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Statement

Notation (p. 230). The adjacency operator of $E_q(n,a)$ acts on $f:\mathbb F_q^n\to\mathbb C$ by $A_af(x)=\sum_{d(x,y)=a}f(y)$ (Eq. (5)). With $e(u)=\exp\{2\pi i\,\mathrm{Tr}(u)/p\}$, where $\mathrm{Tr}$ is the trace from $\mathbb F_q$ to $\mathbb F_p$, the paper sets $e_b(x)=e({}^tb\cdot x)$ for $b,x\in\mathbb F_q^n$ (Eq. (6)). The graph, $d$ and $S_q(n,a)$ are as on the Theorem 1 page.

Proposition 2 (p. 230). For each $b\in\mathbb F_q^n$, $e_b$ is an eigenfunction of $A_a$ with eigenvalue

$$\lambda_b=\sum_{d(s,0)=a}e_b(s).$$

As $b$ runs through $\mathbb F_q^n$ the $e_b$ form a complete set of eigenfunctions, orthogonal for the inner product $(f,g)=\sum_{x\in\mathbb F_q^n}f(x)\overline{g(x)}$, so every eigenvalue of $A_a$ is $\lambda_b$ for some $b$. The paper calls the set "complete orthonormal"; under this unnormalized inner product each $e_b$ has $(e_b,e_b)=q^n$, so orthonormality holds after division by $q^{n/2}$ (an observation of this page). The eigenvalue $\lambda_0=|S_q(n,a)|$ is the degree.

The paper calls the result very old and standard (p. 230). It adds that the combinatorial Laplacian $A_a-kI$, with $k$ the degree, has the same eigenfunctions (p. 230).

Source. A. Medrano, P. Myers, H. M. Stark and A. Terras, Finite analogues of Euclidean space, J. Comput. Appl. Math. 68 (1996), 221-238, doi:10.1016/0377-0427(95)00261-8: the notation and Proposition 2 on p. 230. The edition read is identified on the source card.

Read depth. Claims checked: the statement was read clause by clause on the printed page. Nothing here is independently reviewed.

Proof pointer

P. 230. Substituting $y=s+x$ in $A_ae_b(x)$ and using $e_b(s+x)=e_b(s)e_b(x)$ gives $A_ae_b=\lambda_be_b$; completeness and orthogonality are the standard Fourier analysis on the finite abelian group $\mathbb F_q^n$.