TITLE: Special orthogonal matrices have orthogonal square roots
QUESTION [7 upvotes]: Let $A$ be an orthogonal matrix with $\det (A)=1$. Show that there exists an orthogonal matrix $B$ such that $B^2=A$.
Thank you very much.

REPLY [3 votes]: Depending on what you know, the proof varies from a one-liner to a half page.
A one-line proof is this: every real orthogonal matrix with determinant $1$ can be written as $e^K$ for some skew symmetric matrix $K$; therefore $A=B^2$ for $B=e^{K/2}$.
A longer proof is this: $A$ is real and normal. Hence it is orthogonally similar to its real Jordan form. That is,
$$
A=Q\left(I_p\oplus-I_q\oplus R(\theta_1)\oplus\cdots\oplus R(\theta_m)\right)Q^T
$$
where $Q$ is orthogonal, $\theta_1,\ldots,\theta_m\in(0,\pi)$, $R(\theta)$ denotes the $2\times2$ rotation matrix for angle $\theta$, and $p+q+2m=n$. (See this related question for the reason why such a decomposition is possible.) Since $\det A=1$, $q$ must be an even number. Therefore $A=B^2$, where
$$
B=Q\left(I_p\oplus
\underbrace{R(\frac{\pi}2)\oplus \cdots R(\frac{\pi}2)}_{\frac q2 \text{ blocks}}
\oplus R(\frac{\theta_1}2)\oplus\cdots\oplus R(\frac{\theta_m}2)\right)Q^T.
$$