# Matrix Entrywise Addition is Commutative

## Theorem

Let $\map \MM {m, n}$ be a $m \times n$ matrix space over one of the standard number systems.

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

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

That is:

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

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

## Proof 1

From:

Integers form Ring
Rational Numbers form Ring
Real Numbers form Ring
Complex Numbers form Ring

the standard number systems $\Z$, $\Q$, $\R$ and $\C$ are rings.

Hence we can apply Matrix Entrywise Addition over Ring is Commutative.

$\Box$

The above cannot be applied to the natural numbers $\N$, as they do not form a ring.

However, from Natural Numbers under Addition form Commutative Monoid, the algebraic structure $\struct {\N, +}$ is a commutative monoid.

The result follows from Commutativity of Hadamard Product.

$\blacksquare$

## Proof 2

Let $\mathbf A = \sqbrk a_{m n}$ and $\mathbf B = \sqbrk b_{m n}$ be matrices whose order is $m \times n$.

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 $\ds$ $=$ $\ds \sqbrk {b + a}_{m n}$ Commutative Law of Addition $\ds$ $=$ $\ds \sqbrk b_{m n} + \sqbrk a_{m n}$ Definition of Matrix Entrywise Addition $\ds$ $=$ $\ds \mathbf B + \mathbf A$ Definition of $\mathbf A$ and $\mathbf B$

$\blacksquare$