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Math Help - if I-AB is invertible , is I-BA invertible ?

  1. #1
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    if I-AB is invertible , is I-BA invertible ?

    Hi, everyone ~

    I read Linear Algebra by Hoffman & Kunze.

    At 190pg #8,

    A, B := n*n matrices . Prove that if I-AB is invertible, then I-BA is invertible and (I-BA)^(-1) = I +B(I-AB)^(-1) A.

    Any hint or comment are welcomed !

    Please help !

    Thanks.
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  2. #2
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    Re: if I-AB is invertible , is I-BA invertible ?

    Let X=(I-AB)^{-1}. Then I=X(I-AB)=X-XAB or XAB=X-I. Now use this fact and the distributive law to show

    (I+BXA)(I-BA)=I. (You know this proves the statement, don't you?)
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  3. #3
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    Re: if I-AB is invertible , is I-BA invertible ?

    well assume  I - AB is invertible

    then (I-AB)*A = (A - ABA) = A(I - BA) so  (I-AB)*A = A*(I-BA) using the determinant rule  det(I-AB)A = det(I-AB)det(A) = det(I-BA)*A) = det(I-BA)*det(A) since determinants are just numbers, cancelling det(A) on both sides. det(I-AB) = det(I-BA) = 1 thus I-BA is invertible.
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  4. #4
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    Re: if I-AB is invertible , is I-BA invertible ?

    Hi Jakncoke,

    What if det(A) = 0?. I think your argument fails.
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  5. #5
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    Re: if I-AB is invertible , is I-BA invertible ?

    Quote Originally Posted by johng View Post
    Hi Jakncoke,

    What if det(A) = 0?. I think your argument fails.
    yes you are right! what was i thinking.
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  6. #6
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    Re: if I-AB is invertible , is I-BA invertible ?

    since they actually give the answer, i don't see what the fuss is about:

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

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

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

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

    I - BA + B[(I - AB)^{-1}(I - AB)]A = I - BA + BA = I.
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  7. #7
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    Re: if I-AB is invertible , is I-BA invertible ?

    Yes, i see.
    Thanks everyone !!
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