Exercise 6.5.25

Answers

By the proof of Theorem 6.23 we know that the matrix representations of $T$ and $U$ with respect to the standard basis $α$ are

{[T]}_{α} = (\begin{matrix} \cos 2 ϕ & \sin 2 ϕ \\ \sin 2 ϕ & - \cos 2 ϕ \end{matrix})

and

{[U]}_{α} = (\begin{matrix} \cos 2 ψ & \sin 2 ψ \\ \sin 2 ψ & - \cos 2 ψ \end{matrix}) .

So we have

{[UT]}_{α} = {[U]}_{α} {[T]}_{α} = (\begin{matrix} \cos 2 (ψ - ϕ) & - \sin 2 (ψ - ϕ) \\ \sin 2 (ψ - ϕ) & \cos 2 (ψ - ϕ) \end{matrix}) .

Hence $UT$ is a rotation by the angle $2 (ψ - ϕ)$ .

jephianlin

2011-06-27 00:00