Geometric proof of the Vandermonde determinant?

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), 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 1

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 2

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 3

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 4

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 5

Let $F$ be a field, and $\mathbb{K}$ be the field of fractions of the polynomial ring $R = F[x_{1},x_{2},\ldots x_{n}].$

Take $n$ mutually projections from $\mathbb{K}^{n}$ onto $1$-dimensional spaces, $\{E_{i}: 1 \leq i \leq n \}.$ If you like, let $E_{i}$ be projection onto the $1$-dimensional space of row vectors in $\mathbb{K}^{n}$ with $0$ in place $j$ for $j \neq i.$ Consider the linear transformation $X = \sum_{i=1}^{n} x_{i}E_{i}.$ I claim that $\{ I,X,X^{2},\ldots , X^{n-1} \}$ is linearly independent and has the same linear span as $\{E_{1},E_{2},\ldots, E_{n} \}.$ Clearly the former span is contained in the span of the idempotents. For each $i,$ let $P_{i} = \prod_{j \neq i} \frac{X-x_{j}I}{x_{i}-x_{j}}.$ Now $P_{i} \neq 0$ for any $i$ because $XE_{i} = x_{i}E_{i}$ for each $i.$ In fact, we have $P_{i}E_{j} = 0$ for $j \neq i,$ and $P_{i}E_{i} = E_{i}.$ Since $P_{i}$ is in the linear span of the $E_{k}s,$ we must have $P_{i} = E_{i}$ for each $i.$ Hence each $E_{i}$ is in the linear span of $\{I,X,\ldots, X^{n-1}\}.$ This implies that the rows of the Vandermonde matrix are linearly independent, and that its determinant divides $\prod_{i < j}(x_{i}- x_{j})^{2}.$ In fact, we see that $E_{k} \prod_{i < j}(x_{i}-x_{j}) $ is an $R$-combination of $\{I,X,\ldots ,X^{n-1} \}$ for each $k,$ which implies that the determinant of the Van der Monde matrix divides $\prod_{i < j} (x_{i}-x_{j})$ in $R.$ Consideration of the coefficient of $X^{n-1}$ in each $E_{k}$ shows that this product must also divide the determinant in $R.$

Answer 6

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 7

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?

Answer 8

Ira Gessel used transitive tournaments in graphs to prove Vandermonde’s determinant identity: http://onlinelibrary.wiley.com/doi/10.1002/jgt.3190030315/abstract This proof certainly has some geometric flavour although not in the initial sense of the question.

P.S. The historical process that led to the worldwide adoption of the denomination "Vandermonde determinant" is studied in http://arxiv.org/abs/1204.4716 (A case of mathematical eponymy: the Vandermonde determinant, by Bernard Ycart).

Interestingly, $3\times 3$ Vandermonde determinant turns out to be (up to the square root) the correct physical variable of the odderon due to modular invariance of the odderon: http://arxiv.org/abs/hep-th/9604162 (The Odderon and Invariants of Elliptic Curves, by Romuald A. Janik).

Answer 9

Taking out constants, the Vandermonde determinant is the jacobian of the map $$x\mapsto(x_1+\dots+x_n,\dots,x_1^n+\dots+x_n^n)$$ (not zero since the first $n$ power sums are algebraically independent). By a classical result on Finite Reflection Groups (R. Steinberg, Invariants of finite reflection groups, Canad. J. Math. 12 (1960), 616–618.; for a nice geometric proof see Humphreys' book: Reflection Groups and Coxeter Groups, section 3.13), this determinant is, up to multiplication by a nonzero constant, the product of linear forms corresponding to the hyperplanes of reflection associated to the reflection group with basic invariants: $x_1+\dots+x_n,\dots,x_1^n+\dots+x_n^n$ (in this case it is the symmetric group $S_n$ acting on $\mathbb{R}^n$ with the reflections that send $e_i$ to $e_j$).

Answer 10

I would have posted this as a comment, but it's too long for comment, so I post it here. Here is my version of a sketch a geometric proof for nonsingularity of it when $x_i$'s are distinct, but I don't think that I can improve it to find the determinant:

Two manifolds $P$ and $S$ are called to intersect transversally at a point $A$, if the tangent spaces of $P$ and $S$ together span the whole ambient space. Let $\lambda_1,\ldots, \lambda_n$ be distinct real numbers, and let $A$ be the diagonal matrix $A=\rm{diag}(\lambda_1,\ldots,\lambda_n)$, and let $S$ be the set of all matrices with the same spectrum as $A$. In a small neighborhood around $A$, $S$ becomes a manifold. Let $P$ be the manifold of all the diagonal matrices of size $n$. The tangent space of $S$ and $P$ at $A$ can be computed and shown that the intersection is transversal.

On the other hand, define a function $f$ that maps any diagonal matrix $B$ (with $x_i$'s on its main diagonal) to $(\frac{\rm{tr}B}{1},\frac{\rm{tr}B^2}{2},\dots,\frac{\rm{tr}B^n}{n})$. It can be seen that the Jacobian of $f$ evaluated at $A=\rm{diag}(\lambda_1,\ldots,\lambda_n)$ is the Vandermonde matrix, which is nonsingluar if and only if $\lambda_i$'s are distinct.

Putting the two pieces above together, and with a little bit of discussion, one can show that $\rm{Jac}(f) \big|_A$ being nonsingular is equivalent to having $P$ and $S$ intersect transversally at $A$.

One can see the relations of the above approach to the Terry Tao's answer by noting that the tangent space to $S$ at $A$ is the set $\{[B,A] : B \text{ is a skew-symmetric matrix}\}$.

One relation to the powers of $A$ comes form a way of showing the above Jacobian matrix is nonsingular. Note that $$\rm{Jac}(f)\big|_A = \left[ \begin{array}{} I_{11} & I_{22} & \cdots & I_{nn}\\ A_{11} & A_{22} & \cdots & A_{nn} \\ \vdots & \vdots & \ddots & \vdots\\ A^{n-1}_{11} & A^{n-1}_{22} & \cdots & A^{n-1}_{nn}\end{array} \right]$$ In order to show the nonsingularity above assume that $\left[\begin{array}{} \alpha_1, \ldots, \alpha_n \end{array} \right] \rm{Jac}(f)\big|_A = 0$. This means if you consider the polynomial $p(x) = \sum_{i=1}^{n} \alpha_i x^{i-1}$ and let $X = p(A)$, you want to show if $X\circ I = O$ ($\circ$ is the Schur product) then $X=O$, but that is easy to show, since $A$ is diagonal, hence $p(A)$ is, and so $p(x)$ has $n$ distinct roots, but $\rm{deg}(p(x))=n-1$, thus $p(x)$ is the zero polynomial.

Answer 11

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 $\prod_{i<j} (j-i)=0!1!\dots (n-1)!$.