projection from O(n) to SO(n) Let $A \in \mathbb{O}(n)$. Is there a closed-form expression for the problem:
$$\arg \min_X \ \{\lVert X-A \rVert_F: X \in \mathbb{SO}(n) \}.$$
Simply speaking, I am looking to approximate a general orthogonal matrix by a rotation matrix.
A solution that comes to my mind is the following: Compute determinant of $A$ (which will be $\pm 1$). If its $1$, nothing to do. Else, pick one of the columns of $A$ and multiply it by $-1$. 
Looking forward to your opinion.
 A: If $A\in SO(n)$, the minimiser is obviously $X=A$.
Suppose $A\in O(n)$ and $\det A=-1$. Then $\|X-A\|_F^2=\|A^T X-I\|_F^2=2n-2\operatorname{tr}(A^T X)$. For any skew-symmetric matrix $K$, define
$$
f(t)=2n-2\operatorname{tr}(A^T Xe^{tK}),\quad t\in\mathbb R.
$$
Since $Xe^{tK}$ also belongs to $SO(n)$, if $X$ is a minimiser, we must have $f'(0)=-2\operatorname{tr}(A^TXK)=0$ for every skew-symmetric matrix $K$. Hence $A^TX$ is symmetric.
Now $A^TX$ is a symmetric orthogonal matrix with determinant $-1$. So, an odd number of its eigenvalues are equal to $-1$ and the rest are equal to $1$. Therefore,
$$
\|A^TX-I\|_F=\sqrt{\sum_i\left(\lambda_i(A^TX)-1\right)^2}
$$
is minimised when exactly one eigenvalue of $A^TX$ is equal to $-1$ and the others are equal to $1$. In other words, $X=AV$ for some orthogonal matrix $V$ whose spectrum is $\{1,\ldots,1,-1\}$.
A: Not an answer, but too long to format in a comment.  Note that in the $2 \times 2$ case, all approximations are equally bad.  For instance, take
$$
A = \pmatrix{-1&0\\0&1}
$$
Every element of $SO$ is a rotation by an angle $\theta$.  We find
$$
\|A - X_\theta \|^2 = (\cos \theta - 1)^2 + (\cos \theta + 1)^2 + 2 \sin^2 \theta =\\
2 + 2 \cos^2\theta + 2\sin^2 \theta = 4
$$
