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I know that the nuclear norm $||.||_*$ is defined as the sum of the singular values ($\sigma_i$) of the matrix, that is for an $n\times n$ matrix $L$, the nuclear norm is defined by $$||L||_*=\sum_{i=1}^n \sigma_i(L),$$ and I read in a paper that this $||L||_*-<W,L>$ where $||W||\leq 1$ defines a semi-norm in $\mathbb{R}^{n\times n}$. My question is that how to compare the nuclear norm of $L$ and the nuclear semi-norm of $L$. I mean which one the greater than the other one, and how can I prove it.

Thanks in advance.

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  • $\begingroup$ I cannot understand the sentence "I read in a paper that this $||L||_*-<W,L>$ where $||W||\leq 1$ defines a semi-norm in $\mathbb{R}^{n\times n}$." First, $||L||_*-<W,L>$ is an incomplete mathematical statement, did you mean $||L||_* = <W,L>$? Or maybe $||L||_* - \langle W,L\rangle \geq 0$? Second, I know what a seminorm is, but I don't understand the statement "where $||W||\leq 1$ defines a semi-norm in $\mathbb{R}^{n\times n}$". $\endgroup$ – Ben Grossmann Sep 2 '20 at 5:40
  • $\begingroup$ Sorry Ben, and thank you for your comment, this is because I am not an English speaker. I found in some papers that, this statement $||L||_*-<W,L>$ defines a seminorm in $\mathbb{R}^{n\times n}$ where $||W||\leq 1$. This is the link for the paper link.springer.com/content/pdf/10.1007/s10898-017-0573-2.pdf You can find this statement in page 10. $\endgroup$ – Salma Omer Sep 2 '20 at 8:16
  • $\begingroup$ This doesn't help me because I cannot access the paper without paying for it. Perhaps you can edit your post and write the specific paragraph where you found this statement $\endgroup$ – Ben Grossmann Sep 2 '20 at 8:27
  • $\begingroup$ Ok, this is what is written " Sine $||W^{k-1}||\leq 1$ for each $k$. It is easy to check that $||.||_*-<W^{k-1},.>$ defines a seminorm in the vector space $\mathbb{R}^{n\times n}$". $\endgroup$ – Salma Omer Sep 2 '20 at 8:33
  • $\begingroup$ Ok, so what is made clear in this excerpt that was not clear from your question is that the seminorm being defined is the function $L \mapsto \|L\|_* - \langle W, L \rangle$. That is, the seminorm is $\|\cdot \|_* - \langle W, \cdot \rangle$. $\endgroup$ – Ben Grossmann Sep 2 '20 at 9:37
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As you could see from this post, we have $$ \langle W, L \rangle \leq \|L\|_* \cdot \|W\| \leq \|L\|_* \cdot 1. $$

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  • $\begingroup$ Thank you Ben. As we have $<W,L>\leq ||L||_*$, can we say that the seminorm being defined in this function $|L||_*-<W,L>$ is less than the ordinary nuclear norm that is defined by the sum of the singular values?? I mean can we prove this $$||L||_*-<W,L>\leq ||L||_*$$ $\endgroup$ – Salma Omer Sep 2 '20 at 9:56
  • $\begingroup$ @SalmaOmer No, this statement is not true unless we have some kind of constraint on $L$. For example, if we take $L = W$ then we have $<$, but if $L = -W$ then we have $>$. $\endgroup$ – Ben Grossmann Sep 2 '20 at 11:30
  • $\begingroup$ Ok. Thank you. In my case, $W$ is just a matrix of weights and it has no relation with $L$. $\endgroup$ – Salma Omer Sep 2 '20 at 12:21
  • $\begingroup$ @SalmaOmer Sure, but the point is that for a "random" matrix $L$, we have no reason to believe that $\langle W, L \rangle$ should be always positive or always negative. $\endgroup$ – Ben Grossmann Sep 2 '20 at 13:45
  • $\begingroup$ Thanks Ben. But what if the matrix $L$ is a rank one matrix with one block of ones and all the other entries are zeros? $\endgroup$ – Salma Omer Sep 2 '20 at 16:50

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