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Let K be a field, p1, and AMp(K). If χA(t)=i<p(tλi) in K[t], then the transfer-matrix trace series is i<p(1λix)1

Statement

Let K be a field, let p1, and let AMp(K). Suppose that its characteristic polynomial is displayed as a product of linear factors in the base field,

χA(t)=i<p(tλi)in K[t],

where roots are repeated according to algebraic multiplicity. Then

tr(An)=i<pλin(n0),

and, formally,

n0tr(An)xn=i<p11λix.

When A is a transfer matrix, this is the eigenvalue form of its closed-walk trace series.

Facts & Assumptions

Given: A field K, a positive size p, a matrix AMp(K), and the displayed split factorisation of χA in K[t].

[L1]

For a transfer matrix, the closed-walk series is n0trK(An)xn (Closed walks have trace and logarithmic-derivative generating functions).

[L4]

The trace of an endomorphism equals the trace of any representing matrix (The basis-independent trace of an endomorphism of a finite-dimensional vector space).

[L7]

The case j=1 of the repeated-pole formula is (1λx)1=n0λnxn (Repeated poles expand formally as (1λx)j=n0(n+j1j1)λnxn).

[L8]

Over a field, the commutative-ring matrix trace equals the published matrix trace (For matrices over a field, the commutative-ring trace agrees with the published matrix trace).

Proof

technique · direct
1.1

Let TA be the coordinate endomorphism from [L2]. By [L3], the given factorisation is also χTA(t)=i<p(tλi).

givenL2L3
2.1

Apply [L5] to q(t)=tn. The roots of χTAn are λin with the displayed multiplicities, so [L6] gives tr(TAn)=i<pλin.

step 1.1L5L6
3.1

By [L4] and [L8], the left side in step 2.1 is tr(An) in either trace convention. This proves the coefficient formula, including n=0.

step 2.1L4L8
4.1

Multiply the coefficient formula by xn, sum formally over n0, and apply [L7] to each of the finitely many roots; this yields the displayed rational-series identity.

step 3.1L7algebra
5.1

If A is a transfer matrix, [L1] identifies the left side of step 4.1 with its closed-walk trace series. The split factorisation was assumed at the outset and is not inferred from algebraic closure or diagonalisation.

step 4.1L1

Depends on

Used by

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