TITLE: Validity of my weird proof that $AB$ and $BA$ have the same eigenvalues?
QUESTION [9 upvotes]: On a recent linear algebra exam, I was required to prove that "for every $n \times k$ matrix $A$ and $k \times n$ matrix $B$ over the same field, it holds that $AB$ and $BA$ have the same eigenvalues except for 0."
This is a classic result and I really should have known better, but for some reason the only proof I could come up with is the following. It's kind of uncanny (the examiner rejected it), but I think it might be valid anyway. Am I correct?

Proof:
Let $M_{AB}(t)$ and $M_{BA}(t)$ be the minimal polynomials of $AB$ and $BA$ respectively.
It holds that:
$M_{AB}(AB)=0$.
Therefore:
$B \cdot M_{AB}(AB)=0$
$M_{AB}(BA) \cdot B=0$
$M_{AB}(BA) \cdot BA=0$
So we see that $BA$ is a zero of the polynomial $M_{AB}(t) \cdot t=0$.
Therefore, $M_{BA}(t)\: |\: M_{AB}(t)\,t$, which means that $M_{BA}(t)$ and $M_{AB}(t)$ share the same irreducible factors except possibly for $t$, and the claim follows. $\square$

Please note:
I know that this result has been addressed many times on this site, but I didn't find my proof anywhere, so I think this isn't a duplicate.

REPLY [2 votes]: Your answer looks correct. It's definitely more complicated than the standard proof.
You could spell out the reason that $BM_{AB}(AB)=M_{AB}(BA)B$, but it's actually clever. Not sure I'd call it "uncanny."
Oh, you also left out the symmetry part - that $M_{BA}(t)\mid M_{AB}(t)t$ and $M_{AB}(t)\mid M_{BA}(t)t$. That symmetry is needed to get that they share the same set of non-zero eigenvalues.