Timeline for Does $n^2$ divide $\det[(i+j)^n]_{0\le i,j\le n-1}$ for each integer $n>2$?
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11 events
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| Jun 6, 2018 at 0:10 | comment | added | T. Amdeberhan | @darijgrinberg: Thanks. I saw you comment after I edited my earlier comment but I will leave it there for the reader's benefit. | |
| Jun 6, 2018 at 0:04 | comment | added | Gerhard Paseman | One can subtract the jth column from the (j+1)th column to get n-1 columns of first differences of the nth power function, and then iterate to get fewer columns of second differences all the way down to one column of (n-1)th differences. Not quite a multiple of n!, but close. Gerhard "Someone Take The Baton Now" Paseman, 2018.06.05. | |
| Jun 5, 2018 at 23:58 | comment | added | T. Amdeberhan | A simpler variant is: $\det[(x+i+j)^{n-1}]_{i,j=1}^n=(-1)^{\binom{n}2}(n-1)!^n$. Even more general: $\det[(x_i+x_j)^{n-1}]_{i,j=1}^n=(-1)^{\binom{n}2}\prod_{k=1}^nk^{2k-1-n}V(x_1,\dots,x_n)$ where $V$ is Vandermonde determinant. | |
| Jun 5, 2018 at 23:54 | comment | added | darij grinberg | @T.Amdeberhan: More generally, Exercise 5.17 (c) in my Notes on the combinatorial fundamentals of algebra, version of 26 April 2018 (which is where I got my idea from); it's a fairly well-known result. | |
| Jun 5, 2018 at 23:49 | comment | added | Zhi-Wei Sun | @darij grinberg Great! You are the first one who can answer one of my questions posted in Mathoverflow. It seems that your method does not work for my another similar question which I will pose soon. | |
| Jun 5, 2018 at 23:35 | comment | added | darij grinberg | Minor correction to my first comment: Of course, $A$ should be $\left(\dbinom{n}{k} i^k\right)_{0 \leq i \leq n-1,\ 0 \leq k \leq n}$, not $\left(\dbinom{n}{k} i^k\right)_{0 \leq i \leq n-1,\ 0 \leq j \leq n}$. | |
| Jun 5, 2018 at 23:33 | comment | added | darij grinberg | Note that my proof of $n^2 \mid a\left(n\right)$ generalizes to $\det\left(\left(x_i+y_j\right)^n\right)_{0 \leq i\leq n-1, \ 0 \leq j \leq n-1}$ for any $2n$ integers $x_0, x_1, \ldots, x_{n-1}, y_0, y_1, \ldots, y_{n-1}$. With some more work, the additional $\prod_{k=1}^{n-1}k!$ factor generalizes to this case as well. I actually think we can replace it by an $\left(\prod_{k=1}^{n-1}k!\right)^2$, since the minors of $A$, too, are alternants once the column factors $\dbinom{n}{k}$ are factored out, and thus can be milked for Vandermonde determinants. | |
| Jun 5, 2018 at 23:29 | comment | added | darij grinberg | An additional $\prod_{k=1}^{n-1} k!$ factor can be explained by the fact that every minor of $B$ is divisible by $\prod_{k=1}^{n-1} k!$. (Indeed, every minor of $B$ is, up to sign, an alternant evaluated at the integers $0, 1, \ldots, n-1$, and thus is divisible by the Vandermonde determinant formed from these integers; but the latter Vandermonde determinant is precisely $\prod_{k=1}^{n-1} k!$.) | |
| Jun 5, 2018 at 23:26 | comment | added | darij grinberg | ... the $1$st and $n-1$st columns of $A$ are divisible by $n$, so that each maximal minor of $A$ that contains both of these columns is divisible by $n^2$. It remains to consider the remaining two maximal minors: the one that misses column $1$, and the one that misses column $n-1$. If $n \geq 5$, then these two minors contain both the $2$nd and the $3$rd column of $A$, and thus have a factor of $\dbinom{n}{2}\dbinom{n}{3}$, which is also divisible by $n$ (as a simple case distinction shows). So it remains to solve the $n < 5$ cases, which you have done in your post. | |
| Jun 5, 2018 at 23:22 | comment | added | darij grinberg | Let $A$ be the $n \times \left(n+1\right)$-matrix $\left(\dbinom{n}{k} i^k\right)_{0 \leq i \leq n-1,\ 0 \leq j \leq n}$. Let $B$ be the $\left(n+1\right)\times n$-matrix $\left(j^{n-k}\right)_{0 \leq k \leq n, \ 0 \leq j \leq n-1}$. Then, $AB$ is your $n \times n$-matrix $\left(\left(i+j\right)^n\right)_{0 \leq i \leq n-1, \ 0 \leq j \leq n-1}$ (by the binomial formula). Thus, $a\left(n\right) = \det\left(AB\right)$ can be computed by the Cauchy-Binet formula; you get a sum of several products of a maximal minor of $A$ and a maximal minor of $B$. Moreover, ... | |
| Jun 5, 2018 at 23:03 | history | asked | Zhi-Wei Sun | CC BY-SA 4.0 |