Please help. I don't know where to begin and how to end.
Let $V$ be a finitely-generated vector space over a field $F$ and let $\alpha \in \text{End}(V)$. Show that $\alpha$ is not monic if and only if there exists an endomorphism $\beta \neq \sigma_0$ of $V$ satisfying $\alpha \beta = \sigma_0$.
Thank you.
Suppose $\alpha$ is not monic. Since a linear transformation is monic if and only if its kernel is $\{0\}$, there's a non-zero $v_0 \in V$ such that $\alpha(v_0) =0$. Define $\beta$ as the endomorphism that maps all basis vectors in $V$ to $v_0$. So for any $v \in V$, $\alpha \beta(v) = \alpha(cv_0)= c\alpha(v_0) = 0$ for some $c \in F$.
On the other hand, let $\alpha \beta(v) = 0$ for all $v \in V$, where $\beta \neq \sigma_0$, and further assume that $\alpha$ is monic $\iff \ker(\alpha)=\{0\}$. Then for all $v \in V$, $\alpha(\beta(v))=0 \implies \beta(v)=0$. Since $\beta(v)=0=\sigma_0(v)$ for all $v$, we get $\beta= \sigma_0$, a contradiction. Therefore $\alpha$ is not monic.