Definition:Matrix Entrywise Addition

Theorem
Let $$\mathcal {M}_{S} \left({m, n}\right)$$ be a $m \times n$ matrix space over $$S$$ over an algebraic structure $$\left({S, \circ}\right)$$.

Let $$\mathbf{A}, \mathbf{B} \in \mathcal {M}_{S} \left({m, n}\right)$$.

Then the sum of $$\mathbf{A}$$ and $$\mathbf{B}$$ is written $$\mathbf{A} + \mathbf{B}$$, and is defined as follows.

Let $$\mathbf{A} + \mathbf{B} = \mathbf{C} = \left[{c}\right]_{m n}$$.

Then $$\forall i \in \left[{1 \,. \, . \, m}\right], j \in \left[{1 \,. \, . \, n}\right]: c_{i j} = a_{i j} \circ b_{i j}$$.

Thus $$\left[{c}\right]_{m n}$$ is the $$m \times n$$ matrix whose elements are made by performing the operation $$\circ$$ on corresponding elements of $$\mathbf{A}$$ and $$\mathbf{B}$$.

This operation is called matrix addition.

It follows that matrix addition is defined only when both matrices have the same number of rows and the same number of columns.

The following results can be determined:


 * $$+$$ is closed on $$\mathcal {M}_{S} \left({m, n}\right)$$ iff $$\circ$$ is closed on $$\left({S, \circ}\right)$$.
 * $$+$$ is associative on $$\mathcal {M}_{S} \left({m, n}\right)$$ iff $$\circ$$ is associative on $$\left({S, \circ}\right)$$.
 * $$+$$ is commutative on $$\mathcal {M}_{S} \left({m, n}\right)$$ iff $$\circ$$ is commutative on $$\left({S, \circ}\right)$$.

Closure

 * Let $$\left[{a}\right]_{m n}, \left[{b}\right]_{m n}$$ be elements of $$\mathcal {M}_{S} \left({m, n}\right)$$.

Let $$\left[{c}\right]_{m n} = \left[{a}\right]_{m n} + \left[{b}\right]_{m n}$$.

Then $$\forall i \in \left[{1 \,. \, . \, m}\right], j \in \left[{1 \,. \, . \, n}\right]: c_{i j} = a_{i j} \circ b_{i j}$$.

Thus $$\left({S, \circ}\right)$$ is closed iff $$c_{i j} \in S$$.

As $$\left[{c}\right]_{m n}$$, from the definition of matrix addition, has the same dimensions as both $$\left[{a}\right]_{m n}$$ and $$\left[{b}\right]_{m n}$$, it follows that $$\left[{c}\right]_{m n} \in \mathcal {M}_{S} \left({m, n}\right)$$, and thus $$\left({\mathcal {M}_{S} \left({m, n}\right), +}\right)$$ as it is defined is closed.

Hence the result.

The argument reverses.

Associativity
Let $$\left[{a}\right]_{m n}, \left[{b}\right]_{m n}, \left[{c}\right]_{m n}$$ be elements of $$\mathcal {M}_{S} \left({m, n}\right)$$.

Then let:


 * $$\left[{p}\right]_{m n} = \left({\left[{a}\right]_{m n} + \left[{b}\right]_{m n}}\right) + \left[{c}\right]_{m n}$$;
 * $$\left[{q}\right]_{m n} = \left[{a}\right]_{m n} + \left({\left[{b}\right]_{m n} + \left[{c}\right]_{m n}}\right)$$.

Let $$\circ$$ be associative on $$\left({S, \circ}\right)$$.

$$ $$ $$


 * Now let matrix addition on $$\mathcal {M}_{S} \left({m, n}\right)$$ be associative.

Then it follows trivially that $$\circ$$ is associative on $$\left({S, \circ}\right)$$.

Commutativity
Let $$\left[{a}\right]_{m n}, \left[{b}\right]_{m n}$$ be elements of $$\mathcal {M}_{S} \left({m, n}\right)$$.

Let:
 * $$\left[{c}\right]_{m n} = \left[{a}\right]_{m n} + \left[{b}\right]_{m n}$$.
 * $$\left[{d}\right]_{m n} = \left[{b}\right]_{m n} + \left[{a}\right]_{m n}$$.

Let $$\circ$$ be commutative on $$\left({S, \circ}\right)$$.

$$ $$ $$


 * Now let matrix addition on $$\mathcal {M}_{S} \left({m, n}\right)$$ be commutative.

Then it follows trivially that $$\circ$$ is commutative on $$\left({S, \circ}\right)$$.