Negative Matrix is Inverse for Hadamard Product

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Theorem

Let $\struct {G, \cdot}$ be a group whose identity is $e$.

Let $\map {\MM_G} {m, n}$ be a $m \times n$ matrix space over $\struct {G, \cdot}$.

Let $\mathbf A$ be an element of $\map {\MM_G} {m, n}$.

Let $-\mathbf A$ be the negative of $\mathbf A$.


Then $-\mathbf A$ is the inverse for the operation $\circ$, where $\circ$ is the Hadamard product.


Proof

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

Then:

\(\ds \mathbf A \circ \paren {-\mathbf A}\) \(=\) \(\ds \sqbrk a_{m n} \circ \paren {-\sqbrk a_{m n} }\) Definition of $\mathbf A$
\(\ds \) \(=\) \(\ds \sqbrk a_{m n} \circ \sqbrk {a^{-1} }_{m n}\) Definition of Negative Matrix
\(\ds \) \(=\) \(\ds \sqbrk {a \cdot \paren {a^{-1} } }_{m n}\) Definition of Hadamard Product
\(\ds \) \(=\) \(\ds \sqbrk e_{m n}\) Definition of Inverse Element
\(\ds \leadsto \ \ \) \(\ds \mathbf A \circ \paren {-\mathbf A}\) \(=\) \(\ds \mathbf e\) Definition of Zero Matrix over General Monoid

The result follows from Zero Matrix is Identity for Hadamard Product.

$\blacksquare$


Also see