The relationship between the eigenvalues of matrices $XY$ and $YX$ If $X \in \mathbb{C}^{m \times n}$ and $Y \in \mathbb{C}^{n \times m}$ ($m \geq n$), how to prove the following?
$$\sigma (XY) = \sigma(YX) \cup \underbrace{\left \{ 0, ..., 0 \right \}}_{m-n}$$
Here, $\sigma$ denotes the set of eigenvalues/spectrum.
 A: You could modify the proof of Sylvester's determinant theorem to show that $$\text{det} \left( \lambda I_m - XY \right) = \text{det} \left( \lambda I_n - YX \right)$$ for all $\lambda \neq 0$. This shows equivalence for all nonzero eigenvalues. For the zero eigenvalues, an application of the fundamental theorem of algebra is sufficient. Note that the characteristic polynomial of $XY$ (or $YX$) must have $m$ (or $n$) roots. Since we have examined all roots $\lambda\neq0$, the remaining roots must be zero.
Edit: This proof does not work, see the comments.
A: It is very easy to check that $XY$ and $YX$ have the same nonzero eigenvalues. Just apply $Y$ to the identity $XYv=\lambda v$ and note $Yv\neq 0$ if $\lambda v\neq 0$.
But if you mean $\lambda$ to be the set of eigenvalues counted with multiplicities, then what you are asking is equivalent to 
$$
\chi_{XY}(t)=t^{m-n}\chi_{YX}(t)
$$
where $\chi_A(t)=\det(tI-A)$ is the charateristic polynomial.
Big hint:
$$
\left(\matrix{I&X\\Y&tI}\right)\left(\matrix{tI&-X\\0&I}\right)=\left(\matrix{tI&0\\*&tI-YX}\right)
$$
and
$$
\left(\matrix{I&X\\Y&tI}\right)\left(\matrix{tI&0\\-Y&I}\right)=\left(\matrix{tI-XY&*\\0&tI}\right).
$$
A: Let $\lambda\in\mathbb C-\{0\}$. Partition matrix $M$ as follows:
$$M=\left(\begin{array}\\I_m&A\\\lambda^{-1}B&I_n\end{array}\right).$$
Using determinant factorization property of Schur complements, we have
$$\begin{split}
|M|&=|I_m||M/I_m|\\
&=|I_n-\lambda^{-1}BI_m^{-1}A|\\
&=(-\lambda^{-1})^n|BA-\lambda I_n|\\
&=(-\lambda)^{-n}|BA-\lambda I_n|.\end{split}$$
Using the same factorization, we have
$$\begin{split}
|M|&=|I_n||M/I_n|\\
&=|I_m-AI_n^{-1}\lambda^{-1}B|\\
&=(-\lambda^{-1})^m|AB-\lambda I_m|\\
&=(-\lambda)^{-m}|AB-\lambda I_m|.\end{split}$$
As a result, we have $(-\lambda)^{-m}|AB-\lambda I_m|=(-\lambda)^{-n}|BA-\lambda I_n|$, which in turn implies $|AB-\lambda I_m|=(-\lambda)^{m-n}|BA-\lambda I_n|$. The last equation is called Weinstein–Aronszajn identity.
From Weinstein–Aronszajn identity, it is clear that $AB$ and $BA$ has the same set of non-zero eigenvalues. The remaining eigenvalues, if any, must be zeroes because otherwise it would already have been considered previously.
