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

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.

• You're basically only missing a proof that the multiplicities for the nonzero $\lambda_i$ are the same and for $\lambda = 0$ they are off by the constant $m-n$, right? – AlexR May 24 '15 at 16:38
• @AlexR yes this part seems hard – Vim May 24 '15 at 17:04
• Already answered by Maisam Hedyelloo. – user1551 May 24 '15 at 20:12

## 1 Answer

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.

• Thanks! But I'm still a bit confused. What do you mean by "change of basis" and how do you get B? – Vim May 24 '15 at 16:53
• Oh sorry. Just got it. I was being stupid .... Brilliant answer! – Vim May 24 '15 at 16:55
• One more question, how do you get $\chi (AB)$ and $\chi (BA)$? – Vim May 24 '15 at 17:06
• alright. Got that. – Vim May 24 '15 at 17:28
• I'm impressed by the quality of your questions (I enjoyed reading and thinking about every of them), and by your mastery of English as well. I'm an undergraduate studying mainly math and physics in Paris. I like to crack math problems. – Gabriel Romon May 24 '15 at 17:52