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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.

M.Sina
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kami
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4 Answers4

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$$(I-BA)(I+B(I-AB)^{-1}A)=I$$ here $I$ is the identity matrix

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

jim
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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.

Julien
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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.

copper.hat
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  • Ok, +1, but this do not gives the inverse. –  Mar 07 '13 at 06:51
  • 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. – Julien Mar 07 '13 at 06:55
  • @sbr: I missed that originally, and my only different proof is a block matrix version of jim's solution below. – copper.hat Mar 07 '13 at 07:21
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This should contain all of the information you might want--probably more.

Alex Youcis
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