[Math] How to compute coefficients of the Vandermonde polynomial

abstract-algebracombinatoricslinear algebra

I am trying to find the coefficients of the monomials in the expansion of

$$\prod_{1\le i < j \le n}^n (x_j – x_i)$$

also known as the Vandermonde determinant.

For example, for $n=3$ we have the factorization $(x_3 – x_2)(x_3 – x_1)(x_2 – x_1)$ . The coefficient of $x_3^2 x_2$ is 1, and coefficient of $x_3^2 x_1$ is $-1$. Therefore, when we expand the polynomial we will have

$$\prod_{1\le i < j \le n}^n (x_j – x_i) = x_3^2 x_2 – x_3^2 x_1 +\text{*other stuff*…}$$

Are there any known algorithms/theorems that help with computing these coefficients? Thanks.

Best Answer

Since the Vandermonde Polynomial is the determinant of the Vandermonde Matrix, the coefficient of $$ \prod_{j=1}^nx_j^{\sigma(j)-1} $$ is the parity of $\sigma$, a permutation on $\{1,2,\dots,n\}$. All other terms have a coefficient of $0$. This follows from the Leibniz Formula for the Determinant.


For example,

$x_3^2x_2=x_1^{1-1}x_2^{2-1}x_3^{3-1}$ and $(1,2,3)$ has parity $1$.

$x_3^2x_1=x_1^{2-1}x_2^{1-1}x_3^{3-1}$ and $(2,1,3)$ has parity $-1$.

Note that the parity of a permutation can be computed as $-1$ raised to the the number of pairs that are out of order. In $(1,2,3)$, no pairs are out of order so its parity is $1$. In $(2,1,3)$, only the pair $\{1,2\}$ is out of order so its parity is $-1$.

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