Question

Show that the homomorphism$\mathit{g}$in Example 7 is injective but not surjective.

Short answer

It is proved that$g$ is injective but not surjective.

Step-by-step solution

Prove that g is not surjective

From example 7,$g$is defined as$g:R\to M\left(R\right)$given by $g\left(r\right)=\left(\begin{array}{cc}0& 0\\ -r& r\end{array}\right)$. It is proved thatlocalid="1654012180070" $g$is a homomorphism.

It is clearly observed that there does not exist a real number localid="1654012185303" $r$such that:

$\begin{array}{rcl}g\left(r\right)& =& \left(\begin{array}{cc}0& 0\\ -r& r\end{array}\right)\\ & =& \left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)\end{array}$

where, $\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)\in M\left(R\right)$

This implies thatlocalid="1654012190862" $g$ is not surjective.

Prove that g is injective

Consider $r\ne s$, then

$\begin{array}{rcl}f\left(r\right)& =& \left(\begin{array}{cc}0& 0\\ -r& r\end{array}\right)\\ & \ne & \left(\begin{array}{cc}0& 0\\ -s& s\end{array}\right)\\ & =& f\left(s\right)\end{array}$

Thus, is injective.