Exercise 7.A.1

Answers

Proof. We have

\begin{aligned} ⟨ (z_{1}, \dots, z_{n}), T^{∗} (x_{1}, \dots, x_{n}) ⟩ & = ⟨ T (z_{1}, \dots, z_{n}), (x_{1}, \dots, x_{n}) ⟩ \\ = ⟨ (0, z_{1}, \dots, z_{n - 1}), (x_{1}, \dots, x_{n}) ⟩ \\ = z_{1} \bar{x_{2}} + \dots + z_{n - 1} \bar{x_{n}} \\ = ⟨ (z_{1}, \dots, z_{n}), (x_{2}, \dots, x_{n}, 0) ⟩ . \end{aligned}

Thus $T^{∗} (x_{1}, \dots, x_{n}) = (x_{2}, \dots, x_{n}, 0)$ . □

celiopassos

2017-10-06 00:00