Special orthogonal matrices have orthogonal square roots 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.
 A: Edit: This is indeed true...Thanks @Dirk. 
An orthogonal matrix (see  the "Canonical form" paragraph or this thread exhibited by user1551) $A$ is block diagonalizable in an orthonormal basis with blocks
$$
\left(\matrix {\cos\theta&-\sin\theta\\\sin\theta&\cos\theta}\right)
$$
or $\pm 1$ along the diagonal, i.e. $P^*AP=A'$ with $P$ orthogonal and $A'$ block diagonal of rotations as above and $\pm 1$. I denote $M^*$ the adjoint matrix of $M$, which is nothing but the transpose $M^T$ in the real case.
If $\det A=1$, there are 2$k$ numbers $-1$ on the diagonal, corresponding to $k$ rotations of angle $\pi$. So we only have rotations and $1$'s.
Now, as pointed by @Dirk in the $-1$ case, 

$$
\left(\matrix{\cos\theta&-\sin\theta\\\sin\theta&\cos\theta} \right)=\left(\matrix{\cos\theta/2&-\sin\theta/2\\\sin\theta/2&\cos\theta/2} \right)^2.
$$

The genuine idea is that a rotation, is the square of the half-angle rotation. This works for $-I_2$ in particular, with $\theta=\pi$.
Do that as many times as needed to get $A'=C^2$ with $C$ orthogonal. Then define
$$
B:=PCP^*\qquad\Rightarrow\qquad B^2=PCP^*PCP^*=PC^2P^*=PA'P^*=A
$$
and note that $B^*B=I$, so $B$ is orthogonal.
A: Edit: I missunterstood the question at first thanks @Dirk again. 
Thats  true, $A$ is diagonalizable and with $\pm 1$ on the diagonal. 
you get an even number of $-1$ because else the determinant can't $1$. For this the matrix $B$ will be something like 
$$B= \begin{pmatrix} I_n &0 &0\\ 0 & 0 &1\\ 0& -1 & 0 \\ \end{pmatrix} $$ 
or more of the $$\begin{pmatrix} 0 & 1\\ -1 &0 \\\end{pmatrix}$$ blocks.
So we have $$A=P^{-1} D P = P^{-1} B B P= (P^{-1} B P) (P^{-1} B P) $$
And $B^T B=I$ hence $B$ is orthogonal.
A: 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
$$
B=Q\left(I_p\oplus
\underbrace{R(\frac{\pi}2)\oplus \cdots R(\frac{\pi}2)}_{\frac q2 \text{ blocks}}
\oplus R(\frac{\theta_1}2)\oplus\cdots\oplus R(\frac{\theta_m}2)\right)Q^T.
$$
