Suppose that the equation $A\vec{x} = \vec{b}$ has at least one solution. Prove that the solution is unique if and only if the equation $A\vec{x} = \vec{0}$ has only the trivial solution ($\vec{x} = 0$).
Following is what I believe is a correct proof for the $\Leftarrow$ direction of the statement:
$A\vec{x} = \vec{0} \Rightarrow \vec{x} = \vec{0}$ is the only solution $\Rightarrow$ no free variables.
That and the fact that $A\vec{x} = \vec{b}$ has at least one solution $\Rightarrow$ only one solution $\Rightarrow$ $\vec{x} = \vec{0}$ is unique.
However, it's the $\Rightarrow$ direction I'm having trouble understanding; i.e. if the solution is unique, the equation $A\vec{x} = \vec{b}$ has only the trivial solution $\vec{x} = \vec{0}$. How would I go about proving that way?