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If $u_1+w_1=u_2+w_2$ and $\langle u_1,w_1\rangle=0=\langle u_2,w_2\rangle$, how can we prove that $$\|u_1\|^2+\|w_1\|^2=\|u_2\|^2+\|w_2\|^2$$

I know I can open this to $$\langle u_1,u_1\rangle+\langle w_1,w_1\rangle=\langle u_2,u_2\rangle+\langle w_2,w_2\rangle$$ but from here what can I do with that?

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    What do you want to do? What do you need to prove?2012-12-05
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    No. ${}{}{}{}{}$2012-12-05
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    You're missing either some squares on the norms, or some square roots on the scalar products.2012-12-05
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    @Thibaut Dumont ok because I have that this exist: 1. u1+w1=u2+w2 2. =0= So im trying to find a way to get one of those...2012-12-05
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    You need to *use* those, or you're trying to *get* those? These are not at all the same thing.2012-12-05
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    @ChrisEagle need to use2012-12-05
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    Ok, then you need to actually answer Thibaut's question: what are you trying to do?2012-12-05
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    @ChrisEagle im trying to find a way to use one of those 2 i know exist but im stuck!2012-12-05
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    @DonAntonio i dont understand why nobody helps2012-12-05
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    Ad besides all the above, it is not clear at all in this question what you're given adn what you have to do, @baaa12: what is your question, anyway?2012-12-05
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    @JasperLoy yes sorry2012-12-05
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    Take the square of the norm of both sides of equ (1), use $\| r\|^2=\langle r,r\rangle$ to expand both sides using bilinearity, and then use equ (2) to take out a few terms.2012-12-05

1 Answers 1

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From $u_1+w_1=u_2+w_2$ and $\langle u_1,w_1\rangle=0=\langle u_2,w_2\rangle$, we have

$\|u_1\|^2+\|w_1\|^2$

$=\langle u_1,u_1\rangle+\langle w_1,w_1\rangle+2\langle u_1,w_1\rangle$

$=\langle u_1+w_1,u_1+w_1\rangle$

$=\langle u_2+w_2,u_2+w_2\rangle$

$=\langle u_2,u_2\rangle+\langle w_2,w_2\rangle+2\langle u_2,w_2\rangle$

$=\|u_2\|^2+\|w_2\|^2$