Let $A$ be a $m \times n$ and $B$ be a $n \times m$ such that $AB$ and $BA$ defined then $ch_{BA}(t)=t^{n-m}ch_{AB}(t)$, If $n>m.$
My thought: As we know $tr(AB)=tr(BA)$ when $AB $ and $BA$ is defined so the eigenvalues are same and for that $BA$ has extra eigenvalue as zero. But how do I proof the above thing? Thanks.
Define$$M=\begin{bmatrix}t\operatorname{Id}_m & A \\B & \operatorname{Id}_n \end{bmatrix}\text{ and }N = \begin{bmatrix}\operatorname{Id}_m & 0 \\-B & t\operatorname{Id}_n \end{bmatrix}.$$ We have$$MN=\begin{bmatrix}t\operatorname{Id}_m-AB&tA\\0&t\operatorname{Id}_n\end{bmatrix}\text{ and }NM=\begin{bmatrix}t\operatorname{Id}m&A\\0&t\operatorname{Id}_n-BA\end{bmatrix}$$and therefore $\det(MN)=t^n\operatorname{ch}_{AB}(t)$ and $ \det(NM)=t^m\operatorname{ch}_{BA}(t)$. But $\det(MN)=\det(NM)$.