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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?

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    $\begingroup$ Assume it has two solutions so $Ax_1 = b$ and $Ax_2 = b$ and consider $A(x_1-x_2)$. $\endgroup$
    – Winther
    May 28, 2018 at 18:03
  • $\begingroup$ I'm confused. Remember $A \vec{x} = \vec{b}$ will only have a solution of $\vec{x} = \vec{0}$ when $\vec{b}$ is $\vec{0}$! I'm not 100% sure which direction you've attempted to prove, and which one is giving you trouble. $\endgroup$
    – Theo Bendit
    May 28, 2018 at 18:05
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    $\begingroup$ @imranfat Don't think so, since it's not specified that A is n x n. $\endgroup$
    – eirik-ff
    May 28, 2018 at 18:08
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    $\begingroup$ @eirik-ff Good point. In that case, the first comment is worth looking into. (Proof by contradiction) $\endgroup$
    – imranfat
    May 28, 2018 at 18:09
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    $\begingroup$ You should see the answer below. My first comment was for the direction: $Ax = 0\iff x= 0$ then $Ax=b$ is unique (sorry, I misread the question first time). Because if it wasn't and $Ax_1 = b = Ax_2$ for $x_1\not= x_2$ then $Az = 0$ for $z = x_1-x_2 \not = 0$ giving a contradiction to $Ax = 0$ only having a trivial solution. $\endgroup$
    – Winther
    May 28, 2018 at 18:28

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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.

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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, $A(\vec{x} + \vec{x_0}) = b$.

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