# Matrix Entrywise Addition over Ring is Commutative

## 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.

$\blacksquare$