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Is it possible to satisfy the condition $\sum_{i=1}^N \sum_{j=1}^N\frac{q_i q_j}{a_{ij}}>0$, given that $q_i$ and $q_j$ are non-zero integers, $a_{ij}$ are positive real numbers and $\sum_{i=1}^N q_i =0$? As a special case, allow $N \rightarrow \infty$.

Another formulation of the problem: For a set of integers $A=\left(q_1,q_2,...q_N\right)$, such that $\sum_{i=1}^N q_i =0$, is it possible to find positive weights $a_{ij}$ such that $\sum_{i=1}^N \sum_{j=1}^N\frac{q_i q_j}{a_{ij}}>0$?

Apologies for the edit, I didn't formulate the problem very well earlier. It is probable I am missing a simple proof here, but I can't seem to find it. Thanks!

Nikola Poljak
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1 Answers1

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Let $M$ denote the matrix whose entries are $M_{ij} = \frac{1}{a_{ij}}$. Let $q$ denote the column vector $(q_1,\dots,q_N)$. Then we have $$ \sum_{i=1}^N \sum_{j=1}^N\frac{q_i q_j}{a_{ij}} = q^TMq $$ with that in mind: if $M$ is positive definite, then $q^TMq$ will be positive whenever $q \neq (0,\dots,0)$.

For example, if we take $$ a_{ij} = \begin{cases} 1 & i=j\\ 2 & \text{otherwise} \end{cases} $$ then we will have $\sum_{i=1}^N \sum_{j=1}^N\frac{q_i q_j}{a_{ij}} > 0$ whenever $q$ are not all equal to $0$.

A criterion that should help in the infinite dimensional case: a sufficient condition to make this work is to have $a_{ij} = a_{ji}$ for all $i,j$, and $$ \frac{2}{a_{ii}} > \sum_{j= 1}^N \frac{1}{a_{ij}} \qquad \forall i=1,\dots,N $$ which is to say that $$ a_{ii} < 2\left(\sum_{j= 1}^N \frac{1}{a_{ij}}\right)^{-1}\qquad \forall i $$ Or if you prefer $$ a_{ii} < \left(\sum_{j\neq i} \frac{1}{a_{ij}}\right)^{-1} \qquad \forall i $$

This condition will continue to be sufficient as $N \to \infty$.

Ben Grossmann
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