Regular Quadratic Space - isotrope vector I am currently trying to solve the following exercise:  

Show that every regular quadratic space of finite dimension $E$ that contains at least one isotrope vector, has a basis consisting only of isotrope vectors.

I think my problem with this exercise is that I don't really have a ''feeling'' for the definitions.  
Quadratic form: A quadratic form $q$ over $E$ is an application $E \to K$ such that there exists a bilinear form $f : E\times E \to K$ such that $\forall x\in E$ $q(x)=f(x,x)$.  
$ker(q) = \{x\in E|\forall y\in E: f(x,y)=0\}$  
Regular Quadratic Space: $(E,q)$ is called regular, if $ker(q) = \{0\}$  
Isotrope: $x$ is called isotrope, if $f(x,x)=q(x)=0$ for $x\not=0$ in $E$.  
Is it the right way to work with these definitions? I don't really see how to get from one isotropic vector to the whole basis, and what information it gives me that I have one isotropic vector in the space.  
I would be very happy for a hint to get a start!
All the best, Luca

You don't need the full Witt decomposition, just split off a hyperbolic plane, so
$$V =  W\oplus \langle u, u' \rangle$$ with $f(u,u')=1$, ...
Take any basis $w_1$, $\ldots$, $w_m$ of the subspace $W$. We get the basis of $V$
\begin{eqnarray}
&w_i& - \frac{q(w_i)}{2}(u+u') \text{ for } 1 \le i \le m, \ \text{and}\ 
u, u'
\end{eqnarray}
consisting of isotropic vectors.
