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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? – Thibaut Dumont Dec 5 '12 at 10:14
No. ${}{}{}{}{}$ – Rudy the Reindeer Dec 5 '12 at 10:15
You're missing either some squares on the norms, or some square roots on the scalar products. – Matthew Pressland Dec 5 '12 at 10:15
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? – DonAntonio Dec 5 '12 at 11:19
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. – anon Dec 5 '12 at 11:35
up vote 5 down vote accepted

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


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

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

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

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


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