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Suppose that $w$ is an array of four $m\times n$ (real 0r complex valued) matrices: $w=(w_1, w_2, w_3, w_4) \in \mathbb{R}^{4mn}$. Define $$ ||w||_{\text{nuc},1} = \sum_1^4 ||w_j||_{\text{nuc}} $$ where $||\cdot||_{\text{nuc}}$ is the usual nuclear norm of a matrix (sum of its singular values, or put differently, norm one of the vector $v$ of singular values of that matrix). It is easily seen than the dual of the above norm is $$ ||w||_{\text{nuc}, \infty} = \max_{j=1,2,3,4} ||w_j||_{\text{nuc}}. $$

Our goal is find the orthogonal projection over the $r$-ball of $|| \cdot ||_{\text{nuc}, \infty}$ which is $$ P = \{w\in \mathbb{R}^{4mn} : \, \max_{j=1,2,3,4} ||w_j||_{\text{nuc}} \leq r \} $$ $$= \{w\in \mathbb{R}^{4mn} : \, ||v_j||_1 \leq r,\, j=1,2,3,4 \}.$$ I understand that the projection operator over the norm-one ball comprises a soft thresholding, but I suspect that doesn't work here since $P$ is the norm-one ball of $v_j$'s not $w's$.

Any help will be appreciated.

Edit: A more precise definition for nuclear norm of any matrix $A$ is as follows: $$|| A ||_{\text{nuc}}=\sum_{i=1}^{\min\{m,\,n\}}\!\sigma_{i}(A) $$ where $\sigma_i$ are the singular values.

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  • $\begingroup$ Are you taking the nuclear norm of $n \times m$ matrices? Unless $n = m$, how are you defining this? $\endgroup$ – Theo Bendit Nov 1 '18 at 10:11
  • $\begingroup$ I added more info to the question, and I beg your pardon, I meant "singular values" not eigenvalues. I'd still be grateful if there was help with the case $m=n$. $\endgroup$ – Erfan Nov 1 '18 at 10:25
  • $\begingroup$ Have you solved this? $\endgroup$ – Royi Aug 1 at 8:34

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