Exercise 6.1.15

Answers

1.

If one of

x

y

is zero, then the equality holds naturally and we have

y = 0 x

x = 0 y

. So we may assume that

y

is not zero. Now if

x = cy

, we have

| ⟨ x, y ⟩ | = | ⟨ cy, y ⟩ | = | c | ∥ y ∥^{2}

and

∥ x ∥ \cdot ∥ y ∥ = ∥ cy ∥ \cdot ∥ y ∥ = | c | ∥ y ∥^{2} .

For the necessity, we just observe the proof of Theorem 6.2(c). If the equality holds, then we have

∥ x - cy ∥ = 0,

where

c = \frac{⟨ x, y ⟩}{⟨ y, y ⟩} .

And so $x = cy$ .

2.

Also observe the proof of Theorem 6.2(d). The equality holds only when

ℝe ⟨ x, y ⟩ = | ⟨ x, y ⟩ | = ∥ x ∥ \cdot ∥ y ∥ .

The case $y = 0$ is easy. Assuming $y \neq 0$ , we have $x = cy$ for some scalar $c \in F$ . And thus we have

ℝe (c) ∥ y ∥^{2} = ℝe ⟨ cy, y ⟩ = | ⟨ cy, y ⟩ | = | c | \cdot ∥ y ∥^{2}

and so

ℝe (c) = | c | .

This means $c$ is a nonnegative real number. Conversely, if $x = cy$ for some nonnegative real number, we may check

∥ x + y ∥ = | c + 1 | ∥ y ∥ = (c + 1) ∥ y ∥

and

∥ x ∥ + ∥ y ∥ = | c | ∥ y ∥ + ∥ y ∥ = (c + 1) ∥ y ∥ .

Finally, we may generalize it to the case of $n$ vectors. That is,

∥ x_{1} + x_{2} + \dots + x_{n} ∥ = ∥ x_{1} ∥ + ∥ x_{2} ∥ + \dots + ∥ x_{n} ∥

if and only if we may pick one vector from them and all other vectors are some multiple of that vector using nonnegative real number.

jephianlin

2011-06-27 00:00