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Hilbert matrix

In linear algebra, a Hilbert matrix is a matrix with elements

Hij: = 1 /(i + j − 1)

For example, this is the 5 × 5 Hilbert matrix:

B = \begin{bmatrix}  1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \\[4pt] \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6} \\[4pt] \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & \frac{1}{7} \\[4pt] \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & \frac{1}{7} & \frac{1}{8} \\[4pt] \frac{1}{5} & \frac{1}{6} & \frac{1}{7} & \frac{1}{8} & \frac{1}{9} \end{bmatrix}

The Hilbert matrix can be regarded as derived from the integral

B_{jk} = \int_{0}^{1} x^{j+k} dx,

i.e. as a Gramian matrix for powers of x.

The Hilbert matrices are the canonical examples of ill-conditioned matrices, making them notoriously difficult to use in numerical computation. For example, the determinant of the matrix above is about 3.75\times 10^{-12}. The determinant can be obtained in closed form, as a special case of the Cauchy determinant.

Historical note

In Hilbert's oeuvre, the Hilbert matrix figures in his article Ein Beitrag zur Theorie des Legendreschen Polynoms (published in the journal Acta Mathematica , vol. 18, 155-159, 1894).

That article addresses the following question in approximation theory: "Assume that I = [a, b] is a real interval. Is it then possible to find a non-zero polynomial P with integral coefficients, such that the integral

\int_{a}^b P(x)^2 dx

is smaller than any given bound \varepsilon>0, taken arbitrarily small?" Using the asymptotics of the determinant of the Hilbert matrix he proves that this is possible if the length ba of the interval is smaller than 4.

He derives the exact formula

\det(H_{ij})_{i,j}={{c_n^{\;4}}\over {c_{2n}}}

for the determinant of the n × n Hilbert matrix. Here cn is

\prod_{i=1}^{n-1} i^{n-i}=\prod_{i=1}^{n-1} i!.\,

Hilbert also mentions the curious fact that the determinant of the Hilbert matrix is the reciprocal of an integer which he expresses as the discriminant of a certain hypergeometric polynomial related to the Legendre polynomial. This fact also follows from the identity

{1 \over \det (H_{ij})_{i,j}}={{c_{2n}}\over {c_n^{\;4}}}=n!\cdot \prod_{i=1}^{2n-1} {i \choose [i/2]}

Using Euler-MacLaurin summation on the logarithm of the cn he obtains the raw asymptotic result

\det(H_{ij})_{i,j}=4^{-n^2+r_n}

where the error term rn is o(n2). A more precise asymptotic result (which can be established using Stirling's approximation of the factorial) is

\det(H_{ij})_{i,j}=a_n\, n^{-1/4}(2\pi)^n \,4^{-n^2}

where an converges to some constant a_\infty=0.6450... as n\rightarrow\infty.

Reference

David Hilbert, Collected papers, vol. II, article 21.

01-04-2007 01:16:19
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