How to compare the nuclear seminorm of a matrix with the nuclear norm of the same matrix?

Question

I know that the nuclear norm $\| \cdot \|_*$ 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.

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 $\mathbb{R}^{n\times n}$". — Ben Grossmann, Sep 02 '20 at 05:40
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 https://link.springer.com/content/pdf/10.1007/s10898-017-0573-2.pdf You can find this statement in page 10. — Salma Omer, Sep 02 '20 at 08:16
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 — Ben Grossmann, Sep 02 '20 at 08:27
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 $\mathbb{R}^{n\times n}$". — Salma Omer, Sep 02 '20 at 08:33
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 $|\cdot |* - \langle W, \cdot \rangle$. — Ben Grossmann, Sep 02 '20 at 09:37

score 0 · Accepted Answer · answered Sep 02 '20 at 09:40

0

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

answered Sep 02 '20 at 09:40

Ben Grossmann

225,327

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 $$||L||_-<W,L>\leq ||L||_*$$ – Salma Omer Sep 02 '20 at 09:56
@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 $>$. – Ben Grossmann Sep 02 '20 at 11:30
Ok. Thank you. In my case, $W$ is just a matrix of weights and it has no relation with $L$. – Salma Omer Sep 02 '20 at 12:21
@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. – Ben Grossmann Sep 02 '20 at 13:45
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? – Salma Omer Sep 02 '20 at 16:50
@SalmaOmer Are the entries of $W$ positive? – Ben Grossmann Sep 02 '20 at 16:58
Yes they are all positive. – Salma Omer Sep 02 '20 at 17:00
@Salma Then we can be sure that $\langle W,L \rangle \geq 0$. – Ben Grossmann Sep 02 '20 at 17:51
Exactly, thank you so much. – Salma Omer Sep 02 '20 at 18:06

How to compare the nuclear seminorm of a matrix with the nuclear norm of the same matrix?

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