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Exercise 6.B.11

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Proof. Fix $w \in V$ and, without loss of generality, assume $| | w | |_{2} = 1$ . Then for all $v \in V$ , ${⟨ v - {⟨ v, w ⟩}_{2} w, w ⟩}_{2} = 0$ , which implies also that ${⟨ v - {⟨ v, w ⟩}_{2} w, w ⟩}_{1} = 0$ . Therefore

{⟨ v, w ⟩}_{1} = {⟨ {⟨ v, w ⟩}_{2} w, w ⟩}_{1} = | | w | |_{1}^{2} {⟨ v, w ⟩}_{2} .

Hence, for each fixed $w$ we have ${⟨ v, w ⟩}_{1} = c {⟨ v, w ⟩}_{2}$ for every $v \in V$ for some positive and real constant $c$ ( $c = | | w | |_{1}^{2}$ in the above). Fix $w_{1}, w_{2} \in V$ and let $c_{1}, c_{2} \in 𝔽$ such that

\begin{aligned} {⟨ v, w_{1} ⟩}_{1} & = c_{1} {⟨ v, w_{1} ⟩}_{2}, \\ {⟨ v, w_{2} ⟩}_{1} & = c_{2} {⟨ v, w_{2} ⟩}_{2}, \end{aligned}

for all $v \in V$ . Pluging $v = w_{2}$ in the first equation and $v = w_{1}$ in the second yields

\begin{aligned} {⟨ w_{2}, w_{1} ⟩}_{1} & = c_{1} {⟨ w_{2}, w_{1} ⟩}_{2} \\ {⟨ w_{1}, w_{2} ⟩}_{1} & = c_{2} {⟨ w_{1}, w_{2} ⟩}_{2} . \end{aligned}

Then

c_{1} {⟨ w_{2}, w_{1} ⟩}_{2} = {⟨ w_{2}, w_{1} ⟩}_{1} = \bar{{⟨ w_{1}, w_{2} ⟩}_{1}} = \bar{c_{2} {⟨ w_{1}, w_{2} ⟩}_{2}} = \bar{c_{2}} {⟨ w_{2}, w_{1} ⟩}_{2} .

Hence $c_{1} = \bar{c_{2}}$ . Because both are real, it follows that $c_{1} = c_{2}$ . Therefore, the constant is the same for all $v, w \in V$ . □

celiopassos

2017-10-06 00:00

Exercise 6.B.11

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