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.

Every non-invertible matrix has a matrix which annihilates it. In general it can be constructed with the pseudo-inverse. In the $2\times2$ case the non-invertible:
\begin{bmatrix}
    a & b  \\
    c & d 
\end{bmatrix}
is annihilated by its adjoint:
\begin{bmatrix}
    d & -b  \\
    -c & a 
\end{bmatrix}
