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Math Help - Determinental Identity

  1. #1
    MHF Contributor Drexel28's Avatar
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    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)
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  2. #2
    Super Member girdav's Avatar
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    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.
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