Invert of Matrix I-BA 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.
 A: $$(I-BA)(I+B(I-AB)^{-1}A)=I$$ here $I$  is the identity matrix
so $(I-BA)^{-1}= I+B(I-AB)^{-1}A$
A: 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.
A: 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.
A: This should contain all of the information you might want--probably more.
