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

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Transfer-matrix theorem: weighted-walk generating functions are cofactors of $I - x A$ divided by $\det (I - x A)$

Statement

Let $R$ be a commutative ring, let a finite weighted directed multigraph have $p \geq 1$ vertices and transfer matrix $A$ , and put $M (x) = I_{p} - x A$ . For vertices $u, v$ ,

$\sum_{n \geq 0} (\sum_{W : u ⇝ v, ∣ W ∣ = n} w (W)) x^{n} = (M^{- 1})_{u v} = \frac{(- 1)^{u + v} \det M^{(v, u)}}{\det M} .$

The quotient is a rational formal power series over $R$ because $\det M$ has constant coefficient $1_{R}$ .

Facts & Assumptions

Given: A nonempty finite weighted directed multigraph over $R$ , its $p \times p$ transfer matrix $A$ , vertices $u, v$ , and $M = I_{p} - x A$ .

[L1]

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

[L2]

Formally, $M^{- 1} = \sum_{n \geq 0} A^{n} x^{n}$ over every commutative coefficient ring (Formally, $(I - x A)^{- 1} = \sum_{n \geq 0} A^{n} x^{n}$ over every commutative coefficient ring).

[L3]

The adjugate satisfies $adj (M)_{u v} = (- 1)^{u + v} \det M^{(v, u)}$ (Deleted-row-and-column minors, cofactors, the cofactor matrix and the adjugate over a commutative ring).

[L4]

For a positive-sized square matrix, $M adj (M) = adj (M) M = \det (M) I$ (For every positive-sized square matrix over a commutative ring, $A adj (A) = adj (A) A = \det (A) I$ ).

[L5]

If $\det (M)$ is a unit, then $M^{- 1} = \det (M)^{- 1} adj (M)$ (If $\det (A)$ is a unit, then $A^{- 1} = \det (A)^{- 1} adj (A)$ ).

[L6]

The determinant is the finite Leibniz sum over permutations (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[L7]

A formal series is a unit exactly when its constant coefficient is a unit (A formal power series is a unit exactly when its constant coefficient is a unit).

[L8]

Cauchy multiplication makes $R ⟦ x ⟧$ a commutative ring (Cauchy multiplication makes $R ⟦ x ⟧$ a commutative ring containing $R [x]$ as the finitely supported subring).

Proof

technique · direct

1.1

By [L2], the $(u, v)$ entry of $M^{- 1}$ is $\sum_{n \geq 0} (A^{n})_{u v} x^{n}$ , and [L1] identifies each coefficient with the total weight of the corresponding walks.

givenL1L2

1.2

Setting $x = 0$ gives $M (0) = I_{p}$ . In the Leibniz sum [L6], only the identity permutation contributes to $\det (I_{p})$ , so $\det M$ has constant coefficient $1_{R}$ and is a unit by [L7].

givenL6L7

2.1

Apply [L4] and [L5] over the commutative ring $R ⟦ x ⟧$ from [L8] to obtain $M^{- 1} = \det (M)^{- 1} adj (M)$ .

step 1.2L4L5L8

3.1

Taking the $(u, v)$ entry in step 2.1 and using [L3] gives $(M^{- 1})_{u v} = (- 1)^{u + v} \det M^{(v, u)} / \det M$ .

step 2.1L3

4.1

Combining steps 1.1 and 3.1 proves the formula, with no analytic hypothesis. The assumption $p \geq 1$ is exactly the positive-size domain of [L3] through [L6].

step 1.1step 3.1L3L6∎

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Direct dependencies and their dependencies through the next three levels: 79 results over 18 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)
J. Gallier and J. Quaintance, Linear Algebra for Computer Vision, Robotics, and Machine Learning, Section 6.4 (standard reference, not scraped)

Transfer-matrix theorem: weighted-walk generating functions are cofactors of I−xAI−xA divided by det⁡(I−xA)det(I−xA)