From: hrubin@stat.purdue.edu (Herman Rubin)
Newsgroups: sci.math.research
Subject: Re: Hilbert determinant evaluation
Date: 19 May 1997 21:22:45 -0500

In article <kkedlaya-1705971934270001@kkedlaya.student.princeton.edu>,
Kiran S. Kedlaya <kkedlaya@math.Princeton.EDU> wrote:
>How does one normally compute the determinant of the "Hilbert matrix"
>    A_{ij} = 1/(i + j) (i, j = 1, ..., n)?
>(This matrix, I believe, is a standard example in numerical analysis of a
>matrix whose determinant is much closer to 0 than any of its entries.)
>I recently overheard a problem closely related to this, but my memory
>is a bit rusty concerning the Hilbert matrix. Thanks.

Stated this way, it is not obvious.  But generalized slightly, it is.
Consider instead the determinant of the matrix whose elements are
b_{ij} = 1/(x_i + y_j).  This is a rational function of the x's
and y's, and in lowest terms the denominator can be taken to be
the product of the n^2 b's.  Now the numerator is divisible by
(x_i - x_j) and (y_i - y_j) for i different from j, and thus the
product of all of these divides the numerator.  But this accounts
for degree n^2 - n, and all terms of the product expansion are
of degree -n, so this is it, except possibly for a constant.  The
case where the x's and y's grow rapidly shows the constant is 1.

-- 
This address is for information only.  I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
hrubin@stat.purdue.edu         Phone: (765)494-6054   FAX: (765)494-0558