Exercise 6.1.20

Answers

1.

By Exercise 6.1.19(a), we have the right hand side would be

\frac{1}{4} (4 ℝe ⟨ x, y ⟩) = ⟨ x, y ⟩ .

The last equality is due to the fact $𝔽 = ℝ$ .

2.

Also use Exercise 6.1.19(a). First observe that

∥ x + i^{k} y ∥^{2} = ∥ x ∥^{2} + 2 ℝe ⟨ x, i^{k} y ⟩ + ∥ i^{k} y ∥^{2}

= ∥ x ∥^{2} + 2 ℝe [\bar{i^{k}} ⟨ x, y ⟩] + ∥ y ∥^{2} .

Assuming $⟨ x, y ⟩$ to be $a + bi$ , we have the right hand side would be

\frac{1}{4} [(\sum_{k = 1}^{4} i^{k}) ∥ x ∥^{2} + 2 \sum_{k = 1}^{4} i^{k} ℝe [\bar{i^{k}} ⟨ x, y ⟩] + (\sum_{k = 1}^{4} i^{k}) ∥ y ∥^{2}]

= \frac{1}{2} [bi + (- a) (- 1) + (- b) (- i) + a] = a + bi = ⟨ x, y ⟩ .

jephianlin

2011-06-27 00:00