Timeline for Is the matrix $\left({2m\choose 2j-i}\right)_{i,j=1}^{2m-1}$ nonsingular?
Current License: CC BY-SA 3.0
23 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 2, 2017 at 21:43 | answer | added | Suvrit | timeline score: 21 | |
| Dec 30, 2016 at 3:47 | answer | added | Richard Stanley | timeline score: 14 | |
| Dec 29, 2016 at 17:42 | comment | added | Douglas Zare | See also mathoverflow.net/questions/255869/… and mathoverflow.net/questions/255141/… | |
| S Dec 29, 2016 at 13:45 | history | suggested | user57432 | CC BY-SA 3.0 |
Given mathjax to the math in the title
|
| Dec 29, 2016 at 13:30 | review | Suggested edits | |||
| S Dec 29, 2016 at 13:45 | |||||
| Dec 29, 2016 at 12:36 | answer | added | Mark Wildon | timeline score: 31 | |
| Dec 29, 2016 at 11:53 | comment | added | Johann Cigler | It seems that the eigenvalues of $\left({n+1\choose 2j-i}\right)_{i,j=1}^{n}$ are ${2,2^2,\dots\,2^n}.$ | |
| Dec 29, 2016 at 7:05 | history | edited | Douglas Zare | CC BY-SA 3.0 |
Added information to the title.
|
| Dec 29, 2016 at 7:00 | comment | added | T. Amdeberhan | Just like the matrix of entries $\binom{2i}j$ for $0\leq i, j\leq 2m-1$. | |
| Dec 29, 2016 at 6:59 | answer | added | Douglas Zare | timeline score: 26 | |
| Dec 29, 2016 at 6:51 | comment | added | Douglas Zare | Well, it's pretty exciting that this matrix seems to count domino tilings of an Aztec diamond. | |
| Dec 29, 2016 at 6:36 | comment | added | Pat Devlin | (1) This reminds me of a problem I can't quite remember where some sort of Mobius inversion came in handy. (2) I really want to somehow factor the thing. | |
| Dec 29, 2016 at 6:32 | comment | added | Douglas Zare | There should be a proof using Lindström-Gessel-Viennot. | |
| Dec 29, 2016 at 6:22 | history | edited | Douglas Zare | CC BY-SA 3.0 |
Gave example
|
| Dec 29, 2016 at 6:18 | answer | added | T. Amdeberhan | timeline score: 15 | |
| Dec 29, 2016 at 5:54 | comment | added | user42804 | @AnthonyQuas yes | |
| Dec 29, 2016 at 5:44 | comment | added | Anthony Quas | Are you defining $\binom ab$ to be 0 if $b<0$ or $b>a$? | |
| Dec 29, 2016 at 5:34 | comment | added | Douglas Zare | If $i$ and $j$ start with $0$, numerically the determinant seems to be $2^{A024920} * A063079$ although there are a lot of related sequences, and there might be some pretty simple way to put it. oeis.org/A024920 oeis.org/A063079 | |
| Dec 29, 2016 at 5:14 | comment | added | T. Amdeberhan | Then, the determinant equals $2^{\binom{2m}2}\neq0$ and hence the matrix is regular/invertible. | |
| Dec 29, 2016 at 5:12 | history | edited | user42804 | CC BY-SA 3.0 |
added 10 characters in body
|
| Dec 29, 2016 at 5:11 | comment | added | user42804 | @T.Amdeberhan yes | |
| Dec 29, 2016 at 5:11 | comment | added | T. Amdeberhan | Do $i, j$ start with $1$? | |
| Dec 29, 2016 at 5:02 | history | asked | user42804 | CC BY-SA 3.0 |