The orthogonal matrix nearest to a given matrix

Idea. Minimising the distance is the same as maximising $\operatorname{tr}(Q^TA)$, which becomes $\sum\sigma_iW_{ii}$ for an orthogonal matrix $W$.

Step 1: expansion. With $\langle X,Y\rangle=\operatorname{tr}(X^TY)$,
$$\|A-Q\|_F^2=\|A\|_F^2-2\operatorname{tr}(Q^TA)+\|Q\|_F^2=\sum_i\sigma_i^2-2\operatorname{tr}(Q^TA)+n,$$
using $\|A\|_F^2=\sum\sigma_i^2$ and $\|Q\|_F^2=\operatorname{tr}(Q^TQ)=n$.

Step 2: the trace. By cyclicity of the trace,
$$\operatorname{tr}(Q^TA)=\operatorname{tr}(Q^TU\Sigma V^T)=\operatorname{tr}(V^TQ^TU\,\Sigma)=\operatorname{tr}(W\Sigma)=\sum_i\sigma_iW_{ii},\qquad W:=V^TQ^TU\in O(n).$$

Step 3: bound. Every entry of an orthogonal matrix satisfies $|W_{ii}|\le1$ (columns are unit vectors). Since $\sigma_i\ge0$,
$$\operatorname{tr}(Q^TA)\le\sum_i\sigma_i,$$
with equality when $W=I$, that is $V^TQ^TU=I\iff Q=UV^T$.

Step 4: the minimum.
$$\min_Q\|A-Q\|_F^2=\sum\sigma_i^2-2\sum\sigma_i+n=\sum(\sigma_i-1)^2.$$

Step 5: polar factor. $A=U\Sigma V^T=(UV^T)(V\Sigma V^T)$, with $V\Sigma V^T$ symmetric PSD: the orthogonal factor is $UV^T$.

Step 6: uniqueness for invertible $A$. All $\sigma_i>0$, so equality forces $W_{ii}=1$ for every $i$. A unit column with a $1$ in position $i$ is $e_i$, so $W=I$ and $Q=UV^T$ is the only minimiser.

Numeric check. For a random $4\times4$ matrix, $\|A-UV^T\|_F^2=2.2953=\sum(\sigma_i-1)^2$, and the best of 2000 random orthogonal matrices gave $3.41>2.295$.

Remark. This is the *orthogonal Procrustes problem*. For a singular $A$ the minimiser may be non-unique (e.g. $A=0$: every $Q$ is at distance $\sqrt n$). Consequently the nearest-orthogonal map is continuous on $GL_n$ but not on all of $M_n$.