# If $|u+v| = |u| + |v|$ then $u = \lambda v$. How do I prove $\lambda \ge 0$?

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 $\lambda \ge 0$ ?

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?
• Ahh, so you get a contradiction if $\lambda < 0$ ? – Shuzheng Jun 24 '15 at 10:35
If $\lambda <0$, then
$$|u + v | = |\lambda +1| |v| \neq (|\lambda| +1)|v| = |u| + |v|$$
if $v\neq 0$.