Closure 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 closed on $\map {\MM_S} {m, n}$ $\cdot$ is closed on $\struct {S, \cdot}$.

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

Let $\mathbf A = \sqbrk a_{m n}$ and $\mathbf B = \sqbrk b_{m n}$ be elements of $\map {\MM_S} {m, n}$.

Let $\sqbrk c_{m n} = \sqbrk a_{m n} \cdot \sqbrk b_{m n}$.

Then:
 * $\forall i \in \closedint 1 m, j \in \closedint 1 n: c_{i j} = a_{i j} \circ b_{i j}$

Thus:
 * $\struct {S, \cdot}$ is closed $c_{i j} \in S$.

From the definition of Hadamard product, $\sqbrk c_{m n}$ has the same order as both $\sqbrk a_{m n}$ and $\sqbrk b_{m n}$.

Thus it follows that:
 * $\sqbrk c_{m n} \in \map {\MM_S} {m, n}$

Thus $\struct {\map {\MM_S} {m, n}, \circ}$, as it is defined, is closed.

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

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

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:
 * $a_{i j} \cdot b_{i j} \notin S$

That is:
 * $\mathbf A \circ \mathbf B \notin \map {\MM_S} {m, n}$

because (at least) one element of $\mathbf A \circ \mathbf B$ is not an element of $S$.

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