False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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FALSE: The transfer-matrix identity requires a spectral-radius or convergence hypothesis

Statement

False claim. The identity $(I - x A)^{- 1} = \sum_{n \geq 0} A^{n} x^{n}$ used by the transfer-matrix method requires an analytic convergence or spectral-radius hypothesis.

Facts & Assumptions

Given: A square matrix $A$ over an arbitrary commutative ring $R$ .

[L1]

In $M_{p} (R ⟦ x ⟧)$ , the coefficientwise geometric series is a two-sided inverse of $I - x A$ (Formally, $(I - x A)^{- 1} = \sum_{n \geq 0} A^{n} x^{n}$ over every commutative coefficient ring).

[L2]

For a finite weighted directed multigraph over a commutative ring with $p \geq 1$ vertices and transfer matrix $A$ , the walk generating functions are the entries of $(I_{p} - x A)^{- 1}$ , equal to cofactors of $I_{p} - x A$ divided by its determinant (Transfer-matrix theorem: weighted-walk generating functions are cofactors of $I - x A$ divided by $\det (I - x A)$ ).

Refutation

technique · formal coefficient calculation

1.1

Multiplying $(I - x A)$ by $\sum_{n \geq 0} A^{n} x^{n}$ , the constant coefficient is $I$ and each positive coefficient is $A^{n} - A A^{n - 1} = 0$ ; the same calculation works on the other side. This is the identity in [L1].

givenL1algebra

2.1

Every coefficient uses only finitely many ring operations, and the constant matrix coefficient of $I - x A$ is the invertible matrix $I$ . No topology, norm, absolute value, or limiting operation occurs.

step 1.1

3.1

In the transfer-matrix setting itself — a finite weighted digraph over $R$ with $p \geq 1$ vertices and transfer matrix $A$ — [L2] reads the walk generating functions off this same formal inverse, again with no analytic hypothesis. So the transfer-matrix identity remains valid over every commutative coefficient ring regardless of spectral radius. This refutes the claim.

step 1.1step 2.1L2∎

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 35 results over 9 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

R. P. Stanley, Enumerative Combinatorics, vol. 1, 2nd ed., Theorem 4.7.2 (standard reference, not scraped)