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Hello,

given

Given a symmetric bilinear form $f:V\times V \to K$ , where $V$ is a vectorspace and $K$ is an appropriate field, define the quadratic form $q:V \to K$ as $q(v):= f(v,v)$.

The Polarisation Formula states that $f(x,y) = 1/2\big( q(x+y) - q(x) - q(y)\big)$, which is easily proven.

This means , that any symmetric bilinear form $f:V\times V \to K$ is fully determined by the values $f(v,v)$ for all $v \in V$.

I now want to prove the following theorem: Prove that any symmetric $k$-linear form $M:V\times\cdots \times V \to K$ is determined by the values $M[v]^k := M[v,...,v]$ for all $v\in V$.

How does that work?

2 texified

Hello,

given a bilinear form f:VxV -> K $f:V\times V \to K$ , where V $V$ is a vectorspace and K $K$ is an appropriate field, define the quadratic form $q:V -> K \to K$ as $q(v):= f(v,v) f(v,v)$.

The Polarisation Formula states that f(x,y) $f(x,y) = 1/2 1/2\big( q(x+y) - q(x) - q(y))q(y)\big)$, which is easily proven.

This means, that any symmetric bilinear form f:VxV -> K $f:V\times V \to K$ is fully determined by the values f(v,v) $f(v,v)$ for all $v \in VV$.

I now want to prove the following theorem: Prove that any symmetric k-linear $k$-linear form M:Vx...xV -> K $M:V\times\cdots \times V \to K$ is determined by the values $M[v]^k := M[v,...,v] M[v,...,v]$ for all v in V$v\in V$.

How does that work?

1

Generalization of the Polarisation Formula for Symmetric Bilinear Forms to Symmetric multilinear Forms

Hello,

given a bilinear form f:VxV -> K , where V is vectorspace and K is an appropriate field, define the quadratic form q:V -> K as q(v):= f(v,v) .

The Polarisation Formula states that f(x,y) = 1/2 ( q(x+y) - q(x) - q(y)), which is easily proven.

This means, that any symmetric bilinear form f:VxV -> K is fully determined by the values f(v,v) for all v in V.

I now want to prove the following theorem: Prove that any symmetric k-linear form M:Vx...xV -> K is determined by the values M[v]^k := M[v,...,v] for all v in V.

How does that work?