Exercise 6.1.2

Answers

Compute

⟨ x, y ⟩ = 2 \cdot (2 + i) + (1 + i) \cdot 2 + i \cdot (1 - 2 i) = 8 + 5 i,

∥ x ∥ = {⟨ x, x ⟩}^{\frac{1}{2}} = {(4 + 2 + 1)}^{\frac{1}{2}} = \sqrt{7},

∥ y ∥ = {⟨ y, y ⟩}^{\frac{1}{2}} = {(5 + 4 + 5)}^{\frac{1}{2}} = \sqrt{14},

and

∥ x + y ∥ = ∥ (4 - i, 3 + i, 1 + 3 i) ∥ = {(17 + 10 + 10)}^{\frac{1}{2}} = \sqrt{37} .

We have the Cauchy-Schwarz inequality and the triangle inequality hold since

| 8 + 5 i | = \sqrt{89} \leq \sqrt{7} \sqrt{14} = \sqrt{119}

and

\sqrt{7} + \sqrt{14} \geq \sqrt{37} .

jephianlin

2011-06-27 00:00