question about linear dependent vectors

Question

I have a short question about vectors: ( here $v_i, i=1, ..., n$ are vectors)

If we know that $v_1 + v_2 + ... + v_n = 0$ then for what combination of positive scalars $a_1, ..., a_n$ we have $a_1v_1$+ ... + $a_nv_n=0$ ?

For example it is obvious that if all the scalars $a_i$ are equal the condition is met. I want the condition (if and only if) that generate all such scalars $a_i$.

( initially my question is motivated by the fact that if we regard the vectors from the center O of a regular polygon in the plane pointed to its vertices , then the sum of these vectors is zero, and then if we continue these vectors and choose points on these lines to form another polygon which is irregular in general , what is the condition that still the sum of vectors from the same point O to these new vertices be zero?) thank you .

Answer 1

It depends strongly on the vectors:

• If $v_1,c\dots, v_{n-1}$ are linearly independent and $v_1+\cdots+v_n=0$ then $v_1,\cdots,v_n$ are in an $n-1$ dimensional space. Since $v_1,c\dots, v_{n-1}$ are linearly independent they form a basis. So $v_n$ can be written in terms of $v_1,c\dots, v_{n-1}$ only in a way. So $$a_1v_1+\cdots+a_nv_n=0\iff a_1=\cdots=a_n.$
• If $v_1,c\dots, v_{n-1}$ are linearly dependent and $v_1+\cdots+v_n=0$ then there may have a lot of possibilities. Consider the extreme case $v_1=\cdots =v_{n-1}=-\frac{1}{n-1}v_n.$ Then there are the possibilities $a_1=n-1,a_2=\cdots=a_{n-1}=0,a_n=1.$ Obviously the role of $a_1$ can be played by any other $a_i,$ $i=2,\dots, n-1.$ Even worse, if one of the $v_i$ is zero then $a_i$ can be any number.

To clarify the question, consider the matrix $A$ whose columns are $v_1,\dots, v_n.$ Then, $x=(a_1,\cdots, a_n)^T$ are the solutions of the homogeneous linear system $$Ax=0.$$ If we only know that $(1,\cdots, 1)^T$ is a solution we can't say much about the set of solutions.