TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11

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

Statement

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

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

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 $tr (A)$ 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 $tr (A B) = \sum_{i < m} \sum_{j < n} a_{i j} b_{j i}$ .

givenL1

2.1

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

step 1.1L1L2

3.1

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

step 2.1L1L2∎

Depends on

Used by

Similar matrices have the same trace Corollary
The only scalar multiples of a root that are roots are plus or minus the root Corollary
A 2 by 3 matrix and a 3 by 2 matrix give square products of different sizes but equal traces Example
Hopf trace formula Lemma
For a complex character, χ(1)=dim V, χ is a class function, and |χ(g)|≤χ(1) with equality exactly at scalars Proposition
Real Cartan subalgebras need not be conjugate Proposition
Trace forms are symmetric and invariant Proposition
Cartan's solvability criterion Theorem
Lefschetz fixed-point theorem for finite complexes Theorem
Lie's theorem Theorem
Root spaces of a complex semisimple Lie algebra are one-dimensional Theorem

Dependency tree · two levels

10 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., Theorem 8.49 (standard reference, not scraped)

For A∈Mm×n(F)A∈Mm×n​(F) and B∈Mn×m(F)B∈Mn×m​(F), tr⁡(AB)=tr⁡(BA)tr(AB)=tr(BA)