Positive-definiteness of matrix with $0 \leq a_{ij} \leq \frac{1}{ij}$

Question

Let $A$ be a $N\times N$-matrix with elements $$ a_{ii}=1 \quad\text{and}\quad a_{ij} = \frac{1}{ij} \quad\text{for}~ i\neq j. $$ Then $A$ is positive-definite, as can be easily seen from $$ x^T A x = \sum_i x_i^2 + \sum_{i \neq j} \frac{x_i x_j}{ij} \geq \sum_i \frac{x_i^2}{i^2} + \sum_{i \neq j} \frac{x_i x_j}{ij} = \left(\sum_i \frac{x_i}{i}\right)^2 \geq 0. $$

Assume now that $A$ is a real symmetric $N\times N$-matrix with elements $$ \tag{1} a_{ii}=1 \quad\text{and}\quad 0 \leq a_{ij} \leq \frac{1}{ij} \quad\text{for}~ i\neq j. $$

Is it possible to show that $A$ is also positive-definite (or positive-semidefinite)?

It is quite easy to obtain the result if $N$ is not large (e.g., $N \leq 4$) by estimating the corresponding leading principal minors. However, the case of arbitrary $N$ is unclear.

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This question is a refinement of my previous question, where the assumption $(1)$ was weakened by assuming $|a_{ij}| \leq \frac{1}{ij}$ for $i \neq j$. Under this weaker assumption, a counterexample was presented in a comment.

In another related question, the assumption $(1)$ was weakened by assuming $0 \leq |a_{ij}| \leq 1$ for $i \neq j$, and a counterexample was also given. However, a similar construction does not seem to work in the case of the assumption $(1)$.

Answer 1

Let $E_n=\{A\in S_n;a_{i,i}=1,0\leq a_{i,j}\leq\dfrac{1}{ij} \textbf{ for }i\not= j\}$.

I think that, for every $n$, these matrices are $>0$. Consider the functions

$f_n:A\in E_n\mapsto min(spectrum(A))$, $g_n:A\in E_n\mapsto \det(A)$.

It suffices to show that one of these functions is $>0$ over $E_n$. In particular, it suffices to prove that,

$(*)$ for every $n$, $A\in E_n$ implies that $\det(A)>2-\dfrac{\pi^2}{6}$. The following is easy to prove

$\textbf{Proposition 1}$. Let $B_n\in E_n$ be defined by $b_{i,j}=0$, except $b_{i,i}=1$ and $b_{1,j}=b_{j,1}=\dfrac{1}{j}$. Then $\det(B_n)=1-\dfrac{1}{2^2}-\cdots-\dfrac{1}{n^2}$.

EDIT.

$\textbf{Proposition 2}$. $g_n$ admits a local minimum in $B_n$.

$\textbf{Proof}$. Let $Z_n=\{H\in S_n;h_{i,i}=0,h_{1,j}\leq 0 \text{ for }j>1,h(i,j)\geq 0\text{ for }1<i<j\}$ and $adj(B_n)$ be the adjugate of $B_n$. It suffices to show that, for every $H\in Z_n$, $tr(Hadj(B_n))\geq 0$ and $tr(Hadj(B_n))=0$ implies that $H=0$.

Note that all entries of $U=adj(B_n)$ are positive except the $\{U_{1,j}=U_{j,1};j>1\}$ that are $<0$.

Then, to deduce the required result, it suffices to read the development of $tr(Hadj(B_n))$. For example, when $n=9$,

$\square$

According to some experiments, that follows seems to be true

$\textbf{Conjecture}$. Let $A\in E_n$; then $\det(A)\geq \det(B_n)$.

Clearly, this conjecture implies $(*)$.