# Orthogonal projection on nuclear-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

\begin{aligned} 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 \} \end{aligned}

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? 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 $m=n$. Nov 1 '18 at 10:25
• Have you solved this?
– Royi
Aug 1 '19 at 8:34