Timeline for An two-norm estimate for symmetric $k$-tensors
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Jan 23, 2019 at 23:07 | comment | added | RBega2 | @ZachTeitler Thank you that is very helpful. | |
| Jan 23, 2019 at 23:01 | comment | added | Zach Teitler | Now $\|S\|_*$ is called the nuclear norm of $S$, see for example arxiv.org/abs/1308.3860. I don't know if that helps directly but perhaps searching for "nuclear norm" might turn up something helpful. Good luck. | |
| Jan 23, 2019 at 22:44 | comment | added | RBega2 | @ZachTeitler Good point. I changed the definition to account for this (and to be consistent with the definition of the norm of the sum of spaces). I think this deals with your issue. | |
| Jan 23, 2019 at 22:41 | history | edited | RBega2 | CC BY-SA 4.0 |
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| Jan 23, 2019 at 22:29 | comment | added | Zach Teitler | This may be me being stupid, but I don't get your definition of $\|S\|_*$. If $S_1$ is one of the rank-one summands, why can't you replace the decomposition of $S$ with $(1/m)S_1 + \dotsb + (1/m)S_1$, which in the expression for $\|S\|_*$ replaces $\|S_1\|_2^2$ with $m \|(1/m)S_1\|_2^2 = (1/m)\|S_1\|_2^2$? As $m\to \infty$ this $\to 0$. Do you want to add some condition like the summands $S_i$ are linearly independent, or $r$ is minimal? | |
| Jan 23, 2019 at 0:13 | history | asked | RBega2 | CC BY-SA 4.0 |