Exercise 6.3.14

Answers

It’s linear since

T (c x_{1} + x_{2}) = ⟨ c x_{1} + x_{2}, y ⟩ z

= c ⟨ x_{1}, y ⟩ z + ⟨ x_{2}, y ⟩ z = cT (x_{1}) + T (x_{2}) .

On the other hand, we have

⟨ u, T^{∗} (v) ⟩ = ⟨ ⟨ u, y ⟩ z, v ⟩

= ⟨ u, y ⟩ ⟨ z, v ⟩ = ⟨ u, ⟨ v, z ⟩ y ⟩

for all $u$ and $v$ . So we have

T^{∗} (x) = ⟨ x, z ⟩ y .

jephianlin

2011-06-27 00:00