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Math Help - Linear math help

  1. #1
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    Linear math help

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

    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 R be a ring, a and b two elements in R.
    Prove that if 1-ab is left invertible (resp. right invertible), then 1-ba is left invertible (resp. right invertible).

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

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



    Solution:

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


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

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