Let $A+I$ be invertible. Show that $(A+I)^{-1}$ and $(I-A)$ commute Not really sure where to start with this one. I'm rather terrible with proofs and all, so any help would be greatly appreciated.
 A: \begin{align*}
(A+I)^{-1} (I-A) &= (A+I)^{-1} (I-A) \,I\\
&= (A+I)^{-1} (I-A) \left[ (A+I)(A+I)^{-1} \right] \\
&= (A+I)^{-1} \left[ (I-A)(I+A) \right] (A+I)^{-1} \\
&= (A+I)^{-1} \left[ (I+A)(I-A) \right] (A+I)^{-1} \\
&= \left[ (A+I)^{-1} (I+A)\right](I-A) (A+I)^{-1} \\
&= I\, (I-A) (A+I)^{-1}\\
&= (I-A) (A+I)^{-1}
\end{align*}
A: It is probably best to first prove the general theorem: If $X$ and $Y$ commute and $X$ is invertible, then $X^{-1}$ and $Y$ commute.
Proof:
\begin{align}
X^{-1}Y
&= X^{-1}YI && \text{($I$ is neutral element of multiplication)}\\
&= X^{-1}YXX^{-1} && \text{(definition of $X^{-1}$)}\\
&= X^{-1}XYX^{-1} && \text{($X$ and $Y$ commute)}\\
&= IYX^{-1} && \text{(definition of $X^{-1}$, again)}\\
&= YX^{-1} && \text{(neutrality of $I$ again)}
\end{align}
Note that in the calculation above I implicitly used the associativity of matrix multiplication by not writing any parentheses. Note also that the above calculation is not specific for matrices, but works in any monoid.
Having this general result, for proving that $(A+I)^{-1}$ and $(I-A)$ commute, all that remains to show is that $(A+I)$ and $(I-A)$ commute. But that is easily done by direct calculation.
A: $(A+I)(A-I)=(A-I)(A+I)=A^2-I$ implies that $(A+I)(A-I)(A+I)^{-1}=(A-I)$. This implies that $(A+I)^{-1}(A+I)(A-I)(A+I)^{-1}=(A+I)^{-1}(A-I)$. This implies that $(A-I)(A+I)^{-1}=(A+I)^{-1}(A-I)$
