I'm reading Scharlau's Quadratic and Hermitian forms. In it he defines a quadratic form as
"Let V be a vector space over a field K. A map $q:V\mapsto K$ is called a quadratic form on $V$ if $$q(\alpha x)=\alpha^2q(x)$$ for all $x\in V,\alpha \in K$ and $b_q(x,y)=\frac{1}{2}(q(x+y)-q(x)-q(y))$ is a (necessarily symmetric) bilinear form".
However I'm having trouble verifying it's a bilinear form.Here is my attempt.\
$$b(x+x',y)=\frac{1}{2}((q(x+x'+y)-q(x+x')-q(y))$$ here I get stuck as I don't see how I can get this to split any further so I get $b(x+x',y)=b(x,y)+b(x',y)$
I'm also stuck on the scalar multiplication property of bilinear forms. Here is my attempt. $$b(\alpha x,y)=\frac{1}{2}(q(\alpha x+y)-q(\alpha x)-q(y)= \frac{1}{2}(q(\alpha x+y)-\alpha ^2q(x)-q(y))$$ again I don't see how I can split this any further to show $b(\alpha x,y)=\alpha b(x,y)$
Can anyone give me an idea how to progress into showing this is a bilinear form?
Here's what's going on.
Per the comments under Jose Carlos Santos' answer, in order to be called a quadratic form, the function $q:V\rightarrow K$ must satisfy 2 distinct requirements:
(1) $q(\alpha x) = \alpha^2 q(x)$ for all $\alpha\in K, x\in V$.
(2) The function $b_q:V\times V\rightarrow K$ defined by the equation $b_q(x,y) = \frac{1}{2}(q(x+y) - q(x) - q(y))$ must be a bilinear function.
The second requirement is part of the definition of being called a quadratic form. If $q$ satisfies (1) but not (2), it is not a quadratic form.
So what's going on is just that you were mildly misreading Scharlau's definition. (I was too, at first.)
To see that it is possible to satisfy (1) without satisfying (2), observe that (1) only exerts any control over how $q$ behaves when you scale its input with a scalar factor. It doesn't exert any control on how $q$'s values on a pair of linearly independent vectors are related. (Whereas the bilinearity of $b_q$ clearly requires some sort of relationship of this kind.)
Exploiting this idea to build a concrete example:
Let $V=\mathbb{R}^2$ and let $K=\mathbb{R}$. Construct a candidate $q$ as follows: take an absolutely arbitrary function $g$ on the half-open interval $[0,\pi)$. Now every line through the origin goes through exactly one of the points $(\cos\theta,\sin\theta)$ for $\theta \in [0,\pi)$. This implies that for any nonzero $v\in \mathbb{R}^2$, there is a unique $\beta \in \mathbb{R}$ and a unique $\theta\in [0,\pi)$ such that $v = (\beta\cos\theta,\beta\sin\theta)$. Now define $q(v)$ by the formula $q(v) := \beta^2 g(\theta)$. (And define $q(0)= 0$.)
I claim that the $q$ so defined satisfies $q(\alpha v) = \alpha^2 q(v)$ for any $v\in \mathbb{R}^2$ and any $\alpha\in \mathbb{R}$. Indeed, if $v = (\beta\cos\theta, \beta\sin\theta)$ for some $\beta\in\mathbb{R},\theta\in[0,\pi)$, then for any $\alpha\in\mathbb{R}$ we have
$$q(\alpha v) = q(\alpha\beta\cos\theta,\alpha\beta\sin\theta) = (\alpha\beta)^2 g(\theta) = \alpha^2(\beta^2g(\theta)) =\alpha^2 q(v).$$
Since the original function $g$ was completely arbitrary, $q$ is potentially a mess. (For example, we could let $g$ be some form of the Dirichlet function - $1$ when $\theta$ is a rational fraction of $\pi$, $0$ when it's an irrational fraction of $\pi$.) Then $b_q$ can't hope to be bilinear.