I haven't done any linear algebra for a long time and currently reading about linear algebra in graph theory and had a few queries.
So i'm looking at the definition of a vertex space. Firstly let $G=(V,E)$ be a graph with $n$ vertices and $m$ edges.
It then states that "The vertex space $\mathcal{V}(G)$ of $G$ is the vector-space over the $2$-element sub field $\mathbb{F}_{2}=\{0,1\}$ of all functions $V \rightarrow \mathbb{F}_{2}$."
So just to clarify suppose i 'numbered' vertices $1$ through to $n$. Then all functions $V \rightarrow \mathbb{F}_{2}$ can be represented by a binary vector of length $n$, thus for example $(1,0,0,...,0)$ would represent the set containing just the vertex $1$, so i can represent all functions in this way with each function representing a subset of the vertices.
Now my confusion comes with the scalar multiplication. It says that addition in the vertex space is represented by the 'symmetric difference of sets' which is fine. So suppose i have functions $f,f',f''$ corresponding to subsets $U,U',U'' \subseteq V$. Then according to some standard notes on vector spaces, it follows that $a_{1}U+a_{2}U'+a_{3}U''$ is an element of my vertex space provided $a_{i}$ are elements of the scalar field. So in this case i'm assuming $a_{i} \in \mathbb{F}_{2}$?
If this is correct would we have $1\cdot U=U$ and $0 \cdot U=\emptyset$ (empty vertex set)?
