I was watching a lecture on YouTube on Pseudo-Inverses, and the professor brought up the claim that is the title of this post. I didn't watch the rest of the lecture in hopes of coming up with a proof of this.
I don't know if it's sound, however. Here goes:
Assume that there exists vectors $\vec{x}, \vec{y}$ in the row space of a matrix $A$ such that $A\vec{x} = A\vec{y}.$ Then, $\vec{x} - \vec{y}$ would be in the nullspace of $A$, since $A\vec{x} = A\vec{y} \implies A(\vec{x} - \vec{y}) = \vec{0}.$ Since $\operatorname{Row} A$ is a subspace, then $\vec{x} - \vec{y}$ is also in the row space of $A$. This means that $\vec{x} - \vec{y} \in \operatorname{Row} A \cap \operatorname{Nul} A$. However, $\operatorname{Row} A \cap \operatorname{Nul} A = \{\vec{0}\}$, since those two subspaces are orthogonal complements. (The only vector that lives in a subspace and it's orthogonal complement is the zero vector.) Hence, $\vec{x} - \vec{y} = \vec{0} \implies \vec{x} = \vec{y}.$
We've shown that if $\vec{x}, \vec{y}$ are in the rowspace with matching outputs when left multiplied with $A$, then they're equal. Therefore, the proof by contrapositive is completed.
Is this an ok argument?