Inverse of Hilbert Matrix
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Theorem
Let $H_n$ be the Hilbert matrix of order $n$:
- $\begin{bmatrix} a_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {i + j - 1} \end{bmatrix}$
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:
\(\ds \begin{bmatrix} b_{i j} \end{bmatrix}\) | \(=\) | \(\ds \frac {\paren {\dfrac {\paren {i + n - 1}!} {\paren {i - 1}!} } \paren {\dfrac {\paren {j + n - 1}!} {\paren {j - 1}!} } } {\paren {i + j - 1} \paren {i - 1}! \paren {-1}^{n - i} \paren {n - i}! \paren {j - 1}! \paren {-1}^{n - j} \paren {n - j}!}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {-1}^{i + j} \paren {i + n - 1}! \paren {j + n - 1}!} {\paren {\paren {i - 1}!}^2 \paren {\paren {j - 1}!}^2 \paren {n - i}! \paren {n - j}! \paren {i + j - 1} }\) |
$\blacksquare$
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 $45$