# idempotent in quiver theory

I am studyng quivers and algebras of the text "Elements of the representation theory of associative algebras", author Assem. In the page 46 says: "any idempotent $ε$ of $ε_a(KQ)ε_a$ can be written in the form $ε = λε_a + w$, where $λ$ is in the field $K$ and $w$ is a linear combination of cycles through a of length ≥ 1."

($KQ$ is a path-algebra with $Q$ quiver).

How could the above be demonstrated?

An element in $KQ$ is a linear combination of paths in $Q$ (including length zero paths): $$x=\sum_{p}\lambda_pp$$ If we denote by $s(p)$ and $t(p)$ the source and target of $p$, the multiplication of paths is $$pq=\begin{cases} 0 & \text{if t(p)\ne s(q)} \\[4px] p\sqcup q & \text{if t(p)=s(q)} \end{cases}$$ where $p\sqcup q$ denotes the path obtained by joining $p$ and $q$ at the common target and source. If $a$ is a vertex of $Q$, the zero length path with source and target $a$ is denoted by $\varepsilon_a$, which is clearly idempotent as an element of $KQ$.
Given $x\in KQ$, if we consider $$\varepsilon_a x\varepsilon_a$$ only paths having source and target $a$ survive when we expand the multiplication. Thus an element of $\varepsilon_a(KQ)\varepsilon_a$ has the form $$\lambda_a\varepsilon_a+w$$ where $w$ is a linear combination of paths of length $\ge1$ that have source and target $a$, that is, closed paths starting and ending at $a$. Note that the hypothesis that $\varepsilon_a x\varepsilon_a$ is idempotent has not yet been used.
I don't remember the particular argument in Assem's book, but I guess the next part is proving that $w=0$: if $\lambda_a\varepsilon_a+w$ is idempotent, then either $w^2=w$ or $w^2=-w$, and therefore, by looking at the lengths of paths in $w$, we conclude $w=0$.