Finding the closed form of the determinant of the Hilbert matrix

Question

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.

Answer 1

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.

Answer 2

To find the determinant of the $n\times n$ Hilbert matrix, we will apply a sequence of row operations (multiplication on the left) and column operations (multiplication on the right) that will transform the Hilbert matrix into a diagonal matrix.

The Hilbert matrix is $H_0$ where the element in row $j\in[1,n]$ of column $k\in[1,n]$ of $H_m$ is given by $$ h_m(j,k)=\left\{\begin{array}{cl} \frac1{j+k-1}&\text{if $j=k$ or ($j\gt m$ and $k\gt m$)}\\ 0&\text{otherwise} \end{array}\right.\tag1 $$ for example, for $n=5$, $$ H_0=\begin{bmatrix} 1 & \frac12 & \frac13 & \frac13 & \frac15 \\ \frac12 & \frac13 & \frac14 & \frac15 & \frac16 \\ \frac13 & \frac14 & \frac15 & \frac16 & \frac17 \\ \frac14 & \frac15 & \frac16 & \frac17 & \frac18 \\ \frac15 & \frac15 & \frac17 & \frac18 & \frac19 \end{bmatrix} \quad\text{and}\quad H_2=\begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & \frac13 & 0 & 0 & 0 \\ 0 & 0 & \frac15 & \frac16 & \frac17 \\ 0 & 0 & \frac16 & \frac17 & \frac18 \\ 0 & 0 & \frac17 & \frac18 & \frac19 \end{bmatrix}\tag2 $$ etc.

For a given $m$, we will use element $(m,m)$ in $H_{m-1}$ as a pivot to clear out the rest of column $m$ using row operations and the rest of row $m$ using column operations.

For a given $m$, subtracting $\frac{2m-1}{j+m-1}$ times row $m$ from row $j$ in $H_{m-1}$ multiplies row $j$ by $\frac{j-m}{j+m-1}$ (left multiplication) and column $k$ by $\frac{k-m}{k+m-1}$ (right multiplication); that is, $$ \frac1{j+k-1}-\frac{2m-1}{j+m-1}\frac1{k+m-1}=\frac{j-m}{j+m-1}\frac{k-m}{k+m-1}\frac1{j+k-1}\tag3 $$ Thus, if we define $T_m$ and $D_m$ by $$ \begin{align} t_m(j,k)&=\left\{\begin{array}{cl} 1&\text{if $j=k$}\\ -\frac{2m-1}{j+m-1}&\text{if $j\gt k$ and $k=m$}\\ 0&\text{if $j\gt k$ and $k\ne m$ or $j\lt k$} \end{array}\right.\tag{4a}\\ d_m(j,k)&=\left\{\begin{array}{cl} 1&\text{if $j=k$ and $k\le m$}\\ \frac{j+m-1}{j-m}&\text{if $j=k$ and $k\gt m$}\\ 0&\text{if $j\ne k$}\\ \end{array}\right.\tag{4b} \end{align} $$ $(3)$ yields $$ H_m=D_mT_mH_{m-1}\left(D_mT_m\right)^T\tag5 $$ Since $T_m$ is a lower triangular matrix with ones on the diagonal, $$ \det(T_m)=1\tag{6a} $$ Although we won't use it, it is not difficult to see that $$ \det(D_m)=\binom{n+m-1}{n-m}\tag{6b} $$ What we will use is that since $$ \prod_{m=1}^nd_m(j,k)=\left\{\begin{array}{cl} \binom{2j-2}{j-1}&\text{if $j=k$}\\ 0&\text{if $j\ne k$}\\ \end{array}\right.\tag7 $$ we get $$ \det\left(\prod_{m=1}^nD_m\right)=\prod_{j=1}^n\binom{2j-2}{j-1}\tag8 $$ Iterating the determinant of $(5)$ and applying $\text{(6a)}$, we get $$ \left(\prod_{m=1}^n\det(D_m)\right)\det(H_0)\left(\prod_{m=1}^n\det\left(D_m^T\right)\right)=\det(H_n)\tag9 $$ Applying $(8)$ gives $$ \begin{align} \det(H_0) &=\left(\prod_{m=1}^n\frac1{\det(D_m)}\right)\det(H_n)\left(\prod_{m=1}^n\frac1{\det\left(D_m^T\right)}\right)\tag{10a}\\ &=\prod_{j=1}^n\frac1{\binom{2j-2}{j-1}^2(2j-1)}\tag{10b} \end{align} $$