Exercise 2.5.7

We may let $\beta$ be the standard basis and $\alpha =\{(1,m),(-m,1)\}$ be another basis for ${ℝ}^2$.

1.

We have that ${[T]}_{\alpha }=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -1\hfill \end{array}\right)$ and $Q^{-1}={[I]}_{\alpha }^{\beta }=\left(\begin{array}{cc}\hfill 1\hfill & \hfill m\hfill \\ \hfill -m\hfill & \hfill 1\hfill \end{array}\right)$. We also can calculate that $Q={[I]}_{\beta }^{\alpha }=\left(\begin{array}{cc}\hfill \frac{1}{m^2+1}\hfill & \hfill \frac{m}{m^2+1}\hfill \\ \hfill -\frac{m}{m^2+1}\hfill & \hfill \frac{1}{m^2+1}\hfill \end{array}\right)$. So finally we get

$${[T]}_{\beta }=Q^{-1}{[T]}_{\alpha }Q=\left(\begin{array}{cc}\hfill \frac{1-m^2}{m^2+1}\hfill & \hfill \frac{2m}{m^2+1}\hfill \\ \hfill \frac{2m}{m^2+1}\hfill & \hfill \frac{m^2-1}{m^2+1}\hfill \end{array}\right).$$

That is, $T(x,y)=(\frac{x+2\mathit{ym}-x{m}^2}{m^2+1},\frac{-y+2\mathit{xm}+y{m}^2}{m^2+1})$.

2.

Similarly we have that ${[T]}_{\alpha }=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill \end{array}\right)$. And with the same $Q$ and $Q^{-1}$ we get

$${[T]}_{\beta }=Q^{-1}{[T]}_{\alpha }Q=\left(\begin{array}{cc}\hfill \frac{1}{m^2+1}\hfill & \hfill \frac{m}{m^2+1}\hfill \\ \hfill \frac{m}{m^2+1}\hfill & \hfill \frac{m^2}{m^2+1}\hfill \end{array}\right).$$

That is, $T(x,y)=(\frac{x+\mathit{ym}}{m^2+1},\frac{\mathit{xm}+y{m}^2}{m^2+1})$.