The Vandermonde matrix is the $n\times n$ matrix whose $(i,j)$-th component is $x_j^{i-1}$, where the $x_j$ are indeterminates. It is well known that the determinant of this matrix is $$\prod_{1\leq i < j \leq n} (x_j-x_i).$$

There are many known proofs of this fact, using for example row reduction or the Laplace expansion (here), a combinatorial proof by Art Benjamin and Gregory Dresden (here), and another (slightly less) combinatorial proof by Jennifer Quinn (here, unfortunately not open access). An easy proof follows by noting that the variety of the determinant contains (as a set) the variety of $x_i-x_j$ for all $i < j$ and then by computing the degree of the determinant as a polynomial in the $x_i, x_j$, though I don't know a reference for this proof.

Given that this result is amenable to such a wide variety of proofs (the above list contains three somewhat different flavors of proof---linear algebraic, combinatorial, and algebra-geometric), I have the following question:

Does anyone know a

geometricproof of this result?

For example, one might compute the volume of the parallelepiped whose vertices are given by the rows or columns of this matrix in a clever way. Ideally this would not just boil down to row reduction.

purely geometricallythe powers of $x_j$, the algebraic "addition" in Charles's argument seems to be unavoidable. It's probably a good reason to understand the product $\prod_{1\le i<j\le n}(x_i-x_j)$ from a geometric point of view. It's square, the discriminant, possesses some "hidden" geometry for small $n$ (in the theory of elliptic curves, for example) but probably not the product itself... $\endgroup$ – Wadim Zudilin Jun 24 '10 at 10:47