nLab Vandermonde determinant

The Vandermonde determinant or Vandermonde polynomial Δ(x 1,,x n)\Delta(x_1,\ldots,x_n) is the determinant of the Vandermonde matrix

V(x 1,,x n)=(1 x 1 x 1 n1 1 x 2 x 2 n1 1 x n x n n1) V(x_1,\ldots,x_n) = \left( \array{1 & x_1 & \cdots & x_1^{n-1}\\ 1 & x_2 &\cdots & x_2^{n-1}\\ \cdot &\cdot &\cdot &\cdots\\ 1 & x_n &\cdots &x_n^{n-1}}\right)

Δ(x 1,,x n)= 1i<jn(x jx i)\Delta(x_1,\ldots,x_n)= \prod_{1\leq i\lt j\leq n} (x_j - x_i).


We argue by induction. WLOG, assume x 1,,x n1x_1, \ldots, x_{n-1} are distinct (else the value is zero as claimed), and treat these as constants. Write

det(1 x 1 x 1 n1 1 x 2 x 2 n1 1 x x n1)=f(x)= i=1 n1a ix i.\det \left( \array{1 & x_1 & \cdots & x_1^{n-1}\\ 1 & x_2 &\cdots & x_2^{n-1}\\ \cdot &\cdot &\cdot &\cdots\\ 1 & x &\cdots &x^{n-1}}\right) = f(x) = \sum_{i=1}^{n-1} a_i x^i.

Then the leading coefficient a n1a_{n-1} is 1i<jn1(x jx i)\prod_{1\leq i\lt j\leq n-1} (x_j - x_i) by inductive hypothesis, and f(x)f(x) has simple roots at x=x 1,,x n1x = x_1, \ldots, x_{n-1} since each of those values makes two rows equal (hence with zero determinant). It follows that

f(x)=a n1 i=1 n1(xx i)=( 1i<jn1(x jx i)) i=1 n1(xx i)f(x) = a_{n-1}\prod_{i=1}^{n-1} (x - x_i) = (\prod_{1\leq i\lt j\leq n-1} (x_j - x_i))\prod_{i=1}^{n-1} (x - x_i)

and the inductive step follows by setting x=x nx = x_n.

The Vandermonde determinant appears in many important situations, as a square root of a discriminant, sometimes as a Wronskian, and also in Lagrange interpolation (see wikipedia Lagrange polynomial), Selberg integral etc.

  • wikipedia Vandermonde matrix
  • A. Varchenko, Multidimensional Vandermonde Determinant, Uspekhi Mat. Nauk 43:4 (1988), p. 190.
  • A. Varchenko, Beta-function of Euler, Vandermonde determinant, Legendre equation and critical values of linear functions of configuration of hyperplanes, I. Izv. Akademii Nauk USSR, Seriya Mat., 53:6 (1989), 1206-1235.

Last revised on April 2, 2013 at 18:24:20. See the history of this page for a list of all contributions to it.