I'm following a proof I found online but I get lost halfway through
Suppose $Ax = b$ has more than one solution.
Let $x_1$ and $x_2$ be two possible solutions where $x_1 \neq x_2$
$Ax_1 = Ax_2$
$A(x_1-x_2) = 0$
$x_1 + k(x_1 - x_2)$ is a solution for $k \in R$
Why is this a solution to the linear system? Why are they adding $k(x_1 - x_2)$ to $x_1$?
$Ax = A[x_1 + k(x_1 - x_2)]$
$= Ax_1 + kAx_1 - kAx_2$
$= Ab + kb - kb$
$= b$
On
If the homogeneous system $Ax=0$ has a solution, $k x$ is also a solution, hence there is an infinity of them.
If $Ax=b$ has a solution, let $x_1$, then $A(x-x_1)=0$, and if $x_2$ is also a solution, $k(x_2-x_1)$ are solutions of the homogeneous system.
On
The other answers verify the solution, but don't show how to find the solution. This should be more clear.
Suppose $Ax=b$ has more than one solution.
Let $x_1$ and $x_2$ be two possible solutions where $x1 \neq x2$.
$Ax_1=Ax_2$
$A(x_1−x_2)=0$
$k*A(x_1−x_2)=0$
$Ax_1 + A(k*(x_1−x_2))=b + 0$
$A(x_1 + k*(x_1−x_2))=b$
Therefore $x_1 + k*(x_1−x_2)$ is a solution for each $k$.
Since $x_1 \neq x_2$, $x_1 - x_2 \neq 0$. Therefore $x_1 + k*(x_1 - x_2)$ is different for each $k$.
Therefore there are infinitely many solutions.
Suppose that $Ax = b$. So, we have $x_1$ and $x_2$ as our proposed solutions, with $x_1 \neq x_2$. Then:
$Ax_1 = b$
$Ax_2 = b$
$Ax_1 - Ax_2 = b-b = 0$
$A(x_1-x_2) = 0$
Now, I claim that $x_1 + k(x_1-x_2)$ satisfies $Ax = b$ for any $k \in \mathbb{R}$. This would clearly show that there are infinitely many solutions, since I can pick any value of $k$ that I want and I'd still have a solution.
All that's left is to actually prove that it satisfies the given equation. We have:
$A[x_1+k(x_1-x_2)] = Ax_1 + A(k(x_1-x_2)) = b +k[A(x_1-x_2)] = b +k0 = b$
This proves the desired result. I hope this helps you understand the proof better. The idea is that if you take a linear combination of $x_1$ and $x_2$, then you can show that it does lead to infinitely many solutions.