Re: [obm-l] Matriz Harmonica(e esse onome?)

["Nicolau C. Saldanha" <nicolau@xxxxxxxxxxxxxxxxxxxxx>]

On Fri, Feb 14, 2003 at 03:01:03PM -0300, Johann Peter Gustav Lejeune Dirichlet wrote:
> Turma,ces sabem calcular o determinante de uma matriz n*n onde a(i;j)*(i+j)=1 sempre?Pelo que eu saiba deve ter isso na lista mas de qualquer caso...

Esta se chama uma matriz de Hilbert. Bem, a matriz de Hilbert mais clássica
tem entrada 1/(i+j-1) de tal forma que o canto é 1.

O determinante é:

1^(2(n-1)) * 2^(2(n-2)) * ... * (n-1)^2/
(2^1 * 3^2 * ... * n^(n-1) * (n+1)^n * (n+2)^(n-1) * ... * (2n-1)^2 * 2n

Uma demonstração boa está aqui

http://www.math.niu.edu/~rusin/known-math/97/hilbmat

onde se lê:

> >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.


Procurando isso, acabei de achar uma referência via google:

Man-Duen Choi: Tricks or treats with the Hilbert matrix. Amer. Math.
Monthly, 90:301-312, 1983.

Nunca li, mas pode ser legal. []s, N.
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