How can one characterise the number of linear combinations of m > 2 linearly independent vectors that map onto the same point in the plane? I have m > 2 vectors v in the plane, any two of which are linearly independent to each other. 
Any two of these vectors are enough to fill the plane. 
My question is this:
How can one characterise the number of linear combinations of v that map onto the same point in the plane? 
Also, is this number the same for every point in the plane?
 A: Here is a constructive way to do this.  Let $v_1, ... , v_m$ be your vectors, $m > 0$.  Let's assume they're all nonzero.  Given $c_1, ... , c_m \in \mathbb{R}$, let $$w = c_1v_1 + \cdots + c_mv_m$$ Your question is, what are all the $m$-tuples $r = (r_1, ... , r_m) \in \mathbb{R}^m$ do we have $w = r_1v_1 + \cdots + r_mv_m$?  In other words, for which $r$ do we have $$(r_1-c_1)v_1 + \cdots + (r_m - c_m)v_m = 0$$ From this last point, we can  assume without loss of generality that $c_1 = \cdots = c_m = 0$.  So to answer one of your questions, yes, it is the same for all points in the plane, and you can just take the origin to make life easy.  So this question is what are all the elements $(r_1, ... , r_m)$ for which $$r_1v_1 + \cdots + r_mv_m = 0$$ Now, the function $T: \mathbb{R}^m \rightarrow \mathbb{R}^2$ given by $(r_1, ... , r_m) \mapsto r_1v_1 + \cdots + r_mv_m$ is clearly a linear transformation, and what you want is to compute its kernel.  Now you can use the standard bases for $\mathbb{R}^m, \mathbb{R}^2$, compute the matrix of $T$ (which will be a $2$ by $m$ matrix), and then find a basis for the null space.
A: Having $m$ vectors $v_i \in \mathbb{R}^2$ a linear combination that gives a vector $b \in \mathbb{R}^2$ is 
$$
\lambda_1 v_1 + \dotsb + \lambda_m v_m = b
$$
Each linear combination is characterized by the vector $\lambda = (\lambda_1, \dotsc, \lambda_m)^t$.
This corresponds with the matrix equation
$$
A \lambda=b \quad (*)
$$
with $A = (v_1, \dotsc, v_m)$ and the question of how many linear combinations exist is equivalent to the numbers of solutions $\lambda$ of equation $(*)$.
For this there are three cases:


*

*infinite many solutions

*one solution

*no solution


We can check the case by trying to bring the augmented matrix $(A \mid b)$ into row echelon form:
$$
(A \mid b) = \\
\left(
\begin{array}{rrrr|r}
(v_1)_1 & (v_2)_1 & \dotsb & (v_m)_1 & b_1 \\
(v_1)_2 & (v_2)_2 & \dotsb & (v_m)_2 & b_2 \\
\end{array}
\right)
$$
As $m>2$ there will be always more variables than equations, so a unique solution is not possible. 
What can happen, depending on the choice of the $v_i$ and $b$, are free variables, which will result in infinite many solutions or inconsistent equations, which mean no solution.
