# Orthogonal Projection onto the Nuclear Matrix Norm Ball

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.

• Are you taking the nuclear norm of $n \times m$ matrices? Unless $n = m$, how are you defining this? – Theo Bendit Nov 1 '18 at 10:11
• 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 $m=n$. – Erfan Nov 1 '18 at 10:25
• Have you solved this? – Royi Aug 1 at 8:34