TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

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 $A\in M_{m\times n}(F)$ and $B\in M_{n\times m}(F)$ , $\operatorname{tr}(AB)=\operatorname{tr}(BA)$

Statement

For $A\in M_{m\times n}(F)$ and $B\in M_{n\times m}(F)$ ,

$\operatorname{tr}(AB)=\operatorname{tr}(BA).$

The two products may have different sizes, and the equality includes $m=0$ or $n=0$ .

Facts & Assumptions

Given: A field $F$ and the rectangular matrices in the Statement.

[L1]

Trace is the sum of diagonal entries, and matrix products are row-by-column finite sums (The trace $\operatorname{tr}(A)$ as the sum of the diagonal entries, Rectangular matrix multiplication and the identity matrix $I_n$ , including zero-sized shapes).

[L2]

A finite double sum may be evaluated in either order (Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule).

Proof

technique · direct

1.1

Expanding by [L1] gives $\operatorname{tr}(AB)=\sum_{i<m}\sum_{j<n}a_{ij}b_{ji}$ .

givenL1

2.1

By [L2] and commutativity of multiplication in $F$ , this equals $\sum_{j<n}\sum_{i<m}b_{ji}a_{ij}$ .

step 1.1L1L2

3.1

The final double sum is $\operatorname{tr}(BA)$ by [L1]. If $m=0$ or $n=0$ , both sides are empty double sums and equal $0$ .

step 2.1L1L2∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 40 results over 11 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

S. Axler, Linear Algebra Done Right, 4th ed., Theorem 8.49 (standard reference, not scraped)

For A∈Mm×n(F)A\in M_{m\times n}(F)A∈Mm×n​(F) and B∈Mn×m(F)B\in M_{n\times m}(F)B∈Mn×m​(F), tr⁡(AB)=tr⁡(BA)\operatorname{tr}(AB)=\operatorname{tr}(BA)tr(AB)=tr(BA)