From: Robin Chapman <rjc@noether.ex.ac.uk>
Newsgroups: sci.math.research
Subject: Re: Some questions
Date: Fri, 14 Nov 1997 08:17:53 GMT

Panther wrote:
> 
> 1. I have a square matrix
> 
> [1   1/2  1/3  1/4    ........ 1/N      ]
> [1/2 1/3  1/4  1/5             1/(N+1)  ]
> [.                                      ]
> [.                                      ]
> [.                                      ]
> [.                                      ]
> [                                       ]
> [1/N 1/(N+1)                    1/(2N-1)]
> 
> Does this matrix have a name and does it have an easy inverse ?
> 
This is the Hilbert matrix, and it can be inverted quite easily.
It is also positive definite and a classic example of an ill-conditioned
matrix.

Consider the more general matrix with (i,j)-entry 1/(a_i+b_j).
It's not hard to show that its determinant is the product
of (a_i-a_j)(b_i-b_j) over all i<j divided by the product
of (a_i+b_j) over all i and j. Now observe that the Hilbert
matrix and all its minors have this form. Then with some patience
one can write down the inverse.

All this is in some exercises in Knuth's Fundamental Algorithms.

-- 
Robin Chapman			 	"256 256 256.
Department of Mathematics		 O hel, ol rite; 256; whot's
University of Exeter, EX4 4QE, UK	 12 tyms 256? Bugird if I no.
rjc@maths.exeter.ac.uk             	 2 dificult 2 work out."
http://www.maths.ex.ac.uk/~rjc/rjc.html	 Iain M. Banks - Feersum Endjinn

==============================================================================

[ I get the (i,j) entry of the inverse matrix to be

  (-1)^(i+j)*(i+n-1)!*(j+n-1)!/( (i-1)!^2*(j-1)!^2*(n-i)!*(n-j)!*(i+j-1) )

--djr ]