I am attending a course in foundations in representation theory and I am struggling to grasp the concept of path algebras, more explicitly why paths in a quiver form a $K$-algebra.
Let $Q$ be a quiver and $K$ be a field. Why do paths in a quiver $Q$ form a $K-$algebra, denoted $KQ$?
I understand that the multiplication of paths is associative and that there exists an identity element. Nonetheless no textbook I have studied explains why $KQ$ is a $K$-vector space and how the operation $+: KQ \times KQ \rightarrow KQ$, $(a,b) \mapsto a+b$ of paths is defined. Furthermore they don't explain the scalar multiplication that is required for a $K$-algebra
To construct the path algebra $KQ$, one first forms the $K$-vector space with basis the set of all paths in $Q$. This then tells you what the addition and scalar multiplication are. In particular, a general element of $KQ$ is a finite $K$-linear combination of paths.
We then need to define the multiplication in $KQ$. Since this must be $K$-linear and distributive, it is enough to declare what the multiplication of two basis elements, i.e. paths, is. One proves that this is associative by checking on paths, and that the trivial paths $e_i$ for vertices $i$ form a complete set of orthogonal idempotents.
In fact, the path algebra $KQ$ is naturally graded by path length, so the trivial paths $e_i$ have degree zero and each arrow has degree 1.
As an example, take the quiver $1\to2\to3$. Then the path algebra is isomorphic to the algebra of lower triangular $3\times3$ matrices over $K$ $$ \begin{pmatrix}\ast&0&0\\\ast&\ast&0\\\ast&\ast&\ast\end{pmatrix}$$ where $e_i$ corresponds to the elementary matrix $E_{ii}$, the arrow $a\colon1\to2$ corresponds to $E_{21}$, the arrow $b\colon2\to3$ to $E_{32}$, and the path $ba\colon1\to3$ to $E_{31}$.