A quadratic form is positive-definite iff its set of isotropic vectors is trivial Considering a quadratic form $Q$ in a finite dimensional vector space $V$ can I say that
$\mathscr{I}=\big\{ \vec{o} \big\} \iff Q $ is  definite positive
?
Where $\mathscr{I}$ is the isotropic vectors set.
Are there cases in which 
$\mathscr{I}=\big\{ \vec{o} \big\}$ and $ Q $ is not positive definite?
Thanks in advice
 A: As orangeskid points out, the answer is essentially affirmative on real vector spaces, because of Sylvester's law of inertia. 
More precisely, a real quadratic form without isotropic vectors has a sign: it is either positive or negative definite. 
This, of course, fails on complex vector spaces, see e.g. $Q(z)=i\lvert z\rvert^2$ for $z\in \mathbb{C}$.
P.S.: Re to rschwieb's asnswer. On complex vector spaces one still has the following property: if a vector $v\ne 0$ exists such that $Q(v)>0$ and a vector $w\ne 0$ exists such that $Q(w)<0$ then there must exist a vector $u\ne 0$ such that $Q(u)=0$. 
(Here $Q(x)=B(x, x)$ and $B$ is a sesquilinear form.)
A: 
can I say that $\mathscr{I}=\big\{ \vec{o} \big\} \iff Q $ is  definite positive?

No: consider the quadratic form on $\Bbb R^2$ given by $Q((x,y))=-x^2-y^2$. There are no isotropic vectors, and it is negative definite.
There is a proposition you can prove to show that without nonzero isotropic vectors, you get a definite form:
If there are vectors $v,w$ such that $Q(v)>0$ and $Q(w)<0$, then there is a nonzero isotropic vector.
Sketch: Convert $Q$ to a bilinear form $B$ such that $B(z,z)=Q(z)$ for all z. 
Use Gram-Schmidt to replace $w$ with another negative value vector that is perpendicular to $v$. (The result cannot have a positive Q value.)
Now normalize both by their length so that v has length 1 and w has length -1.  Then $Q(v+w)=B(v+w,v+w)= Q(v)+Q(w)=0$. 
V+w can't be zero since the two vectors are linearly independent.
