TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

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The $(i, j)$ entry of $A (G)^{k}$ counts walks of length $k$

Statement

Let $G$ be a finite simple graph with ordered vertex set $V (G) = {v_{1}, \dots, v_{n}}$ , let $A = A (G)$ , and let $k \geq 1$ be an integer. Then for every $1 \leq i, j \leq n$ , the entry $(A^{k})_{i j}$ equals the number of walks of length $k$ from $v_{i}$ to $v_{j}$ in $G$ .

Facts & Assumptions

Given: A finite simple graph $G$ with ordered vertices $v_{1}, \dots, v_{n}$ , its adjacency matrix $A$ , and an integer $k \geq 1$ .

[F1]

The entry $A_{i j}$ is $1$ exactly when $v_{i}$ and $v_{j}$ are adjacent, and is $0$ otherwise (The adjacency matrix of a finite simple graph).

[F2]

A walk of length $k + 1$ from $v_{i}$ to $v_{j}$ is exactly a walk of length $k$ from $v_{i}$ to some intermediate vertex $v_{r}$ , followed by the edge $v_{r} v_{j}$ (Walks, closed walks, trails, paths and cycles, with length equal to the number of traversed edges).

Proof

technique · direct

1.1

We argue by induction on $k$ . For $k = 1$ , the number of walks of length $1$ from $v_{i}$ to $v_{j}$ is $1$ when $v_{i}$ and $v_{j}$ are adjacent and $0$ otherwise, so it is exactly $A_{i j}$ by [F1].

F1base

2.1

Assume the claim for some $k \geq 1$ . Then $(A^{k + 1})_{i j} = \sum_{r = 1}^{n} (A^{k})_{i r} A_{r j}$ . By the inductive hypothesis, $(A^{k})_{i r}$ counts the walks of length $k$ from $v_{i}$ to $v_{r}$ , while [F1] makes $A_{r j}$ the indicator that the final edge $v_{r} v_{j}$ exists. Therefore the summand counts the length- $(k + 1)$ walks from $v_{i}$ to $v_{j}$ whose last intermediate vertex is $v_{r}$ , and summing over $r$ counts all such walks exactly once by [F2].

step 1.1F1F2induction

3.1

The base case and inductive step prove the claim for every $k \geq 1$ .

step 1.1step 2.1discharge-induction∎

Depends on

Used by

The trace of A(G)ᵏ counts closed walks of length k Corollary

Dependency tree · two levels

10 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Richard P. Stanley, Enumerative Combinatorics, Volume 1, Theorem 4.7.1 (standard reference, not scraped)

The (i,j)(i,j) entry of A(G)kA(G)k counts walks of length kk