Can an adjacency matrix A of a graph $G $(of $100$ connected vertices) have at least $1$ $0$ entry in all of its powers $(A_1, A_2, A_3, ...)$?

Question

Conceptually, this means for any $A^k$ (with $k$ being a natural number obviously), there's going to be a pair $(i, j)$ with $0$ walks of length k from $i$ to $j$.

How do we go about proving such thing could or couldn't exist?

Answer 1

Let $G$ just be a straight line graph with $100$ vertices numbered from $1$ to $100$ in order along the line.

Then in $(A^{2n})_{12}=0$ and $(A^{2n+1})_{11}=0$.