0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

As you could see from this post, we have $$ \langle W, L \rangle \leq \|L\|_* \cdot \|W\| \leq \|L\|_* \cdot 1. $$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.