TITLE: Prove the positive definiteness of Hilbert matrix
QUESTION [9 upvotes]: This is so called Hilbert matrix which is known as a poorly conditioned matrix. 
$$
A =        \left(\begin{matrix}
        1 & \frac{1}{2} & \frac{1}{3} & ... & \frac{1}{n} \\
        \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & ... & \frac{1}{n + 1} \\
        \vdots & \vdots & \vdots & \ddots & \vdots \\
        \frac{1}{n} & \frac{1}{n + 1} & \frac{1}{n + 2} & ... & \frac{1}{2n - 1} \\
        \end{matrix}\right)
$$
The task is to prove that matrix A is positively definite. 
A possible way to go is to look at the scalar product: 
$$(f, g) = \int_0^1f(x)g(x)dx$$ 
within the space of polynomials of degree not higher than $(n - 1)$. There our form $(\cdot, \cdot)$ is bilinear, symmetric and positively defined thus our polynomials' space with that form is, in fact, an euclidean space of degree $n$. 
Now let's take the following basis: ${1, x, x^2, ..., x^\left(n - 1\right)}$. It's easy to see that matrix A is a Gram matrix for this basis. 
But is there another approach?

REPLY [11 votes]: Let $X=(x_i)_{1\leq i\leq n} \in \cal{M}_{n,1}(\mathbb{R}).$  We have
$$
{}^tXAX=\sum_{1\leq i,j\leq n}\frac{x_ix_j}{i+j-1}=\sum_{1\leq i,j\leq n}x_ix_j\int_0^1t^{i+j-2}dt=\int_0^1\left(\sum_{i=1}^nx_it^{i-1}\right)^2dt>0
$$ for $X\neq0$, giving the announced result since $A$ is symmetric.