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This question already has an answer here:

Suppose $A$ and $B$ are two square Matrix. Let $I-AB$ be invertible. I would like to know why $I-BA$ is also invertible? Also what is invert of $I-BA$? Thanks.

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marked as duplicate by JSchlather, Asaf Karagila, Stefan Hansen, Hagen von Eitzen, Jim Mar 7 '13 at 8:16

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  • $\begingroup$ this is a duplicate $\endgroup$ – Rustyn Mar 7 '13 at 6:43
  • $\begingroup$ usually you link the duplicate not just say. $\endgroup$ – user45099 Mar 7 '13 at 6:47
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    $\begingroup$ @user1709828 Sometimes it's difficult to find the other question. $\endgroup$ – JSchlather Mar 7 '13 at 6:54
  • $\begingroup$ Thank you very much for all answers. $\endgroup$ – kami Mar 7 '13 at 7:10
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$$(I-BA)(I+B(I-AB)^{-1}A)=I$$ here $I$ is the identity matrix

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

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    $\begingroup$ This is a special case of the Sherman–Morrison–Woodbury-Justin-Bieber formula. $\endgroup$ – copper.hat Mar 7 '13 at 6:42
  • $\begingroup$ how did Justin-Bieber come into formula? $\endgroup$ – jim Mar 7 '13 at 6:45
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    $\begingroup$ Damn, that guy is everywhere nowadays... $\endgroup$ – copper.hat Mar 7 '13 at 6:49
  • $\begingroup$ I'm just impressed he gets to have his first and last name in the formula. Never seen that before. $\endgroup$ – JSchlather Mar 7 '13 at 7:45
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Wait, here is another approach:

The non-zero eigenvalues of $AB$ and $BA$ are the same (If $ABv= \lambda v$, then $B(AB)v = (BA) (Bv) = \lambda (Bv)$, and since $Bv \neq 0$, we see that $\lambda$ is an eigenvalue of $BA$).

Since $I-AB$ is invertible, this means that $1$ is not an eigenvalue of $AB$, and hence not of $BA$ either. Hence $I-BA$ is invertible.

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  • $\begingroup$ Ok, +1, but this do not gives the inverse. $\endgroup$ – user63181 Mar 7 '13 at 6:51
  • $\begingroup$ That's funny. Usually, I use that $1-ab$ is invertible if and only if $1-ba$ is invertible to conclude that the spectum of $ab$ minus $0$ is the same as the spectrum of $ba$ minus $0$ in a general Banach algebra. $\endgroup$ – Julien Mar 7 '13 at 6:55
  • $\begingroup$ @sbr: I missed that originally, and my only different proof is a block matrix version of jim's solution below. $\endgroup$ – copper.hat Mar 7 '13 at 7:21
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Edit: well, jim just gave you the formula. But if you want to play the game, the hint I give you is a fun way to manipulate expressions which do not a priori make sense to compute a formula which really works. And that's probably like that that the fact and the formula were discovered.

Hint: pretend that the Neumann series of $(I-BA)^{-1}$ converges and use it to make the Neumann series of $(I-AB)^{-1}$ appear. You'll get a formula giving you a candidate for $(I-BA)^{-1}$. Just check that it works.

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This should contain all of the information you might want--probably more.

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