When do $n$ vectors add up to give a resultant of zero? 
When do $n$ vectors add up to give a resultant of zero?

My take on this questions was this :
So I took up few pencils for the $n=4$,also $n=3$ it has to lie on a plane, whereas two they should be opposite, so coming back, with the $n=4$ I observed that if they formed a closed loop of segments, they would be zero. But the polygon law of addition, had stated that would be the the case only when they are coplanar, but the pencils I arranged up were not coplanar.
P.S. I am honestly sorry if this is not phrased properly, this is my first question, Kindly excuse me.
 A: I'll assume that you are confortable with a matrix notation. Let $x_1,\cdots, x_n \in \mathbb{R}^k$ be your $k$-dimensional vectors. We want to sum then and get zero, then we want to know numbers $\alpha_1, \cdots, \alpha_n$ such that
$$ \alpha_1 x_1 + \cdots + \alpha_nx_n = 0. $$
Writing $M = \left[\begin{array}{c:c:c:c}x_1 & x_2 & \cdots & x_n \end{array}\right]$ a matrix with $x_1, \cdots, x_n$ as columns we can rewrite your equation as the system
$$ M\alpha = 0 $$
where $\alpha = \begin{bmatrix}\alpha_1\\ \alpha_2 \\ \vdots \\ \alpha_n\end{bmatrix}$. Now we need to know how many solutions we have for $M\alpha = 0$. You can find the path to answer that at any Linear Algebra book or course. But here is the answer:
If $k < n$ then you have infinitely many ways to do that.
If $k = n$ and $M$ has non-zero determinant then the only way is to set $\alpha_1=\cdots=\alpha_n = 0$.
If $k=n$ and $M$ has zero determinant, then you have infinitely many ways to do that.
If $k > n$ and you can find other $n-k$ vectors $x_{k+1}, \cdots, x_{n}$ such that $M$ has non-zero determinant then the only way is to set $\alpha_1=\cdots=\alpha_n = 0$.
If $k > n$ and you can't find other $n-k$ vectors $x_{k+1}, \cdots, x_{n}$ such that $M$ has non-zero determinant then you have infinitely many ways to do that.
