Hereditary Algebras I recently began to study representation theory of algebras and I found this problem:

Suppose $\Lambda$ is a finite dimensional algebra over an arbitrary field. If $\Lambda$ is hereditary, basic and connected and if $e$ is an idempotent of $\Lambda$, then $e\Lambda e$ is hereditary.

I would like that someone give me a simple proof of it because I have no idea of what to do.
 A: I will write $A$ instead of $\Lambda$...
If $A$ is hereditary, there is a projective resolution $$0\to P_1\to P_0\to A\to0$$ of $A$ as an $A$-bimodule. The right $A$-module $eA$ is projective, as is the left $A$-module $Ae$, so the functor $$M\longmapsto eA\otimes_AM\otimes_AAe$$ from the category of $A$-bimodules to the category of $eAe$-bimodules is exact. We have, therefore a short exact sequence of $eAe$-bimodules $$0\to eA\otimes_AP_1\otimes_AAe\to eA\otimes_AP_0\otimes_AAe\to eA\otimes_AA\otimes_AAe\to0$$ The $eAe$-bimodule $eA\otimes_AA\otimes_AAe$ is isomorphic to $eAe$, and the $eAe$-bimodules $eA\otimes_AP_1\otimes_AAe$ are projective. This means that the projective dimension of $eAe$ as an $eAe$-bimodule is at most $1$. This implies that $eAe$ is hereditary.

If you know that the algebra is the path algebra $kQ$ of an acyclic quiver and you suppose (as you may) that $e$ is the sum of a set of vertices $S\subseteq Q_0$, then it is easy to construct a quiver $Q'$ such that $ekQe\cong kQ'$. Indeed, the vertices of $Q'$ are the elements of $S$ and the arrows in $Q'$ are the paths in $Q$ from a vertex of $S$ to another which cannot be factored as a product of two such paths.
For example, if $Q$ is 

and $S=\{1,3,4\}$, the quiver $Q'$ is

