0
$\begingroup$

In an induction proof of a lemma I would like to prove the following statement.

Let $U$ be a non-empty finite set and let $X_{i,j}$ for all $i\not=j \in U$ be real numbers. Assume for each $j\in U$ we have real numbers $\beta^j_i>0$ for each $j\not= i \in U$ such that $\sum_{i\in U\setminus\{j\}}\beta^j_i X_{i,j}>0$. Can we construct real numbers $\beta_i>0$ such that $ \sum_{j\in U} \sum_{i\in U\setminus \{j\}}\beta_i X_{i,j}>0$.

Obviously $\sum_{j\in U} \sum_{i\in U\setminus \{j\}}\beta_i^j X_{i,j}>0$ and more generally $\sum_{j\in U} a_j\sum_{i\in U\setminus \{j\}}\beta_i^j X_{i,j}>0$ for non-negative number $a_j$, but I need $\beta_i^j$ to be equal for all $j$.

Inspired by above remark I have tried finding positive weights $a_j$ and let $\beta_i = \sum_{j\in U\setminus \{i\}} a_j \beta_i^j$ to no avail.

Defining the matrix $X = \{X_{i,j}\}$ we can possibly rephrase it as some statement about matrices.

This is possibly already known but it is hard to find good terms to search for. If anyone have any feedback I would be very pleased.

$\endgroup$
2
  • $\begingroup$ So, for $U = \left\{1,2,3\right\}$, you are given that $\beta_2^1 X_{2,1} + \beta_3^1 X_{3, 1} > 0$ and cyclically, and you want to find $\beta_1, \beta_2, \beta_3$ (all $\beta$s are positive) such that $\beta_1\left(X_{1,2} + X_{1,3}\right) + \beta_2\left(X_{2,1} + X_{2,3}\right) + \beta_3\left(X_{3,1} + X_{3,2}\right) > 0$, right? $\endgroup$ Commented Sep 20, 2021 at 20:31
  • $\begingroup$ Pretty sure it's impossible. Just take $X_{2,1} = X_{3,2} = X_{1,3} = -2$ and $X_{1,2} = X_{2,3} = X_{3,1} = 1$. The $\beta_j^i$s are easily found, but the $\beta_j$s cannot exist. $\endgroup$ Commented Sep 20, 2021 at 20:32

0

You must log in to answer this question.

Browse other questions tagged .