# Proving that $\|u_1\|^2+\|w_1\|^2=\|u_2\|^2+\|w_2\|^2$

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. ${}{}{}{}{}$ –  Matt N. Dec 5 '12 at 10:15
You're missing either some squares on the norms, or some square roots on the scalar products. –  Matt Pressland Dec 5 '12 at 10:15
Have you read this site's FAQ, @baaa12 ? People get discouraged when they see someone questions by someone with accpet rate equal to 0%: this means you haven't accepted any answer to any of your questions and, thus, it looks like you don't like the answers you've received here. You should take care of this, imfho. –  DonAntonio Dec 5 '12 at 11:18
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
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## 1 Answer

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

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

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

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

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

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

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

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