Phase transition in matrix

Question

Playing around with Matlab I noticed something very peculiar:

Take the symmetric matrix $A \in \mathbb R^{n \times n}$ defined by

$$A_{ij}= i \delta_{ij} - \frac{\varepsilon}{\sqrt{i}\sqrt{j}}\,.$$

Here $\delta_{ij}$ is the Kronecker delta.

We first note that this matrix is not diagonally dominant if $n$ is large enough.

This is because $\lim_{n \rightarrow \infty} \sum_{i=1}^{n} \vert A_{i,1}\vert=\infty >\vert A_{1,1} \vert.$

It is obvious that we require $\varepsilon<1$ in order for $A$ to be positive definite, since otherwise $A_{1,1}\le 0.$

However, I noticed that for let's say $\varepsilon=0.1$ one can make the dimension as large as one wants and the matrix remains positive definite.

Question: How can one show that $A$ is positive definite independent of the dimension if $\varepsilon$ is sufficiently small but fixed ?

Answer 1

The claim is true with $\epsilon=\frac6{\pi^2}\,$.

To see this, remark that by changing variable $x_i=y_i\sqrt i\,$, this is equivalent to proving that $$\epsilon\left(\left(\frac1{ij}\right)\right)_{1\le i,j}\le I_\infty.$$ The first (infinite) matrix is $V\otimes V$ with $V=(1,\frac12\,,\ldots,\frac1n\,,\ldots)$. It is symmetric, rank-one, with eigenvalues $0$ (infinitely many times) and ${\rm Tr}(V\otimes V)=\frac{\pi^2}6$ (simple).