Timeline for Prove that the matrix $[\Gamma(\lambda_{i}+\mu_{j})]$ is nonsingular
Current License: CC BY-SA 4.0
17 events
when toggle format | what | by | license | comment | |
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Jul 22, 2018 at 19:46 | vote | accept | VSP | ||
Jul 22, 2018 at 19:03 | vote | accept | VSP | ||
Jul 22, 2018 at 19:46 | |||||
Jul 22, 2018 at 19:03 | vote | accept | VSP | ||
Jul 22, 2018 at 19:03 | |||||
Jul 22, 2018 at 18:58 | vote | accept | VSP | ||
Jul 22, 2018 at 19:03 | |||||
Jul 22, 2018 at 18:58 | vote | accept | VSP | ||
Jul 22, 2018 at 18:58 | |||||
Jul 21, 2018 at 9:25 | answer | added | Mateusz Kwaśnicki | timeline score: 6 | |
Jul 20, 2018 at 18:49 | history | edited | Abdelmalek Abdesselam |
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Jul 20, 2018 at 15:39 | comment | added | Abdelmalek Abdesselam | @prince: thank you for this beautiful question. May I ask about the motivation or context for it? | |
Jul 19, 2018 at 15:34 | answer | added | Abdelmalek Abdesselam | timeline score: 18 | |
Jul 19, 2018 at 12:56 | comment | added | VSP | I have checked it many times on Mathematica, the matrix is getting singular only if any two $\alpha_{i}$'s or $\beta_{j}$'s are becoming same, which is not allowed in our problem. | |
Jul 19, 2018 at 12:42 | comment | added | VSP | Bullet51, can you please tell me, why did you mention $\Gamma(2)=\Gamma(1)=1$? | |
Jul 19, 2018 at 12:15 | comment | added | VSP | I don't think, this matrix can get all it's entries equal to 1. | |
Jul 19, 2018 at 12:10 | comment | added | VSP | thanks for replying. Here what are your $ \lambda_{i}$ and $\mu_{j}$?? I have checked it by Mathematica. And I couldn't find any factorization of this matrix too. I have tried it to prove it in many ways. But no success. | |
Jul 19, 2018 at 11:11 | comment | added | LeechLattice | $\Gamma(2)=\Gamma(1)=1$. | |
Jul 19, 2018 at 10:34 | comment | added | Alex M. | Can you give some motivation for why you expect it to be non-singular? | |
Jul 19, 2018 at 10:29 | review | First posts | |||
Jul 19, 2018 at 10:33 | |||||
Jul 19, 2018 at 10:28 | history | asked | VSP | CC BY-SA 4.0 |