LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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.

The trace of a matrix commutator is zero

Statement

Let $R$ be a commutative ring, let $d \geq 0$ , and let $A, B \in M_{d} (R)$ . Then

$tr (A B - B A) = 0.$

Facts & Assumptions

Given: A commutative ring $R$ , an integer $d \geq 0$ , and square matrices $A = (a_{i j}), B = (b_{i j}) \in M_{d} (R)$ .

[F1]

The trace of a square matrix over a commutative ring: for a square matrix $C = (c_{i j})$ , $tr (C) = \sum_{i < d} c_{i i}$ , with the empty sum equal to $0_{R}$ when $d = 0$ .

[F2]

Entrywise ring-matrix operations, rectangular matrix products, identity matrices and transpose: for square matrices $A = (a_{i j})$ and $B = (b_{i j})$ , $(A B)_{i i} = \sum_{j < d} a_{i j} b_{j i}$ .

Proof

1.1givenF1F2algebra

By the matrix-product formula [F2] and trace definition [F1], $tr (A B) = \sum_{i < d} \sum_{j < d} a_{i j} b_{j i}$ .

2.1givenF1F2step 1.1algebra

By [F1] and the matrix-product formula [F2], $tr (B A) = \sum_{i < d} \sum_{j < d} b_{i j} a_{j i}$ . Since $R$ is commutative, reordering each product and exchanging the finite dummy indices gives $tr (B A) = \sum_{i < d} \sum_{j < d} a_{i j} b_{j i} = tr (A B)$ .

3.1givenF1step 2.1algebra∎

The diagonal entries of $A B - B A$ are the differences of the corresponding diagonal entries, so [F1] and step 2.1 give $tr (A B - B A) = tr (A B) - tr (B A) = 0$ . The same calculation covers $d = 0$ because both sums are empty.

Depends on

Used by

No nonzero finite-dimensional Weyl modules at nonzero central character in characteristic zero Theorem

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

Sheldon Axler, Linear Algebra Done Right, 4th ed., §8D, Theorem 8.49 (standard reference, not scraped)