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Problem 52: On $M_2(\R)$, define the operations of addition and multiplication by a real number $\oplus$ and $\cdot$ , respectively as follows:
$A \oplus B = −(A + B)$,
$k\cdot A = −kA$,

Problem 52: On $M_2(\R)$, define the operations of addition and multiplication by a real number $\oplus$ and $\cdot$ , respectively as follows:
$A \oplus B = −(A + B)$,
$k\cdot A = −kA$,
where the operations on the right-hand sides of these equations are the usual ones associated with $M_2(\R)$. Determine which of the axioms for a vector space are satisfied by $M_2(\R)$ with the operations $\oplus$ and $\cdot$.

Solution:
The Axioms of a Vector Space
The following properties must hold for all $ u,v,w \in \V$ and $a,b \in \R$:
Closure Properties
$(A1) u+v \in \V$.
$(A2) av \in \V$.
Properties of Addition
$(A3) u+v=v+u$.
$(A4) u+(v+w)=(u+v)+w$.
$(A5)$ There is an element $\mathbf{0} \in \V$ such that $\mathbf{0}+v=v$ for all $v \in \V$.
$(A6)$ Given an element $v \in \V$, there is an element $−v \in \V$ such that $v+(−v)=\mathbf{0}$.
Properties of Scalar Multiplication
$(A7) a(bv)=(ab)v$.
$(A8) a(u+v)=au+av$.
$(A9) (a+b)v=av+bv$.
$(A10) 1v=v$ for all $v \in \V$.

Prove A1:
Let $A,B \in M_2(\R)$
\begin{align*} A\oplus B = −(A + B) \in M_2(\R) \end{align*} A1 holds.

Prove A2:
Let $r \in \R$ and $A \in M_2(\R)$
\begin{align*} r\cdot A = =rA\in M_2(\R) \end{align*} A2 holds.

Prove A3:
Let $A,B \in M_2(\R)$
\begin{align*} A\oplus B = -(A+B)\\ = -(B+A)\\ = B\oplus A \end{align*} A3 holds.

Prove A4:
Let $A,B,C \in M_2(\R)$
\begin{align*} (A\oplus B)\oplus C = -((A\oplus B)+C) = -((-(A+B)+C) = -(-A-B+C) = A+B-C A\oplus (B\oplus C) = -(A+(B\oplus C)) = -(A+(-(B+C))) = -(A-B-C) = -A+B+C \end{align*} $\therefore (A\oplus B)\oplus C \ne A\oplus (B\oplus C)$
A4 not holds

Prove A5:
An element $B$ is needed such that $A \oplus B = A$ for all $A\in M_2(\R$),
but $−(A + B) = A \Rightarrow B = −2A$.
Since this depends on $A$, there is no zero vector.
A5 not holds

Prove A6:
since there is no zero vectro, we cannot define the additive inverse.
A6 not holds.

Prove A7:
Let $A \in M_2(\R)$ and $r,s \in \R$
\begin{align*} (rs)\cdot A = −(rs)A\\ s\cdot (t \cdot A) = s\cdot (-tA) = -s((-tA)) = (st)A\\ \end{align*} so it follows that $(rs)\cdot A \ne s\cdot (t \cdot A)$
A7 not holds.

Prove A8:
Let $A,B \in M_2(\R)$ and $r \in \R$
\begin{align*} r\cdot(A\oplus B) = -r((A\oplus B)) = -r(-(A+B)) = r(A+B) = rA+ rB\\ r\cdot A \oplus r\cdot B = -(r\cdot A + r\cdot B) = -((-rA) + (-rB)) = rA-rB \end{align*} Thus $r\cdot(A\oplus B)) \ne r\cdot A \oplus r\cdot B$.
A8 not holds.

Prove A9:
Let $A\in M_2(\R)$ and $r,s \in \R$
\begin{align*} (r+s)\cdot A = -(r+s)A = -rA+(-sA)\\ r\cdot A\oplus s\cdot A = (-rA)\oplus(-sA) = -(-(rA)+(-sA))= rA+(-sA) \end{align*} Thus $(r+s)\cdot A \ne r\cdot A\oplus s\cdot A$.
A9 not holds.

Prove A10:
Let $A\in M_2(\R)$ and $1 \in \R$
\begin{align*} 1\cdot A =-1A = -A \ne A \end{align*} A10 not holds.

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