Question: 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](http://www.proofwiki.org/wiki/Vandermonde_Determinant)), a combinatorial proof by Art Benjamin and Gregory Dresden ([here](http://home.wlu.edu/~dresdeng/papers/VDM.pdf)), and another (slightly less) combinatorial proof by Jennifer Quinn ([here](http://www.sciencedirect.com/science/article/pii/S0378375810000406), 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:

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> Does anyone know a *geometric* proof of this result?
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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.

Answer: 
I have what looks like the first half of an answer, but for some strange reason I can't see how to finish the job.

Let $A$ be an $n \times n$ real matrix with eigenvalues $x\_1,...,x\_n$. On the one hand, the adjoint operator $ad(A): B \mapsto [A,B]$, acting from the space $C(A)^\perp$ of symmetric matrices orthogonal to centraliser of A to the space of skew-symmetric matrices, has determinant $\prod\_{1 \leq i < j \leq n} (x\_i - x\_j)$ (as can be seen by diagonalising $A$, and after fixing some sign conventions).

On the other hand, generically the centraliser $C(A)$ is an $n$-dimensional space spanned by $1, A, \ldots, A^{n-1}$, and the determinant of this basis is $\det(x\_j^{i-1})\_{1 \leq i < j \leq n}$ (again up to some normalisations). 

So presumably there must be some special property of the basis $1,A,\ldots,A^{n-1}$ of the kernel of the adjoint operator $ad(A)$ which would connect the two quantities and finish the job, but I can't see it yet, though it looks very close; I have a vague feeling one wants to work somehow in the non-commutative polynomial ring $M\_n({\bf R})(X)$ formed by adjoining an non-commutative indeterminate X to the matrix ring $M\_n({\bf R})$, and then specialise X to A, but my algebra is not good enough to push this through immediately. The fact that $\det(T^\*) = \det(T)$ for any linear transformation T also seems relevant somehow.
My favorite proof is the following (I am not sure that it is geometric, but however). At first, replace each $x\_i^k$ to $x\_i(x\_i-h)(x\_i-2h)\dots (x\_i-(k-1)h)$. I claim that determinant still equals $\prod (x\_j-x\_i)$. Then for $h=0$ we get what we need, while the trick is that we prove it for any other $h$ instead. Say, for $h=1$, it suffices by homogenuity. Note that both determinant and product of differences are polynomials of degree (say, at most) $n(n-1)/2$. It suffices to check that their values coincide on the set of points $\{(x\_1,x\_2,\dots,x\_n)=(c\_1,c\_2,\dots,c\_n)\}$, where $c\_i$ are non-negative integers and $\sum c\_i\leq n(n-1)/2$. This is easy and useful lemma. And in those points everything is clear. Both sides do vanish unless $(c\_1,\dots,c\_n)$ is a permutation of $0,\dots,n-1$. In this latter case the determinant contains unique non-zero summand, and it just equals the product of the sign of above permutation and $\prod\_{i<j} (j-i)=0!1!\dots (n-1)!$.