For a finite graph $G$, we say a matrix $B\in \mathbb{R}^{n\times n}$ represents $G$ if $B$ belongs to this set: $$\{ B=(b_{ij}) \mid B\textrm{ is positive semidefinite with trace at most 1 and } b_{ij}=0 \textrm{ if } i \textrm{ and } j \textrm{ are adjacent vertices of } G \textrm{ } (i\neq j) \}.$$ In the above set, considering only the elements of the upper triangle and ignoring those entries which correspond to $i$ and $j$ being adjacent, prove this set is full dimensional in the $n(n+1)/2-|E|$-dimensional space.
A set $S\in \mathbb{R}^n$ is full dimensional if it contains an $n$-dimensional ball. Alternatively, any element $z\in \mathbb{R}^n$ can be written as $z=x-y$ for some $x,y\in S$.
I have not dealt with full dimensionality before. Do we try to create a spanning set/basis?
It is proved here that the SDP cone is full dimensional: Let $E_{ij}$ denote the matrix with $1$'s in the positions $(i,i),(i,j),(j,i),(j,j)$ and zeros everywhere else. The set $\{E_{ij}\mid 1\leq i,j\leq n\}$ is a basis for the $S^{n}$ space of symmetric matrices: any $A=(a_{ij})\in S^n$ can be written as $$A= \sum_{i\leq j}a_{ij}E_{ij} - \sum_i \left(\sum_{i<j}a_{ij}\right) E_{ii}.$$ It is easy to see that each $E_{ij}$ is PSD: $$x^TE_{ij}x=(x_i+x_j)^2\geq 0$$ for $i\neq j$ and for $i=j$ we have $x^TE_{ii}x=x_i^2\geq 0$. Hence the PSD cone is full dimensional.
In the case of the set in the question, we similarly reduce the $E_{ij}$ set to the $(n(n+1)/2-|E|)$th dimension and prove that it is a basis. Note that in this case $tr(E_{ij})$ should be at most $1$, hence we can take the new $E_{ij}$ to have $1/2$ in place of $1$.