Proof If $AB-I$ Invertible then $BA-I$ invertible. I have these problems :


*

*Proof If $AB-I$ invertible then $BA-I$ invertible.

*Proof If $I-AB$ invertible then $I-BA$ invertible.


I think I solve it correctly, But I'm not so sure, I'll be glad to receive feedback.


*

*If $AB-I$ invertible then :  $$\det|AB-I| \neq 0 \implies \\ \det|A-I||B| \neq 0 \implies \\ \det|B||A-I| \neq 0 
\implies\\ \det|BA-I| \neq 0$$ 


Therefore $BA-I$ invertible.


*

*If $I-AB$ invertible then :


$$\det|I-AB| \neq 0 \implies \\
\det|I-B||A| \neq 0 \implies \\
\det|I-BA| \neq 0$$
Therefore $I-BA$ invertible.
 A: Here's a way to prove this statement:


*

*You want to prove that $BA-I$ is invertible if $AB-I$ is invertible. This is equivalent to proving that $AB-I$ is not invertible if $BA-I$ is not invertible.

*Can you relate non-invertibility to some condition on the eigenvalues of $AB$ and $BA$? Hint: $0$ is an eigenvalue of $AB-I$ if and only if $1$ is an eigenvalue for $AB$.

*Finally, what is the relation between the eigenvalues of $AB$ and those of $BA$?


If you want more hints, just ask, but please try to solve it by yourself first.
A: If $AB-I$ is invertible with inverse $C$ then look at $-I+BCA$ as inverse of $BA-I$.
\begin{align}
(BA-I)(-I+BCA) &=-BA+BABCA+I-BCA\\
&= -BA+B(AB-I)CA+I\\
&=-BA+BA+I=I
\end{align}
and for the other side
\begin{align}
(-I+BCA)(BA-I)&=-BA+I+BCABA-BCA\\
&=-BA+BC(AB-I)A+I\\
&= -BA+BA+I=I
\end{align}
For the second part you can do something similar. Assume $I-AB$ has inverse $D$, which is $-C$, then $I-BA$ has inverse $I+BDA=-(-I+BCA)$.
A: There's a slick way to discover the inverse by first solving the problem for (formal) power series.  
$$\begin{eqnarray} \rm (1-ab)^{-1} &=&\rm 1+ ab + a\color{#c00}{ba}b + a\color{#0a0}{baba}b +\,\cdots\\
&=&\rm 1+ a (1\, +\, \color{#c00}{ba}\ \ +\ \ \color{#0a0}{baba}\,\ +\,\cdots)b\\
&=&\rm 1+ a (1\,-\,ba)^{-1}b\end{eqnarray}\qquad\qquad$$
Simple algebra proves that this formula is universally correct (as in Kaladin's answer).  
At first glance, it seems highly remarkable that such a method should work. Halmos posed the challenge of explaining why this works in one of his popular expositions in Math. Intelligencer. Some explanations are known - see here for more (see also here).
A: Your result will be true if $A$ or $B$  is supposed to be invertible.
Since we have $\det(ABA-A)=\det(A(BA-I))$:
$$ \det((AB-I)A)=\det (A(BA-I))\implies
 \det A\cdot\det (AB-I)=\det A \cdot\det(BA-I).$$
result follows if $\det A\neq0$.
Moreover $$\begin{align}\det(BA-I)\cdot\det B &= \det((BA-I)B)\\ &=\det(BAB-B)\\&=\det(B(AB-I))\\&=\det B\cdot \det(AB-I).\end{align}$$
Result follows if $\det B \neq 0$.
