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Consider $E$ and $F$ two vector spaces over $\mathbb{R}$, and $f : E^n \longmapsto F$ a $n$-linear map.

Assume that $f$ is symmetric, i.e. for all $(v_1,...,v_n) \in E^n$, for all permutation $\sigma \in S_n$, $$f(v_1,...,v_n) = f(v_{\sigma(1)},...,v_{\sigma_n}).$$ Then show that $f$ is uniquely defined by the values of the $f(v,v,...,v)$ for $v \in E$.


I tried showing the result using induction, but had some problems deriving $n+1$ from $n$. I mainly used the following base case.

Proof for $\textbf{n=2}$: for all $(u, v) \in E^2$, denote $a = \frac{u+v}{2}$ and $b= \frac{u-v}{2}$. Then the multilinearity yields $f(u,v) = f(a+b,a-b) = f(a,a)+f(b,a)-f(a,b)-f(b,b)$ and as $f$ is symmetric, $f(u,v) = f(a,a)-f(b,b)$.

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  • $\begingroup$ I suppose you need to assume the field is not of characteristic $2$, and probably not of any finite characteristic either. $\endgroup$ – edm Sep 24 '18 at 12:22
  • $\begingroup$ @edm that's right ! I am going to define the space vectors over R $\endgroup$ – charmd Sep 24 '18 at 12:41
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    $\begingroup$ See math.stackexchange.com/questions/481167/polarization-formula/… for example. $\endgroup$ – Jens Schwaiger Sep 24 '18 at 13:09

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