Commutativity of Hadamard Product
Theorem
Let $\struct {S, \cdot}$ be an algebraic structure.
Let $\map {\MM_S} {m, n}$ be a $m \times n$ matrix space over $S$.
For $\mathbf A, \mathbf B \in \map {\MM_S} {m, n}$, let $\mathbf A \circ \mathbf B$ be defined as the Hadamard product of $\mathbf A$ and $\mathbf B$.
The operation $\circ$ is commutative on $\map {\MM_S} {m, n}$ if and only if $\cdot$ is commutative on $\struct {S, \cdot}$.
Proof
Necessary Condition
Let the operation $\cdot$ be commutative on $\struct {S, \cdot}$.
Let $\mathbf A = \sqbrk a_{m n}$ and $\mathbf B = \sqbrk b_{m n}$ be elements of the $m \times n$ matrix space over $S$.
Then:
\(\ds \mathbf A \circ \mathbf B\) | \(=\) | \(\ds \sqbrk a_{m n} \circ \sqbrk b_{m n}\) | Definition of $\mathbf A$ and $\mathbf B$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \sqbrk {a \cdot b}_{m n}\) | Definition of Hadamard Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \sqbrk {b \cdot a}_{m n}\) | as $\cdot$ is commutative | |||||||||||
\(\ds \) | \(=\) | \(\ds \sqbrk b_{m n} \circ \sqbrk a_{m n}\) | Definition of Hadamard Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \mathbf B \circ \mathbf A\) | Definition of $\mathbf A$ and $\mathbf B$ |
That is, $\circ$ is commutative on $\map {\MM_S} {m, n}$.
$\Box$
Sufficient Condition
Suppose $\struct {S, \cdot}$ is such that $\cdot$ is not commutative.
Then there exists $a$ and $b$ such that:
- $a \cdot b \ne b \cdot a$
Let $\mathbf A$ and $\mathbf B$ be elements of $\map {\MM_S} {m, n}$ such that:
- $a_{i j} = a$, $b_{i j} = b$
where:
- $a_{i j}$ is the $\tuple {i, j}$th element of $\mathbf A$
- $b_{i j}$ is the $\tuple {i, j}$th element of $\mathbf B$
Then:
- $a_{i j} \cdot b_{i j} \ne b_{i j} \cdot a_{i j}$
That is:
- $\mathbf A \circ \mathbf B \ne \mathbf B \circ \mathbf A$
because they differ (at least) at indices $\tuple {i, j}$.
That is, $\circ$ is not commutative on $\map {\MM_S} {m, n}$.
$\blacksquare$