenvalue of $AB^{-1}$, then $ \lambda$ is an eigenvalue of $ B^{-1}A$ and vice versa.
One way of the proof.
We have  $B(B^{-1}A ) B^{-1} = AB^{-1}. $ Assuming $ \lambda$ is an eigenvalue of $AB^{-1}$ then we have, 
$$\begin{align*}
\det(\lambda I - AB^{-1})  &= \det( \lambda I - B( B^{-1}A ) B^{-1} )\\
 &=  \det( B(\lambda I - B^{-1}A ) B^{-1})\\
 &= \det(B) \det\bigl( \lambda I - B^{-1}A  \bigr) \det(B^{-1})\\ 
&= \det(B) \det\bigl( \lambda I - (B^{-1}A )\bigr)   \frac{1}{ \det(B) }\\ \
&= \det( \lambda I - B^{-1}A ). 
\end{align*}$$ 
It follows that $ \lambda$ is an eigenvalue of $ B^{-1}A.$ The other side of the lemma can also be proved similarly.
Is there another way how to prove the statement?

A shorter way of seeing this would be to observe that if
$$
(AB^{-1})x=\lambda x
$$
for some non-zero vector $x$, then by multiplying that equation by $B^{-1}$ (from the left) we get that
$$
(B^{-1}A)(B^{-1}x)=\lambda (B^{-1}x).
$$
In other words $(B^{-1}A)y=\lambda y$ for the non-zero vector $y=B^{-1}x$. This process is clearly reversible.
