Closure 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 closed on $\map {\mathcal M_S} {m, n}$ $\circ$ is closed on $\struct {S, \circ}$.

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

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

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

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

From the definition of matrix entrywise addition, $\sqbrk c_{m n}$ has the same dimensions as both $\sqbrk a_{m n}$ and $\sqbrk b_{m n}$.

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

Thus $\left({\map {\mathcal M_S} {m, n}, +}\right)$, as it is defined, is closed.

Hence the result.

The argument reverses.