Matrix Space Semigroup under Hadamard Product

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Theorem

Let $\map {\MM_S} {m, n}$ be the matrix space over a semigroup $\struct {S, \cdot}$.

Then the algebraic structure $\struct {\map {\MM_S} {m, n}, \circ}$, where $\circ$ is the Hadamard product, is also a semigroup.

If $\struct {S, \cdot}$ is a commutative semigroup then so is $\struct {\map {\MM_S} {m, n}, \circ}$.

If $\struct {S, \cdot}$ is a monoid then so is $\struct {\map {\MM_S} {m, n}, \circ}$.

Proof

$\struct {S, \cdot}$ is a semigroup and is therefore closed and associative.

As $\struct {S, \cdot}$ is closed, then so is $\struct {\map {\MM_S} {m, n}, \circ}$ from Closure of Hadamard Product‎.

As $\struct {S, \cdot}$ is associative, then so is $\struct {\map {\MM_S} {m, n}, \circ}$ from Associativity of Hadamard Product.

Thus if $\struct {S, \cdot}$ is a semigroup then so is $\struct {\map {\MM_S} {m, n}, \circ}$.

If $\struct {S, \cdot}$ is commutative, then so is $\struct {\map {\MM_S} {m, n}, \circ}$ from Commutativity of Hadamard Product.

Thus if $\struct {S, \cdot}$ is a commutative semigroup then so is $\struct {\map {\MM_S} {m, n}, \circ}$.

Let $\struct {S, \cdot}$ be a monoid, with identity $e$.

Then from Zero Matrix is Identity for Hadamard Product, $\struct {\map {\MM_S} {m, n}, \circ}$ also has an identity and is therefore also a monoid.

$\blacksquare$