# A variant of the Vandermonde determinant

A very hard proof that $\sum_{i=0}^n i = \frac{n(n+1)}2$ (in comparison with the elementary level of the identity) is to compare degrees in the Vandermonde identity, which you prove playing around with the columns and use induction. Using Laplace expansion, the determinant of the Vandermonde matrix is a polynomial of degree $0 + 1+2+ \cdots + n$ and the result is $\prod_{0 \le i < j \le n} (x_j - x_j)$, a polynomial of degree $\binom{n+1}2$.

Knowing that there are formulas for the sum of $k^{\text{th}}$ powers, this begs the following question : is there a neat formula for the "$k$-Vandermonde" polynomials $$\det \left( \begin{bmatrix} 1 & x_0^{1^k} & x_0^{2^k} & \cdots & x_0^{n^k} \\ 1 & \vdots & \vdots & \ddots & \vdots \\ 1 & \vdots & \vdots & \ddots & \vdots \\ 1 & x_n^{1^k} & x_n^{2^k} & \cdots & x_n^{n^k} \end{bmatrix} \right).$$ I conjecture in the case $k=2$ that $$\det \left( \begin{bmatrix} 1 & x_0^{1^2} & x_0^{2^2} & \cdots & x_0^{n^2} \\ 1 & \vdots & \vdots & \ddots & \vdots \\ 1 & \vdots & \vdots & \ddots & \vdots \\ 1 & x_n^{1^2} & x_n^{2^2} & \cdots & x_n^{n^2} \end{bmatrix} \right) = \left( \prod_{0 \le i < j \le n} (x_j - x_i) \right) \left( \sum_{0 \le i \le j \le n} x_i x_j \right)( \cdots )$$ since the formula holds for $n=2$ (checked with Wolfram Alpha, in which case the $(...)$ factor equals $1$ ; the conjecture just suggests that these polynomials divide the rest). I know that the first factor divides the determinant ; this is just because the determinant vanishes when $x_i = x_j$, so in the UFD $\mathbb Z[x_0,\cdots,x_n]$, the distinct irreducibles $x_i - x_j$ divide the determinant. Also note that the $k$-Vandermonde divided by the $1$-Vandermonde is therefore a symmetric polynomial.

EDIT : I checked $n=3$ as the comments suggested, and indeed my conjecture is wrong, but the claims I made during the conjecture (about the product of the linear factors) is true.

Any ideas?

• Your conjecture fails for $n = 3$ (or $n=4$, depending on how you correct the LHS: as it stands, it has $n$ columns and $n+1$ rows, so one of these should be changed). I don't think you can say much about the general case. An alternant always is the Vandermonde determinant times a Schur polynomial (see the first Corollary in combinatorics.org/ojs/index.php/eljc/article/view/v9i1n5 or Corollary 2.54 in arxiv.org/abs/1409.8356v2 ), and the Schur polynomial doesn't really simplify usually. – darij grinberg Jun 23 '15 at 20:17
• @darijgrinberg : the row/column issue was just a typo. – Patrick Da Silva Jun 23 '15 at 20:18
• Yeah, but the $\sum_{i \leq j} x_i x_j$ factor isn't there for $n=3$. – darij grinberg Jun 23 '15 at 20:19

## 1 Answer

If you divide both sides of your conjecture by the Vandermonde, then the l.h.s. is exactly the Schur polynomial $s_{\lambda}(x_1,\cdots,x_n)$ where $\lambda_i+n-i=(n-i)^k$. Thus the $(\cdots)$ part is quite complicated if you now refer to the fact that $s_\lambda=\sum_T x^T$, the sum over semistandard young tableau of shape $\lambda$. To see that your conjecture doesn't quite work out, pick something like $n=4$ and find a semi-standard young tableau of shape $\lambda$ which has more than two distinct numbers.

• At least I know that this determinant has a nice well-known structure! Thanks a lot! And if one wants to overkill the elementary result, perhaps counting the number of semistandard tableaux and working a bit could give back the original identity, for instance using the Littlewood-Richarson rule to derive the inductive formula for the sum of $k^{\text{th}}$ powers? What do you think? Does this sound do-able or is it too complicated? – Patrick Da Silva Jun 23 '15 at 20:54
• This trick shows that $$\det \left( \begin{bmatrix} 1 & x_0 & x_0^{2^k} \\ 1 & x_1 & x_1^{2^k} \\ 1 & x_2 & x_2^{2^k} \end{bmatrix} \right) = \left(\prod_{0 \le i < j \le 2} (x_j - x_i) \right) \left( \underset{t_0 + t_1 + t_2 = 2^k-2}{\sum_{t_0,t_1,t_2 \ge 0}} x_0^{t_0} x_1^{t_1} x_2^{t_2} \right),$$ which is pretty cool! – Patrick Da Silva Jun 23 '15 at 21:03