Suppose your are given an antipodal graph $G=(V,E)$, that is, for every vertex $v\in V$ there is a unique maximally distant vertex $v'\in V$.
Under which condistions does the following hold:
If $\theta_2$ denotes the second-largest eigenvalue of $G$ (i.e., of its adjacency matrix), then for every $\theta_2$-eigenvector $u\in\Bbb R^V$ we have $u_v=-u_{v'}$ for all $v\in V$.
For example, is this true if $G$
- is walk-regular, (resp. vertex-transitive)
- is 1-walk-regular (resp. vertex- and edge-transitive, or arc-transitive),
- is distance-regular (resp. distance-transitive), or
- has a symmetry $\phi\in\mathrm{Aut}(G)$ mapping $v\mapsto v'$ for all $v\in V$.
Especially in the last case I can imagine that we have $u_v = \pm u_{v'}$ for all $v\in V$ and eigenvectors $u\in \smash{\Bbb R^V}$ of $G$ (not necessarily to $\theta_2$). But I am specifically interested in the case $\theta_2$ and whether we then always have the negative sign.
