The eigenvalue $\lambda$ is said to be a main eigenvalue if $\mathcal{E}(e)\not \subseteq \textbf{j}^{\perp}$, where $\mathcal{E}(e)$ is the eigenspace of $\lambda$ and $\textbf{j}$ is the all-1 vector in $\mathbb{R}^n$.
The line graph $L(G)$ of a graph $G$ is the graph whose vertices are the edges of $G$, with two vertices in $L(G)$ adjacent whenever the corresponding edges in $G$ have exactly one vertex in common.
There is this fact that no line graph has $-2$ as a main eigenvalue. I tried to prove that, but I couldn't. I was wondering if someone could help me about it. Thanks in advance.
Let $B$ be the vertex-edge incidence matrix of the graph $X$ and let $L(X)$ be the line graph of $X$. Each column of $B$ sums to $2$, and so the sum of rows of $B$ is twice the all-ones vector.
Next $B^TB = 2I + A(L(X))$, and therefore the kernel of $B$ is the $-2$-eigenspace of $L(X)$. Since the row sum of $B$ is twice $\mathbf{j}$, each element of the eigenspace is orthogonal to $\mathbf{j}$. Hence $-2$ cannot be a main eigenvalue.