Null vector space in Minkowski space

Let us consider a Minkowski space of the form: $$ds^2 = -dt^2 + dx^2 + dy^2 +dz^2.$$

What would the linearly independent null vectors of this space be?

I am aware this is a trivial question but is something that has not been made clear to me and so causes me some confusion when reading some of the GR literature.

• The set of null vectors is not a vector space (it's the light cone), so it doesn't really make sense to ask for linearly independent vectors; there are infinitely many of them. – Javier Oct 20 at 22:48

The light cone consists of all vectors $$\mathbf V$$ such that $$\boldsymbol \eta(\mathbf V,\mathbf V) = 0$$ (where $$\boldsymbol \eta$$ is the Minkowski metric). In Cartesian coordinates $$(t,x,y,z)$$, this means that
$$\boldsymbol \eta(\mathbf V,\mathbf V) = -(V^t)^2 + (V^x)^2+(V^y)^2+(V^z)^2 = 0$$
or $$V^t = \pm \sqrt{(V^x)^2+(V^y)^2+(V^z)^2}$$