# Let $A,B\in \Bbb{K}^{n^2}, B\neq0$ prove that if $AB=0$ then $A$ is not invertible.

Let $A,B\in \Bbb{K}^{n^2}, B\neq0$ prove that if $AB=0$ then $A$ is not invertible.

I'm having trouble proving this, I tried saying that $|AB|=|A||B|=0 \implies |A|=0 \text{ or } |B|=0$ but that got me nowhere.

Besides a regular proof using matrix properties, I would like to see if this can be proved in a more ring theory-ish way, saying that as $A$ is a zero divisor, it can't have an inverse, but I don't know how.

If $AB=0$ and $B\ne 0$, there must be a nonzero column of $B$, call this $b$ and see that $$Ab = 0$$ i.e. $b\ne 0$ is an eigenvector of $A$ for the eigenvalue $0$. Thus $A$ must be singluar.
Let $A$ invertible. Then $AB=0\Rightarrow A^{-1}(AB)=A^{-1}0\Rightarrow B=0$ - a contradiction. $A,B$ can be elements of an arbitrary ring with unity.