I don't fully understand the proof in this link on Springer of Lemma 2.2. Why is it that the $i$th and $j$ components of $x$ are equal when $i\sim j$? Also, what does $i\sim j$ even mean? Does it mean there is a path from $i$ to $j$?
I get that if $G$ is connected and components of $x$ are equal if their corresponding vertices are joined by a path, then $x$ must have all components equal. But is there a formal justification for why the components have to be equal if their corresponding vertices are joined by a path (e.g. there's a justification that the diagonal entries of the Laplacian matrix of a graph $G$ are the vertex degrees)?
The notation $i\sim j$ means that vertices $i$ and $j$ are connected by an edge, say $e_k$. Let $q_k$ be the $k$-th column of $Q$. If $i\sim j$, the only two non-zero entries in $q_k$ are those in rows $i$ and $j$; call them $q_{k,i}$ and $q_{k,j}$, respectively. One of them is $1$, the other is $-1$, and
$$[0]=x'q_k=x_iq_{k,i}+x_jq_{k,j}=q_{k,i}(x_i-x_j)\,,$$
so $x_i=x_j$.
Now suppose that there is a path $i=i_0\sim i_1\sim\ldots\sim i_\ell=j$ from $i$ to $j$. Then
$$x_i=x_{i_0}=x_{i_1}=\ldots=x_{i_\ell}=x_j$$
by repeated applications of the previous result. $G$ is connected, so there is a path between any two vertices, and therefore $x_1=x_2=\ldots=x_n$. Thus, the left null space of $Q$ is either trivial or one-dimensional, and hence $\operatorname{rank}Q=n$ or $\operatorname{rank}Q=n-1$, respectively. On the other hand, the $n$ rows of $Q$ are sum to a zero vector and are therefore linearly dependent, so $\operatorname{rank}Q\le n-1$. Combining results, we see that $\operatorname{rank}Q=n-1$.