Let $Q$ be a wild quiver without oriented cycles and let $V$ be an indecomposable representation of $Q$. Assume that $V\_i\neq 0$ for each vertex $i$ of $Q$. The base field $k$ is algebraically closed. If $V$ is not a Schur representation, $\operatorname{End}V$ is a local $k$-algebra different from $k$, so there is a non-trivial nilpotent endomorphism. What I would like to know is the following: does there exist a nilpotent endomorphism $\phi$ such that for each vertex $i$ $\phi$ is not zero at $V\_i$? 


I have a large body of examples in which this question has positive answer, however I still can't prove it in full generality... am I missing some key example?


This question arised while solving problems for a homework sheet for a course on quiver representations, however it does not help me in any way to solve that problem, which was simply to prove that $V$ is indecomposable iff $End V$ is local.


Edit: forgot the key word, I want $\phi$ to be nilpotent.



Endomorphism algebras of indecomposable quiver representations
Take the quiver $1 \rightarrow 2 \rightrightarrows 3$.


Let $V$ be the representation with $V\_1=k$, $V\_2=k^2$, $V\_3=k^2$, with arrows acting by $\pmatrix{0&1}$, $\pmatrix{0&1\\0&0}$ and $\pmatrix{1&0\\0&1}$.


Then, up to scalar multiplication, the only nilpotent endomorphism is zero at vertex $1$ and $\pmatrix{0&1\\0&0}$ at vertices $2$ and $3$.


