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I'm trying to prove that if $|u+v| = |u| + |v|$ implies $u = \lambda v$ for $\lambda \ge 0$.

To this end I have $|u + v|^2 = (|u| + |v|)^2 \Rightarrow |u|^2 + |v|^2 + 2|u||v| = |u|^2 + |v|^2 + 2u\cdot v \Rightarrow |u||v| = u \cdot v \Rightarrow u = \lambda v$, where the last implication follows by Cauchy-Schwarz.

However, how do I prove that $\lambda \ge 0$ ?

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If $\lambda <0$ then for $v\neq 0$ $$u\cdot v=\lambda v\cdot v=\lambda |v|^2<0.$$ Can you take it from here?

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  • $\begingroup$ Ahh, so you get a contradiction if $\lambda < 0$ ? $\endgroup$ – Shuzheng Jun 24 '15 at 10:35
  • $\begingroup$ @user111854, thats right. $\endgroup$ – Ofir Schnabel Jun 24 '15 at 10:37
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If $\lambda <0$, then

$$|u + v | = |\lambda +1| |v| \neq (|\lambda| +1)|v| = |u| + |v|$$

if $v\neq 0$.

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