The inverse of the matrix $\{1/(i+j-1)\}$

Question

Let $n$ be a positive integer. Show that the matrix

$$\begin{pmatrix} 1 & 1/2 & 1/3 & \cdots & 1/n \\ 1/2 & 1/3 & 1/4 & \cdots & 1/(n+1) \\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1/n & 1/(n+1) & 1/(n+2) & \cdots & 1/(2n-1) \end{pmatrix}$$

is invertible and all the entries of its inverse are integers. This is an exercise in Hoffman and Kunze's linear algebra book. Any hints will be appreciated!

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http://math.stackexchange.com/questions/430060/why-does-the-inverse-of-the-hilbert-matrix-have-integer-entries?noredirect=1&lq=1 — StubbornAtom, Feb 04 '17 at 11:40

score 4 · Answer 1 · answered Jun 19 '13 at 19:21

4

The matrix you have is a Hilbert matrix. The $(i,j)$ entries of its inverse are given by $$(-1)^{i+j}(i+j-1){n+i-1 \choose n-j}{n+j-1 \choose n-i}{i+j-2 \choose i-1}^2$$which are clearly integers.

answered Jun 19 '13 at 19:21

The inverse of the matrix $\{1/(i+j-1)\}$

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