# Quadratic forms…

Let V be an F-vector space (F $= \mathbb{R}$ or $\mathbb{C}$)

My notes define a quadratic form as:

A map $q:V \rightarrow F$ s.t. $q(v)=\beta(v,v)$ for some (symmetric) bilinear form $\beta:V\times V \rightarrow F$.

Later on the notes define an Hermitian form $\gamma:V\times V \rightarrow F$ to be a conjugate-symmetric sesquilinear form. It then says that given an Hermitian form $\gamma,$ we can define a quadratic form $q:V \rightarrow F$ by $q(v)=\gamma(v,v).$

Does this fit with the definition of "quadratic form" given above - I'm quite confused?

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## 1 Answer

Given a quadratic form $q(v)$, one can find the associated bilinear form by: $$\beta_q(u,v)=\frac12\left( q(u+v)-q(u)-q(v)\right)$$ So to check the validity of the statement, it is sufficient to check that given a Hermitian form $\gamma$, define $q_\gamma(v)=\gamma(v,v)$, you will get $\beta_{q_\gamma}$ to be a bilinear form. This is a simple check.

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When I do this I get $\beta_q(u,v)=\frac{1}{2}(\gamma(x,y)+\gamma(y,x)) = \Re[\gamma(x,y)]$ where the last equality follows as $\gamma$ is Hermitian. So doesn't this mean $\beta_q$ is bilinear if and only if F $= \mathbb{R}$? – J1779 Oct 11 '12 at 12:26