Elements of Inverse of Hilbert Matrix are Integers

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Theorem

Let $H_n$ be the Hilbert matrix of order $n$:

$\begin{bmatrix} a_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {i + j - 1} \end{bmatrix}$


Consider its inverse $H_n^{-1}$.


All the elements of $H_n^{-1}$ are integers.


Proof

From Inverse of Hilbert Matrix, $H_n^{-1} = \begin {bmatrix} b \end{bmatrix}_n$ can be specified as:

$\begin{bmatrix} b_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac {\paren {-1}^{i + j} \paren {i + n - 1}! \paren {j + n - 1}!} {\paren {\paren {i - 1}!}^2 \paren {\paren {j - 1}!}^2 \paren {n - i}! \paren {n - j}! \paren {i + j - 1} } \end{bmatrix}$


Thus:

\(\ds b_{i j}\) \(=\) \(\ds \frac {\paren {-1}^{i + j} \paren {i + n - 1}! \paren {j + n - 1}!} {\paren {\paren {i - 1}!}^2 \paren {\paren {j - 1}!}^2 \paren {n - i}! \paren {n - j} ! \paren {i + j - 1} }\)
\(\ds \) \(=\) \(\ds \paren {\frac {\paren {-1}^{i + j} } {i + j - 1} } \paren {\frac {\paren {i + n - 1}!} {\paren {i - 1}! \, n!} } \paren {\frac {\paren {j + n - 1}!} {\paren {j - 1}! \, n!} } \paren {\frac {n! \, i} {i! \, \paren {n - i}!} } \paren {\frac {n! \, j} {j! \, \paren {n - j}! } }\)
\(\ds \) \(=\) \(\ds \frac {\paren {-1}^{i + j} i j} {i + j - 1} \binom {i + n - 1} n \binom {j + n - 1} n \binom n i \binom n j\) Definition of Binomial Coefficient
\(\ds \) \(=\) \(\ds \paren {-1}^{i + j} j \binom {i + n - 1} {i - i} \binom {j + n - 1} {n - 1} \binom {i + j - 2} {n - i} \binom n j\)

All of the factors of the above expression are integers, from Binomial Coefficient is Integer.

$\blacksquare$


Sources

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