Let $c_i>0$ be a scalar, $v_i\in R^d$ be a unit-length vector, $i=1,...,n$. In a previous question, it is proved that
Fact: There exist $v_i$ such that $\sum_{i=1}^n c_i v_i=0$ if and only if $c_j \le \frac{1}{2}\sum_{i=1}^n c_i, \forall j$.
We can find a solution when $c_j \le \frac{1}{2}\sum_{i=1}^n c_i, \forall j$. The method is to reorder $c_i$ and form a triangle (please refer to here for details).
My question is: whether the solution to $\sum_{i=1}^n c_i v_i=0$ is unique? If not, can you show any other approaches to construct a different solution?
Many thanks.
Thanks to Xiaochuan, I think I have the answer to this question. In many cases the solution to $\sum_{i=1}^n c_i v_i=0$ is not unique. One general case is: if the vectors $v_i$ can be divided into some subgroups, and in each subgroup $\sum_{i\in g_k} c_i v_i=0$ holds, where $i\in g_k$ means the $i$th vector belongs to the $k$th subgroup. Then for the whole group of $v_i$, we have $\sum_{i=1}^n c_i v_i=\sum_{k=1}^{subgroupnum}\sum_{i\in g_k} c_i v_i= 0$. Xiaochuan's example is a special case. If $c_i$ is equal to each other, any two $v_i$ can form a subgroup and make $\sum_{i=1}^2 v_i=0$ hold.
However, another interesting question arises, what constraints on $c_i$ can make the solution unique? Obviously, the solution as well as its uniqueness is determined by $c_i$.