# Parallelogram law, dot product [duplicate]

Prove that if $||\cdot||$ satisfies $||u-v||^2 + ||u+v||^2 = 2(||u||^2 + ||v||^2)$ , then $u \cdot v = \frac{1}{2} (||u+v||^2 - ||u||^2 - ||v||^2)$ is dot product and $||u||^2 = u \cdot u$.

I've already shown that $(u+w)\cdot v = u \cdot v + w \cdot v$, but I have serious troubles showing that $(\lambda v)\cdot(w) = \lambda (v\cdot w)$.

Could you help me with that?

## marked as duplicate by Norbert, TMM, Davide Giraudo, Sasha, Lord_FarinApr 23 '13 at 20:17

• Presumably $\|\cdot \|$ satisfies all the norm axioms? – rschwieb Apr 23 '13 at 19:34
• You can use $\|u-v\|^2=(u-v)\cdot(u-v)$ and $\|u+v\|^2=(u+v)\cdot(u+v)$ to get $u\cdot v$ easily. – xpaul Apr 23 '13 at 19:40