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Let $r>0$, and let $A\notin\mathcal{B}_r(0)\triangleq\{Y\in\mathbb{R}^{n\times n}:\|Y\|_2\leq r\}$, where $\|\cdot\|_2$ is the induced 2-norm. Let $\bar A$ be the orthogoanl projection of $A$ on the boundary of $\mathcal{B}_r(0)$ defined by $$\bar A\triangleq{\rm argmin}_{X\in\partial \mathcal{B}_r(0)}\|X- A\|_{\rm F},$$ where $\partial \mathcal{B}_r(0)$ denotes the boundary of $\mathcal{B}_r(0)$, and $\|\cdot\|$ denotes the Frobinius norm.

How can I orthogonally project $A$ to the boundary of $\mathcal{B}_r(0)$?

I know that if, in the definition of $\mathcal{B}_r(0)$, we replace $\|\cdot\|_2$ by $\|\cdot\|_{\rm F}$ , then $\bar A=\dfrac{Ar}{\|A\|_{\rm F}}$ is the orthogoanl projection of $A$ on the boundary of $\mathcal{B}_r(0)$. Can we show that $\bar A=\dfrac{Ar}{\|A\|_{2}}$ also works here?

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  • $\begingroup$ What is $\|\cdot \|_2$ defined as? $\endgroup$ – BananaCats Category Theory App Jan 3 '18 at 22:26
  • $\begingroup$ $\|A\|_2$ is defined as the maximum singular value of $A$. $\endgroup$ – krms Jan 3 '18 at 22:29
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    $\begingroup$ If you want to project an $n$-dimensional vector $v$ onto the hypercube $[-r,r]^n$, it suffices to "clip" each component $v_i$ so that it lies in the interval $[-r,r]$. (Explicitly, $\max(\min(v_i,r),-r)$.) For your question (assuming "projection" means "element of $B_r(0)$ closest to $A$ in Frobenius distance), I suspect the answer is to write the SVD and "clip" each singular value so it lies in $[0,r]$. $\endgroup$ – angryavian Jan 3 '18 at 22:35
  • $\begingroup$ @angryavian Seems reasonable, but how can I show that rigorously? $\endgroup$ – krms Jan 3 '18 at 22:37
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    $\begingroup$ For the question to make sense, you would have to define what "project orthogonally" means in the operator norm. The operator norm is not an inner product norm. Also, you would have to say why $rA/\|A\|$ does not work for you. $\endgroup$ – Martin Argerami Jan 3 '18 at 23:24

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