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Given a matrix $X$, let $\mathbf{proj}(X)=\underset{\|Y\|_2\le 1}{\arg\min} \|X-Y\|_F$. Now the question is to solve $\mathbf{proj}(X)$.

Proposition: Suppose $X=U \,\mathbf{diag}(\sigma)\,V^T$ is the SVD of $X$, then $\mathbf{proj}(X)=U\, \mathbf{diag}\bigl( \min(\sigma_1,1),\cdots,\min(\sigma_n,1) \bigr)\,V^T$.

How to prove this proposition? My difficulty focus on proving that $X$ and $\mathbf{proj}(X)$ have the same singular vector.

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  • $\begingroup$ $\mathbf{proj}(X) = aX$ for some scalar $a$. $\endgroup$ – Paul Sinclair Apr 21 at 20:26

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