TITLE: If the determinant of an real orthogonal matrix $Q$ is 1, prove that there exists a real skew-symmetric $K$ such that $Q=e^K$.
QUESTION [1 upvotes]: If the determinant of an real orthogonal matrix $Q$ is 1, prove that there exists a real skew-symmetric $K$ such that $Q=e^K$.

Every complex orthogonal matrix $Q$ can be represented in the form
$$Q=Re^{iK}$$
where $R$ is real orthogonal matrix and $K$ a real skew-symmetric matrix.
Should I use that theorem to prove it?

REPLY [0 votes]: One approach is as follows: first, note (or show as a consequence of the spectral theorem for normal matrices) that if $Q$ is a real orthogonal matrix with determinant $1$, then there exists an orthogonal matrix $P$ such that $Q = PDP^{T}$, where $D$ is the block diagonal matrix with blocks $D_1,\dots,D_k$ where
$$
D_j = \pmatrix{a_j & -b_j\\b_j & a_j}, \quad 1 \leq j \leq k
$$
for some real numbers $a_j,b_j$.
Now, find a skew-symmetric matrix $K_j$ such that $e^{K_j} = D_j$.  Define $K$ to be the block diagonal matrix with blocks $K_1,\dots,K_k$.  Finally, note that
$$
\exp[PKP^T] = P\exp(K)P^T = PDP^T = Q
$$
and that $PKP^T$ is skew-symmetric.
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