Cayley transformation of a skew-symmetric matrix is orthogonal? If $S$ is skew-symmetric ($S^{T} = -S$), how do I show that $Q$ is orthogonal where 
$$Q = (I + S)(I - S)^{-1}$$ which is the Cayley transformation of $S$.
 A: Just compute
$$QQ^T=(I+S)(I-S)^{-1}(I+S)^{-1}(I-S)$$
and since $(I+S)=-(I-S)+2I$ commutes with $(I-S)^{-1}$ then the result follows easily.
Remarks


*

*We used the result $(A^{-1})^T=(A^T)^{-1}$

*Prove for a skew-symmetric matrix S that $\pm1$ are not eigenvalues for it so that the inverse of $(S-I)$ and $(S+I)$ exist.

A: First we show that Q can also be written as:
$$Q=(I-S)^{-1}(I+S).-------(*)$$
Since, $I+S=-(I-S)+2I$
$(I+S)(I-S)^{-1}=(-(I-S)+2I)(I-S)^{-1}=-I+2(I-S)^{-1}=(I-S)^{-1}(-(I-S)+2I)=(I-S)^{-1}(I+S).$
Now, since $S$ is skew  symmetric, $S^{T}=-S.$ Hence, 
$Q^{T}=[(I+S)(I-S)^{-1}]^{T}=(I-S^{T})^{-1}(I+S^{T})=(I+S)^{-1}(I-S).$
Hence, $QQ^{T}=(I+S)(I-S)^{-1}(I+S)^{-1}(I-S)=(I-S)^{-1}(I+S)(I+S)^{-1}(I-S)=I.$ (Using (*))
Similarly, $Q^{T}Q=I.$ Hence Q is orthogonal.
A: I was referred to this question from this one, which concerns itself with unitary and skew-hermitian matrices.  We thus assume that $S$ is skew-hermitian, i.e., that
$S^\dagger = -S; \tag 0$
Then the adoint of $(I + S)(I - S)^{-1}$ is given by
$((I + S)(I - S)^{-1})^\dagger = ((I - S)^{-1})^\dagger (I + S)^\dagger$
$= (I - S^\dagger)^{-1}(I + S^\dagger) = (I + S)^{-1}(I - S), \tag 1$
where we have exploited the fact that
$((I - S)^{-1})^\dagger = ((I - S)^\dagger)^{-1}, \tag{1.1}$
which follows from the general formula
$AA^{-1} = I \tag{1.2}$
by taking adjoints:
$(A^{-1})^\dagger A^\dagger = I^\dagger = I, \tag{1.3}$
whence
$(A^{-1})^\dagger = (A^\dagger)^{-1}. \tag{1.4}$
Then
$((I + S)(I - S)^{-1})^\dagger (I + S)(I - S^{-1}) =  (I + S)^{-1}(I - S)(I + S)(I - S)^{-1}$
$= (I + S)^{-1}(I + S)(I - S)(I - S)^{-1} = I, \tag 2$
and
$(I + S)(I - S)^{-1}(I + S)^{-1}(I - S)$
$= (I + S)(I + S)^{-1}(I - S)^{-1}(I - S) = I, \tag 3$
where in (3) we have used the fact that
$(I - S)^{-1}(I + S)^{-1} = (I + S)^{-1}(I - S)^{-1}, \tag 4$
which follows from
$(I + S)(I - S) = (I - S)(I + S) \tag 5$
by taking inverses on both sides.  (2) and (3) together show that $(I + S)(I - S)^{-1}$ is a unitary operator.
