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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?

Martin Sleziak
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mozharovsky
  • 233

2 Answers2

12

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.

Olivier Oloa
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6

Let $H_n$ be the n-th order Hilbert matrix. To prove $H_n$ is positive defined, it suffices to show all the principal minor determinant of $H_n$ are positive. Say, $\det(H_m)>0$ for all $0\leq m\leq n$. This is true by the properties of Hilbert matrix. (see Hilbert Matrix).

Lion
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