Geometric proof of the Vandermonde determinant? - MathOverflow

Geometric proof of the Vandermonde determinant?

2010-06-23T16:01:55Z

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.

Answer by Charles Siegel for Geometric proof of the Vandermonde determinant?

2010-06-23T17:06:44Z

First, I'll prove a lemma: on the rational normal curve of degree $n-1$ in $\mathbb{P}^{n-1}$, any collection of $n$ points spans the whole projective space. This is basically a consequence of the notion of degree: assume not. Then all $n$ points are contained in a hyperplane, but the curve is degree $n-1$, which means that no hyperplane can contain MORE than $n-1$ points, contradiction.

Thus, $n$ points on the rational normal curve are linearly independent, and so the determinant will vanish if and only if two or more of the points coincide. Using this, we can see that the determinant is a scalar multiple of the desired polynomial. Then, to see that the scalar is one, we just pick a coefficient on each side and compare.

Answer by Pietro Majer for Geometric proof of the Vandermonde determinant?

2010-06-24T00:00:36Z

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).

Answer by Terry Tao for Geometric proof of the Vandermonde determinant?

2010-06-24T05:34:07Z

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.

Answer by Agol for Geometric proof of the Vandermonde determinant?

2010-06-25T03:18:45Z

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?

Answer by Tom Goodwillie for Geometric proof of the Vandermonde determinant?

2010-06-25T11:03:26Z

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 $(\Lambda^{top}Sym)\otimes (\Lambda^{top}Skew)^{-1}$.

Answer by Ben Krause for Geometric proof of the Vandermonde determinant?

2010-12-02T12:01:36Z

A professor of mine brought this question up during a seminar the other day, and I offered a "peudo-geometric" answer, interpreting the Gaussian elimination proof in terms of volume preserving shear maps $\tau_{\lambda} \in End(R^n)$ of the form $(x_1,x_2,\dots,x_n) \mapsto (x_1-\lambda,x_2-\lambda x_1,\dots,x_n-\lambda x_{n-1})$ for $\lambda \in R$ scalar. Essentially, this argument boiled down to the fact that Gaussian row-reduction operations are "well-behaved" with respect to volume. I wonder if this cuts it?