Problem 51: On $M_2(\R)$, define the operation of addition by $A + B = AB$, and use the usual scalar multiplication operation. Determine which axioms for a vector space are satisfied by$M_2(\R)$ with the above operations

Problem 51: On $M_2(\R)$, define the operation of addition by $A + B = AB$, and use the usual scalar multiplication operation. Determine which axioms for a vector space are satisfied by$M_2(\R)$ with the above operations

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+B = AB \in M_2(\R) \end{align*}

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

Prove A3:
Let $A,B \in M_2(\R)$
\begin{align*} A+B = AB \ne BA = B+A \end{align*} A3 not holds.

Prove A4:
Let $A,B,C \in M_2(\R)$
\begin{align*} (A+B)+C = (AB)+C\\ = (AB)C \\ = ABC \\ = A(BC) \\ = A(B+C) \\ = A+(B+C). \end{align*} A4 holds

Prove A5:
Let $A\in M_2(\R)$ and $I \in M_2(\R)$
\begin{align*} A+I = AI\\ = A\\ \end{align*}

Prove A6:
We wish to determine whether for each matrix $A\in M_2(\R)$ we can find a matrix $B\in M_2(\R)$
such that $A+B = I$, remember that we have shown in A5 that the zero vector is $I$,
equivalently, such that $AB = I$.
However, this equation can be satisfied only if $A$ is nonsingular, therefore the axiom fails. A6 not holds.

Prove A7:
Let $A \in M_2(\R)$ and $r,s \in \R$
\begin{align*} (rs)A = r(sA) \end{align*}

Prove A8:
Let $A,B \in M_2(\R)$ and $r \in \R$
\begin{align*} r(A+B) = rAB\\ rA+rB = (rA)(rB) = r^2AB \end{align*} Thus $r(A+B) \ne rA+rB$.
A8 not holds.

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

Prove A10:
Let $A\in M_2(\R)$ and $1 \in \R$
\begin{align*} 1\cdot A =A \in M_2(\R) \end{align*} A10 holds