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 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 $$V^t = \pm \sqrt{(V^x)^2+(V^y)^2+(V^z)^2}$$
As mentioned by Javier in the comments, the set of null vectors is not a vector space, so it's not clear what you mean when you ask for the linearly independent null vectors.
The question is:
What would the linearly independent null vectors of this space be?
Since the null vectors do not form a vector space, the question only makes sense if "this space" is understood to be the whole Minkowski space.
First, there is no need for differentials in a vector space (Minkowski space is a vector space). Choosing an orthonormal basis, the scalar square of a vector $v=(t,x,y,z)$ is $$v^2=-t^2+x^2+y^2+z^2.$$ Let $M$ denote the Minkowski space, and $L$ the set of null vectors, i.e. the subset of $M$ having scalar square $0$. Since for any $\lambda\in\mathbb R^+$, $(\lambda v)^2=\lambda^2 v^2$, if $v$ is in $L$, then $\lambda v$ is also in $L$, that is, $L$ (the set of null vectors) is a linear cone. Since the dimension of $M$ is $4$, any set of linearly independent vectors in $L$ can have at most $4$ elements. An example of a $4$-element set of linearly independent elements in $L$ is $$\{(1,-1,0,0),\ (1,1,0,0), (1,0,1,0), (1,0,0,1)\}$$ This set is a null basis, that is, a basis of $M$ consisting of null vectors.
Note that the fact that two light-like vectors are linearly independent is equivalent to the fact that their scalar product is not $0$.
