[Math] Geometric proof of the Vandermonde determinant

alternative-proofeuclidean-geometrylebesgue-measurelinear algebramg.metric-geometry

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 geometric proof 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.

Best Answer

A proof that may be called "geometric" with some good will from your side is as follows. Denote $V(a_1,a_2,\dots,a_n)$ the Vandermonde matrix with entries $a_j^{\ i-1}$, and consider the function $$f(t):=\det V(a_1+t, a_2+t,\dots, a_n+t \,).$$ One easily computes the derivative of the matrix and notices that $$\partial_t V:= NV,$$ for a certain matrix $N$ with null trace (actually, a nilpotent matrix with at most $n-1$ non-zero entries). Therefore by the Liouville formula, $$f^{\\ '}(t)=\mathrm{tr}(N) \ f(t)=0,$$ that proves that the Vandermonde determinant is translation-invariant. In particular for $t:=-a_1$ one has a reduction step that leads plainly to the product formula. (Alternatively, one can solve the above linear ODE getting $V=\exp(tN)V(0)$ and conclude as above, for $\exp(tN)$ is a triangular matrix with unit diagonal entries, hence with determinant equal to 1).