Exercise 6.11.5

Let $\alpha =\left\{e_1,e_2\right\}$ be the standard basis in ${ℝ}^2$.

1.

We may check that the rotation $T_{\varphi }$ is a linear transformation. Hence it’s enough to know

$$\{\begin{array}{c}T(e_1)=(\cos <!--\; FUNCTION\; APPLICATION\; -->\varphi ,\sin <!--\; FUNCTION\; APPLICATION\; -->\varphi ),\hfill \\ T(e_2)=(-\sin <!--\; FUNCTION\; APPLICATION\; -->\varphi ,\cos <!--\; FUNCTION\; APPLICATION\; -->\varphi )\hfill \end{array}$$

by directly rotate these two vectors. Hence we have ${[T_{\varphi }]}_{\alpha }=A$.

2.

Denote that

$$A_{\varphi }=\left(\begin{array}{cc}\hfill \cos <!--\; FUNCTION\; APPLICATION\; -->\varphi \hfill & \hfill -\sin <!--\; FUNCTION\; APPLICATION\; -->\varphi \hfill \\ \hfill \sin <!--\; FUNCTION\; APPLICATION\; -->\varphi \hfill & \hfill \cos <!--\; FUNCTION\; APPLICATION\; -->\varphi \hfill \end{array}\right).$$

Directly compute that $A_{\varphi }{A}_{\psi }=A_{\varphi +\psi }$. So we have

$${[T_{\varphi }{T}_{\psi }]}_{\alpha }={[T_{\varphi }]}_{\alpha }{[T_{\psi }]}_{\alpha }=A_{\varphi }{A}_{\psi }=A_{\varphi +\psi }={[T_{\varphi +\psi }]}_{\alpha }.$$

3.

By the previous argument we kow that

$$T_{\varphi }{T}_{\psi }=T_{\varphi +\psi }=T_{\psi +\varphi }=T_{\psi }{T}_{\varphi }.$$