Solution to $A \vec{x} = \vec{b}$ is unique if and only if the equation $A \vec{x} = \vec{0}$ has the trivial solution $\vec{x} = \vec{0}$ 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? 
 A: Let it be that the equation $Ax=b$ has a solution and let it be that this solution is unique. Further let us denote this solution by $x_0$.
If $y$ is a solution of $Ax=0$ then $A(x_0-y)=Ax_0-Ay=b-0=b$ and we are allowed to conclude that $x_0-y=x_0$ or equivalently $y=0$ on base of the uniqueness. So $Ax=0$ only has the trivial solution $x=0$.

Let it be that the equation $Ax=b$ has a solution that we denote by $x_0$ and let it be that the equation $Ax=0$ only has the trivial solution $x=0$. 
Now if $y$ is a solution for $Ax=b$ then $A(x_0-y)=Ax_0-Ay=b-b=0$ and we are allowed to conclude that $x_0-y=0$ or equivalently $y=x_0$. This is because $Ax=0$ only has the trivial solution (from our assumption). So apparantly the solution of $Ax=b$ is unique.
A: For the sake of contradiction, assume the negation: the solution to $A\vec{x} = \vec{b}$ is unique and $A\vec{x} = \vec{0}$ has a non-trivial solution, we'll call it $\vec{x_0}$. Then, consider the equation $A\vec{x} + A\vec{x_0} = b$. This must be equivalent to $A\vec{x}$ since $A\vec{x_0} = 0$. But this means that $A(\vec{x} + \vec{x_0}) = b$. Thus, we have a contradiction, since we assumed $A\vec{x} = b$  is unique, but we found another solution, $A(\vec{x} + \vec{x_0}) = b$.
