Sum of Elements in Inverse of Hilbert Matrix
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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.
The sum of all the elements $b_{i j}$ of $H_n^{-1}$ is:
- $\ds \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j} = n^2$
Proof
From Hilbert Matrix is Cauchy Matrix, $H_n$ is a special case of a Cauchy matrix:
- $\begin{bmatrix} c_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {x_i + y_j} \end{bmatrix}$
where:
- $x_i = i$
- $y_j = j - 1$
Then:
\(\ds \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j}\) | \(=\) | \(\ds \sum_{k \mathop = 1}^n x_k + \sum_{k \mathop = 1}^n y_k\) | Sum of Elements in Inverse of Cauchy Matrix | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^n k + \sum_{k \mathop = 1}^n \paren {k - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \sum_{k \mathop = 1}^n k - n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n \paren {n + 1} - n\) | Closed Form for Triangular Numbers | |||||||||||
\(\ds \) | \(=\) | \(\ds n^2\) |
$\blacksquare$
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 $45$