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Show that the matrix

$A = \begin{bmatrix} 1 & \frac{1}{2} & \ldots & \frac{1}{n}\\ \frac{1}{2} & \frac{1}{3} & \ldots & \frac{1}{n+1}\\ \vdots & \vdots & & \vdots \\ \frac{1}{n} &\frac{1}{n+1} &\ldots &\frac{1}{2n-1} \end{bmatrix}$

is invertible and $A^{-1}$ has integer entries.

This problem appeared in Chapter one of my linear algebra textbook so I assume nothing more is required to prove it than elementary row operations. I've been staring at it for a while and considered small values of $n$ but there doesn't seem to be a clever way of easily reducing it to the identity matrix. Could someone give me a hint or a poke in the right direction rather than a full solution as I'm still hoping to work it out myself.

Alp Uzman
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Herman Chau
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  • Very nice situation. Today I asked the same question and got this page as a link. I saw the question many times previously but today I became very much interested to know the solution. Many student has faced the same several times. – Supriyo Dec 26 '13 at 16:28

1 Answers1

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If your linear algebra textbook is Kenneth Hoffmann and Ray Kunze's Linear Algebra book then I was in the same situation you're right now a few years ago, but I was adviced at the moment that this particular exercise wasn't as easy as one might expect at first.

The matrix you have is called the Hilbert matrix and the question you have was already asked a couple of times in math overflow here and here. They have excellent answers so I will just point you to them.

Community
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Adrián Barquero
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    Thanks! I am using Hoffman and Kunze and according to the comments in your second link, it seems that a simple proof is possible with the concept of a determinants, but they're not presented yet in Hoffman and Kunze. It's nice to know what the matrix is called though! – Herman Chau Jun 22 '11 at 11:05
  • Thanks, those two links were very helpful. Also, it's nice to know what the matrix is called! – Herman Chau Jun 22 '11 at 11:06
  • @Herman Chau I'm glad it helped you. And yes, it is good to know that the matrix has a name, since Hoffmann and Kunze's book didn't say anything about it. – Adrián Barquero Jun 22 '11 at 11:10
  • @AdriánBarquero And did you find any solution just using the tools provided for the first exercise? I'm curious. – Bijesh K.S May 07 '17 at 18:53