Zero Matrix is Identity for Hadamard Product
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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:
| \(\ds \mathbf A \circ \mathbf e\) | \(=\) | \(\ds \sqbrk a_{m n} \circ \sqbrk e_{m n}\) | Definition of $\mathbf A$ and $\mathbf e$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sqbrk {a \cdot e}_{m n}\) | Definition of Hadamard Product | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sqbrk a_{m n}\) | Definition of Identity Element | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \mathbf A \circ \mathbf e\) | \(=\) | \(\ds \mathbf A\) | Definition of Zero Matrix over General Monoid |
Similarly:
| \(\ds \mathbf e \circ \mathbf A\) | \(=\) | \(\ds \sqbrk e_{m n} \circ \sqbrk a_{m n}\) | Definition of $\mathbf A$ and $\mathbf e$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sqbrk {e \cdot a}_{m n}\) | Definition of Hadamard Product | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sqbrk e_{m n}\) | Definition of Identity Element | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \mathbf e \circ \mathbf A\) | \(=\) | \(\ds \mathbf A\) | Definition of Zero Matrix over General Monoid |
$\blacksquare$