Question

Show that $\mathit{M}\mathbf{\left(}\mathbf{R}\mathbf{\right)}$ is a vector space over $\mathit{R}$.

Short answer

It is shown that, $M\left(R\right)$ is a vector space over $R$.

Step-by-step solution

Definition of a vector space

A space consisting of vectors, together with the associative and commutative operations of addition of vectors and the associative and distributive operation of multiplication of vectors by scalars.

Showing that  x(v1+v2)=xv1+xv2

let the additive Abelian group $M\left(R\right)$ set of all $2\times 2$ matrices over the field $R$ of real numbers.

Let role="math" localid="1654750517501" $v_1=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right),v_2=\left(\begin{array}{cc}e& f\\ g& h\end{array}\right)\in M\left(R\right)$

And let $x,y\in R$

Then,

role="math" localid="1654750697288" $$\begin{gathered} x(v_1+v_2)=x\left(\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)+\left(\begin{array}{cc}e& f\\ g& h\end{array}\right)\right) \\ =\left(x\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)+x\left(\begin{array}{cc}e& f\\ g& h\end{array}\right)\right) \\ =x{v}_1+x{v}_2 \end{gathered}$$

So,$x(v_1+v_2)=x{v}_1+x{v}_2$

Showing that x(yv1)=(xy)v1

Now, find the product of $xy$ and $v_1$.

$$\begin{gathered} xy\left(v_1\right)=xy\left(\begin{array}{cc}a& b\\ c& d\end{array}\right) \\ =\left(\begin{array}{cc}xya& xyb\\ xyc& xyd\end{array}\right) \end{gathered}$$

As $x,y\in R$ so,$xy=yx$ . Thus

role="math" localid="1654751171682" $$\begin{gathered} x\left(y{v}_1\right)=x\left(\begin{array}{cc}ya& yb\\ yc& yd\end{array}\right) \\ \begin{array}{c}\\ =xy\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\\ =xy\left(v_1\right)\end{array} \end{gathered}$$

Hence $x\left(y{v}_1\right)=\left(xy\right)v_1$.

Showing that identity of R  is I  .

As the given field is a real number. Therefore, the identity of $R$ is $I$ . So,

$\begin{array}{c}I{v}_1=I\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\\ =\left(\begin{array}{cc}Ia& Ib\\ Ic& Id\end{array}\right)\\ =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\\ =v_1\end{array}$

Showing that M(R)   is a vector space over R .

Therefore, by the definition if $F$ is an additive abelian group role="math" localid="1654754091516" $V$ equipped with scaler multiplication such that for all $x,y\in F$and $v_1,v_2\in V$. So, hence prove that $M\left(R\right)$ is a vector space over$R$.