Matrix Entrywise Addition over Ring is Commutative/Proof 1
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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
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 $\mathbf B$ |
$\blacksquare$