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I want to prove (or disprove) the following:

Given an arbitrary square matrix, $A$, of size $(n \times n)$, and the matrix $A_S$ defined by $$A_S= \frac{A+A^T}{2}$$ prove that $A_S$ is the nearest symmetric matrix to $A$ with respect to the spectral norm.

That is, prove that:

$$A_S = \arg\min_{\tilde{A}\in S} \|A-\tilde{A}\|_2$$

Where $S$ is the set of all $(n \times n)$ symmetric matrices.

But I am unsure where to start... does anyone have any suggestions on how to tackle this problem? In addition, if the above is not true, what would be the nearest symmetric matrix to $A$ in the above sense?

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    $\begingroup$ Just to be clear, the norm you are using is $\|B\|_2 := \sup_{\|v\| = 1} \|Bv\|_2$? $\endgroup$
    – angryavian
    Commented Oct 30, 2019 at 19:52
  • $\begingroup$ Apologies @Mason I am unsure I understand your question. $\endgroup$
    – JDoe2
    Commented Oct 30, 2019 at 19:59
  • $\begingroup$ @angryavian, yes that is the norm I was using. $\endgroup$
    – JDoe2
    Commented Oct 30, 2019 at 20:00
  • $\begingroup$ Did you try some numerical experiments? $\endgroup$ Commented Oct 31, 2019 at 14:44

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As shown in Rahul's answer, the statement as it stands is false. $A_S$ is not necessarily the nearest symmetric matrix to $A$, because there can be other symmetric matrices that are equally close to $A$.

However, $A_S$ is always a global minimiser of $\|A-X\|$ over all (real or complex) symmetric matrices $X$. Let $K=(A-A^T)/2$. For any symmetric matrix $X$, we have \begin{aligned} 2\|K\|&=\|(K-X)+(K+X)\|\\ &\le\|K-X\|+\|K+X\|\\ &=\|K-X\|+\|-K-X\|\\ &=\|K-X\|+\|(K-X)^T\|\\ &=2\|K-X\|. \end{aligned} Therefore, given any symmetric matrix $S$, if we put $X=S-A_S$, we obtain $$ \|A-A_S\|=\|K\|\le\|K-X\|=\|A-S\|. $$ Note that this proof applies as long as $\|M\|=\|M^T\|$. In particular, the inequality still holds if the norm in question is replaced by Frobenius norm or any Ky Fan $k$-norm. It doesn't even have to be a submultiplicative norm. Any vector $p$-norm $\|M\|:=\|\operatorname{vec}(M)\|_p$ can be used as well.

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  • $\begingroup$ I see! Thank you! $\endgroup$
    – JDoe2
    Commented Nov 1, 2019 at 14:32
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The minimizer is not necessarily unique. For example take $A=\begin{bmatrix}&1&\\-1&&\\&&0\end{bmatrix}$, $A_S=0$, $\tilde A=\begin{bmatrix}0&&\\&0&\\&&1\end{bmatrix}$. Then $\|A-A_S\|_2=1=\|A-\tilde A\|_2$.

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  • $\begingroup$ Ah yes, Thank you @Rahul ! $\endgroup$
    – JDoe2
    Commented Nov 1, 2019 at 14:33

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