Exercise 2.1.4

It’s a linear transformation. And $N(T)=\left\{\left(\begin{array}{ccc}\hfill a_{11}\hfill & \hfill 2a_{11}\hfill & \hfill -4a_{11}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill & \hfill a_{23}\hfill \end{array}\right)\right\}$ with basis

$R(T)=\left\{\left(\begin{array}{cc}\hfill s\hfill & \hfill t\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right)\right\}$ with basis $\left\{\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right),\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right)\right\}$.