$(E-A)$ and $(E+A)^{-1}$ commute Suppose $E$ is the identity matrix and $A$ is an orthogonal matrix, and that $E+A$ is invertible. Prove:
1)$(E-A)$ and $(E+A)^{-1}$ commute
2)$(E-A)(E+A)^{-1}$ is an anti-symmetric matrix.
My attempted solution:

A is 0 matrix/ or Det(A)=1?


For2 there is a wrong deduction
\begin{align}\left[(E-A)(E+A)^{-1}\right]^t=\left((E+A)^{-1}\right)^t(E-A)^t=(E+A)^{-1}(E-A)=(E-A)(E+A)^{-1}\end{align}
Where is the $-1$?
 A: Hint: For (1) note, that by direct computation 
$$ (E+A)(E-A) = E - A^2 = (E-A)(E+A)$$
now multiply wisely by $(E+A)^{-1}$.
For (2), we have to prove
$$ \bigl( (E+A)^{-1}(E-A)\bigr)^t = -(E-A)(E+A)^{-1} \iff 
   (E-A^t)(E+A) = -(E+A^t)(E-A) $$
A: Suppose that $X$ and $Y$ commute and that $X$ is inversible. Automatically, this implies that  $X^{-1}$ and $Y$ commute because
$$YX^{-1} = (X^{-1} X) Y X^{-1}= X^{-1}(YX)X^{-1} = X^{-1}Y(XX^{-1}) = X^{-1}Y.$$
Here, is obvious that $X = E+A$ and $Y = E-A$ commute because they are polynomials of $A$. You can also check directly that $(E+A)(E-A) = E - A^2 = (E-A)(E+A)$.
A: *

*Note that by Cayley-Hamilton theorem, $(E+A)^{-1}$ is a polynomial in $E+A$ and therefore also a polynomial in $A$. Since polynomials in $A$ commute, $(E+A)^{-1}$ commutes with $E-A$, regardless of whether $A$ is orthogonal or not.

*In your proof attempt, you wrote $\left((E+A)^{-1}\right)^t(E-A)^t=(E+A)^{-1}(E-A)$ without justifying it. Actually, since $A$ is orthogonal, we have $A^tA=AA^t=I$ and hence you should have
\begin{align*}
\left((E+A)^{-1}\right)^t(\color{red}{E-A})^t
&=(E+A^t)^{-1}(E-A)^t\\
&=\left(A^t(A+E)\right)^{-1}\left(\color{red}{(A^t-E)A}\right)^t.
\end{align*}
You may continue your proof from here. Note that, by pulling out $A$ from $E-A$ in the above, we produce the difference $A^t-E$. Taking transpose, we get $A-E$ and this is the source of $-1$.


As a remark, the mapping $A\mapsto(E+A)^{-1}(E-A)$ is known as the Cayley transform.
