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 , Kiran S. Kedlaya 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