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Exercise 6.A.6

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Proof. Suppose $⟨ u, v ⟩ = 0$ . Note that $u$ is orthogonal to any multiple of $v$ . Then, by the Pythagorean Theorem (6.13), we have

| | u + av | |^{2} = | | u | |^{2} + | a | | | v | |^{2} \leq | | u | |^{2} .

Taking the square root of both sides completes the forward direction.

For the converse, we will prove the contrapositive, that is, if $⟨ u, v ⟩ \neq 0$ , then $| | u | | > | | u + av | |$ for some $a \in 𝔽$ .

Suppose $⟨ u, v ⟩ \neq 0$ . Note that neither $u$ nor $v$ can equal $0$ . We have

\begin{aligned} | | u + av | |^{2} & = ⟨ u + av, u + av ⟩ \\ = ⟨ u, u ⟩ + ⟨ u, av ⟩ + ⟨ av, u ⟩ + ⟨ av, av ⟩ \\ = | | u | |^{2} + ⟨ u, av ⟩ + ⟨ av, u ⟩ + | a |^{2} | | v | |^{2} \\ < | | u | |^{2} \end{aligned}

where the last line follows provided that $⟨ u, av ⟩ + ⟨ av, u ⟩ + | a |^{2} | | v | |^{2} < 0$ . By 6.14, we can write $u = cv + w$ for some $c \in 𝔽$ and $w \in V$ such that $⟨ v, w ⟩ = 0$ . Note that $c \neq 0$ , because $⟨ v, v ⟩ \neq 0$ and

0 \neq ⟨ u, v ⟩ = ⟨ cv + w, v ⟩ = ⟨ cv, v ⟩ + ⟨ w, v ⟩ = c ⟨ v, v ⟩

Choose $a = - c$ , then

\begin{aligned} ⟨ u, av ⟩ + ⟨ av, u ⟩ + | a |^{2} | | v | |^{2} & = ⟨ cv + w, av ⟩ + ⟨ av, cv + w ⟩ + | a |^{2} | | v | |^{2} \\ = ⟨ cv, av ⟩ + ⟨ w, av ⟩ + ⟨ av, cv ⟩ + ⟨ av, w ⟩ + | a |^{2} | | v | |^{2} \\ = cā ⟨ v, v ⟩ + a \bar{c} ⟨ v, v ⟩ + | a |^{2} | | v | |^{2} \\ = (cā + a \bar{c} + | | c | |^{2}) | | v | |^{2} \\ = (- c \bar{c} - c \bar{c} + | | c | |^{2}) | | v | |^{2} \\ = - | c |^{2} | | v | |^{2} \\ < 0 \end{aligned}

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celiopassos

2017-10-06 00:00

Exercise 6.A.6

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