 Solution

Use the Characteristic polynomials of AB and BA, which also account for the zero eigenvalue and algebraic multiplicities. The block determinantequals both $\det (I+tAB)$ and $\det (I+tBA)$, by taking the two Schur complements. For $\lambda \ne 0$, set $t=-1/\lambda$ and multiply by ${\lambda }^n$ to obtainBoth sides are polynomials, so agreement away from zero proves equality there as well. Their roots, with multiplicity, are the same. The familiar eigenvector argument $ABv=\mu v\Rightarrow BA(Bv)=\mu Bv$ works directly only when $\mu \ne 0$, because then $Bv\ne 0$; it should not be used to discard the zero case.