Left annihilator of $2\times2$ matrices. If we let $R$ be the ring of $2\times2$ complex matrices.
When is the left annihilator just equal to $\{0 \} $?
I see that if $A$ is invertible $\text{Ann}_{R} (A)$ is trivial since if $M \in \text{Ann}_R (A)$ then $MA =0 $ so we can just multiply on the right by $A^{-1} $ and so $M=0$.
But what about when $A$ is not invertible. Essentially I’m looking for zero divisors of the ring of $2\times 2$ matrices.

If $A$ is not invertible, then rank is 1 or $A=0$. If rank is 1, then columns are a multiple of each other. Then you can classify $2 \times 2$ non invertible matrices in the following way
$$\begin{bmatrix} 
a & ka \\
b & kb\end{bmatrix} \hspace{1cm} \text{or} \hspace{1cm} \begin{bmatrix} 
ka & a \\
kb & b\end{bmatrix} $$
Thus left annihilator can be
$$\begin{bmatrix} 
b & -a \\
lb & -la\end{bmatrix} \hspace{1cm} \text{or} \hspace{1cm} \begin{bmatrix} 
lb & -la \\
b & -a\end{bmatrix} $$
