# Matrix Entrywise Addition over Ring is Closed

## 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 closed on $\map {\MM_R} {m, n}$.

That is:

$\mathbf A + \mathbf B \in \map {\MM_R} {m, n}$

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 $\map {\MM_R} {m, n}$.

Let $\sqbrk c_{m n} = \sqbrk a_{m n} + \sqbrk b_{m n}$.

By definition of matrix entrywise addition:

$\forall i \in \closedint 1 m, j \in \closedint 1 n: a_{i j} + b_{i j} = c_{i j}$

By Ring Axiom $\text A0$: Closure under Addition, $R$ is closed under addition.

Hence:

$\forall i \in \closedint 1 m, j \in \closedint 1 n: c_{i j} \in R$

From the definition of matrix entrywise addition, $\sqbrk c_{m n}$ has the same order as both $\sqbrk a_{m n}$ and $\sqbrk b_{m n}$.

Thus it follows that:

$\sqbrk c_{m n} \in \map {\MM_R} {m, n}$

Thus $\struct {\map {\MM_R} {m, n}, +}$, as it is defined, is closed.

$\blacksquare$

## 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 $\text A0$: Closure under Addition that ring addition is closed.

The result then follows directly from Closure of Hadamard Product.

$\blacksquare$