Exercise 2.5.6

Let $\alpha$ be the standard basis (of ${𝔽}^2$ or ${𝔽}^3$). We have that $A={[L_A]}_{\alpha }$ and hence ${[L_A]}_{\beta }={[I]}_{\alpha }^{\beta }{[L_A]}_{\alpha }{[I]}_{\beta }^{\alpha }$. So now we can calculate ${[L_A]}_{\beta }$ and $Q={[I]}_{\beta }^{\alpha }$ and $Q^{-1}={[I]}_{\alpha }^{\beta }$.

1.

${[L_A]}_{\beta }=\left(\begin{array}{cc}\hfill 6\hfill & \hfill 11\hfill \\ \hfill -2\hfill & \hfill -4\hfill \end{array}\right)$ and $Q=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right)$.

2.

${[L_A]}_{\beta }=\left(\begin{array}{cc}\hfill 3\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -1\hfill \end{array}\right)$ and $Q=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -1\hfill \end{array}\right)$.

3.

${[L_A]}_{\beta }=\left(\begin{array}{ccc}\hfill 2\hfill & \hfill 2\hfill & \hfill 2\hfill \\ \hfill -2\hfill & \hfill -3\hfill & \hfill -4\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right)$ and $Q=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 2\hfill \end{array}\right)$.

4.

${[L_A]}_{\beta }=\left(\begin{array}{ccc}\hfill 6\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 12\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 18\hfill \end{array}\right)$ and $Q=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill -2\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)$.