TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16

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The $(u, v)$ entry of $A^{n}$ is the total weight of length- $n$ walks from $u$ to $v$

Statement

Let $A$ be the transfer matrix of a finite weighted directed multigraph over a commutative ring $R$ . For every $n \in N$ and vertices $u, v$ ,

$(A^{n})_{u v} = \sum_{W : u ⇝ v, ∣ W ∣ = n} w (W) .$

The sum is over all length- $n$ walks from $u$ to $v$ . Powers of a square matrix are the ones given by the recursion $A^{0} : = I_{p}$ and $A^{n + 1} : = A^{n} A$ , where $p$ is the number of vertices; the cited matrix laws supply the product and the identity but no power notation, so the recursion is fixed here. At $n = 0$ , both sides are $1_{R}$ when $u = v$ and $0_{R}$ otherwise.

Facts & Assumptions

Given: A finite weighted directed multigraph over $R$ , its transfer matrix $A$ , vertices $u, v$ , and a length $n \in N$ .

[L1]

The transfer entry $A_{u v}$ is the sum of the weights of all edges from $u$ to $v$ , and the unique empty walk at a vertex has weight $1_{R}$ (Finite weighted directed multigraphs, weighted walks and their transfer matrices).

[L2]

Matrix multiplication is $(A B)_{u v} = \sum_{z} A_{u z} B_{z v}$ and the identity matrix has diagonal entries $1$ and off-diagonal entries $0$ (Entrywise ring-matrix operations, rectangular matrix products, identity matrices and transpose).

[L3]

For matrices over a commutative ring of compatible shapes, $A (B C) = (A B) C$ and $I_{m} A = A = A I_{n}$ , and the entrywise additive and distributive laws hold, including all zero-sized shapes (Matrix arithmetic over a commutative ring is associative, unital and distributive, and transpose reverses products).

Proof

technique · induction

1.1baseL1L2

For $n = 0$ , the recursion in the Statement gives $A^{0} = I_{p}$ , so $(A^{0})_{u v} = (I_{p})_{u v}$ , which is $1_{R}$ when $u = v$ and $0_{R}$ otherwise by [L2]; this agrees with the empty-walk convention in [L1], the empty walk being the unique walk of length $0$ from $u$ to $v$ and existing only when $u = v$ .

1.2ihL2

Assume the formula at length $n$ . The recursion gives $A^{n + 1} = A^{n} A$ , so $(A^{n + 1})_{u v} = \sum_{z < p} (A^{n})_{u z} A_{z v}$ by the product formula of [L2].

2.1step 1.2L1L3algebra

Substitute the induction formula and the edge-sum definition [L1] into step 1.2. The distributive laws of [L3] expand the result into one product for each length- $n$ walk from $u$ to $z$ followed by one edge from $z$ to $v$ .

3.1step 2.1L1

Every length- $(n + 1)$ walk has a unique penultimate vertex and last edge, so the expansion in step 2.1 is exactly the total weight of all length- $(n + 1)$ walks from $u$ to $v$ .

4.1step 1.1step 3.1discharge-induction∎

The base case and induction step prove the formula for all $n \in N$ .

Depends on

Used by

Dependency tree · two levels

8 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

R. P. Stanley, Enumerative Combinatorics, vol. 1, 2nd ed., Theorem 4.7.1 (standard reference, not scraped)

The (u,v)(u,v) entry of AnAn is the total weight of length-nn walks from uu to vv