Matrix Space Semigroup under Hadamard Product

Theorem
Let $$\mathcal {M}_{S} \left({m, n}\right)$$ be the matrix space over a semigroup $$\left({S, \circ}\right)$$.

Then the algebraic structure $$\left({\mathcal {M}_{S} \left({m, n}\right), +}\right)$$, where $$+$$ is matrix addition, is also a semigroup.

If $$\left({S, \circ}\right)$$ is a commutative semigroup then so is $$\left({\mathcal {M}_{S} \left({m, n}\right), +}\right)$$.

If $$\left({S, \circ}\right)$$ is a monoid then so is $$\left({\mathcal {M}_{S} \left({m, n}\right), +}\right)$$.

Proof

 * $$\left({S, \circ}\right)$$ is a semigroup and is therefore closed and associative.


 * As $$\left({S, \circ}\right)$$ is closed, then so is $$\left({\mathcal {M}_{S} \left({m, n}\right), +}\right)$$ from the definition of matrix addition.


 * As $$\left({S, \circ}\right)$$ is associative, then so is $$\left({\mathcal {M}_{S} \left({m, n}\right), +}\right)$$ from the definition of matrix addition.

Thus if $$\left({S, \circ}\right)$$ is a semigroup then so is $$\left({\mathcal {M}_{S} \left({m, n}\right), +}\right)$$.


 * If $$\left({S, \circ}\right)$$ is commutative, then so is $$\left({\mathcal {M}_{S} \left({m, n}\right), +}\right)$$ from the definition of matrix addition.

Thus if $$\left({S, \circ}\right)$$ is a commutative semigroup then so is $$\left({\mathcal {M}_{S} \left({m, n}\right), +}\right)$$.


 * Let $$\left({S, \circ}\right)$$ be a monoid, with identity $$e$$.

Then from zero matrix, $$\left({\mathcal {M}_{S} \left({m, n}\right), +}\right)$$ also has an identity and is therefore also a monoid.