Exercise 6.9.1

Answers

(b)

It comes from that $T_{v} (e_{i}) = e_{i}$ for $i = 2, 3$ .

(c)

By the axiom (R4), we know that

T_{v} (\begin{matrix} a \\ 0 \\ 0 \\ d \end{matrix}) = (\begin{matrix} a^{'} \\ 0 \\ 0 \\ d^{'} \end{matrix}) .

(d)

We compute that, for $i = 2, 3$ and $j = 1, 4$ ,

⟨ T_{v}^{∗} (e_{i}), e_{j} ⟩ = ⟨ e_{i}, T_{v} (e_{j}) ⟩ = 0

by the fact that $T_{v} (e_{j}) \in span ({e_{1}, e_{4}})$ . Hence we know that $span ({e_{2}, e_{3}})$ is $T_{v}^{∗}$ -invariant.

On the other hand, we compute that, for $i = 2, 3$ and $j = 1, 4$ ,

⟨ T_{v}^{∗} (e_{j}), e_{i} ⟩ = ⟨ e_{j}, T_{v} (e_{i}) ⟩ = ⟨ e_{j}, e_{i} ⟩ = 0 .

So $span ({e_{1}, e_{4}})$ is $T_{v}^{∗}$ -invariant.

jephianlin

2011-06-27 00:00