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:

> 
> 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.

Answer: 
This is an interpretation of Terry Tao's answer (and BCnrd's comment).

If $A$ is $n\times n$ symmetric then $ad\_A:X\mapsto AX-XA$ maps symmetric matrices to skew matrices. Generically this is surjective, and generically its kernel has the first $n$ powers of $A$ as a basis. Choosing bases for the symmetric matrices and for the skew matrices (independent of $A$!), you then have a determinant to be computed, which appears to depend on the generic matrix $A$. However, we have -

Funny Fact: This is independent of $A$.

Proof of FF: If $A$ is diagonal, then a computation using the first bases you think of shows that this determinant is the quotient of a Vandermonde determinant and the usual expression for the same. Over the real numbers, you can use conjugation by orthogonal matrices to reduce to diagonal case. The real version implies the general version.

If you want to turn this into a proof of the Vandermonde identity, then you have to find an independent reason for FF. I do not have one to offer.

A cool restatement of FF is:

Although the basis $1, A, \dots , A^{n-1}$ for $ker(ad\_A)$ is (of course) dependent on $A$, the generator which it gives you for the top exterior power of this vector space does not depend on $A$. Here I am using the short exact sequence $0\to ker(ad\_A) \to Sym\to Skew\to 0$ to identify the $1$-dimensional vector spaces $\Lambda^n ker(ad\_A)$ and $(\Lambda^{top}Sym)\otimes (\Lambda^{top}Skew)^{-1}$.
This isn't the answer you're looking for, but I discovered it while attempting to find a geometric approach. 

Consider the polynomials $$p\_j(x) = (x-x\_1)\cdots (x-x\_j) = \sum\_{i=0}^{n-1} a\_{i,j} x^i,$$
where $p\_0(x)=1$ by convention, $0\leq j\leq n-1$. 
Of course, $a\_{j,j}=1$ and $a\_{i,j}=0$ if $i>j$, and $a\_{i,j}$ is the $i$th symmetric polynomial in $x\_1,\ldots, x\_j$ up to sign. 
If we multiply the Vandermonde matrix $[x\_i^{j-1}]$ by the upper unipotent matrix $[a\_{i-1,j-1}]$, 
we get the matrix $[p\_{j-1}(x\_i)]$. This is a lower triangular matrix, since $p\_{j-1}(x\_i) =0$ if $i \leq j-1$, and the diagonal entries are $p\_{i-1}(x\_i)$. 
Clearly, 
$$\prod\_{i=1}^n p\_{i-1}(x\_i) = \prod\_{i=1}^n (x\_i-x\_1) \cdots (x\_i-x\_{i-1}) = \prod\_{1\leq i < j\leq n} (x\_j-x\_i).$$

Maybe there's a way of seeing this product geometrically as a natural affine transformation?