Prove that $UV^T$ is the nearest orthogonal matrix to $A$ given $||A||=\max_{x\neq 0}\dfrac{||Ax||}{||x||}=\sigma_1$

Question

Prove that $UV^T$ is the nearest orthogonal matrix to $A$ ie., $||A-Q||\geq ||A-UV^T||$ for all orthogonal matrices $Q$ given the norm of the matrix is defined as $||A||=\max_{x\neq 0}\dfrac{||Ax||}{||x||}=\sigma_1$

The SVD of $A$ is $A=U\Sigma V^T=\sum_{i=1}^{r}\sigma_i u_i v_i^T$

$$ ||A-Q||=||U\Sigma V^T-Q||=||U^T\Big(U\Sigma V^T-Q\Big)V||=||\Sigma-U^TQV||=||\Sigma-Q'|| $$

So we need to solve for $Q'$ when $||\Sigma-Q'||$ is minimized for the orthogonal matrix $Q'$.

How do I prove it ?

Note:

Similar question has been asked at Showing that matrix $Q=UV^T$ is the nearest orthogonal matrix to A., but it is for the Frobenius norm

Answer 1

\begin{aligned} \|\Sigma-Q'\| &=\max_{\|u\|=1}\|\Sigma u-Q'u\|\\ &\ge\max_{\|u\|=1}\big|\,\|\Sigma u\|-\|Q'u\|\,\big|\\ &=\max_{\|u\|=1}\big|\,\|\Sigma u\|-1\,\big|\\ &=\max(|\sigma_1-1|,|\sigma_n-1|)\\ &=\|\Sigma-I\|. \end{aligned}