= Solution
First, $I-A$ is invertible. If $(I-A)x=0$ for a real <vector>, then $Ax=x$, whereas the <skew-symmetric matrix> identity gives $x^TAx=0$; hence $x^Tx=0$ and $x=0$. Thus its real <kernel> is trivial, which suffices for invertibility. The same argument applies to $I+A$.
Using $A^T=-A$ and the rule for the <transpose> of a product,
$$
N^T=[(I-A)^{-1}]^T(I+A)^T=(I+A)^{-1}(I-A).
$$
The factors $I+A$ and $I-A$ commute, as do the required inverses. Consequently
$$
N^TN=(I+A)^{-1}(I-A)(I+A)(I-A)^{-1}=I.
$$
Thus $\boxed{N\text{ is orthogonal}}$. This is the <Cayley transform of a real skew-symmetric matrix>.
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