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Let $V=\mathbb{R}^n$. Given a norm $|\cdot |$, find the inner product for which $\langle v,v\rangle=|v|$.

So basically I need to find an inner product such that $\forall v\in V. \langle v,v\rangle=1$, right? How do I do that?

I was also given a hint: "Find a way to write $\langle u,v\rangle$ as a function of norms." I think they mean $\langle u,v\rangle= \dfrac{|u+v|^2-|u|-|v|}{2}$?

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  • $\begingroup$ Yes. Or $\frac14(\lVert u+v\rVert-\lVert u-v\rVert)$. $\endgroup$ – Bernard Jun 5 '15 at 16:57
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The parallelogram law says for a norm from an inner product, $2 || x||^2 + 2 ||y||^2 = ||x-y||^2 + ||x+y||^2$.

You can get from this the polarization identity $$\langle x,y\rangle = \frac{||x+y||^2-||x-y||^2}{4}$$

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  • $\begingroup$ There is also a polarization identity for complex norms. $\endgroup$ – copper.hat Jun 5 '15 at 17:20
  • $\begingroup$ Yeah. There's a lot of them -- you can get a different one from considering the $n$-th primitive roots of unity for each $n\geq 3$. $\endgroup$ – Batman Jun 5 '15 at 17:22

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