If $S^T = -S$ show that $I-S$ is nonsingular and that $(I-S)^{-1}(I+S)$ is orthogonal. Let $S$ be an $n\times n$ matrix with $S^T = -S$. Show that $I-S$ is nonsingular and that $(I-S)^{-1}(I+S)$ is orthogonal (the question can be found here).
$\mathbf{Hint}:$ First show that $(I-S)^{T}(I-S)$ is nonsingular. Then show and use the fact that
$$(I+S)(I-S) = (I-S)(I+S)$$
Have no idea even with the hint! 
 A: Use the hint and the relation $S^T=-S$ to get
$$(I-S)^T(I-S) = I^TI - I^TS-S^TI+S^TS = I+S^TS.$$
Now you need to show that $I+S^TS$ is nonsingular. One approach is to prove that $x^T(I+S^TS)x > 0$ for any $x\neq 0$.
Now use the result from your previous question to conclude $I-S$ is nonsingular.
For the last part, simplify $[(I-S)^{-1}(I+S)]^T(I-S)^{-1}(I+S)$ as much as you can using $S^T=-S$, then use the hint.
A: Hint: Work with complex matrices. $S+S^*=0$  means $S = i H$ where $H$ is hermitian. Write $H = \tan \frac{T}{2}$ with $T$ hermitian and $\sigma(T) \subset (-\pi,\pi)$. Then we have 
$$\frac{1+S}{1-S}= \frac{1 + i \tan \frac{T}{2}}{1 - i \tan \frac{T}{2}}= \exp(i T)$$
is a unitary matrix. 
A: First, note that
$(I - S)^T(I - S) = (I - S^T)(I - S)$
$= I - (S + S^T) + S^TS = I + S^TS \tag{1}$
since
$S + S^T = S - S = 0. \tag{2}$
Next, realize and/or recall that $I + S^TS$ is always nonsingular, for any real matrix $S$; this follows from the fact that, for any vector $v \ne 0$,
$\langle v, (I + S^TS)v \rangle = \langle v, v \rangle + \langle v, S^TSv \rangle$
$= \langle v, v \rangle + \langle Sv, Sv \rangle \ge \langle v, v \rangle > 0, \tag{3}$
by virtue of the facts that $\langle Sv, Sv \rangle \ge 0$ and $\langle v, v \rangle > 0$ for all $v \ne 0$.  Now (3) implies that
$(I + S^TS)v \ne 0 \tag{4}$
for all non-zero $v$; hence, $I + S^TS$ is nonsingular.  This in turn implies both $I - S$ and $I - S^T = I + S$ are nonsingular, since their product is the invertable matrix $I + S^TS$.  So that's done.
As for $(I - S)^{-1}(I + S)$, recall that for any invertable matrix $R$ we have
$(R^{-1})^T = (R^T)^{-1} \tag{5}$
which may be seen as follows:  we have
$RR^{-1} = I, \tag{6}$
and transposing we obtain
$(R^{-1})^TR^T = I \tag{7}$
as well; it is easily seen that (7) implies (5).  We thus have
$((I - S)^{-1})^T = (I - S^T)^{-1} = (I + S)^{-1}, \tag{8}$
whence
$(I - S)^{-1}(I + S)((I - S)^{-1}(I + S))^T$
$= (I - S)^{-1}(I + S)(I + S)^T((I - S)^{-1})^T = (I - S)^{-1}(I + S)(I - S)(I + S)^{-1}$
$= (I -S)^{-1}(I - S)(I + S)(I + S)^{-1} = I, \tag{9}$
as was to be shown.
A: First part: note that
$$
(I-S)^T(I-S) = I - S-S^T + S^TS = I+S^TS
$$
Now, we need to show that $I + S^TS$ is invertible.  There are a few options here: I'd say that you should either use positive definiteness or SVD.
Once you show that $(I+S)(I-S) = (I-S)(I+S)$, we have
$$
[(I-S)^{-1}(I+S)][(I-S)^{-1}(I+S)]^T = \\
(I - S)^{-1}(I+S)(I+S)^T[(I - S)^{-1}]^T = \\
(I - S)^{-1}(I+S)(I-S)[(I - S)^{-1}]^T = \\
(I - S)^{-1}(I-S)(I+S)[(I - S)^{-1}]^T = \\
\Big[(I - S)^{-1}(I-S)\Big] \Big[(I-S)^T[(I - S)^{T}]^{-1}\Big]
$$
