Zero Matrix is Identity for Hadamard Product

Theorem
Let $\struct {S, \cdot}$ be a monoid whose identity is $e$.

Let $\map {\MM_S} {m, n}$ be an $m \times n$ matrix space over $S$.

Let $\mathbf e = \sqbrk e_{m n}$ be the zero matrix of $\map {\MM_S} {m, n}$

Then $\mathbf e$ is the identity element for Hadamard product.

Proof
Let $\mathbf A = \sqbrk a_{m n} \in \map {\MM_S} {m, n}$.

Then:

Mutatis mutandis, the same proof can be used to show that $\mathbf e \circ \mathbf A = \mathbf A$.

Also see

 * Zero Matrix is Identity for Matrix Entrywise Addition