How to prove this identity involving characteristic polynomials on both sides?

Question

Suppose $A\in \Bbb C^{m\times n},B\in \Bbb C^{n\times m},m\ge n$, prove: $$\det(\lambda I_m-AB)=\lambda^{m-n}\det(\lambda I_n-BA)$$ I don't want to get into nasty determinant calculation. Instead, I think comparing the polynomial factors on both sides might help. My attempt so far shows that $AB$ and $BA$ share the same non-zero eigenvalues, and that if $BA$ has 0 as eigenvalue, so does $AB$. I guess I'm on the right track but I can't proceed. The multiplicities of $(\lambda-\lambda_i)$s on both sides seem to be big trouble. I cannot prove they are equal. Can you help me? Thanks in advance.

Answer 1

I don't think the following could be caracterized as "nasty determinant calculation". I don't know how one can prove the equality without indulging in some computation.

Let $r=\operatorname{rank}(A)$

From a well-known theorem, derive that there exists $P,Q$ invertible $m\times m$ and $n \times n$ matrices such that $$A=P\begin{bmatrix}I_r& 0\\ 0 &0\end{bmatrix}Q $$

where $I_r$ denotes the $r\times r$ identity matrix.

By changes of basis, $$B=Q^{-1}\begin{bmatrix}E& F\\ G &H\end{bmatrix}P^{-1}$$

For some submatrices $E,F,G,H$.

---

Note that $AB=P\begin{bmatrix}E& F\\ 0&0\end{bmatrix}P^{-1}$ and $BA=Q^{-1}\begin{bmatrix}E& 0\\ G&0\end{bmatrix}Q$.

Hence $\chi_{AB}:=\det(XI_m-AB)=\det(XI_r-E)(X)^{m-r}$ and $\chi_{BA}:=\det(XI_n-BA)\det(XI_r-E)(X)^{n-r}$

Hence $\chi_{BA}=(X)^{n-m}\chi_{AB}$, as desired.