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

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The trace of $A (G)^{k}$ counts closed walks of length $k$

Statement

Let $G$ be a finite simple graph with adjacency matrix $A (G)$ , and let $k \geq 1$ . Then

$tr (A (G)^{k})$

equals the number of closed walks of length $k$ in $G$ .

Facts & Assumptions

Given: A finite simple graph $G$ with adjacency matrix $A$ , and an integer $k \geq 1$ .

[L1]

The entry $(A^{k})_{i i}$ counts the walks of length $k$ from $v_{i}$ back to $v_{i}$ (The $(i, j)$ entry of $A (G)^{k}$ counts walks of length $k$ ).

[F1]

The trace of a square matrix is the sum of its diagonal entries (The trace of a square matrix over a commutative ring).

Proof

technique · direct

1.1

For each vertex $v_{i}$ , a walk of length $k$ from $v_{i}$ to itself is exactly a closed walk of length $k$ whose chosen start vertex is $v_{i}$ , so [L1] makes $(A^{k})_{i i}$ the number of such closed walks.

2.1

Summing step 1.1 over all vertices counts every closed walk of length $k$ exactly once for each of its chosen start vertices, and by [F1] that sum is $tr (A^{k})$ . Hence the trace counts the closed walks of length $k$ .

step 1.1F1∎

Depends on

Used by

A finite simple graph is bipartite if and only if its adjacency spectrum is symmetric about 0 Theorem

Dependency tree · two levels

6 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, Corollary 4.7.3 (standard reference, not scraped)

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