Let $\mathbb{K}$ be a Field and $A\in M_2(\mathbb{K})$, $A \not = 0.$ Show that $A$ is a unit iff $A$ is a left non-zero divisor in $M_2(\mathbb{K}).$ If $\mathbb{K}$ is not a field, does this hold ?
My Attempt: The forward direction is straightforward. Suppose that $A$ is a left non-zero divisor in $M_2(\mathbb{K}).$ Then $\forall X\in M_2(\mathbb{K})$, $AX=0\implies X=0.$ Consider the map $f:M_2(\mathbb{K})\to M_2(\mathbb{K})$ such that $f(X)=AX.$ Observe that this map is injective since $f(X_1)=f(X_2)\implies A(X_1-X_2)=0\implies X_1=X_2.$ Since the dimensions of the vector spaces is the same in the domain and the co-domain we have that $f$ is also surjective. Hence $\exists B$ such that $f(B)=I$ or $AB=I\implies BA=I$ and so $A$ is a unit. Is this proof correct?
Your proof looks good. Perhaps, for completeness, you should state that $f$ is a linear transformation (so that your reasoning about dimensions applies to it).
For $\Bbb{K}$ not a field the result does not hold in general. E.g., if $\Bbb{K} = \Bbb{Z}$, $\left(\matrix{2\;0 \\0\;2}\right)$ is neither a zero divisor nor a unit