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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_Farin Apr 23 '13 at 20:17

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  • $\begingroup$ It follows from definition of inner product, or dot product, specially if given that your field is real. $\endgroup$ – user45099 Apr 23 '13 at 19:33
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    $\begingroup$ Presumably $\|\cdot \|$ satisfies all the norm axioms? $\endgroup$ – rschwieb Apr 23 '13 at 19:34
  • $\begingroup$ @user57 I think he's showing that the inner product can be retrieved from the norm without assuming you have defined the norm with the inner product. $\endgroup$ – rschwieb Apr 23 '13 at 19:35
  • $\begingroup$ 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. $\endgroup$ – xpaul Apr 23 '13 at 19:40
  • $\begingroup$ @rschwieb Ok, great. overlooked it. $\endgroup$ – user45099 Apr 23 '13 at 19:40