Problem
Hilbert Matrices
A deceptively simple matrix: can you prove it is always nonsingular?
Prove that the matrix given by
\[\begin{bmatrix} 1 & \frac 12 & \dots & \frac 1n\\ \frac 12 & \frac 13 & \dots & \frac 1{n+1}\\ \vdots & \vdots & \ddots & \vdots\\ \frac 1n & \frac 1{n+1} & \dots & \frac 1{2n-1}\\ \end{bmatrix}\]is non-singular.
Explore the solution
Observe that if we denote the matrix above with \(H = [H_{ij}]\), then we have
\[H_{ij} = \int_{0}^{1} t^{i+j-2}\,dt.\]Now let \(\vec{x} = (x_1, x_2, \dots, x_n)^\intercal \neq \vec 0\) be an arbitrary column vector. Some straightforward calculations give:
\[\begin{align} \vec{x}^\intercal H \vec{x} &= \sum_{i, j=1}^{n}x_i x_j\int_{0}^{1} t^{i+j-2}\,dt\\ &= \int_{0}^{1}\left(\sum_{i=1}^{n} x_it^{i-1}\right)^2\,dt > 0. \end{align}\]Therefore, the only solution to the equation \(H\vec x = \vec 0\) is the trivial solution and the conclusion follows.