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