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$(\mathcal{M}_{m\times n}(\mathbb{K}),+)$ is an abelian group. That is, the sum of matrices verifies the following properties

  1. Associativity: $(A+B)+C=A+(B+C)\qquad\forall\,A,B,C\in \mathcal{M}_{m\times n}(\mathbb{K})$
  2. Commutativity: $A+B=B+A\qquad\forall\,A,B\in \mathcal{M}_{m\times n}(\mathbb{K})$
  3. Identity Element: $\exists\, 0\in\mathcal{M}_{m\times n}(\mathbb{K})\,|\,A+0=0+A=A\qquad\forall\,A\in \mathcal{M}_{m\times n}(\mathbb{K})$
  4. Inverse Element: $\forall\,A\in\mathcal{M}_{m\times n}(\mathbb{K})\quad\exists\,-A\in\mathcal{M}_{m\times n}(\mathbb{K})\,|\,A+(-A)=(-A)+A=0$

Let $A=(a_{ij})_{ij}$, $B=(b_{ij})_{ij}$ and $C=(c_{ij})_{ij}$ be any three matrices in $\mathcal{M}_{m\times n}(\mathbb{K})$
  • Associativity: $$ \begin{split} (A+B)+C & = [(a_{ij})_{ij}+(b_{ij})_{ij}]+(c_{ij})_{ij} \\ & = (a_{ij}+b_{ij})_{ij}+(c_{ij})_{ij} \\ & = ([a_{ij}+b_{ij}]+c_{ij})_{ij} \\ & = (a_{ij}+[b_{ij}+c_{ij}])_{ij} \\ & = (a_{ij})_{ij}+(b_{ij}+c_{ij})_{ij} \\ & = (a_{ij})_{ij}+[(b_{ij})_{ij}+(c_{ij})_{ij}] \\ & = A+(B+C) \end{split} $$
  • Commutativity: $$ \begin{split} A+B & = (a_{ij})_{ij}+(b_{ij})_{ij} \\ & = (a_{ij}+b_{ij})_{ij} \\ & = (b_{ij}+a_{ij})_{ij} \\ & = (b_{ij})_{ij}+(a_{ij})_{ij} \\ & = B+A \end{split} $$
  • Identity Element: The identity element suitable for this sum is the matrix filled up with zeros, that is $$0 = (0)_{ij} = \left(\begin{array}{cccc} 0 & 0 & \cdots & 0\\ 0 & 0 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0\\ \end{array}\right)$$ and it does indeed meet the requirements of identity element: $$ \begin{split} A+0 & = (a_{ij})_{ij}+(0)_{ij} \\ & = (a_{ij}+0)_{ij} \\ & = (a_{ij})_{ij} \\ & = A \end{split} $$ $$ \begin{split} 0+A & = (0)_{ij}+(a_{ij})_{ij} \\ & = (0+a_{ij})_{ij} \\ & = (a_{ij})_{ij} \\ & = A \end{split} $$
  • Opposite Element: Each matrix $A$ has an opposite matrix - namely $-A=(-a_{ij})_{ij}$, that is, the matrix whose elements are exactly the opposite of those of $A$. Let's check:

    $$ \begin{split} A+(-A) & = (a_{ij})_{ij}+(-a_{ij})_{ij} \\ & = (a_{ij}+(-a_{ij}))_{ij} \\ & = (0)_{ij} \\ & = 0 \end{split} $$ $$ \begin{split} (-A)+A & = (-a_{ij})_{ij}+(a_{ij})_{ij} \\ & = ((-a_{ij})+a_{ij})_{ij} \\ & = (0)_{ij} \\ & = 0 \end{split} $$