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Thread: Linear math help

  1. #1
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    Linear math help

    I'm trying to show that if $\displaystyle AB - I_n$ is invertible then:

    $\displaystyle B(AB-I_n)^{-1}A - I_n = (BA-I_n)^{-1}$

    I have no idea where to start...?
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  2. #2
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    Yes that's an interesting exercise:
    Let $\displaystyle R$ be a ring, $\displaystyle a$ and $\displaystyle b$ two elements in $\displaystyle R$.
    Prove that if $\displaystyle 1-ab$ is left invertible (resp. right invertible), then $\displaystyle 1-ba$ is left invertible (resp. right invertible).

    Here, $\displaystyle AB-I$ is invertible, and the inverse of $\displaystyle BA-I$ is given, but you have to verify it is correct.

    So just multiply $\displaystyle B(AB-I)^{-1}A-I$ by $\displaystyle BA-I$.



    Solution:

    $\displaystyle (B(AB-I)^{-1}A-I)(BA-I)$
    $\displaystyle =B(AB-I)^{-1}ABA-BA-B(AB-I)^{-1}A+I$
    $\displaystyle =B(AB-I)^{-1}ABA-B(AB-I)^{-1}A-BA+I$
    $\displaystyle =B((AB-I)^{-1}AB-(AB-I)^{-1})A-BA+I$
    $\displaystyle =B((AB-I)^{-1}(AB-I))A-BA+I$
    $\displaystyle =BA-BA+I$
    $\displaystyle =I$


    $\displaystyle B(AB-I)^{-1}A-I$ is a left inverse of $\displaystyle BA-I$, therefore it's the inverse of $\displaystyle BA-I$.

    Conclusion: $\displaystyle B(AB-I_n)^{-1}A - I_n = (BA-I_n)^{-1}$
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