# Inverse of Hilbert Matrix

## Theorem

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

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

Then its inverse $H_n^{-1} = \sqbrk b_n$ can be specified as:

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

## 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:

$x_i = i$
$y_j = j - 1$

From Inverse of Cauchy Matrix, the inverse of the square Cauchy matrix of order $n$ is:

$\begin{bmatrix} b_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac {\ds \prod_{k \mathop = 1}^n \paren {x_j + y_k} \paren {x_k + y_i} } {\ds \paren {x_j + y_i} \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne i} } \paren {y_i - x_k} } } \end{bmatrix}$

Thus $H_n^{-1}$ can be specified as:

$\begin{bmatrix} b_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac {\ds \prod_{k \mathop = 1}^n \paren {i + k - 1} \paren {j + k - 1} } {\ds \paren {i + j - 1} \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne i} } \paren {i - k} } \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {j - k} } } \end{bmatrix}$

First, from Product of Products:

$\ds \prod_{k \mathop = 1}^n \paren {i + k - 1} \paren {j + k - 1} = \prod_{k \mathop = 1}^n \paren {i + k - 1} \prod_{k \mathop = 1}^n \paren {j + k - 1}$

We address in turn the various factors of this expression for $b_{i j}$.

 $\ds \prod_{k \mathop = 1}^n \paren {i + k - 1}$ $=$ $\ds \prod_{k \mathop = 0}^{n - 1} \paren {i + k}$ Translation of Index Variable of Product $\ds$ $=$ $\ds i^{\overline n}$ Definition of Rising Factorial $\ds$ $=$ $\ds \frac {\paren {i + n - 1}!} {\paren {i - 1}!}$ Rising Factorial as Quotient of Factorials

and similarly:

$\ds \prod_{k \mathop = 1}^n \paren {j + k - 1} = \frac {\paren {j + n - 1}!} {\paren {j - 1}!}$

Then:

 $\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne i} } \paren {i - k}$ $=$ $\ds \paren {\prod_{1 \mathop \le k \mathop < i} \paren {i - k} } \paren {\prod_{i \mathop < k \mathop \le n} \paren {i - k} }$ $\ds$ $=$ $\ds \paren {i - 1}! \paren {\prod_{i \mathop < k \mathop \le n} \paren {i - k} }$ Definition of Factorial $\ds$ $=$ $\ds \paren {i - 1}! \paren {\prod_{0 \mathop < k \mathop \le n - i} \paren {-k} }$ Translation of Index Variable of Product $\ds$ $=$ $\ds \paren {i - 1}! \paren {-1}^{n - i} \paren {\prod_{0 \mathop < k \mathop \le n - i} k}$ $\ds$ $=$ $\ds \paren {i - 1}! \paren {-1}^{n - i} \paren {n - i}!$ Definition of Factorial

and similarly:

$\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {j - k} = \paren {j - 1}! \paren {-1}^{n - j} \paren {n - j}!$

Thus we can write:

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

$\blacksquare$