Negative Matrix is Inverse for Hadamard Product
Jump to navigation
Jump to search
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.
$\blacksquare$