2
$\begingroup$

The inverse of the matrix

$A=\left( \matrix{1 & \frac{1}{2} & \frac{1}{3} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} }\right)$

is

$A^{-1}=\left( \matrix{ 9 & -36 & 30 \\ -36 & 192 & -180 \\ 30 & -180 & 180 } \right)$.

Then, perhaps the matrix

$B=\left( \matrix{1 & \frac{1}{2} & ... & \frac{1}{n} \\ \frac{1}{2} & \frac{1}{3} & ... & \frac{1}{n+1} \\ ... & ... & & ...\\ \frac{1}{n} & \frac{1}{n+1} & ... & \frac{1}{2n-1} }\right)$

is invertible and $B^{-1}$ has integer entries. How can I prove it?

  • 2
    This is a Hilbert matrix and the elements of its inverse can be represented as the product of binomial coefficients. This might help you get started.2012-07-01

1 Answers 1

0

Here you will find your answer and many other things about Hilbert matrices : http://www.jstor.org/stable/2975779

  • 0
    [Google](http://www.google.com/#q=%22Tricks+or+Treats+with+the+Hilbert+Matrix) spits out quite a few additional links for that paper: [1](http://vigo.ime.unicamp.br/HilbertMatrix.pdf) [2](http://www.cecm.sfu.ca/~jborwein/Expbook/Updates/Already%20Included/choi.pdf) and also a related [MO thread](http://mathoverflow.net/questions/60000/elementary-proof-that-the-hilbert-matrix-is-invertible-with-integer-entries).2012-07-01