Let $Q$ be a quiver over defined as follows
Then $KQ\cong$ $\begin{pmatrix}K&K&K\\0&K&K\\0&0&K\end{pmatrix}$, where $KQ$ is just the path algebra. What the professor did was he just took
$e_1 =\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix}$, $e_1 =\begin{pmatrix}0&0&0\\0&1&0\\0&0&0\end{pmatrix}$, $e_1 =\begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix}$
$\alpha =\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix}$, $\beta =\begin{pmatrix}0&0&0\\0&0&1\\0&0&0\end{pmatrix}$,
to get $\alpha \beta =\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}$,
hence $KQ\cong$ $\begin{pmatrix}K&K&K\\0&K&K\\0&0&K\end{pmatrix}$.
I did not understand what was going on, and I asked him, but he just repeated what he wrote on the board. So could you guys help me to understand what he did?
Thanks,
What he did is the following recipe, which works for all quivers whose underlying graph is a tree:
Label the vertices such that the arrows are increasing. Then put $e_i \mapsto E_{ii}$ and the unique path (since underlying graph is a tree) from $i$ to $j$ mapsto $E_{ij}$. One can check easily that this defines an isomorphism. It maps a basis to a basis (of the subalgebra generated by those matrices). And multiplication works as it should: The unique path from $i$ to $j$ to $k$ is mapped to $E_{ij}\cdot E_{jk}=E_{ik}$.
More generally if your quiver is acyclic you can put a vector space of dimension = number of arrows form $i$ to $j$ on position $ij$ and just define multiplication as it should behave.