Matrix Entrywise Addition over Ring is Commutative

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Theorem

Let $\struct {R, +, \circ}$ be a ring.

Let $\map {\MM_R} {m, n}$ be a $m \times n$ matrix space over $R$.

For $\mathbf A, \mathbf B \in \map {\MM_R} {m, n}$, let $\mathbf A + \mathbf B$ be defined as the matrix entrywise sum of $\mathbf A$ and $\mathbf B$.


The operation $+$ is commutative on $\map {\MM_R} {m, n}$.

That is:

$\mathbf A + \mathbf B = \mathbf B + \mathbf A$

for all $\mathbf A$ and $\mathbf B$ in $\map {\MM_R} {m, n}$.


Proof 1

Let $\mathbf A = \sqbrk a_{m n}$ and $\mathbf B = \sqbrk b_{m n}$ be elements of the $m \times n$ matrix space over $R$.

Then:

\(\ds \mathbf A + \mathbf B\) \(=\) \(\ds \sqbrk a_{m n} + \sqbrk b_{m n}\) Definition of $\mathbf A$ and $\mathbf B$
\(\ds \) \(=\) \(\ds \sqbrk {a + b}_{m n}\) Definition of Matrix Entrywise Addition over Ring
\(\ds \) \(=\) \(\ds \sqbrk {b + a}_{m n}\) Ring Axiom $\text A2$: Commutativity of Addition
\(\ds \) \(=\) \(\ds \sqbrk b_{m n} + \sqbrk a_{m n}\) Definition of Matrix Entrywise Addition over Ring
\(\ds \) \(=\) \(\ds \mathbf B + \mathbf A\) Definition of $\mathbf A$ and $\mathbf B$

$\blacksquare$


Proof 2

By definition, matrix entrywise addition is the Hadamard product of $\mathbf A$ and $\mathbf B$ with respect to ring addition.

We have from Ring Axiom $\text A2$: Commutativity of Addition that ring addition is commutative.

The result then follows directly from Commutativity of Hadamard Product.

$\blacksquare$


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