# If $Ax=b$, for $b\ne 0$, has more than one solution, then $Ax=0$ does as well. T or F. Prove this.

I get that this is true, because there's one free variable, so no matter what the augmented matrix is, there always will be an infinite amount of solutions. Right? But how to I explain this as a proof?

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If $x_1$ and $x_2$ are both solutions of $Ax=b$ then what can you say about $A(x_2-x_1)$?
Which means $x_2-x_1$ is a solution and so is any multiple of it. So $k(x_2-x_1)$ is a solution for any $k$. – Maesumi Feb 1 '13 at 20:30
So final answer is Yes. If there are two distinct solutions $x_1,x_2$, then the solution set of $Ax=0$, the so-called null space, contains all multiples of $x_2-x_1$, so it has infinitely many solutions. – Maesumi Feb 1 '13 at 22:45
$$\rm\begin{eqnarray} ax = &\rm b&\rm = ax',\ \ x\ne x'\\ \rm \Rightarrow\ a(x\!-\!x') = &0&\rm = a\,0,\ \ \,x -x'\ne 0\end{eqnarray}$$