If AB is diagonalisable, then so is $(BA)^2$ I have shown that the minimal polynomials of $XY$ and $YX$ differ by at most a factor of $t$ for any square complex matrices $X$ and $Y$ and that their characteristic polynomials are equal. Using this, I argued as follows:
$AB$ diagonalisable so $(AB)^2$ diagonalisable so we can write down its minimal polynomial. $(BA)^2=B(ABA)$ so the minimal polynomial of $(BA)^2$ differs from the minimal polynomial of $(AB)^2$ by at most a factor of $t$. So if $t$ doesn't divide $m_{AB}(t)$ then we're done since $m_{(BA)^2}$ then a product of distinct linear factors.
I'm having trouble figuring out what to do if $t$ divides $m_{AB}$, in other words if $AB$ has $0$ as an eigenvalue. If $m_{(BA)^2}$ drops the factor of $t$ or is $m_{(AB)^2}$ then we're done, but how can I show it doesn't gain a factor of $t$?
 A: noting that both matrices are square, here's a simple rank based argument
$\text{rank}\Big(\big(AB\big)\Big)\geq \text{rank}\Big(\big(BA\big)^2\Big)\geq \text{rank}\Big(A\big(BA\big)^2B\Big)=\text{rank}\Big(\big(AB\big)^3\Big)=\text{rank}\Big(\big(AB\big)\Big)$
$\implies \dim \ker \Big(\big(AB\big)^2\Big)=\dim \ker \Big(\big(AB\big)\Big)= \dim \ker\Big(\big(BA\big)^2\Big)$
$\implies \text{alg mult }_{\lambda =0} \big((BA)^2\big)=\text{alg mult }_{\lambda =0} \big((AB)^2\big)= \text{geo mult}_{\lambda =0} \Big(\big(BA\big)^2\Big)$
A: View $A$ as a linear map between two vector spaces $V$ and $W$, and view $B$ as a linear map from $W$ to $V$. Since $AB$ is diagonalisable, by picking two appropriate bases of $V$ and $W$, the two linear maps have matrix representations
$$
A=\pmatrix{\Delta&0\\ 0&0},\ B=\pmatrix{X&Y\\ Z&W}
$$
where $\Delta$ is nonsingular and $\Delta X=D\oplus0$ for some nonsingular diagonalisable matrix $D$. It follows that
$$
BA=\pmatrix{X\Delta&0\\ Y\Delta&0}\sim\pmatrix{\Delta X&0\\ Y&0}=:
\left(\begin{array}{cc|c}
D&0&0\\ 0&0&0\\ \hline F&G&0
\end{array}\right)
$$
where the symbol $\sim$ denotes similarity and $Y=\pmatrix{F&G}$. Consequently
$$
(BA)^2\sim
\left(\begin{array}{cc|c}
D&0&0\\ 0&0&0\\ \hline F&G&0
\end{array}\right)^2
=\left(\begin{array}{cc|c}
D^2&0&0\\ 0&0&0\\ \hline FD&0&0
\end{array}\right)
\sim\left(\begin{array}{cc|c}
D^2&0&0\\ 0&0&0\\ \hline 0&0&0
\end{array}\right).
$$
