Subtle conditions that imply linear independence

Question

Suppose you have a set of $2 < k < n$ non-zero, distinct vectors in $\mathbb{R}^n$ with non-negative integer coefficients such that the sum of the entries of each vector is equal to the same positive integer $c$. I want to ensure the vectors in this set are linearly independent. My hope is that since I already have two assumptions on these vectors, I do not have to add full blown linear independence as a further assumption, but there is a weaker one (no matter if convoluted or complicated) that would ensure they are. Any ideas?

Answer 1

No, $2<k<n$ is not enough. Let $n=4$, $k=3$, and let $c=6$. Consider the following example: $$ u=\begin{pmatrix} 1 \cr 1 \cr 1 \cr 3\end{pmatrix},\; v=\begin{pmatrix} 1\cr 1 \cr 3 \cr 1\end{pmatrix},\; w=\begin{pmatrix} 1\cr 1 \cr 2 \cr 2\end{pmatrix},\; $$ They are not linearly independent, since $u+v=2w$. One can extend this example by adding further entries $1$ in $u,v,w$. So three vectors are already too many. So we need $k=2$ or $k=1$.