# Determinental Identity

• Aug 5th 2011, 11:02 AM
Drexel28
Determinental Identity
The challenge section has been pretty dead as of late. I think one of the reasons is that some of the problems (or at least the problem's subject matter) is somewhat inaccessible. So, I propose the following question which has the threefold benefit of a) having an easy to understand statement, b) having both a difficult but elementary and an easier but less elementary solution, and c) having real applications in matrix based mathematics. So:

Problem: Let $A$ and $B$ be a $n\times k$ and $k\times n$ complex matrices. Then,

$\det\left(I_n+AB\right)=\det\left(I_k+BA\right)$
• Aug 5th 2011, 12:04 PM
girdav
Re: Determinental Identity
We have a more general result: if we denote by $\chi_{AB}:=\det(xI_n-AB)$ (resp. $\chi_{BA}$) the characteristic polynomial of $AB$ (resp $BA$), we have $\chi_{AB}(x)x^k =\chi_{BA}(x)x^n$. To see that, we notice that
\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 $x=-1$ to show the asked result.