TITLE: Proof that any linear system cannot have exactly 2 solutions.
QUESTION [17 upvotes]: How would you go about proving that for any system of linear equations (whether all are homogenous or not) can only have either (if this is true):

One solution
Infinitely many solutions
No solutions

I found this a bit difficult to prove (even though its a very fundamental thing about any linear equation). The intuitive geometric explanation is that a line can only intersect at one point, and if it intersects at a later point, it can't be a linear equation, but I don't think this is a convincing proof. 
I thought of if you assume that there are two (or more, but I picked two) solutions for some linear system, then for the points in between 
Solution Set 1: X1, X2....., Xn

Solution Set 2: X1, X2....., Xn
Then (I think), the points between S1 and S2, must be infinitely many points (and thus infinitely many solutions) such that these points can also satisfy the linear system, which would mean the system has 2 infinite solutions. 
However, I don't think this is rigorous enough and nor do I understand completely why its true. Can anyone help in explaining (correcting) and elaborating on the intuition and proof of this?

REPLY [8 votes]: In general, in a vector space over a general field $F$, the number of solutions $\vec{x}$ to the system
$$
A \vec{x} = \vec{w}
$$
is either $0$ or $|\ker A|$. (Because if $\vec{x}_0$ satisfies the equation, then the set of solutions is $\vec{x}_0 + \ker A$.)
And $|\ker A| = |F|^k$ for some $k \ge 0$.
Assuming the base field $F$ has infinite cardinality $\alpha$ and the vector space is finite-dimensional, it follows that the number of solutions is $0, 1,$ or $\alpha$ (since $|F|^0 = 1$ and $|F|^k = \alpha$ for $k \ge 1$).