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I think I've got the two ideas needed to solve this, but it feels like they're not tied together properly. I'm not sure if I'm allowed to do something like this:

Let $A$ be an invertable $n\times n$ matrix, and $b$ be an n-dimensional vector.

\begin{align} Ax=b&\Longrightarrow A^{-1}Ax=A^{-1}b\\ &\Longrightarrow x=A^{-1}b \end{align}

Therefore, there exists at least one solution to the equation $Ax=b$. Additionally, for the equation $Ay=b$:

\begin{align} Ay=b&\Longrightarrow A^{-1}Ay=A^{-1}b\\ &\Longrightarrow y=A^{-1}b\\ &\Longrightarrow y=x \end{align}

Therefore, for any two unique combinations of $A$ and $b$, there is a unique $x$ for $Ax=b$.

The problem I feel exists with this is that I'm doing two separate proofs and referencing one in the other, when I feel like I can only do that if they're combined into one single proof. Am I mistaken?

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Curiously when I visited this question the only upvoted (and accepted) answer was the unique (sic) one that does not correctly address the interrogation that OP expressed. The point is that unique existence really has two different aspects, and that showing them separately is quite normal. Although one can present them in a combined fashion (see the answer by copper.hat) as an equivalence between two equations (but the equivalence still involves separate implications in two directions). – Marc van Leeuwen May 14 '15 at 11:07
up vote 2 down vote accepted

A handy way to deal with uniqueness proofs is to assume by contradiction that there exist distinct solutions.

Assume that $x_1$ and $x_2$ are distinct solutions to $Ax=b$.

Then, $Ax_1 = b$ and $Ax_2 = b$. Since $A$ is invertible, we have $x_1 = A^{-1}b$ and $x_2 = A^{-1}b$. Thus, because $A^{-1}b = A^{-1}b$, we have by transitivity $x_1 = x_2$, but we assumed they are distinct.

Therefore, the solution must be unique.

This is essentially what you're trying to do, but it is not two different proofs.

Instead, we leverage the power of transitivity and reflexivity of equality to show that distinct solutions cannot exist.

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Basically, you're just showing that if $x_1$ and $x_2$ are solutions of the system, they must be equal. Note that this actually is not a proof by contradiction, since the assumption that they are distinct is unnecessary. That of course does not take away the fact that it is a handy way of thinking about it. – Eric Spreen Jul 1 '13 at 19:01
@EricSpreen Of course, which is why I worded it like that. It is a natural thing to think "well, what if there were two solutions?" The remainder of the proof follows a bit more naturally from there, and it removes some of the uneasiness that comes from just stating equality and hoping it works. – Emily Jul 1 '13 at 19:10
This is just doing half the work. Showing unique existence requires showing uniqueness and showing existence. You did not do the latter, whereas the proof OP presents properly does both aspects. Therefore I find this as an answer to the question OP posed quite misleading. – Marc van Leeuwen May 14 '15 at 11:03
@Marcvanleeuwen The implication here was more to clean up the OP's second part, not that there didn't need to be two parts. My comments were maybe misleading. "Not two different proofs" wasn't meant to imply there weren't two steps, but rather that it wasn't necessarily the existence proof applied twice. I'll clean up the wording. – Emily May 14 '15 at 12:39

If $A$ is invertible and $b$ is given, then $Ax=b$ iff $x = A^{-1}b$.

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If $A$ is invertible, left multiplication by $A$ is an isomorphism on $\mathbb{R}^n$. An isomorphism is a bijective linear map. For the linear system $Ax = b$, surjectivity tells us that a solution exists, and by injectivity the solution is unique.

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Actually, your first calculation shows uniqueness, as starting from $Ax=b$ you infer that $x=A^{-1}b$. But by simply plugging in the value $A^{-1}b$ for $x$ you also get existence as for this choice of $x$ you get $Ax=AA^{-1}b=b$.

Note that the first step used the existence of a left inverse (i.e. you made use of the fact that $A^{-1}A$ is th eidentity), whereas the existence made use of the right inverse property (i.e. that $AA^{-1}$ is the identity).

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