Exercise 6.9.4

We know that $B_v^{\text{∗}}A{B}_v={[T_v^{\text{∗}}{L}_A{T}_v]}_{\beta }$ and ${[L_A]}_{\beta }$. So (a) and (b) in the Corollary is equivalent. We only prove (a) by the steps given by Hints. For brevity, we write $U=T_v^{\text{∗}}{L}_A{T}_v$ and $C=B_v^{\text{∗}}A{B}_v$.

1.

We have $U(e_i)=e_i$ for $i=2,3$ by Theorem 6.39. By Theorem 6.41 we have

$$\{\begin{array}{c}\hfill U(e_1)+U(e_4)=U(e_1+e_4)=U(w_1)=a{w}_2=a{e}_1-a{e}_4,\hfill \\ \hfill U(e_1)-U(e_4)=U(e_1-e_4)=U(w_2)=b{w}_1=b{e}_1+b{e}_2.\hfill \end{array}$$

Solving this system of equations we get that

$$\{\begin{array}{c}\hfill U(e_1)=p{e}_1-q{e}_4,\hfill \\ \hfill U(e_4)=q{e}_1-p{e}_2,\hfill \end{array}$$

where $p=\frac{a+b}{2}$ and $q=\frac{a-b}{2}$. Write down the matrix representation of $U$ and get the result.

2.

Since $C$ is self-adjoint, we know that $q=-q$ and so $q=0$.

3.

Let $w=e_2+e_4$. Then we know that $⟨L_A(w),w⟩=0$. Now we calculate that

$$U(w)=U(e_2+e_4)=e_2-p{e}_4.$$

By Theorem 6.40 we know that

$$⟨U(w),w⟩=⟨e_2-p{e}_4,e_2+e_4⟩=1-p=0.$$

Hence we must have $p=0$.