For $p$ a prime, $p>2$, my graph $G$ is defined as follows:
The vertices of $G$ are all the non self-orthogonal 1-dimensional subspaces in $\mathbb{F}_p^3$, with 2 vertices adjacent if and only if their corresponding 1-dimensional vector spaces are orthogonal.
For example, if $p = 5$: $$u = \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix} v= \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix} z = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$$ $u$ is self orthogonal because $u^\top u = 0$ in $\mathbb{F}_5$. $v$ and $z$ are adjacent because $v^\top z = 0$.
This graph $G$, for any $p$, has order $p^2$. Vertices in $G$ have degrees of either $p-1$ or $p+1$.
Now for my question, I suspect the adjacency matrix of $G$ has the following eigenvalues (not counting multiplicities): $$(-\sqrt{p}, \, -1, \, 0, \, \sqrt{p}, \, p)$$ The algebraic multiplicities follow an interesting pattern as well: $$ \big( \frac{p^2-p-2}{2}, \, p, \, 1, \, \frac{p^2-p-2}{2}, \,1 \big)$$ but I have not been able to find a proof for this.
I have verified this property for primes up to and including $p=37$. Is there perhaps a class of matrices for which the eigenvalues satisfy this relation?