TITLE: Linear Transformation Question $ \operatorname{Hom}(V,W)$ and $ \operatorname{Hom}(W,V)$ satisfying conditions
QUESTION [5 upvotes]: Can I please get help on this questions. I'm not sure how to start it or end it.

Let $V$ and $W$ be vector spaces over a field $F$. Let $\alpha$ be an element of $ \operatorname{Hom}(V,W)$ and $\beta$ be an element of $ \operatorname{Hom}(W,V)$ satisfy the condition  $\alpha\cdot\beta\cdot\alpha=\alpha$. 
If $w \in  \operatorname{Im}(\alpha)$, show that $\alpha^{-1}(w) = \{ \beta(w)+v - \beta \alpha(v)  \mid v \in V \}$.

REPLY [2 votes]: It's generally true that, for $w$ in the image of $\alpha$, $\alpha^{-1}(w)=\{x+y:y\in\ker\alpha\}$, where $x\in V$ is a fixed vector such that $\alpha(x)=w$.
Since $w=\alpha(x)$, then $w=\alpha(x)=\alpha\beta\alpha(x)=\alpha\beta(w)$, so we can take $x=\beta(w)$.
Now it's generally true that $\alpha^{-1}(w)=\{\beta(w)+y:y\in\ker\alpha\}$, so we just need to determine $\ker\alpha$.
Let $v\in V$; then
$$
\alpha(v-\beta\alpha(v))=\alpha(v)-\alpha\beta\alpha(v)=0
$$
so $v-\beta\alpha(v)\in\ker\alpha$. Conversely, if $y\in\ker\alpha$, then $y=y-\beta\alpha(y)$. Thus
$$
\ker\alpha=\{v-\alpha\beta(v):v\in V\}
$$