Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

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.

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.