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Math Help - Invertible matrices proofs

  1. #1
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    Invertible matrices proofs

    Hey, i am having with this question. Would appreciate any help:

    A and B are matrices, with (I-AB) an invertible matrix.

    Show that the inverse of (I-BA) = I+BCA

    Where C is the inverse of (I-AB)
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  2. #2
    AMI
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    \bullet (I+BCA)(I-BA)=I-BA+BCA-BCABA
    C=(I-AB)^{-1}\Rightarrow C(I-AB)=I \Rightarrow C-CAB=I\Rightarrow B(C-CAB)A=BIA \Rightarrow BCA-BCABA=BA\Rightarrow(I+BCA)(I-BA)=I.
    \bullet (I-BA)(I+BCA)=I+BCA-BA-BABCA
    C=(I-AB)^{-1}\Rightarrow (I-AB)C=I \Rightarrow C-ABC=I\Rightarrow B(C-ABC)A=BIA \Rightarrow BCA-BABCA=BA\Rightarrow(I-BA)(I+BCA)=I.

    \Longrightarrow(\exists)\text{ }(I-BA)^{-1}=I+BCA.
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  3. #3
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    Quote Originally Posted by AMI View Post
    \bullet (I+BCA)(I-BA)=I-BA+BCA-BCABA
    C=(I-AB)^{-1}\Rightarrow C(I-AB)=I \Rightarrow C-CAB=I\Rightarrow B(C-CAB)A=BIA \Rightarrow BCA-BCABA=BA\Rightarrow(I+BCA)(I-BA)=I.
    \bullet (I-BA)(I+BCA)=I+BCA-BA-BABCA
    C=(I-AB)^{-1}\Rightarrow (I-AB)C=I \Rightarrow C-ABC=I\Rightarrow B(C-ABC)A=BIA \Rightarrow BCA-BABCA=BA\Rightarrow(I-BA)(I+BCA)=I.

    \Longrightarrow(\exists)\text{ }(I-BA)^{-1}=I+BCA.
    thanks
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  4. #4
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    Just for additional information: It is a simple application of the more general Woodbury formula. Trying to prove that identity is fun
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