The spectral theorem for a real $n \times n$ symmetric matrix $A$ says that $A$ is diagonalizable with all eigenvalues real. If $A$ happens to have non-negative integer entries, it can be interpreted as the adjacency matrix of an undirected graph $G$, and the spectral theorem gives us information about how the sequences $A_{ij}^n$ behave, which count the number of walks of length $n$ from vertex $i$ to vertex $j$. In particular, it says that $A_{ij}^n$ has the form $\sum_{k=1}^{n} a_{ijk} \lambda_k^n$ for some real $\lambda_k$ and some $a_{ijk}$.

If $A$ is not symmetric, on the other hand, two things can happen that don't happen in the above case:

The $\lambda_k$ may fail to be real. In other words, there can be "periodicity" of period greater than $2$ in the sequences $A_{ij}^n$. This happens, for example, if $A$ is a directed cycle graph.

The coefficients $a_{ijk}$ may be polynomials in $n$. This happens, for example, if $A$ is a directed path graph with loops based at each vertex.

Is it possible to prove "combinatorially" that neither of these things can happen when $A$ is symmetric? (What I mean is that, given only that you know what $A_{ij}^n$ looks like in terms of eigenvalues, what can you prove just by looking at $G$?) For example, it is straightforward to prove that the coefficient of the largest positive eigenvalue is constant by a path-counting argument which I described here. I think a path-counting argument can also in principle prove that the $\lambda_k$ are real via some kind of mixing argument which could show that the only constraint on the length of a long walk from a vertex to itself is its length mod 2 (due to the presence of even cycles). I think such an argument can at least show that the eigenvalues of maximum absolute value must be real, but I don't know how easy it is to deduce information about the other eigenvalues.

If you have a solution with "adjacency matrix" replaced by "Laplacian," I'd also be interested in that.