Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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. –  mathematician1975 Jul 1 '12 at 10:32

1 Answer 1

up vote 0 down vote accepted

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

share|improve this answer
2  
Especially for those who don't access to the article through JSTOR, will you please indicate how the article gives an answer to this question? –  Jonas Meyer Jul 1 '12 at 10:48
    
Google spits out quite a few additional links for that paper: 1 2 and also a related MO thread. –  Martin Sleziak Jul 1 '12 at 14:16

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.