Properties of Hadamard Product

Theorem
Let $\mathcal M_S \left({m, n}\right)$ be a $m \times n$ matrix space over $S$ over an algebraic structure $\left({S, \circ}\right)$.

Let $\mathbf A, \mathbf B \in \mathcal M_S \left({m, n}\right)$.

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

The operation of matrix entrywise addition satisfies the following properties.


 * $+$ is closed on $\mathcal M_S \left({m, n}\right)$ $\circ$ is closed on $\left({S, \circ}\right)$.
 * $+$ is associative on $\mathcal M_S \left({m, n}\right)$ $\circ$ is associative on $\left({S, \circ}\right)$.
 * $+$ is commutative on $\mathcal M_S \left({m, n}\right)$ $\circ$ is commutative on $\left({S, \circ}\right)$.

Closure
Let $\left[{a}\right]_{m n}, \left[{b}\right]_{m n}$ be elements of $\mathcal M_S \left({m, n}\right)$.

Let $\left[{c}\right]_{m n} = \left[{a}\right]_{m n} + \left[{b}\right]_{m n}$.

Then:
 * $\forall i \in \left[{1 \,.\,.\, m}\right], j \in \left[{1 \,.\,.\, n}\right]: c_{i j} = a_{i j} \circ b_{i j}$

Thus:
 * $\left({S, \circ}\right)$ is closed iff $c_{i j} \in S$.

As $\left[{c}\right]_{m n}$, from the definition of matrix addition, has the same dimensions as both $\left[{a}\right]_{m n}$ and $\left[{b}\right]_{m n}$, it follows that $\left[{c}\right]_{m n} \in \mathcal M_S \left({m, n}\right)$.

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

Hence the result.

The argument reverses.

Associativity
Let $\left[{a}\right]_{m n}, \left[{b}\right]_{m n}, \left[{c}\right]_{m n}$ be elements of $\mathcal M_S \left({m, n}\right)$.

Then let:


 * $\left[{p}\right]_{m n} = \left({\left[{a}\right]_{m n} + \left[{b}\right]_{m n}}\right) + \left[{c}\right]_{m n}$
 * $\left[{q}\right]_{m n} = \left[{a}\right]_{m n} + \left({\left[{b}\right]_{m n} + \left[{c}\right]_{m n}}\right)$

Let $\circ$ be associative on $\left({S, \circ}\right)$.

Now let matrix addition on $\mathcal M_S \left({m, n}\right)$ be associative.

Then it follows trivially that $\circ$ is associative on $\left({S, \circ}\right)$.

Commutativity
Let $\left[{a}\right]_{m n}, \left[{b}\right]_{m n}$ be elements of $\mathcal M_S \left({m, n}\right)$.

Let:
 * $\left[{c}\right]_{m n} = \left[{a}\right]_{m n} + \left[{b}\right]_{m n}$
 * $\left[{d}\right]_{m n} = \left[{b}\right]_{m n} + \left[{a}\right]_{m n}$

Let $\circ$ be commutative on $\left({S, \circ}\right)$.

Now let matrix addition on $\mathcal M_S \left({m, n}\right)$ be commutative.

Then it follows trivially that $\circ$ is commutative on $\left({S, \circ}\right)$.