Does the sum $\sum_{j=1}^n a_{ij}(x_i-x_j)=0$, for all $i=1,...,n$, where $a_{ji}=a_{ij}\neq 0$ implies that the $x_i=x_j$ for all $i,j$?

Question

Assume we have $$ \sum_{j=1}^n a_{ij}(x_i-x_j)=0, \qquad \forall i=1,...,n, $$ where $a_{ij}$ are complex coordinates satisfying $a_{ji}=a_{ij}$ and are not zeros. Can we conclude that $x_i=x_j$ for all $i,j$ ?

In fact, for $n=2$, one can observe that the result follows immediately for $n=2$. For n=3, we can solve the linear system and obtain $x_1=x_2=x_3$.

How can we prove it for any $n\in \mathbb{N}$ ?

Answer 1

Let ${A= \begin{pmatrix} 1 & -1 & 2 & -2 \\ -1 & 1& -2 & 2\\ 2 & -2& 5 & -5 \\ -2 & 2& -5 & 5 \end{pmatrix}}$

It is clear that $A$ is symmetric and no elements are zero while the rows (and columns) sum to $0$. So if we find an eigenvector $(a,b,c,d)$ for the eigenvalue $0$ (determinant is zero by design as some rows are proportional) with not all elements equal we are done. But by writing the equation for that we get $a=b, c=d$ so we can take $a=b=1, c=d=2$ and get a counterexample to the claim.