Trace of a product is symmetric

Statement

Let $R$ be a commutative ring. Let $A$ be an $m \times n$ matrix and $B$ an $n \times m$ matrix over $R$. Then $\tr(AB) = \tr(BA)$, where $\tr$ is the trace.

Remarks

Both sides equal $\sum_{i=1}^{m} \sum_{j=1}^{n} a_{ij} b_{ji}$. The matrices $AB$ and $BA$ can have different sizes, and the identity still holds. A consequence is $\tr(P^{-1} A P) = \tr A$, so similar matrices have the same trace. For three matrices, only cyclic permutations are allowed in general: $\tr(ABC) = \tr(BCA)$, but $\tr(ABC)$ can differ from $\tr(BAC)$.

Explanations

L1 human

Although $AB$ and $BA$ are usually different matrices, they have the same trace. Each diagonal sum adds the same products $a_{ij} b_{ji}$, only in another order. This fact makes the trace a property of a linear map, not of the basis that we choose.