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?
Best Answer
Suppose that $\vec v$ and $\vec w$ are distinct solutions for the system $A\vec x = \vec b$ so that $A \vec v = A \vec w = \vec b$. Then $\frac{1}{2}(\vec v + \vec w)$ must be distinct from both $\vec v$ and $\vec w$ and must also solve the system since: $$ A(\tfrac{1}{2}(\vec v + \vec w)) = \tfrac{1}{2}(A\vec v + A\vec w) = \tfrac{1}{2}(\vec b + \vec b) = \vec b $$ We can then apply the same argument to $\vec v$ and $\frac{1}{2}(\vec v + \vec w)$ in order to get infinitely many distinct solutions.