Proving that an matrix have integers entries

Asked Sep 04 '13 at 18:07

Active Sep 04 '13 at 18:54

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Considering the Matrix below with $n \in \mathbb{N}$

$$ A = \left( \begin{array}{ccc} 1 & \frac{1}{2} & \cdots & \frac{1}{n} \\ \frac{1}{2} & \frac{1}{3} & \cdots & \frac{1}{n+1} \\ \vdots & \vdots & & \vdots \\ \frac{1}{n} & \frac{1}{n+1} & \cdots & \frac{1}{2n-1} \\ \end{array} \right) $$

How i show that this matrix is invertible and $A^{-1}$ has only integers entries using some elimination method , like for example Gauss-Jordan? I tried to prove that $A$ is invertible, but i consider very difficult to solve the problem on this way and using this approach i have no idea how to prove that the entries are integers.

edited Sep 04 '13 at 18:24

asked Sep 04 '13 at 18:07

Mike

4

Hilbert matrix (http://en.wikipedia.org/wiki/Hilbert_matrix) – Sep 04 '13 at 18:11
possible duplicate of Why does the inverse of the Hilbert matrix have integer entries? – rschwieb Sep 04 '13 at 18:14
I see this question, but that use the idea of a Cauchy matrix, and a want to see something more easy, maybe the calculation using the Gauss-Jordan – Mike Sep 04 '13 at 18:18
@ExampleMo OK, if you want to have these restrictions then you must make sure they're clear in the problem statement, or else someone else is going to cast closing votes. (I've retracted mine at the moment.) – rschwieb Sep 04 '13 at 18:20
Ok i will reformulate the question – Mike Sep 04 '13 at 18:21

Proving that an matrix have integers entries

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