Exercise 6.5.32

Answers

1.

We pick

𝜃

to be the value

⟨ x, y ⟩

. Thus we have

\bar{⟨ x, 𝜃y ⟩} = \bar{⟨ x, ⟨ x, y ⟩ y ⟩}

= \bar{⟨ y, x ⟩ ⟨ x, y ⟩} = ⟨ x, y ⟩ ⟨ y, x ⟩

= ⟨ x, ⟨ x, y ⟩ y ⟩ = ⟨ x, 𝜃y ⟩

and so the value is real. Now pick $u$ to be the normalized vector $\frac{x - 𝜃y}{∥ x - 𝜃y ∥}$ . And compute

H_{u} (x) = x - 2 ⟨ x, u ⟩ u = x - 2 ⟨ x, \frac{x - 𝜃y}{∥ x - 𝜃y ∥} ⟩ \frac{x - 𝜃y}{∥ x - 𝜃y ∥}

= x - \frac{2}{∥ x - 𝜃y ∥^{2}} ⟨ x, x - 𝜃y ⟩ (x - 𝜃y)

= x - \frac{2 ∥ x ∥^{2} - 2 ⟨ x, 𝜃y ⟩}{∥ x - 𝜃y ∥^{2}} (x - 𝜃y)

= x - \frac{∥ x ∥^{2} - 2 ⟨ x, y ⟩ + ∥ 𝜃y ∥^{2}}{∥ x - 𝜃y ∥^{2}} (x - 𝜃y)

= x - x + 𝜃y = 𝜃y .

2.

Pick

u

to be the normalized vector

\frac{x - y}{∥ x - y ∥}

. Compute

H_{u} (x) = x - 2 ⟨ x, u ⟩ u = x - 2 ⟨ x, \frac{x - y}{∥ x - y ∥} ⟩ \frac{x - y}{∥ x - y ∥}

= x - \frac{2}{∥ x - y ∥^{2}} ⟨ x, x - y ⟩ (x - y)

= x - \frac{2 ∥ x ∥^{2} - 2 ⟨ x, y ⟩}{∥ x - y ∥^{2}} (x - y)

= x - \frac{∥ x ∥^{2} - 2 ⟨ x, y ⟩ + ∥ y ∥^{2}}{∥ x - y ∥^{2}} (x - y)

= x - x + y = y .

jephianlin

2011-06-27 00:00