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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 $c_i$.

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Where is the "here" you asked us to refer for details? Broken hyperlink? –  Alex Becker Mar 28 '11 at 1:44
I did respond to your comment to my answer in the question you linked. Did you find a mistake in that? –  Aryabhata Mar 28 '11 at 14:08
Hello Moron. Sorry I didn't notice your reply because it was hidden by default. Do you mean there is a mistake in the example you given? I think $c_1=1,c_2=2,c_3=3,c_4=4$ can give three solutions. In addition to the two you mentioned, the other one is $1+4 and 2+3$. The uniqueness does bother me. I also noticed that if $v_i$ satisfy $\sum_{i=1}^nc_iv_i=0$, then any transformation on $v_i$ will preserve the equation, that is $\sum_{i=1}^nc_iAv_i=0$ where $A$ is an arbitrary matrix. Hence when we consider the uniqueness, this kind of solutions should not be accounted as different ones. –  Shiyu Mar 28 '11 at 15:09
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I think the answer to this question is no. Just take $R^2$ for example. We consider the complex number $\Bbb C$ for convenience. let $v_1=a+bi,v_2=a-bi;v_3=-a+bi,v_4=-a-bi$. We also ask these $v_i$ to be with absolute value 1. Take all $c_i$ as $1$.

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Hi Xiaochuan: In this example, you assume all $c_i$ are equal. Thanks. At least in this case, there are infinite number of solutions. It reminds me that the solution to $\sum_{i=1}^n c_i v_i=0$ is also not unique if: the vectors $v_i$ can be divided into some subgroups. In each subgroup, $\sum_{i=1}^{n_k}c_i v_i =0$ hold. Hence the whole group will have $\sum_{i=1}^n c_i v_i =0$. –  Shiyu Mar 28 '11 at 3:42
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