Orthogonal matrix representation If $\mathbf{M}$ is anti-symmetric, then $\mathbf{U}=(\mathbf{I}-\mathbf{M})(\mathbf{I}+\mathbf{M})^{-1}$ is orthogonal with $\det\mathbf{U}=1$. This is just manipulation and noticing that $\mathbf{I}-\mathbf{M}$ and $\mathbf{I}+\mathbf{M}$ commute. A sort-of converse is that every orthogonal $\mathbf{U}$ with $\det\mathbf{U}=1$ and without $-1$ as an eigenvalue can be written in this form, for some anti-symmetric $\mathbf{M}$.
Is there a clever expression for $\mathbf{M}$ in terms of $\mathbf{U}$ that one can write to prove this fact, or should I be thinking about another approach? The first thing I tried was $\mathbf{M}=\mathbf{U}-\mathbf{U}^{\text{T}}$ which would've been nice, but doesn't quite work.

Let us pretend for a minute that we are working with functions. Then $u = \frac{1 - m}{1 + m}$ and we can express $m$ in terms of $u$ as
$$ (1 + m)u = (1-m) \implies m(1 + u) = 1 - u \implies m = \frac{1-u}{1+u} $$
so the expression for $m$ in terms of $u$ is the same as the expression of $u$ in terms of $m$. Hence, a good guess will be to take $M = (1-U)(1+U)^{-1}$ (which is well-defined as $-1$ is not an eigenvalue of $U$) and indeed, this will work.
