How to show two matrices commute? How do I show that $(I_n+M)$ and $(I_n-M)^{-1}$ commute where $I_n$ is the $n\times n$ identity matrix and $M$ is an $n\times n$ matrix.?
I have been trying to figure this out for ages, I think I must be missing something simple, any help would be much appreciated. 
 A: You have
$$
I=(I-M)(I-M)^{-1}=(I-M)^{-1}-M(I-M)^{-1},
$$
and
$$
I=(I-M)^{-1}(I-M)=(I-M)^{-1}-(I-M)^{-1}M.
$$
Comparing the two equalities, you get
$$
(I-M)^{-1}M=M(I-M)^{-1}. 
$$
A: For brevity denote $I_n+M=A,$ and $I_n-M=B.$ Assuming that $B$ is invertible, we want to show that $AB^{-1}=B^{-1}A.$ To do so, consider that $A+B=2I_n,$ so multiplying both sides by on the right by $B^{-1},$ we get
$$B^{-1}(A+B)=B^{-1}2I_n=2B^{-1},$$
So $B^{-1}(A+B)=2B^{-1},$ a similar argument by multiplying on the left gives, 
$$(A+B)B^{-1}=2I_nB^{-1}=2B^{-1},$$ 
so we get
$$B^{-1}(A+B)=(A+B)B^{-1}$$
$$B^{-1}A +I_n= AB^{-1}+I_n.$$
Subtracting $I_n$ from both sides yields the desired result.
A: It is trivial to see that if $X$ and $Y$ commute, and $Y^{-1}$ exists then $X$ and $Y^{-1}$ commute: just pre- and post-multiply $XY=YX$ by $Y^{-1}$.
Showing $I+M$ and $I-M$ commute is left as an exercise for the reader. 
A: $I+M,(I-M)^{-1},\exp(2I+M^2-M^3),\sin(I-M^4),(I-3M+M^6)^{-7},\arcsin(M-M^2)$ 
pairwise commute because all are polynomials in $M$ (when they are defined).
