Regular Quadratic Space - isotrope vector

Question

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

Answer 1

First of all, I will assume that:

1) The ground field is of characteristic $\neq 2$, as otherwise this statement is just not true. The space $(F_2)^2$ with the form $q(x) = x_1^2 + x_2^2$ is a counterexample: it has a single non-zero isotropic vector, $(1,1)$.

2) In your definition, $f$ has to be symmetric. Otherwise $ker(q)$ is not well-defined, as it would depend on $f$.

Now let's prove that if $V$ is a proper subset of $E$ and $x\in V$ is a non-zero isotropic vector, then there is an isotropic vector $y\not\in V$. Using this, we will be able to do the following: given a non-zero isotropic vector $x_1$, find isotropic vector $x_2\not\in Span(x_1)$; then find $x_3\not\in Span(x_1, x_2)$; and so on, until we construct a basis of $E$.

In order to prove this, we first need to find a vector $z\not\in V$ such that $f(z,x)\neq 0$. By definition of a regular space, $\exists~z_1\in E: f(z_1,x)\neq 0$; if $z_1\not\in V$, then take $z=z_1$, and if $z_1\in V$, then let $z_2$ be an arbitrary vector from $E\backslash V$; if $f(z_2,x)\neq 0$, then take $z=z_2$, and if $f(z_2, x) = 0$, then take $z = z_1+z_2$.

Now set $y=z+cx$, where $c\in K$. Then $q(y) = f(z+cx,z+cx) = f(z,z) + 2cf(z,x)$; if we choose $c = -\frac{f(z,z)}{2f(z,x)}$, then $q(y) = 0$, so $y$ is an isotropic vector and $y\not\in V$.

Answer 2

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.