Non-singular matrix Assuming $B$, $I+B$, $I+B^{-1}$ are all non-singular, show that 
$$(I+B)^{-1}+(I+B^{-1})^{-1}=I$$
All I know is that determinants not equal to $0$ and that the inverse of $B$ exists.
 A: i think you can establish $A = I$ by manipulating what you have  $$A = (I+B)^{-1}+(I+B^{-1})^{-1}=I \tag 1$$
post and pre multiplying by $(I+B)$ and  $(1)$ gives you
$$\begin{align}(I+B)A(I+B)&=(I+B)A(I+B^{-1})B=(I+B^{-1}+I+B)B \\&=I+2B+B^2\\
&=(1+B)^2\\
\end{align}\\$$
now cancel $(I+B)$ on the left and right to give $$A = I. $$
A: Consider $I+B^{-1}=B^{-1}(B+I)$
Hence, $(I+B^{-1})^{-1}=[B^{-1}(I+B)]^{-1}$
$(I+B^{-1})^{-1}=(I+B)^{-1}B$
Hence, l.h.s becomes 
$(I+B)^{-1}+(I+B^{-1})^{-1}=(I+B)^{-1}+(I+B)^{-1}B$
$(I+B)^{-1}+(I+B^{-1})B=(I+B)^{-1}(I+B)=I$, hence shown
A: HINT


*

*Assume first $B$ is diagonal. Can you prove your result holds?

*Now assume $B$ is diagonalizable, so $B = VDV^{-1}$ and $I = V I V^{-1}$ and $B^{-1} = VD^{-1}V^{-1}$. Can you use (1) and prove your result holds?

*Last generalization is to take care of the case when $B$ has Jordan form...

A: Hint
Call $X$ your matrix. It is enough to find an invertible matrix $A$ s.t. 
$$ XA=A.$$
Try with $A=(I+B)(I+B^{-1})$, that is invertible by assumption.
A: Verify that $(I+B^{-1})(I+B)=(I+B^{-1})+(I+B)$ and multiply from left by $(I+B^{-1})^{-1}$ and from right by $(I+B)^{-1}$.
