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This question already has an answer here:

Assume $I+AB$ is invertible, prove then that $I+BA$ is invertible and $(I+BA)^{-1} = I-B(I+AB)^{-1}A$.

My work:

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

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

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

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

Thus $(I+BA)$ is invertible and $B(I+AB)^{-1}B^{-1}$ is its inverse. But I have no clue how to arrive to the given inverse formula. I feel like I'm missing something. Can anyone help?

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marked as duplicate by Brahadeesh, Pierre-Guy Plamondon, Cesareo, mrtaurho, Namaste linear-algebra Jan 4 at 20:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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$$ \begin{equation} \begin{split} (I+BA)(I-B(I+AB)^{-1}A) &= I+ BA-B(I+AB)^{-1}A-BAB(I+AB)^{-1}A \\ &=I+B(I-(I+AB)^{-1}-AB(I+AB)^{-1})A \\ &= I+B(I-(I+AB)(I+AB)^{-1}) \\ &= I+B(I-I) \\ &= I \end{split} \end{equation} $$ $$ \begin{equation} \begin{split} (I-B(I+AB)^{-1}A)(I+BA) &= I+ BA-B(I+AB)^{-1}A-B(I+AB)^{-1}ABA \\ &=I+B(I-(I+AB)^{-1}-(I+AB)^{-1}AB)A \\ &= I+B(I-(I+AB)^{-1}(I+AB)) \\ &= I+B(I-I) \\ &= I \end{split} \end{equation} $$

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  • $\begingroup$ This can be discovered very slickly using power series (a famous problem of Kaplansky). $\endgroup$ – Bill Dubuque Dec 11 '16 at 20:43
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Because $I+AB$ is invertible the matrix $I-B(I+AB)^{-1}A$ is well defined.

We can try this suggested inverse out and get: \begin{align} (I+BA)(I-B(I+AB)^{-1}A) &= (I+BA-(I+BA)B(I+AB)^{-1}A) \\ &= (I+ BA - B(I+AB)(I+AB)^{-1}A) \\ &= (I+BA-BA) \\ &= I \end{align} So $I-B(I+AB)^{-1}A$ is a right inverse to $I+BA$.

Now we try from the left: \begin{align} (I-B(I+AB)^{-1}A)(I+BA) &= I+BA - B(I+AB)^{-1}A(I+BA) \\ &= I+BA - B(I+AB)^{-1}(I + AB)A \\ &= I+BA - BA \\ &= I \end{align} And it turns out to be a left inverse as well. So $I-B(I+AB)^{-1}A$ is the inverse to $I+BA$ and $I+BA$ is invertible.

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  • $\begingroup$ But, how would we discover this formula? $\endgroup$ – littleO Dec 11 '16 at 18:00
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    $\begingroup$ @littleO For a very slick,simple way to discover it see here. $\endgroup$ – Bill Dubuque Dec 11 '16 at 22:59

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