Trace of Product of Matrices

Theorem

Let $\struct {R, +, \circ}$ be a commutative ring.

Let $\mathbf A = \sqbrk a_{m n}$ be an $m \times n$ matrix over $R$.

Let $\mathbf B = \sqbrk b_{n m}$ be an $n \times m$ matrix over $R$.

Then:

$\map \tr {\mathbf A \mathbf B} = \map \tr {\mathbf B \mathbf A}$

where $\map \tr {\mathbf A}$ denotes the trace of $\mathbf A$.

Let $\mathbf A \mathbf B = \mathbf C = \sqbrk c_m$.

Let $\mathbf B \mathbf A = \mathbf D = \sqbrk d_n$.

$\ds \forall i, j \in \closedint 1 m: \, $

$\ds c_{i j}$

$=$

$\ds \sum_{k \mathop = 1}^n a_{i k} \circ b_{k j}$

$\ds \forall i, j \in \closedint 1 n: \, $

$\ds d_{i j}$

$=$

$\ds \sum_{k \mathop = 1}^m b_{i k} \circ a_{k j}$

Therefore:

$\ds \map \tr {\mathbf A \mathbf B}$

$=$

$\ds \sum_{j \mathop = 1}^m c_{jj}$

$\ds $

$=$

$\ds \sum_{j \mathop = 1}^m \sum_{k \mathop = 1}^n a_{j k} \circ b_{k j}$

$\ds $

$=$

$\ds \sum_{k \mathop = 1}^n \sum_{j \mathop = 1}^m a_{j k} \circ b_{k j}$

$\ds $

$=$

$\ds \sum_{k \mathop = 1}^n \sum_{j \mathop = 1}^m b_{k j} \circ a_{j k}$

$\circ$ is commutative

$\ds $

$=$

$\ds \sum_{k \mathop = 1}^n d_{kk}$

$\ds $

$=$

$\ds \map \tr {\mathbf B \mathbf A}$

$\blacksquare$