I have the following question:
Assume $A \in M_{2\times2}(\mathbb{R})$ has the property that "if $A\vec{x}=A\vec{y}$, then $\vec{x}=\vec{y}$. Prove that $A$ is invertible.
My attempt:
Consider $A\vec{x}=A\vec{y}$
$A\vec{x} - A\vec{y}= \vec{0}$
$A(\vec{x}-\vec{y}) = \vec{0}$
Since $\vec{x}=\vec{y}$, this implies that $\vec{x} - \vec{y}= \vec{0}$
Thus, $\operatorname{Null}(A) = \{\vec{0} \}$
By the invertible matrix theorem, A is invertible.
What I'm worried about is if I can conclude that $\operatorname{Null}(A) = \{\vec{0} \}$.
If not, I was wondering from the hypothesis, if I could say that $A\vec{x} = \vec{b}$ has a unique solution for some $b \in \mathbb{R}^n$, and so $\operatorname{rank}(A) = n$...
Here's a slightly different approach which might be clearer. Suppose $A x = 0$. Then take $y = 0$, so $Ax = Ay$. But now $x = y = 0$, so the only solution to $Ax = 0$ is $x = 0$. In particular, Null($A$) is just the zero vector, so $A$ is invertible.