isotropic vector-space of a quadratic form Consider the quadratic form $q(x) = x_1x_2 + x_3x_4$. Find a subvector space $V$ of dimension 2 such that $\left.q\right|_V = 0$. What are the possible values for rank and signature of $(V,\left.q\right|_V)$?
After completing it into squares we see that $q$ is of rank $4$ and $(2,2)$ as signature.
I know that such a vector space $V$ is called isotropic vector space. The vectors of $V$ have to satisfy the equation
$$ x_1x_2 = - x_3x_4.
$$
So we could for example choose $V = \{ (1,-1,-1,-1), (-1,0,-1,0)\}$, then $V$ is isotropic of dimension 2 because the two vectors are independent and they satisfy $x_1x_2 = x_3x_4$.
But can we say in general the vectors that generates $V$ are the vectors $(-\frac{x_3x_4}{x_2}, x_2, x_3, x_4)$ with $x_2 \neq 0$, but this is not of dimension $2$ is it?
I know also that maximal isotropic sub vector spaces are all of the same dimension.
 A: You first have to distinguish two things: the (affine) quadric $Q = \{x\in \mathbb{R}^4\,|\, q(x)=0\}$ defined by $q$ (I assume you are working over the real numbers, since you mention the signature, but this is not really relevant for the problem), and the totally isotropic subspaces $V$ (there are infinitely many of them). By definition of "totally isotropic", you always have $V\subset Q$, but $Q$ is not a vector space. It is, by construction, defined by a quadratic equation, not a linear one. So unless $Q=\{0\}$ (we say that $q$ is anisotropic in that case), $Q$ is never a vector space. Make sure it is absolutely clear to you that $(-\frac{x_3x_4}{x_2}, x_2, x_3, x_4)$ is not the formula for a vector space (multiplying or dividing variables is forbidden).
So we need to find a vector space $V$ of dimension $2$ which is included in $Q$. Here $2$ is the maximal possible dimension; in general, for a real quadratic form, if the signature of the form is $(r,s)$, the dimension of a maximal totally isotropic subspace is $\min(r,s)$. The thing is, it is not enough to find two independent isotropic vectors; they also have to be orthogonal. For instance, the vectors you suggest are not suitable: their sum
$$(1,-1,-1,-1)+(-1,0,-1,0) = (0,-1,-2,-1)$$
is not isotropic, so the subspace they generate is not totally isotropic. This is because those two vectors are not orthogonal for the quadratic form $q$ (or rather, for its polar bilinear form).
In the case of this form, it is rather easy to see that the space $V=\{(a,0,b,0)\,|\, a,b\in \mathbb{R}\}$ is totally isotropic, and it clearly has dimension $2$.
