$A$ is a real orthogonal matrix, prove that $ (I+A)^{-1}(I-A)$ is skew-symmetric $A$ is a real orthogonal matrix and $(I + A)$ is non-singular. Prove that $
(I+A)^{-1}(I-A)$ is a skew-symmetric matrix.
Attempt:
$[(I+A)^{-1}(I-A)]^t=(I-A)^t((I+A)^{-1})^t=(I-A)^t((I+A)^{t})^{-1}=(I-A^t)   ((I+A^t))^{-1}$ 
Please help me further.
 A: $A$ is a real orthogonal matrix means $A^T=A^{-1}$:
$$
(I-A^T)(I+A^T)^{-1} = (I-A^{-1})(I+A^{-1})^{-1} =
A(I-A^{-1})A^{-1}(I+A^{-1})^{-1}=A(I-A^{-1})\bigl((I+A^{-1})A\bigr)^{-1}=(A-I)(A+I)^{-1} =-(I-A)(I+A)^{-1}
$$
UPDATE
Ok, let's show that
$$
(I-A^{-1}) = A(I-A^{-1})A^{-1}
$$
and
$$
(I+A)^{-1} (I-A) = (I-A)(I+A)^{-1}.
$$
It's obvious for me, but it's needed. It's simple for first:
$$
A(I-A^{-1})A^{-1} = (A-AA^{-1})A^{-1}=(A-I)A^{-1}=(AA^{-1}-A^{-1})=(I-A^{-1}).
$$
And it's not too hard for second:
$$
(I-A)(I+A)=(I+A)(I-A)
$$
(I think it's clear; if not, expand and substract), so
$$
(I+A)^{-1}(I-A)(I+A)=(I+A)^{-1}(I+A)(I-A) = (I-A)
$$
and
$$
(I+A)^{-1}(I-A)(I+A)(I+A)^{-1}=(I-A)(I+A)^{-1},
$$
or
$$
(I+A)^{-1}(I-A)=(I-A)(I+A)^{-1}.
$$
Note: we used that $I+A$ is non-singular (and have not use $(I-A)^{-1}$, for example).
A: \begin{align}
\bigl((I+A)^{-1}(I-A)\bigr)^t
&=(I-A^t)(I+A^t)^{-1}\\[4px]
&=(I-A^{-1})(I+A^{-1})^{-1}\\[4px]
&=(I-A^{-1})\bigl((A+I)A^{-1}\bigr)^{-1}\\[4px]
&=(I-A^{-1})A(A+I)^{-1}\\[4px]
&=(A-I)(A+I)^{-1}\\[4px]
&=-(I-A)(A+I)^{-1}
\end{align}
Now we're almost done: let's compare what we get upon multiplying on the left by $I+A$ the given matrix and its transpose in simplified form as obtained above:
\begin{align}
&(I+A)(I+A)^{-1}(I-A) &&-(I+A)(I-A)(A+I)^{-1}\\[4px]
&(I-A) &&-(I-A)(I+A)(A+I)^{-1}\\[4px]
&(I-A) &&-(I-A)
\end{align}
so the two matrices are indeed the negative of one another.
Just note that $(I-A)(I+A)=I-A^2=(I-A)(I+A)$.
A: First note that
$$
       (I+A)^{-1}(I-A)=(I+A)^{-1}\{-(I+A)+2I\}=-I+2(I+A)^{-1}
$$
And $A^{t}=A^{-1}$. Therefore,
\begin{align}
       ((I+A)^{-1})^{t} & =(I+A^{t})^{-1} \\
                        & =(I+A^{-1})^{-1} \\
                        & =\{A^{-1}(I+A)\}^{-1} \\
                        & = (I+A)^{-1}A \\
                        & = (I+A)^{-1}\{(A+I)-I\} \\
                        & = I-(I+A)^{-1}.
\end{align}
Finally,
$$
   \{(I+A)^{-1}(I-A)\}^{t}=-I+2I-2(I+A)^{-1}=-(I+A)^{-1}(I+A).
$$
