Timeline for Formula expressing symmetric polynomials of eigenvalues as sum of determinants
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27 events
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Aug 2, 2021 at 9:53 | vote | accept | Jules | ||
Oct 17, 2020 at 22:29 | history | edited | Jules | CC BY-SA 4.0 |
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S Oct 2, 2020 at 8:01 | history | bounty ended | CommunityBot | ||
S Oct 2, 2020 at 8:01 | history | notice removed | CommunityBot | ||
Sep 24, 2020 at 14:57 | answer | added | Carlo Beenakker | timeline score: 11 | |
S Sep 24, 2020 at 6:51 | history | bounty started | Jules | ||
S Sep 24, 2020 at 6:51 | history | notice added | Jules | Authoritative reference needed | |
Sep 23, 2020 at 1:38 | history | edited | Jules | CC BY-SA 4.0 |
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Sep 23, 2020 at 1:35 | comment | added | Jules | I put a concise version of my proof in the question. | |
Sep 23, 2020 at 1:33 | history | edited | Jules | CC BY-SA 4.0 |
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Sep 23, 2020 at 0:25 | comment | added | Fedor Petrov | @TimothyChow done, and I found an issue preparing this answer. | |
Sep 23, 2020 at 0:23 | answer | added | Fedor Petrov | timeline score: 11 | |
Sep 22, 2020 at 16:48 | comment | added | Timothy Chow | @FedorPetrov : Perhaps you could collect your comments into an answer even though they don't directly answer the "reference request" part of the question. | |
Sep 22, 2020 at 14:31 | comment | added | Jules | Ah, now I understand. I misunderstood your initial comment and thought there was a particular known family of functions $H$ that you were talking about, but you are talking about the $p_i$ that are of that form for any $H$. Sorry for the confusion and thank you for the explanation. | |
Sep 22, 2020 at 14:20 | comment | added | Fedor Petrov | because the product of two such functions is again such a function. | |
Sep 22, 2020 at 13:15 | comment | added | Jules | Thanks! I think I understand all the steps except one. I could not understand why the linear span is the same as the algebra. The degree of these special $p_i$ seems to be $\leq n$ so I'd expect that the same holds true for their span, whereas in general the $p_i$ can have any degree. Could you help me understand where my reasoning went wrong? By the way, it is easy to prove the reverse implication, that any symmetric polynomial can be written in elementary symmetric polynomials, by taking $A$ to be the companion matrix of a polynomial in the identity :) | |
Sep 21, 2020 at 23:44 | comment | added | Fedor Petrov | At first, the linear span of such functions is the same as the algebra generated by them. Next, it suffices to consider $p$ with a finite support $A$. Then for $H(m)=m+\alpha$ for $m\in A$ (hereafter: and 0 otherwise) we get a function of the form $p_i=(i_1+\alpha)...(i_n+\alpha)$. Varying $\alpha$ and taking linear combinations we get any elementary symmetric polynomial in $i_1,..., i_n$. They generate an algebra of all symmetric polynomials, and any symmetric function on $A$ is represented by a symmetric polynomial. | |
Sep 21, 2020 at 23:36 | history | edited | Jules | CC BY-SA 4.0 |
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Sep 21, 2020 at 23:25 | history | edited | Jules | CC BY-SA 4.0 |
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Sep 21, 2020 at 23:18 | history | edited | Jules | CC BY-SA 4.0 |
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Sep 21, 2020 at 23:13 | comment | added | Jules | That sounds cool! What is the name of these functions $H$ and where could I find a proof that they generate all symmetric functions of natural number tuples? (which I think means that any symmetric $p$ can be written as a linear combination of those special $p$'s?) | |
Sep 21, 2020 at 22:49 | comment | added | Fedor Petrov | That's nice. I do not know the reference. When $p_i=\prod_{k=1}^n H(i_k)$ for certain function $H$ defined on $\{0,1,\ldots\}$, both parts factorize and are equal to the determinant of $\sum_m H(m)A^m$. The functions of such type generate all symmetric functions $i=(i_1,\ldots,i_n)\mapsto p_i$, so the result follows. | |
Sep 21, 2020 at 22:33 | history | edited | Jules | CC BY-SA 4.0 |
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Sep 21, 2020 at 22:24 | history | edited | Jules | CC BY-SA 4.0 |
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Sep 21, 2020 at 22:18 | history | edited | Jules | CC BY-SA 4.0 |
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Sep 21, 2020 at 21:45 | history | edited | Jules | CC BY-SA 4.0 |
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Sep 21, 2020 at 21:30 | history | asked | Jules | CC BY-SA 4.0 |