# Hilbert matrices are positive definite

It can be seen by using its integral form:

$\displaystyle H_{ij} = \int_0^1 dx\,x^{i+j-2}$

Then we can express the positive definite condition as

$\displaystyle \sum_{i,j} x_i H_{ij} x_j > 0$

for $\mathbf{x} \ne 0$.

Replacing and operating:

$\displaystyle \sum_{i,j} x_i \int_0^1 du\,u^{i+j-2} x_j = \int_0^1 du \sum_{i,j} x_i u^{i+j-2} x_j$

$\displaystyle = \int_0^1 du \sum_{i,j} x_i u^{i-1} x_j u^{j-1}$

$\displaystyle = \int_0^1 du \sum_i x_i u^{i-1} \sum_j x_j u^{j-1}$

$\displaystyle = \int_0^1 du \left( \sum_i x_i u^{i-1} \right) \left( \sum_j x_j u^{j-1} \right)$

$\displaystyle = \int_0^1 du \left( \sum_i x_i u^{i-1} \right)^2$

The term inside the parentheses is just a (polynomial) function of $u$:

$\displaystyle = \int_0^1 du\,p(u)^2$

The only way the integral can be zero is if $p(u)$ is identically zero:

$\displaystyle \sum_i x_i u^{i-1} = 0 \implies \forall i: x_i = 0$.

QED