In my studies of matrix theory I came across the famous Hilbert matrix, which is a square $ n \times n $ matrix $ H $ with entries given by: $ h_{ij} = \frac{1}{i+j-1} $ and this is an example of a Cauchy matrix, which is a matrix $ C_n $ of the form $ c_{ij} = \frac{1}{x_i+y_j} $ and for this matrix there is the well known formula for the determinant:
$$\det(C) = \dfrac {\displaystyle \prod_{1 \mathop \le i \mathop < j \mathop \le n} \left({x_j – x_i}\right) \left({y_j – y_i}\right)} {\displaystyle \prod_{1 \mathop \le i, \, j \mathop \le n} \left({x_i + y_j}\right)}$$
Now I think I can substitute the sequences for the Hilbert matrix but I cannot see how to get the closed form they got here (under Properties):
$$\det(H) = \frac{c_n^4}{c_{2n}}$$
where
$$c_n = \prod_{i=1}^{n-1} i^{n-1} =\prod_{i=1}^{n-1} i!$$
and I was hoping someone would please help me obtain the closed form. Thanks.
Best Answer
If you are willing to accept the use of special functions, then there is an explicit closed form for the determinant of the $n\times n$ Hilbert matrix $\mathbf H_n$. Using the expression for the $\mathbf D$ factor of the $\mathbf L\mathbf D\mathbf L^\top$ decomposition shown in this MO answer (and originally derived by Hitotumatu), one can multiply together the diagonal elements of $\mathbf D$ (and then recall that $\det \mathbf L=1$) to get
$$\begin{align*} \det \mathbf H_n&=\det\mathbf D=\prod_{k=1}^n \frac1{(2k-1)\tbinom{2k-2}{k-1}^2}\\ &=\frac{G(n+1)^4}{G(2n+1)} \end{align*}$$
where $G(n)=\prod\limits_{k=1}^{n-1}\Gamma(k)$ is the Barnes $G$-function.