Commutativity of Hadamard Product

Theorem
Let $\map {\mathcal M_S} {m, n}$ be a $m \times n$ matrix space over $S$ over an algebraic structure $\struct {S, \circ}$.

Let $\mathbf A, \mathbf B \in \map {\mathcal M_S} {m, n}$.

Let $\mathbf A + \mathbf B$ be defined as the matrix entrywise sum of $\mathbf A$ and $\mathbf B$.

The operation $+$ is commutative on $\map {\mathcal M_S} {m, n}$ $\circ$ is commutative on $\struct {S, \circ}$.

Proof
Let $\sqbrk a_{m n}, \sqbrk b_{m n}$ be elements of $\map {\mathcal M_S} {m, n}$.

Let:
 * $\sqbrk c_{m n} = \sqbrk a_{m n} + \sqbrk b_{m n}$
 * $\sqbrk d_{m n} = \sqbrk b_{m n} + \sqbrk a_{m n}$

Let $\circ$ be commutative on $\struct {S, \circ}$.

Then:

Now let matrix addition on $\map {\mathcal M_S} {m, n}$ be commutative.

Then it follows trivially that $\circ$ is commutative on $\struct {S, \circ}$.