# Invert of Matrix I-BA [duplicate]

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.

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

• this is a duplicate – Rustyn Mar 7 '13 at 6:43
• usually you link the duplicate not just say. – user45099 Mar 7 '13 at 6:47
• @user1709828 Sometimes it's difficult to find the other question. – JSchlather Mar 7 '13 at 6:54
• Thank you very much for all answers. – kami Mar 7 '13 at 7:10

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

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

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

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.

• Ok, +1, but this do not gives the inverse. – user63181 Mar 7 '13 at 6:51
• 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 7 '13 at 6:55
• @sbr: I missed that originally, and my only different proof is a block matrix version of jim's solution below. – copper.hat Mar 7 '13 at 7:21

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.

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