PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

For matrices over a field, the commutative-ring trace agrees with the published matrix trace

Statement

Let $K$ be a field, let $p \in N$ , and let $A \in M_{p} (K)$ . Then the commutative-ring trace ${tr}_{K} (A)$ equals the published field-matrix trace $tr (A)$ . This includes $p = 0$ .

Facts & Assumptions

Given: A field $K$ , a size $p \in N$ , and a matrix $A = (a_{i j}) \in M_{p} (K)$ .

[L1]

The commutative-ring trace is ${tr}_{K} (A) = \sum_{i < p} a_{i i}$ , with empty sum zero when $p = 0$ (The trace of a square matrix over a commutative ring).

[L2]

The published field trace is $tr (A) = \sum_{i < p} a_{i i}$ , with empty sum zero when $p = 0$ (The trace $tr (A)$ as the sum of the diagonal entries).

Proof

technique · direct

1.1givenL1

By [L1], the commutative-ring trace of $A$ is the finite diagonal sum $\sum_{i < p} A_{i i}$ .

1.2givenL2

By [L2], the published field trace of $A$ is the same finite diagonal sum.

2.1step 1.1step 1.2∎

Comparing steps 1.1 and 1.2 proves equality; for $p = 0$ both are the same empty sum $0_{K}$ .

Depends on

Used by

Let K be a field, p≥1, and A∈ Mₚ(K). If χ_A(t)=∏_i<p(t-λᵢ) in K[t], then the transfer-matrix trace series is ∑_i<p(1-λᵢ x)⁻¹ Corollary

Dependency tree · two levels

5 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

S. Axler, Linear Algebra Done Right, 4th ed., Definition 8.47 (standard reference, not scraped)