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May 31, 2014 at 6:20 comment added Tom Copeland S. Winitzki in Linear Algebra via Exterior Products (pages 126-7) gives a derivation of the product formula that seems to combine some of the observations of Algol and Terry Tao.
May 19, 2014 at 1:23 comment added Tom Copeland Similarly, with $e_k(x_1,...,x_n)$ the elementary symmetric polynomials, $de_1 \wedge de_2 \; ...\wedge de_n= |V_n|\; dx_1 \wedge dx_2 \; ...\wedge dx_n.$
Oct 31, 2010 at 17:46 comment added Terry Tao One can phrase this a little bit more geometrically as follows. The $n$-dimensional space $V$ of polynomials of degree less than $n$ has two bases: the standard basis $1,x,\ldots,x^{n-1}$, and Agol's basis $1, (x-x_1), (x-x_1)(x-x_2),\ldots,(x-x_1)\ldots(x-x_n)$. It also has the map $\Phi: P \mapsto (P(x_1),\ldots,P(x_n))$ from $V$ to $R^n$. The Vandermonde det is det of $\Phi$ with respect to the standard basis, while the product $\prod_{1 \leq i < j \leq n} (x_i-x_j)$ is the det of $\Phi$ wrt Agol's basis (where it becomes triangular). But the two bases are linked by a unipotent map.
Jun 25, 2010 at 16:35 comment added Ian Agol @Daniel: yes, if you look at the row reduction argument given in your link, it's actually what you get by doing only the column operations, which is how I found it (since I wanted to see what kind of affine transformation this produced). The matrix can be factored into block unipotent matrices which have $n−1−j$ copies of $−x_j$ just above the diagonal in the lower right corner, for $j=1,\ldots,n−1$.
Jun 25, 2010 at 13:05 comment added Daniel Litt I think this is exactly the same as row reduction if you write it out, no? That said, it's a great way of doing the computation.
Jun 25, 2010 at 4:41 comment added Wadim Zudilin It sounds like en.wikipedia.org/wiki/Companion_matrix.
Jun 25, 2010 at 3:23 history made wiki Post Made Community Wiki by Ian Agol
Jun 25, 2010 at 3:18 history answered Ian Agol CC BY-SA 2.5