 Solution

If $A$ is an invertible matrix, thenIf instead $B$ is an invertible matrix, thenThus in either case $AB$ and $BA$ are similar matrices. Explicitly, whenever $Y=P^{-1}XP$ and $Xu=\lambda u$ with $u\ne 0$, one has $Y(P^{-1}u)=\lambda P^{-1}u$ and $P^{-1}u\ne 0$. Applying the same argument with $P^{-1}$ in place of $P$ proves equality of the sets of eigenvalues, including zero.