Let $l$ be a positive integer. Does the matrix $$ M_l \ := \ \left( \binom{l-(2p+1)}{j} \right)_{0\leq p,j \leq[(l-1)/2]} $$ have nonzero determinant?

  • $\begingroup$ Have you searched for "Vandermonde determinant"? $\endgroup$ – Jason Starr Feb 21 '15 at 16:32
  • $\begingroup$ Yes, but Vandermonde determinant didn't solve the problem. I found a paper that solves the problem but to a much more general case. I would like a more simpler solution, I believe it exists. $\endgroup$ – Liliam Feb 21 '15 at 16:40
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    $\begingroup$ @JasonStarr How do you reduce this to Vandermonde? Just curious :) $\endgroup$ – Alex Degtyarev Feb 21 '15 at 16:50
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    $\begingroup$ @AlexDegtyarev. For elements $a_0,\dots,a_{n-1}$ in a commutative ring $R$, the Vandermonde determinant $\text{det}(a_i^j)_{0\leq i,j\leq n-1}$ equals $0!\cdot 1!\cdots (n-1)!$ times the determinant $\text{det}(\binom{a_i}{j})_{0\leq i,j\leq n-1}$. Thus, the determinant above is, up to a product of factorials, a Vandermonde determinant. $\endgroup$ – Jason Starr Feb 21 '15 at 17:53
  • $\begingroup$ @JasonStarr: thanks, this is nice. For proof, I guess, this is the interpolation by Newton "monomials" rather than conventional ones, right? $\endgroup$ – Alex Degtyarev Feb 21 '15 at 18:52

There is a nice result of Gessel and Viennot that computes your determinant in terms of NE lattice paths. The original paper is available here. Aigner and Ziegler also give a nice exposition of this result in their Proofs From the Book.

  • $\begingroup$ Thank you Chris. The paper has a very general result, I wanted some more direct thing. I'll look in the book. $\endgroup$ – Liliam Feb 21 '15 at 17:33
  • $\begingroup$ @Liliam, I think if you take $k=\left\lfloor\frac{\ell-1}{2}\right\rfloor$, $a_i=2i$ and $b_j=j$ in their Theorem 1, then you get something pretty close to your determinant. $\endgroup$ – Chris McDaniel Feb 21 '15 at 18:12
  • $\begingroup$ Chris, the book solved my problem, thank you so much. $\endgroup$ – Liliam Feb 26 '15 at 0:02

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