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: 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.
A: Even if $A$ is $n\times m$ and $B$ is $m\times n$ with $m\le n$, we have
$$
\det(\lambda I_n-AB)=\lambda^{n-m}\det(\lambda I_m-BA)\tag{1}
$$
Proof:
Drawing from an answer of julien's,
$$
\begin{bmatrix}I_n&-A\\0&\lambda I_m\end{bmatrix}
\begin{bmatrix}\lambda I_n&A\\B&I_m\end{bmatrix}
=\begin{bmatrix}\lambda I_n-AB&0\\\lambda B&\lambda I_m\end{bmatrix}\tag{2}
$$
$$
\begin{bmatrix}I_n&0\\-B&\lambda I_m\end{bmatrix}
\begin{bmatrix}\lambda I_n&A\\B&I_m\end{bmatrix}
=\begin{bmatrix}\lambda I_n&A\\0&\lambda I_m-BA\end{bmatrix}\tag{3}
$$
Since the determinants on the left sides of $(2)$ and $(3)$ are equal, the determinants on the right side prove
$$
\lambda^m\det(\lambda I_n-AB)=\lambda^n\det(\lambda I_m-BA)\tag{4}
$$
In the case of square matrices, since the characteristic polynomials are the same, the eigenvalues are the same.
