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 ?
> 
> 
> 



Phase transition in matrix
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).


