Inverse of Hilbert matrix
Dong-Ping Deng
and mathematica (I can access to both). The first problem I'd like to solve
was to find the inverse of a NxN Hilbert matrix, the matrix looks like
H(N,N) =
1/1 1/2 1/3 1/4 ... 1/N
1/2 1/3 1/4 1/5 ... 1/(N+1)
1/3 1/4 1/5 1/6 ... 1/(N+2)
...
...
...
1/N 1/(N+1) ... 1/(N+N)
Of course, for a give value of N, I can find inverse of H. Could anyone give
a hint to find the inverse as a function of N ? Thanks in advance.
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dp
Bruce Ikenaga
I saw this long ago as a problem in Hoffman and Kunze ("Linear
Algebra"). I *think* there is a formula for this in Abramowitz
and Stegun ("Handbook of Mathematical Tables ... ").
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Bruce Ikenaga
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US mail: Dept. of Math, CWRU, Cleveland, Ohio 44106
E-mail : b...@po.CWRU.edu
Deane Yang
I first saw this when I was grading a linear algebra course which used
Hoffman and Kunze. I asked many people if they knew how to solve this
problem. Finally, while I was a graduate student at Harvard, Takahiro Shiota
kindly provided a very elegant solution.
The idea is to observe that the Hilbert matrix represents the L^2
inner product for polynomials of degree N-1 on the interval [0,1] with
respect to the basis 1, x, x^2,..., x^{N-1}.
On the other hand, it is well-known, especialy to physicists what
an orthonormal basis is. It's given by the Legendre polynomials.
Using the formulas for these, you can easily
derive the matrix that represents the change of basis. Squaring this
matrix yields the inverse. You also get a formula for the determinant.
I think it's something like 1/N!!, where N!! means the product of the
first N odd integers.
I'll leave the details to you.
Deane Yang
Department of Mathematics
Bruce Balden
>In article <DENG.91Fe...@max.bnl.gov> de...@bnl.gov (Dong-Ping Deng) writes:
>>I am recently introduced into the field of symbolic computation with macsyma
>>and mathematica (I can access to both). The first problem I'd like to solve
>>was to find the inverse of a NxN Hilbert matrix, the matrix looks like
As pointed out by Donald Knuth and others, the Hilbert matrix is a
special case of a Cauchy matrix
a(i,j)=1/(x(i)+y(j))
The inverse is then found by computing the determinants in the
explicit inverse formula. This turns out to be feasible.
The resulting formulas are a little bit complicated but simplify
greatly for the special case of the Hilbert matrix (x(i)=i,y(j)=j-1)
The general case is given on p36 of Knuth's Art of Computer Programming
Volume 1, Chapter 1.
The specialization to the Hilbert case is given in a later exercise.
Knuth cites as a reference
J. Todd
J. Res Nat. Bur Stand. 65(1961) pp19-22.
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Bruce Balden, Department of Thaumaturgy, Xenophon Consulting
Usenet: uunet!van-bc!balden Internet: bal...@wimsey.bc.ca