Adacency matrix of a line digraph

Question

Let $G$ be directed graph with adjacency matrix $A$ and let us assume that $A$ is primitive, i.e. there exists $N\in\mathbb{N}$ such that $(A^N)_{i,j}>0$ for all $i,j$.

Let now $L(G)$ be the line digraph of $G$ and denote by $B$ its adjacency matrix.
(Two vertices representing directed edges from $u$ to $v$ and from $w$ to $x$ in $G$ are connected by an edge from $uv$ to $wx$ in the line digraph when $v = w$).

My guess is that the matrix $B$ is again primitive and this seems to fit with some examples I calculated. However, I am not sure how to prove it rigorously. Can anybody share some good ideas?

I have tried to look for ways of expressing $B$ in terms of $A$, but for digraph I have found no connection…

Thank you very much for your help!

Answer 1

I think you're looking for what is close to what is defined here as $W_1$:

We define the 0,1 edge matrix $W_1$ by orienting the $m$ edges of $G$ and labeling them as in formula (2.1). Then $W_1$ is the $2m×2m$ matrix with $ij$ entry $1$ if edge $e_i$ feeds into $e_j$ provided that $e_j\neq e^{−1}_i$, and $ij$ entry $0$ otherwise. By “$e_i$ feeds into $e_j$,” we mean that the terminal vertex of edge $e_i$ is the same as the initial vertex of edge $e_j$.

Let $u,w,x$ be labels of vertices of $G$ and not caring about the no-backtracking thing for now, the following should give an $n^2$ dimensional representation of $B$: $$ B=\sum_u \sum_w \langle w|Au\rangle \sum_{|x\rangle\in A|w\rangle} E_{u,w}\otimes E_{w,x} $$

Setting $E_{u,w}\otimes E_{w,u}=0$ we can remove the backtrackers, to get a $n^2$ dimensional representation of $W_1$...