<s>[INST] Prove that if the columns of the $m\times n$ matrix $A$ are linearly independent, then $Ax=b$ has at most one solution. 
Prove that if the columns of the $m\times n$ matrix $A$ are linearly independent, then $Ax=b$ has at most one solution.

I was thinking that if the $n$ columns are linearly independent, then the dimension of the column space is $n$, which implies the dimension of the row space is $n$ also. Hence $A$ is $n\times n$, but wouldn't this imply there's exactly one solution of $Ax=b$ rather than at most one?
Is this logic correct, or am I missing something?
 [/INST] Let $A = \begin{bmatrix}\mathbf{a}_1 & \mathbf{a}_2 & \cdots & \mathbf{a}_n\end{bmatrix}$ and $\mathbf{x} = \begin{bmatrix}x_1\\x_2\\\vdots\\x_n\end{bmatrix}$ and $\mathbf{b} = \begin{bmatrix}b_1\\b_2\\\vdots\\b_m\end{bmatrix}$.
Now, $A\mathbf{x} = x_1 \mathbf{a}_1 + x_2 \mathbf{a}_2 + \cdots + x_n\mathbf{a}_n$.
Let $\mathbf{u}$ and $\mathbf{v}$ be two solutions of $A \mathbf{x} = \mathbf{b}$. Then, we have:
$$\begin{cases}
u_1 \mathbf{a}_1 + u_2 \mathbf{a}_2 + \cdots + u_n\mathbf{a}_n &=& \mathbf{b} & (1)\\
v_1 \mathbf{a}_1 + v_2 \mathbf{a}_2 + \cdots + v_n\mathbf{a}_n &=& \mathbf{b} & (2)
\end{cases}$$
Subtract $(2)$ from $(1)$ to get:
$$(u_1-v_1) \mathbf{a}_1 + (u_2-v_2) \mathbf{a}_2 + \cdots + (u_n-v_n) \mathbf{a}_n = 0$$
Using the fact that $\mathbf{a}_1$ through $\mathbf{a}_n$ are linearly independent, we get $u_1-v_1 = u_2-v_2 = \cdots = u_n-v_n = 0$, i.e. $u_i=v_i$, i.e. $\mathbf{u} = \mathbf{v}$.
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