Vektorraum/R/Skalarprodukt/Normgleichheit/2/Aufgabe/Lösung

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Es ist

{\displaystyle {}{\begin{aligned}\mid \mid \!2u\!\mid \mid ^{2}&=\mid \mid \!(u+v)+(u-v)\!\mid \mid ^{2}\\&=\left\langle (u+v)+(u-v),(u+v)+(u-v)\right\rangle \\&=\left\langle (u+v),(u+v)\right\rangle +\left\langle (u+v),(u-v)\right\rangle +\left\langle (u-v),(u+v)\right\rangle +\left\langle (u-v),(u-v)\right\rangle \end{aligned}}}

und

{\displaystyle {}{\begin{aligned}\mid \mid \!2v\!\mid \mid ^{2}&=\mid \mid \!(u+v)-(u-v)\!\mid \mid ^{2}\\&=\left\langle (u+v)-(u-v),(u+v)-(u-v)\right\rangle \\&=\left\langle (u+v),(u+v)\right\rangle -\left\langle (u+v),(u-v)\right\rangle -\left\langle (u-v),(u+v)\right\rangle +\left\langle (u-v),(u-v)\right\rangle .\end{aligned}}}

Somit ist

${\displaystyle {}\mid \mid \!2u\!\mid \mid ^{2}-\mid \mid \!2v\!\mid \mid ^{2}=4\left\langle (u+v),(u-v)\right\rangle \,.}$

Also ist

${\displaystyle {}\mid \mid \!u\!\mid \mid =\mid \mid \!v\!\mid \mid \,}$

genau dann, wenn

${\displaystyle {}\left\langle (u+v),(u-v)\right\rangle =0\,}$
ist.
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