Matrix Entrywise Addition is Commutative

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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.

By definition, matrix entrywise addition is the Hadamard product with respect to addition of numbers.

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 $\mathbf B$

$\blacksquare$


Also see


Sources

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