# Proof $f(u,v)=\frac{1}{4}(q(u+v)-q(u-v))+\frac{i}{4}(q(u+iv)-q(u-iv))$

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.
• You want $f$ to be Hermitian, not symmetric: $f(\alpha u,v)=\alpha f(u,v)=f(u,\overline\alpha v)$ etc. – 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." 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$?$ – emonHR Nov 14 '19 at 8:25
• 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... – Mnifldz Nov 14 '19 at 8:29