The Hilbert matrix is
symmetric and
positive definite. The Hilbert matrix is also
totally positive (meaning that the determinant of every
submatrix is positive). The Hilbert matrix is an example of a
Hankel matrix. It is also a specific example of a
Cauchy matrix. The determinant can be expressed in
closed form, as a special case of the
Cauchy determinant. The determinant of the
n ×
n Hilbert matrix is : \det(H) = \frac{c_n^4}{c_{2n}}, where : c_n = \prod_{i=1}^{n-1} i^{n-i} = \prod_{i=1}^{n-1} i!. Hilbert already mentioned the curious fact that the determinant of the Hilbert matrix is the reciprocal of an integer (see sequence in the
OEIS), which also follows from the identity : \frac{1}{\det(H)} = \frac{c_{2n}}{c_n^4} = n! \cdot \prod_{i=1}^{2n-1} \binom{i}{[i/2]}. Using
Stirling's approximation of the
factorial, one can establish the following asymptotic result: : \det(H) \sim a_n\, n^{-1/4}(2\pi)^n \,4^{-n^2}, where
an converges to the constant e^{1/4}\, 2^{1/12}\, A^{-3} \approx 0.6450 as n \to \infty, where
A is the
Glaisher–Kinkelin constant. The
inverse of the Hilbert matrix can be expressed in closed form using
binomial coefficients; its entries are : (H^{-1})_{ij} = (-1)^{i+j}(i + j - 1) \binom{n + i - 1}{n - j} \binom{n + j - 1}{n - i} \binom{i + j - 2}{i - 1}^2, where
n is the order of the matrix. It follows that the entries of the inverse matrix are all integers, and that the signs form a checkerboard pattern, being positive on the
principal diagonal. For example, : \begin{bmatrix} 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & \frac{1}{7} \\ \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & \frac{1}{7} & \frac{1}{8} \\ \frac{1}{5} & \frac{1}{6} & \frac{1}{7} & \frac{1}{8} & \frac{1}{9} \end{bmatrix}^{-1} = \left[\begin{array}{rrrrr} 25 & -300 & 1050 & -1400 & 630 \\ -300 & 4800 & -18900 & 26880 & -12600 \\ 1050 & -18900 & 79380 & -117600 & 56700 \\ -1400 & 26880 & -117600 & 179200 & -88200 \\ 630 & -12600 & 56700 & -88200 & 44100 \end{array}\right]. The condition number of the
n ×
n Hilbert matrix grows as O\left(\left(1 + \sqrt{2}\right)^{4n}/\sqrt{n}\right). ==Applications==