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Show that if $A$ and $I+AB$ are invertible, then $I+BA$ is also invertible with $$(I+BA)^{-1} = A^{-1}(I+AB)^{-1}A$$

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Multiply $I+BA$ on the left with $A^{-1}(I+AB)^{-1}A$ try to simplify it using the distributive property. Then do the same on the right. –  Tim Seguine Jan 26 '13 at 22:13
    
Three answers, none from me, and yet mine is the only vote for the question so far. –  Michael Hardy Jan 26 '13 at 22:16
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3 Answers

up vote 11 down vote accepted

Hint: Just do it. Use the fact that

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

to show that

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

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Is it also necessary to show $A^{-1} (I+AB)^{-1} A (I + BA)= I$? –  mez Jan 26 '13 at 22:14
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@mezhang For general square matrices $M, N$, if we know that $M N = I$, then $N M = I$. –  Calvin Lin Jan 26 '13 at 22:16
    
@mezhang if a square matrix has a one sided inverse, then it has an inverse. CalvinLin: Your comment is only true for square matrices. –  Tim Seguine Jan 26 '13 at 22:19
    
@Tim Could you show me a proof for your statement? I don't want to raise a separate question. Thanks. –  mez Jan 26 '13 at 22:21
    
@mezhang Check the lemma in my answer here: math.stackexchange.com/questions/280884/… –  Git Gud Jan 26 '13 at 22:22
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$$ (I + BA) = A^{-1}(I + AB)A \Rightarrow (I + BA)^{-1} = A^{-1}(I + AB)^{-1}A $$

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Since $\,A^{-1}C^{-1}A=(A^{-1}CA)^{-1}\,$ , we get with $\,C=I+AB\,$ :

$$(I+BA)\left(A^{-1}(I+AB)^{-1}A\right)=(I+BA)(A^{-1}(I+AB)A)^{-1}=$$

$$(I+BA)(I+BA)^{-1}=I$$

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