1. ## Linear math help

I'm trying to show that if $AB - I_n$ is invertible then:

$B(AB-I_n)^{-1}A - I_n = (BA-I_n)^{-1}$

I have no idea where to start...?

2. Yes that's an interesting exercise:
Let $R$ be a ring, $a$ and $b$ two elements in $R$.
Prove that if $1-ab$ is left invertible (resp. right invertible), then $1-ba$ is left invertible (resp. right invertible).

Here, $AB-I$ is invertible, and the inverse of $BA-I$ is given, but you have to verify it is correct.

So just multiply $B(AB-I)^{-1}A-I$ by $BA-I$.

Solution:

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

$B(AB-I)^{-1}A-I$ is a left inverse of $BA-I$, therefore it's the inverse of $BA-I$.

Conclusion: $B(AB-I_n)^{-1}A - I_n = (BA-I_n)^{-1}$