Isotropic cone of a quadratic form Let $E$ be a finite dimensional vector space over $K$. Let q be a quadratic form on $E$. Let $C(q)$ denote the isotropic cone of $q$ (the isotropic cone of a quadratic form $q$ is the set of all isotrops of $E$ under $q$). In other words $C(q) = \{x \in E$ such that $q(x) = 0\}$.
We know $ker(q)$ is a subset of $C(q)$. In case $ker(q) = C(q)$, $C(q)$ become a vector subspace of $E$. If not, then $C(q)$ is not necessarily a vector subspace.
Question: Is it true that $C(q)$ is a subspace of $E$ if and only if $C(q) = Ker(q$)? If not, what should $C(q)$ satisfy to become a vector subspace?
 A: Edit: in three introductory books I have on quadratic forms, your subspace $\ker q$ is called the radical of the space. This topic is usually mentioned in the sections on Witt Cancellation. The books, which I certainly recommend, are Cassels, Gerstein, and Lam. Any one of these would be helpful supplementary material for your course. 
(Positive) Semi-definite means that the field is the real numbers and we always have $q(x) \geq 0.$ Positive definite means $q(x) > 0$ unless the vector $x=0.$ 
Could vary by field but is rare in any case, amounts to a certain type of degeneracy. Over the reals, the null cone is a vector subspace only if the quadratic form is semidefinite but not definite; this follows from Sylvester's Law of Inertia.  
For example, if
$$ g(x_1,x_2,x_3,x_4,x_5,x_6,x_7) = x_1^2 + x_7^2  $$
the null cone is
$$ x_1 = x_7 = 0 $$
which is a 5-dimensional subspace. 
Edit, Thursday:
Here is a different example, shows how important the field is.
This time the field is the rational numbers $\mathbb Q.$
$$  q(v,w,x,y,z) = v^2 - 2 w^2.  $$
Since $\sqrt 2$ is irrational, the only way to get $v^2 - 2 w^2 = 0$ is to have both $v,w=0.$ The radical $\ker q$ is no different, in particular no smaller; ignoring a possible factor of $2,$ the induced bilinear form
$$  (v,w,x,y,z) \cdot (v,-w,x,y,z) = v^2 + 3 w^2.  $$ Therefore something is in the radical only when $v,w=0.$
