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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 .

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ "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

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