Suppose $G$ is a $k-$regular graph and $A$ is its adjacency matrix, and let $[ \ 1 \ ]$ denote "all ones" square matrix of size $|V(G)|$. After some computations, I believe the following holds:
$$\lim_{n\to\infty} A^n/k^n=\frac{1}{|V(G)|}[ \ 1 \ ]$$
I sort of understand why, in terms of walks, but this intuition is not aiding me in the proof that this limit would require. Could someone point me in the right direction on how to prove this statement? Thanks.
Assume the graph is connected. First of all, it's easy to check that the constant vector $v_{const}=\textbf{1}/\sqrt{\mid V(G)\mid}$ is an eigenvector of A/k with eigenvalue 1 (by regularity). By the Perron-Frobenius theorem, this is the largest eigenvalue, and it has multiplicity 1 (this is where I use the connectivity, see the section on Non-Negative matrices in the wikipedia page).
Thus, $A/k=v_{const}v_{const}^T+\sum_i \lambda_i v_iv_i^T$ where $\lambda_i$ have magnitude strictly less than 1, and the $v_i$ are orthogonal to each other and to $v_{ones}$ (because A is symmetric).
In particular, $(A/k)^n=v_{const}v_{const}^T+\sum_i \lambda_i^n v_iv_i^T$, and in the limit, we get $v_{const}v_{const}^T=\textbf{1}\textbf{1}^T/\mid V(G)\mid$. Finally, note that $\textbf{1}\textbf{1}^T$ coincides with your matrix $[1]$