Commutativity of Hadamard Product

Theorem
Let $\struct {S, \cdot}$ be an algebraic structure.

Let $\map {\MM_S} {m, n}$ be a $m \times n$ matrix space over $S$.

For $\mathbf A, \mathbf B \in \map {\MM_S} {m, n}$, let $\mathbf A \circ \mathbf B$ be defined as the Hadamard product of $\mathbf A$ and $\mathbf B$.

The operation $\circ$ is commutative on $\map {\MM_S} {m, n}$ $\cdot$ is commutative on $\struct {S, \cdot}$.

Necessary Condition
Let the operation $\cdot$ be commutative on $\struct {S, \cdot}$.

Let $\mathbf A = \sqbrk a_{m n}$ and $\mathbf B = \sqbrk b_{m n}$ be elements of the $m \times n$ matrix space over $S$.

Then:

That is, $\circ$ is commutative on $\map {\MM_S} {m, n}$.

Sufficient Condition
Suppose $\struct {S, \cdot}$ is such that $\cdot$ is not commutative.

Then there exists $a$ and $b$ such that:
 * $a \cdot b \ne b \cdot a$

Let $\mathbf A$ and $\mathbf B$ be elements of $\map {\MM_S} {m, n}$ such that:
 * $a_{i j} = a$, $b_{i j} = b$

where:
 * $a_{i j}$ is the $\tuple {i, j}$th element of $\mathbf A$
 * $b_{i j}$ is the $\tuple {i, j}$th element of $\mathbf B$

Then:
 * $a_{i j} \cdot b_{i j} \ne b_{i j} \cdot a_{i j}$

That is:
 * $\mathbf A \circ \mathbf B \ne \mathbf B \circ \mathbf A$

because they differ (at least) at indices $\tuple {i, j}$.

That is, $\circ$ is not commutative on $\map {\MM_S} {m, n}$.

Also see

 * Closure of Hadamard Product
 * Associativity of Hadamard Product