Something really basic but I have to ask it: I was taught that the formula of symmetric bi linear form of the quadratic form if: $f(v,w) = 1/2(q(v+w)-q(v)-q(w))$
but $q$ is linear so what did we get here? if I open the bracelets it is $q(v)+q(w)-q(v)-q(w) =0 $
Therefore I don't understand. In addition there are more formulas such as:
$f(v,w)= \frac{1}{4}(q(v+w)-q(v-w))$ and I don't understand how they are related
Thanks
How do you define a quadratic form? I think that nearly all definitions of a quadratic form $q : X\rightarrow \mathbb{R}$ on a linear space $X$ over $\mathbb{R}$ require the Parallelogram Law $$ q(u+v)+q(u-v)=2q(u)+2q(v),\;\;\; u, v \in X. $$ Some require an additional scalar property $$ q(\alpha u) =\alpha^{2}q(u),\;\;\; u \in X,\; \alpha \in \mathbb{R}. $$ A quadratic form $q$ is not linear, as you suggest in your question.
Every bilinear form $f : X\times X\rightarrow\mathbb{R}$ over a real linear space $X$--whether or not $f$ is symmetric--generates a quadratic form $q(u)=f(u,u)$ which satisfies both of the above. To verify this, assume $f$ is bilinear, and notice that the Parallelogram Law for $q$ and the scalar identity are implied by these two identities for $f$: $$ \begin{align} f(u+v,u+v)+f(u-v,u-v) & =f(u,u)+f(u,v)+f(v,u)+f(v,v) \\ & +f(u,u)-f(u,v)-f(v,u)+f(v,v) \\ & =2f(u,u)+2f(v,v),\;\;\; u,v \in X. \end{align}\\ f(\alpha u,\alpha u) = \alpha^{2}f(u,u),\;\;\; \alpha \in \mathbb{R},\;\; u \in X. $$ On the other hand, it is not necessarily true that every function $q : X\rightarrow\mathbb{R}$ which satisfies the Parallelogram Law comes from a bilinear form $f$ through $q(u)=f(u,u)$; I seem to recall that it's not true even if you add the second identity $q(\alpha u)=\alpha^{2}q(u)$ (it is true, though, if you require both identities and require that $q$ be non-negative.) However, if $q(u)=f(u,u)$ does hold for all $u\in X$ for some symmetric bilinear form $f$, then the following must hold: $$ f(u,v) = \frac{1}{4}\{q(u+v)-q(u-v)\},\;\;\; u,v \in X. $$ Assuming the Parallelogram Law, the above is equivalent to $$ f(u,v)=\frac{1}{4}q(u+v)-\frac{1}{4}\{-q(u+v)+2q(u)+2q(v)\}=\frac{1}{2}\{q(u+v)-q(u)-q(v)\}. $$