Eigenvalues of $AB$ and $BA$ where $A$ and $B$ are rectangular matrices

Let $\lambda\neq 0$ be an eigenvalue of $AB$

Then, for some non-zero $v$, $ABv=\lambda v$

Hence $BABv=\lambda Bv$

Equivalently $(BA)(Bv)=\lambda (Bv)$

Note that $Bv \neq 0$. Otherwise, $ABv=\lambda v=0$, hence $\lambda=0$

Hence $\lambda$ is a non-zero eigenvalue of $BA$

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Switching $A$ and $B$ in the previous proof, it also holds that a non-zero eigenvalue of $BA$ is a non-zero eigenvalue of $AB$

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Conclusion: $AB$ and $BA$ have the same non-zero eigenvalues.

This is a totally different approach, but way more powerful.

I'm going to prove that $\chi_{BA}=(-X)^{n-m}\chi_{AB}$ by elementary means.

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Let $r=\operatorname{rank}(A)$

From a well-known theorem, derive that there exists $P,Q$ invertible $m\times m$ and $n \times n$ matrices such that $$A=P\begin{bmatrix}I_r& 0\\ 0 &0\end{bmatrix}Q $$

where $I_r$ denotes the $r\times r$ identity matrix.

By changes of basis, $$B=Q^{-1}\begin{bmatrix}E& F\\ G &H\end{bmatrix}P^{-1}$$

For some submatrices $E,F,G,H$.

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Note that $AB=P\begin{bmatrix}E& F\\ 0&0\end{bmatrix}P^{-1}$ and $BA=Q^{-1}\begin{bmatrix}E& 0\\ G&0\end{bmatrix}Q$.

Hence $\chi_{AB}=\det(E-XI_r)(-X)^{m-r}$ and $\chi_{BA}=\det(E-XI_r)(-X)^{n-r}$

Hence $\chi_{BA}=(-X)^{n-m}\chi_{AB}$.

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The results in the two other answers are now a simple consequence of the formula.

Also, note that $BA$ will have an eigenvalue of $0$, since it's $n\times n$, but the maximum rank of each of $A$ and $b$ is $m<n$.