quadratic and bilinear forms Does a quadratic form always come from symmetric bilinear form ?
We know when $q(x)=b(x,x)$ where $q$ is a quadratic form and $b$ is a symmetric bilinear form.
But when we just take a bilinear form and $b(x,y)$ and write $x$ instead of $y$,does it give us a quadratic form ?
 A: Yes, every quadratic form (over a finite-dimensional $\mathbb{R}$-space) can be expressed in terms of a symmetric bilinear form, because if your quadratic form $Q(x)$ (for $x \in \mathbb{R}^n$) is written as
$$ Q(x) = \sum_{i\le j} c_{ij} x_i x_j $$
then $Q(x) = x^T Ax$, where $A$ is a symmetric matrix given by $A = (a_{ij})$ with
$$a_{ij} = a_{ji} = \frac{c_{ij}}{2}, i < j, \quad a_{ii} = c_{ii}$$
A: Sure. In finite dimension over the real numbers you do not even need the bilinear form symmetric, because any matrix is the sum of a symmetric and a skew-symmetric matrix, and these are easy to find. 
Meanwhile, as long as characteristic of the underlying field is not $2,$ you can go back with $$ c(x,y) = \frac{1}{2} \left( q(x+y) - q(x) - q(y) \right).     $$
Since this recipe gives a symmetric bilinear form, those are preferred.
Even in infinite dimension, I suppose you can symmetrize with 
$$  f(x,y) = \frac{1}{2} \left( h(x,y) + h(y,x) \right).  $$
There won't be any matrices.
A: If we have b symmetric bilinear form we can get q quadratic form
$q\colon V \to \mathbb{R}$
q(v)=b(v,v)
conversely if q is a quadratic form
$q\colon V \to \mathbb{R}$
we can define
$\frac 12$(q(v+w)-q(v)-q(w)):=b(v,w)
the vital answer is you just get a bilinear form not always a symmetric bilinear
form.
because the definition $\frac 12$(q(v+w)-q(v)-q(w)) leads us to $\frac 12$(b(v,w)+b(w,v))  
