Isotropic Subspace Implying the Existence of a Linear Equation Issue Given a quadratic form in $n = 2 \nu + 1$ complex dimensions written as
$$F = (x^0)^2 + x^1 x^{1'} + ... + x^{\nu} x^{\nu'}$$
a vector $x = (x^0,x^1,...,x^{\nu},x^{1'},...,x^{\nu'})$, with $x^0,x^i,x^{i'} \in \mathbb{C}$, $i = 1,...,\nu$, is isotropic if it satisfies
$$F = 0.$$
Given this setup, I would like to understand the following passage here (from this book) which I reproduce in the following quote:

We have seen in Chapter I (Section 10) that any isotropic subspace (i.e., a subspace in which all the vectors are isotropic) has dimension at most $\nu$. If the equations defining an isotropic subspace do not include an equation connecting $x^0,x^1,...,x^{\nu}$ it would be possible to express the $x^{i'}$ components of each vector in the subspace as the same linear combinations of the $x^0,x^1,...,x^{\nu}$ components which are arbitrary: such vectors do not satisfy $F = 0$. We shall establish the equations of an isotropic $\nu$-plane assuming the general case where there is no linear relation between $x^1,x^2,...,x^{\nu}$.

I believe the first sentence can be understood as follows: Any $p$-dimensional isotropic subspace generated by $p$ basis vector $\mathbf{e}_1,...,\mathbf{e}_p$, and the space orthogonal to this space has dimension $n-p$, however since the $\mathbf{e}_i$, $i=1,...,p$, are orthogonal to themselves they live among these $n-p$ vectors so there are only $n-2p$ basis vectors remaining in the total space. Clearly we must have $n - 2p \geq 0$ which means $2p \leq 2 \nu + 1$ so that $p \leq \nu$.
The remaining sentences are trying to justify the fact that a relation of the form
$$\eta_0 := \xi_0 x^0 + \xi_1 x^1 + ... + \xi_{\nu} x^{\nu} = 0$$
should exist (where the $\xi_0,\xi_i$, $i=1,...,\nu$, are complex coefficients). The abstract existence of this relation is what I am mainly trying to understand from this passage.
In the case of $n = 3$ it can be inferred directly by taking $F = (x^0)^2 + x^1 x^{1'} = 0$ and writing $x^0 = \sqrt{-x^1} \sqrt{x^{1'}} = \xi_0 \xi_1$ so that $\xi_0 x^0 = - x^1 \xi_1$ implies $\eta_0 = \xi_0 x^0 + \xi_1 x^1 = 0$, however I have no idea what he's trying to say regarding what would happen if we assumed this wasn't possible and what it would imply about $x^{1'}$ being expressible in terms of 'the same linear combination' of $x^0,x^1$ (which are arbitrary and so do not satisfy $F=0$...?), and so cannot see how to directly infer the existence of $\eta_0 = \xi_0 x^0 + \xi_1 x^1 = 0$ without any computations (so that it can easily generalize to higher dimensions).
A potentially ridiculous idea is to re-write $F$ as
$$F = (x^0,x^1) \cdot (x^0,x^{1'}) = 0$$
which implies (?) the existence of other vectors $(\xi_0,\xi_1)$ orthogonal to $(x^0,x^1)$ so that
$$\eta_0 = (x^0,x^1) \cdot (\xi_0,\xi_1) = \xi_0 x^0 + \xi_1 x^1 = 0$$
holds, it's doubtful he means something like this, but it's worth determining how valid this is. Indeed from it we find
$$F = \frac{1}{\xi_0}[\xi_0 x^0 x^0 + \xi_0 x^1 x^{1'}] = \frac{x^1}{\xi_0}[-\xi_1 x^0 + \xi_0 x^{1'}] = \frac{x^1}{\xi_0} (x^{1'},-x^0) \cdot (\xi_0,\xi_1) := \frac{x^1}{\xi_0} \eta_1 = 0$$
which is important later on in this discussion.
Any idea what's going on?
 A: I think this is basically about Gaussian elimination. Consider a subspace $V$ of $\mathbb C^{2\nu+1}$ with $\dim(V) \leq \nu$ (eg. any isotropic subspace). Then $V$ can be defined by a system $(\Sigma)$ of $m \geq \nu + 1$ independent linear equations:
$$(\Sigma) : \forall 1 \leq i \leq m, \quad \quad \quad a_{i,0}x^0 + \sum_{j=1}^{\nu} (a_{i,j}x^j + a'_{i,j} x^{j'}) = 0.$$
Since we have at least $\nu+1$ independant equations, we may combine them to eliminate all the $\nu$ variables $x^{1'}, \ldots, x^{\nu'}$ in one equation by the process of Gaussian elimination. This equation will give you a non trivial relation of the form $\eta_0$.

Edit: (on request of the comment)
In the quote, the author intends to prove that the vectors of any isotropic subspace satisfy a non trivial linear equation in the coordinates $x^0,\ldots ,x^{\nu}$. They prove it by contraposition: if the vectors of my subspace does not satisfy any such equation, then it wouldn't be isotropic. I personally find their justification a little bit difficult to grasp, so I suggest proving the statement directly, without contraposition.
On the other hand, for $d\geq 0$, any vector subspace $V$ of $F^n$ of dimension $k$ can be determined by $n-k$ independant linear equations in the coordinates of the vectors. Equivalently, in a more abstract way, any $k$-dimensional subspace of a vector space of dimension $n$ is the intersection of $n-k$ hyperplanes. Thus, in the context of the question above, since $V$ is assumed to by isotropic, it has dimension $k \leq \nu$. Thus it is determined by $m := (2\nu+1) - k \geq \nu + 1$ independent linear equations.
I claim that from now, the isotropy of $V$ is not needed anymore. In fact, the vectors inside any vector subspace of $F^{2\nu+1}$ of dimension $\leq \nu$ must satisfy a non trivial equation in the coordinates $x^0,\ldots ,x^{\nu}$ only. To prove this, we use Gaussian elimination. Let's go back to the system of equation $(\Sigma)$ as above. We want to make linear combinations of the equations $(\Sigma_i)$ for $1\leq i \leq m$,
$$(\Sigma_i):\quad a_{i,0}x^0 + \sum_{j=1}^{\nu} (a_{i,j}x^j + a'_{i,j} x^{j'}) = 0,$$
in order to cancel all the variables $x^{1'},\ldots,x^{\nu'}$. This is possible because there are $m \geq \nu + 1$ equations, strictly more than the number $\nu$ of variables we want to eliminate. Thus, there exists a non trivial linear combination
$$b_1(\Sigma_1) + \ldots + b_m(\Sigma_m)$$
of the equations $(\Sigma_i)_{1\leq i \leq m}$ such that no variable $x^{1'},\ldots,x^{\nu'}$ occurs in it. Observe that the non triviality of this linear combination is guaranteed by the independence of the equations $(\Sigma_i)$'s. Thus, we have obtained a relation of the form $\eta_0$ as expected.
A: The idea behind the passage is to establish the existence of a specific linear relation between x0,x1,…,xν components of a vector in an isotropic subspace. The passage explains that if there is no such relation, it would be possible to express the xi′ components of each vector in the subspace as the same linear combinations of the x0,x1,…,xν components. This would mean that the vectors in the subspace do not satisfy F=0, which contradicts the definition of isotropic subspace. Therefore, the existence of a linear relation of the form η0=ξ0x0+ξ1x1+…+ξνxν=0 is established.
The author does not go into details about how to directly infer the existence of such a relation, but it is clear that the idea is to use the orthogonality property of vectors in an isotropic subspace. The validity of your potentially ridiculous idea to rewrite F as F=(x0,x1)⋅(x0,x1′)=0 is not clear, as it is not specified in the passage.
