Timeline for Efficient way to express a symmetric tensor in terms of rank one elements
Current License: CC BY-SA 4.0
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Jan 23, 2019 at 23:05 | comment | added | Zach Teitler | Right, I forgot about that $1/C$. Anyway the obvious source of extra decompositions is by scaling each $x_i$ on the right hand side by some factor $t_i$, so you end up with $(t_1 x_1)^{k_1} \dotsm (t_n x_n)^{k_n}$; as long as the $t$'s multiply out to $1$, you got a different decomposition. I haven't thought about how to get $t$'s that minimize the $c_i$'s (or $\sum |c_i|$), hmm. And unfortunately there are other decompositions, these scalings don't hit all possible decompositions. I'm not sure how to describe the other ones, let alone minimize them. Interesting question. | |
Jan 23, 2019 at 22:59 | comment | added | RBega2 | @ZachTeitler One question about how you go the bounds on $c_i$ from the article you linked to. If I am understanding equation (8) correctly, then there is a coefficient $1/C$ that should be pretty large, possibly making the bounds one gets on on $c_i$ smaller than what you wrote. | |
Jan 23, 2019 at 22:51 | comment | added | RBega2 | @ZachTeitler Yes I mean monomial--this is very far from my area of research so I am a bit out of my element even with terminology. Thank you for the reference! It certainly looks like the sort of result I am interested in. | |
Jan 23, 2019 at 22:37 | comment | added | Zach Teitler | When you say $P = x_1^{k_1} \dotsm x_n^{k_n}$ is a homogeneous polynomial, do you actually intend that $P$ is a monomial? If so, then you might wish to look at arxiv.org/abs/1201.2922 (sorry for self-promoting :( ). The paper gives a simple explicit decomposition of $P$ in which each $\mathbf{v}_i$ has roots of unity as entries, so I guess length $\sqrt{n}$, and each $c_i$ is also a root of unity; if you scale to use unit vectors then I guess $|c_i|=n^{k/2}$ if I'm understanding right. And $N = \left(\prod(k_i+1)\right)/(\min\{k_i+1\})$. The decompositions are not unique though. | |
Jan 23, 2019 at 21:41 | history | asked | RBega2 | CC BY-SA 4.0 |