Re: Determinental Identity

We have a more general result: if we denote by $\displaystyle \chi_{AB}:=\det(xI_n-AB)$ (resp. $\displaystyle \chi_{BA}$) the characteristic polynomial of $\displaystyle AB$ (resp $\displaystyle BA$), we have $\displaystyle \chi_{AB}(x)x^k =\chi_{BA}(x)x^n$. To see that, we notice that

$\displaystyle \begin{align*}\begin{pmatrix} xI_n &A\\ xB&xI_k\end{pmatrix}&=\begin{pmatrix}xI_n-AB &A\\0&xI_k\end{pmatrix}\begin{pmatrix} I_n&0\\B&I_k\end{pmatrix}\\&=\begin{pmatrix} I_n&0\\B&I_k\end{pmatrix}\begin{pmatrix} xI_n&A\\0&xI_k-BA\end{pmatrix}\end{align*}$

and take the determinant.

Now, we take $\displaystyle x=-1$ to show the asked result.