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
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.3$: Sums and Products: Exercise $45$