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

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Closed walks have trace and logarithmic-derivative generating functions

Statement

Let $R$ be a commutative ring, let $p \geq 1$ , let $A \in M_{p} (R)$ be a transfer matrix, and put $Q (x) = \det (I_{p} - x A)$ . Then

$\sum_{n \geq 0} {tr}_{R} (A^{n}) x^{n} = \frac{{tr}_{R [x]} (adj (I_{p} - x A))}{Q (x)},$

and

$\sum_{n \geq 1} {tr}_{R} (A^{n}) x^{n} = - \frac{x Q^{'} (x)}{Q (x)} .$

The coefficient ${tr}_{R} (A^{n})$ is the total weight of all closed walks of length $n$ . Each trace subscript names the commutative ring its argument's entries lie in: $A^{n} \in M_{p} (R)$ , while $adj (I_{p} - x A)$ has entries in $R [x]$ , so that numerator is ${tr}_{R [x]}$ . The defining formula $\sum_{i < p} a_{i i}$ is the same in each case. The second expression is a formal logarithmic derivative; no logarithm or convergence is required.

Facts & Assumptions

Given: A positive-sized transfer matrix $A$ over a commutative ring $R$ , $M = I_{p} - x A$ , and $Q = \det M$ .

[L1]

The transfer-matrix formula identifies each entry of $M^{- 1}$ with the corresponding weighted-walk series and with an adjugate entry divided by $Q$ (Transfer-matrix theorem: weighted-walk generating functions are cofactors of $I - x A$ divided by $\det (I - x A)$ ).

[L2]

The determinant derivative satisfies $Q^{'} = - {tr}_{R [x]} (adj (M) A)$ , the trace being over $R [x]$ where those entries lie ( $\frac{d}{d x} \det (I - x A) = - tr (adj (I - x A) A)$ ).

[L3]

The ring trace is the finite sum of diagonal entries, including the empty-sum convention (The trace of a square matrix over a commutative ring).

Proof

technique · direct

1.1givenL1L3

Sum the diagonal instances of [L1]. By [L3], the left side is $\sum_{n \geq 0} {tr}_{R} (A^{n}) x^{n}$ and the numerator on the right is ${tr}_{R [x]} (adj (M))$ , proving the first identity and the closed-walk interpretation.

1.2L1L3algebra

From [L1], $M^{- 1} = \sum_{n \geq 0} A^{n} x^{n} = adj (M) / Q$ . Multiplying by $A$ , taking the trace over $R ⟦ x ⟧$ where $M^{- 1} A$ has its entries, and then multiplying by $x$ gives $x {tr}_{R ⟦ x ⟧} (M^{- 1} A) = \sum_{n \geq 1} {tr}_{R} (A^{n}) x^{n}$ , the right-hand coefficients being the traces over $R$ of the matrices $A^{n}$ .

2.1step 1.2L2algebra

Dividing [L2] by the unit $Q$ gives $- x Q^{'} / Q = x {tr}_{R ⟦ x ⟧} (M^{- 1} A)$ . Combine this with step 1.2 to obtain the second identity.

3.1step 1.1step 2.1∎

Both calculations take place in $R ⟦ x ⟧$ , so they require no topology or spectral-radius assumption.

Depends on

Used by

Dependency tree · two levels

14 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., Corollary 4.7.3 (standard reference, not scraped)