Exercise 6.8.25

Answers

Let $A = ψ_{β} (H)$ for some orthonormal basis $β$ . And let $T$ be the operator such that ${[T]}_{β} = A$ . By Exercise 6.8.5 we have

H (x, y) = {[ϕ_{β} (x)]}^{t} A [ϕ_{β} (y)] = {[ϕ_{β} (x)]}^{t} {[T]}_{β} [ϕ_{β} (y)] = {[ϕ_{β} (x)]}^{t} [ϕ_{β} (T (y))] .

Also, by Parseval’s Identity in Exercise 6.2.15 we know that

⟨ x, T (y) ⟩ = \sum_{i = 1}^{n} ⟨ x, v_{i} ⟩ ⟨ T (y), v_{i} ⟩ = {[ϕ_{β} (x)]}^{t} [ϕ_{β} (T (y))]

since $β$ is orthonormal.

jephianlin

2011-06-27 00:00