Inverse of a particular matrix

[preda]

I have a very special matrix, square n*n, where the lines 0..(n-1) are:
line(i)={1/(i+1), 1/(i+2), ... , 1/(i+n)}

for example, for n==2:

1   1/2
1/2 1/3

for n==3:

1   1/2 1/3
1/2 1/3 1/4
1/3 1/4 1/5

I would like to find the inverse (as a function of "n").

(I would also be interested in *how* to solve the question (not only in the answer), if it's not too complicated)

Thank you!

[preda]

The above matrix originated from attempting to do a least-squares polynomial fit to a function evaluated over a (large) set of equidistant points. (the order of the matrix corresponds to the degree of the polynomial.)

[masser]

Have you heard about Hilbert matrices?

[preda]

Originally Posted by masser

Have you heard about Hilbert matrices?

Thanks! I thought that must be some well-known matrix, but I didn't know its name :)

[masser]

As the wiki article states, those matrices are useful for testing linear algebra solvers; that's how I first learned about them.

[preda]

In pari/gp:

1 / mathilbert(n)

[Dr Sardonicus]

Hilbert matrices are notorious for being "numerically bad." It is however ridiculously easy to prove they're "positive definite," hence invertible.

If n is a positive integer, f = f(x) is a polynomial of degree n-1,

f = a₀ + a₁*x + .... + a_n-1x^n-1, we can write [f] as the product of a 1xn and an nx1 matrix (Pari-GP notation),

[f] = [a₀, a₁, ..., a_n-1]*[1;x;...;x^n-1]. Taking the transpose,

[f] = [1,x,...,x^n-1]*[a₀; a₁; ...; a_n-1] Multiplying,

[f²] = [a₀, a₁, ..., a_n-1]*[1;x;...;x^n-1]*[1,x,...,x^n-1]*[a₀; a₁; ...; a_n-1]

Multiplying the middle two matrices gives the nxn matrix whose i, j entry is x^i+j-2. Now, integrate from 0 to 1, and we find:

The (1x1 matrix whose entry is) the integral from 0 to 1 of f²(x) dx may thus be expressed

[a₀, a₁, ..., a_n-1]*H_n*[a₀; a₁; ...; a_n-1].

The integral is positive unless f is identically zero, so H_n is positive-definite.

All you need, then, is a set of polynomials of degree 0, 1, 2, ..., n-1 which are orthogonal WRT integration from 0 to 1, (starting, obviously, with the constant polynomial 1). The "standard" set is the "shifted" Legendre polynomials. (The usual Legendre polynomials are orthogonal WRT integration from -1 to 1.)

[Batalov]

I want to muck with forum's tex software. (by posting straight from Wiki.)

The inverse of the Hilbert matrix can be expressed in closed form using binomial coefficients; its entries are

\((H^{-1})_{ij} = (-1)^{i+j}(i + j - 1) \binom {n + i - 1}{n - j} \binom {n + j - 1}{n - i} \binom {i + j - 2}{i - 1}^2\)

[retina]

Originally Posted by Batalov

I want to muck with forum's tex software. (by posting straight from Wiki.)


\((H^{-1})_{ij} = (-1)^{i+j}(i + j - 1) \binom {n + i - 1}{n - j} \binom {n + j - 1}{n - i} \binom {i + j - 2}{i - 1}^2\)

That isn't tex, that is relying on some ugly JS interpreter.

This is tex:


At least I can see the tex, although binom isn't defined.

[Nick]

Originally Posted by retina

That isn't tex, that is relying on some ugly JS interpreter.

This is tex:


At least I can see the tex, although binom isn't defined.

Use {n \choose r}.

[retina]

Originally Posted by Nick

Use {n \choose r}.