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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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    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

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

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

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

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

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

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