TITLE: Vandermonde matrix is totally positive
QUESTION [14 upvotes]: A totally positive matrix $M\in \mathcal{M}_{n\times m}(\mathbb R)$ is such that all of its minors of all sizes are positive. It is true that any Vandermonde matrix (with well-ordered positive entries) is totally positive. It seems that this fact should be classic. Although I can prove it by a variational argument, I cannot find a reference (in books I can think of or on the Internet) and I would like to know whether this is the "standard" way of proving the result, or if there is another (algebraic?) method known to the community.
Any pointer to a reference or direct proof would be very much appreciated !

REPLY [9 votes]: The minors of a Vandermonde matrix are known as alternants. Their positivity follows from the fact that they are products of the Vandermonde determinant of the variables involved with a Schur function in these variables. For a short and self-contained proof of this fact, see "Corollary (The Bi-Alternant Formula)" in John R. Stembridge, A Concise Proof of the Littlewood-Richardson Rule, The Electronic Journal of Combinatorics 9 (2002), Note #N5. This note itself is the distillate of several years of algebraic combinatorics (ideas of Lindstrom, Gessel, Viennot, Gasharov, Bender and Knuth are all in there), and longer proofs have been found before (e.g., in Macdonald's book).