# Proof

The proof is very easy but long since it only requires the verification of the rules for the matrices $M_{m\times n}(F)$ over a field $F$ component by component. We will use two different symbols to distinguish between the addition of the components "$+$" in the field $F$ and the addition of matrices $\oplus$ in $M_{m\times n}(F)$.

### Ad $(1)$

• Let any three matrices $A,B,C\in M_{m\times n}(F)$ be given.
• Let $A=(a_{ij})$, $A=(b_{ij})$ and $C=(c_{ij})$, $i=1,\ldots,m$, $j=1,\ldots,n$ be the components of $A,B,C$.
• Since $(F,+)$ is itself an Abelian group, we have $$a_{ij}+(b_{ij}+c_{ij})=(a_{ij}+(b_{ij})+c_{ij}$$ for all $i=1,\ldots,m$, $j=1,\ldots,n$.
• Thus, by the definition of matrix addition, it follows $$A\oplus(B\oplus C)=(A\oplus B)\oplus C.$$

### Ad $(2)$

• Let any two matrices $A,B\in M_{m\times n}(F)$ be given.
• Let $A=(a_{ij})$, $A=(b_{ij})$, $i=1,\ldots,m$, $j=1,\ldots,n$ be the components of $A,B$.
• Since $(F,+)$ is itself an Abelian group, we have $$a_{ij}+b_{ij}=b_{ij}+a_{ij}$$ for all $i=1,\ldots,m$, $j=1,\ldots,n$.
• Thus, by the definition of matrix addition, it follows $A\oplus B=B\oplus A.$

### Ad $(3)$

• Let $A,O\in M_{m\times n}(F)$ be given, where $O$ is the zero matrix, with the components $A=(a_{ij})$, $O=(b_{ij})$, b_{ij}=0, $i=1,\ldots,m$, $j=1,\ldots,n$ where $0\in F.$
• It follows $a_{ij}+0=0+a_{ij}=a_{ij}$ for all $i=1,\ldots,m$, $j=1,\ldots,n$.
• Thus, $A\oplus O=O\oplus A=A.$

### Ad $(4)$

• Let $A\in M_{m\times n}(F)$ be given with the components $A=(a_{ij})$, $i=1,\ldots,m$, $j=1,\ldots,n.$
• In $(F,+)$ the element $-a_{ij}$ is the inverse element of the addition $+$.
• Thus, $a_{ij}+(-a_{ij})=-a_{ij}+a_{ij}=0$ for all $i=1,\ldots,m$, $j=1,\ldots,n$.
• By setting $-A$ as the matrix with the components $(-a_{ij})$, $i=1,\ldots,m$, $j=1,\ldots,n,$ we get $$A\oplus -A=-A\oplus A=O$$ with $O$ being the zero matrix in $M_{m\times n}(F)$.

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