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Let $A\in \mathbb{R}^{n\times n}$. Find $\overline O$, orthogonal matrix, to minimize $\|A-O\|_F$. That is;

$$\min_{O\in O(n)} \|A-O\|_F$$ Where $O(n)$ are the set of orthogonal matrices of $\mathbb{R}^{n\times n}$.

$\|\cdot \|_F $ is the Frobenius Norm

So I think it has something to do with SVD. We know that we can write $A$ as $$A = \sum_{j=1}^r \sigma_j u_j v_j^t$$

Where $rank(A) = r$

Also, if we denote $A_k = \sum_{j=1}^k \sigma_j u_j v_j^t$ then we know that $$ \min_{B\in Mat_{n\times n}(\mathbb{R}) \\ rank(B)\le k} \|A-B\| = \|A-A_k\| = \sigma_{k+1} $$

I've tried to utilize those theorem but couldn't figure it out.

I'd be glad for help!

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    What is $F$? Is it a matrix? – 2017-02-26
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    https://en.wikipedia.org/wiki/Matrix_norm#Frobenius_norm – 2017-02-26
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    [Orthogonal Procrustes Problem](https://en.wikipedia.org/wiki/Orthogonal_Procrustes_problem) – 2017-02-26
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    The [polar decomposition](https://en.wikipedia.org/wiki/Polar_decomposition) gives such a minimum, check if it is also the minimum wrt. the Frobenius norm. – 2017-02-26

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We know that $$ {\displaystyle \|B\|_{\rm {F}}={\sqrt {\operatorname {trace} (B^{*}B)}}}, $$ so we need to minimize

$$\min \operatorname {trace} (A^*-O^{*})(A-O) = \min\operatorname {trace} (A^*A-O^{*}A-A^*O + I)$$ or equivalently $$\max \operatorname {trace} O^{*}A$$

This problem was already arised in the question.