Question

The entries of an $n\times n$ array of numbers are denoted by $x_{ij},$ with $1\le i,j\le n.$ For all i, j, k, $i,j,k,1\le i,j,k\le n$ the following equality holds $x_{ij}+x_{jk}+x_{ki}=0.$
Prove that there exist numbers $t_1,t_2,...,t_n$ such that $x_{ij}=t_i-tj$ for all $i,j,k,1\le i,j,k\le n.$

Answer 1

Setting i= j= k in the given condition yields $3x_{ii}=0,$ hence $x_{ii}=0,$

for all $i,1\le i\le n.$ For k =j we obtain $x_{ij}+x_{jj}+x_{ji}=0,$

hence $x_{ij}=-x_{ji},$ for all i, j. Now fix i and j and add up the equalities $x_{ij}+x_{jk}+x_{ki}=0,$ for k = 1, 2, ... , n.

It follows that $n{x}_{ij}+\sum _{k=1}^n{x}_{jk}+\sum _{k=1}^n{x}_{jk}=0$ or $n{x}_{ij}+\sum _{k=1}^n{x}_{jk}-\sum _{k=1}^n{x}_{ik}=0.$

If we define $t_i=\frac{1}{n}\sum _{k=1}^n{x}_{jk}$

we obtain $x_{ij}=t_i-t_j,$ for all $i,j$ as desired.