Timeline for Nuclear norm as minimum of Frobenius norm product
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Oct 27, 2020 at 4:09 | history | edited | Hans | CC BY-SA 4.0 |
Explicate the inequalities.
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| Sep 17, 2020 at 14:58 | vote | accept | Hans | ||
| Sep 17, 2020 at 14:53 | history | edited | Hans | CC BY-SA 4.0 |
Added the definition of $U$ and $V$.
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| Apr 19, 2018 at 5:49 | comment | added | Hans | @DenisSerre: You are right. I have now revamped and corrected my proof. Please review. | |
| Apr 18, 2018 at 9:43 | history | edited | Hans | CC BY-SA 3.0 |
Revamped my proof.
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| Apr 18, 2018 at 6:37 | comment | added | Denis Serre | @Hans. Your answer is wrong, because you start from a false identity. The trace of $\sqrt{VU^TUV^T}$ is not equal to that of $\sqrt{U^TU}\,\sqrt{V^TV}$ in general, it is only larger. Your mistake comes from the fact that square root does not behave well under product. Consider the case where $V=D>0$ is diagonal and set $S=\sqrt{U^TU}$. Then $A:=SD$ is diagonalisable with real eigenvalues and arbitrary otherwise. You are comparing the traces of $\sqrt{A^TA}$ and that of $A$. The first one is obviously larger. | |
| Apr 17, 2018 at 7:56 | history | edited | Hans | CC BY-SA 3.0 |
added 34 characters in body
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| Apr 17, 2018 at 6:45 | comment | added | Tobias Fritz | no, the downvoter was not me. | |
| Apr 17, 2018 at 1:29 | comment | added | Hans | @TobiasFritz: Did you happen to downvote my answer? | |
| Apr 16, 2018 at 23:52 | history | edited | Hans | CC BY-SA 3.0 |
Added a simplifying proof.
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| Apr 16, 2018 at 23:29 | comment | added | Hans | @TobiasFritz: Good that we agree. As a matter of fact, I now think both our methods, my use of Von Neumann's trace inequality and your citing of general matrix Holder inequality, are a bit circuitous. It is just a simple application of Cauchy-Schwartz inequality. I will write the simplified proof out. | |
| Apr 16, 2018 at 22:28 | comment | added | Tobias Fritz | well, the most important thing is that we've answered your question, so let's leave it at that. I certainly don't doubt that you've come up with the argument yourself! | |
| Apr 16, 2018 at 22:21 | comment | added | Hans | @TobiasFritz: Which part you do you not agree, that I did not repost your answer or that our methods are distinct? If it is the latter, are you saying I proved the special case $p=q=2$ of the matrix Holder inequality, which you quoted rather than proved, without knowing it? Is it not even more in my favour? Does it not imply further that there was no reposting except the difference in posting time? As I said before, I had checked the source arxiv.org/pdf/1106.6189v2.pdf of the matrix Holder inequality, the proof there was distinct from mine for $p=q=2$. Do you not agree? | |
| Apr 16, 2018 at 20:53 | comment | added | Tobias Fritz | I'm not sure I agree, because your argument uses the trace inequality only to derive the $p=q=2$ case of the matrix Hölder inequality, which is precisely your $\|UV^T\|_\sigma \leq \|U\|\, \|V\|$. | |
| Apr 16, 2018 at 20:00 | history | edited | Hans | CC BY-SA 3.0 |
added 99 characters in body
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| Apr 16, 2018 at 19:54 | comment | added | Hans | @TobiasFritz: To give a definitive statement to the second sentence of my last comment, I just checked the source of the matrix Holder inequality and found it to be very much different from the Von Neumann's trace inequality. So our approaches are distinct, and thus one is not a repost of another at all. | |
| Apr 16, 2018 at 19:33 | comment | added | Hans | @TobiasFritz: You have a wrong assumption. I was close to finishing mine when you posted yours. Also, I am using Von Neumann's trace inequality, while I do not know whether or not the matrix Holder inequality you cited is the same as Von Neumann's trace inequality. | |
| Apr 16, 2018 at 19:05 | comment | added | Tobias Fritz | That's essentially the same as my answer. Why do you repost it? | |
| Apr 16, 2018 at 19:01 | history | answered | Hans | CC BY-SA 3.0 |