<s>[INST] Geometric or matrix intuition on $A(A + B)^{-1}B = B (A + B)^{-1} A$ I am curious about a seemingly simple identity in matrix algebra. Though matrix multiplication is not commutative (the classic example of noncommutativity, it does allow a commutativity of sorts around a very specific third matrix: 
$$
\color{blue}{A (A + B)^{-1} B} = \color{blue}{B (A + B)^{-1} A}
$$
There is a very simple algebraic proof:
$$
\begin{eqnarray}
\color{blue}{A (A + B)^{-1} B} + \color{red}{B (A + B)^{-1} B} &=& (A + B)(A + B)^{-1} B\\
                                   &=& B\\
                                   &=& B (A + B)^{-1} (A + B) \\
                                   &=& \color{blue}{B (A + B)^{-1} A} + \color{red}{B (A + B)^{-1} B}\\
\end{eqnarray}
$$
because matrix addition is commutative. The identity follows by cancelling $B (A + B)^{-1} B$. (If there is a simpler linear proof, say, without needing cancellation, please say so. That extra weird matrix, while obviously looking right and obviously doing the job, just pops out of nowhere. Or does it?)
Algebra is blind manipulation of symbols. The identity holds in any abstract ring with multiplicative inverses. But for a given model, the identity says something about that model. I've only seen the identity in the context of matrices, but I don't see what's so special about it there.
What does this identity do for matrices? 
Is there something, in matrix theory, for which this is special? Does it ever really come up in proofs? $(A+B)^{-1}$ can't be the only matrices that allow such quasi-commutativity, can they? Is there a visualization or geometric interpretation or a meaningful anything interpretation? 
 [/INST] Just put
$$
A = \left( {A + B} \right) - B
$$
to get
$$
\eqalign{
  & \left( {\left( {A + B} \right) - B} \right)\left( {A + B} \right)^{\, - \,1} B = B\left( {A + B} \right)^{\, - \,1} \left( {\left( {A + B} \right) - B} \right)  \cr
 & \quad  \Downarrow   \cr  
  & B - B\left( {A + B} \right)^{\, - \,1} B = B - B\left( {A + B} \right)^{\, - \,1} B \cr} 
$$
Maybe a better insight on what is going on can be obtained by putting
$$
\left\{ \matrix{
  A = {1 \over 2}\left( {A + B} \right) - {1 \over 2}\left( {B - A} \right) \hfill \cr 
  B = {1 \over 2}\left( {A + B} \right) + {1 \over 2}\left( {B - A} \right) \hfill \cr}  \right.
$$
which then gives
$$
\eqalign{
  & {1 \over 4}\left( {\left( {A + B} \right) - \left( {B - A} \right)} \right)\left( {A + B} \right)^{\, - \,1} \left( {\left( {A + B} \right) + \left( {B - A} \right)} \right) =   \cr 
  &  = {1 \over 4}\left( {\left( {A + B} \right) + \left( {B - A} \right)} \right)\left( {A + B} \right)^{\, - \,1} \left( {\left( {A + B} \right) - \left( {B - A} \right)} \right) =   \cr 
  & \quad  \Downarrow   \cr 
  & \left( {I - \left( {B - A} \right)\left( {A + B} \right)^{\, - \,1} } \right)\left( {\left( {A + B} \right) + \left( {B - A} \right)} \right) =   \cr 
  &  = \left( {I + \left( {B - A} \right)\left( {A + B} \right)^{\, - \,1} } \right)\left( {\left( {A + B} \right) - \left( {B - A} \right)} \right)  \cr 
  & \quad  \Downarrow   \cr 
  & \left( {A + B} \right) - \left( {B - A} \right)\left( {A + B} \right)^{\, - \,1} \left( {B - A} \right) =   \cr 
  &  = \left( {A + B} \right) - \left( {B - A} \right)\left( {A + B} \right)^{\, - \,1} \left( {B - A} \right) \cr} 
$$
</s>