Timeline for Higher order generalization of Cauchy-Schwarz?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Aug 1, 2020 at 3:24 | history | became hot network question | |||
| Jul 31, 2020 at 20:24 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title, added tag
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| Jul 31, 2020 at 20:09 | vote | accept | Malkoun | ||
| Jul 31, 2020 at 20:08 | answer | added | Robert Bryant | timeline score: 16 | |
| Jul 31, 2020 at 20:02 | comment | added | Malkoun | @RobertBryant, could you please write it as an answer? The answer turned out to be simple (and I should have thought about it), but it is guiding me in the right direction (for the problem I am interested in, which inspired this post). Is the post too trivial? Should I delete it? | |
| Jul 31, 2020 at 19:55 | comment | added | Malkoun | @RobertBryant, ah yes true! The famous sum of squares trick, when working over $\mathbb{R}$. Thank you. How can one obtain all such polynomials? Can one use some form of the positivstellensatz perhaps? | |
| Jul 31, 2020 at 19:52 | comment | added | Robert Bryant | Actually, there's an octic polynomial: $$Q(v_1,v_2,\ldots,v_n) = \sum_{1\le i<j\le n} ((v_i,v_i)(v_j,v_j)-(v_i,v_j)^2)^2.$$ | |
| Jul 31, 2020 at 19:49 | comment | added | Malkoun | @KevinCasto, yes but, the determinant of the Gramian vanishes iff the vectors are linearly dependent. What I would like is though, a polynomial which vanishes iff the vectors lie in the same $1$-dimensional subspace. | |
| Jul 31, 2020 at 19:35 | comment | added | Kevin Casto | Indeed the Gramian is positive semi-definite, so its determinant is always nonnegative, and is positive just when the vectors are linearly independent. See en.wikipedia.org/wiki/Gramian_matrix#Gram_determinant | |
| Jul 31, 2020 at 19:31 | history | edited | Malkoun | CC BY-SA 4.0 |
added 198 characters in body
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| Jul 31, 2020 at 19:22 | history | asked | Malkoun | CC BY-SA 4.0 |