Let $q$ be the Hermitian quadratic form associated with a symmetric bilinear form $f$, on a vector space $V$ over the field $F$. Prove that $$f(u,v)=\frac{1}{4}(q(u+v)-q(u-v))+\frac{i}{4}(q(u+iv)-q(u-iv))$$

I have no idea how to start$?$ All I know by definition is

A mapping $q:V\rightarrow F$ is a quadratic form if $q(v)=f(v,v)$ for some symmetric bilinear form $f$ on $V$

Then if $1+1\neq 0$ in $F$, then bilinear form $f$ can be obtained from the quadractic form $q$ by the following polar form of $f$: $$f(u,v)=\frac{1}{2}(q(u+v)-q(u)-q(v))$$ But I don't think this help me to solve that proof. Any kind of help will be appreciated.
Thanks for your time. Thanks in advance .

  • $\begingroup$ You want $f$ to be Hermitian, not symmetric: $f(\alpha u,v)=\alpha f(u,v)=f(u,\overline\alpha v)$ etc. $\endgroup$ – Angina Seng Nov 14 '19 at 8:15

You basically approach this by starting on the right side with the expressions for $q$. Since we have $q(v) = f(v,v)$ we can then start computing each of these quantities one at a time: \begin{eqnarray*} q(u+v) & = & f(u+v, u+v) \\ & = & f(u, u+v) + f(v,u+v) \\ & = & f(u,u) + f(u,v) + f(v,u) + f(v,v). \end{eqnarray*}

Repeat this with the others and collect like terms.

  • $\begingroup$ "Repeat this with the others and collect like terms." What I get is $$\frac{1}{2}(f(u,v)+if(u,v))$$ I think It should be $\frac{1}{2}(f(u,v)+f(u,v))$ which make the proof complete. Where I did wrong$?$ $\endgroup$ – emonHR Nov 14 '19 at 8:25
  • $\begingroup$ There might be some missing information. Is $f$ a real symmetric bilinear form or is it complex? I would anticipate $f(u,v) = \overline{f(v,u)}$. If it is Hermitian then that's a different story... $\endgroup$ – Mnifldz Nov 14 '19 at 8:29
  • $\begingroup$ I write same thing which was written in my Question paper @Mnifldz Sir. Ok let take it Hermitian I think that piece of information is missing. $\endgroup$ – emonHR Nov 14 '19 at 8:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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