Orbits of vectors under the action of $\mathrm{GL}_n(\mathbb Q)$ Context.
While working on a larger proof, I would love to have the following lemma, but I can't even decide if it's true or not.
The question.
We consider the action of $\mathrm {GL}_n(\mathbb Q)$ on $\mathbb R^n$ such that $\varphi\cdot x=\varphi(x)$ for $\varphi\in \mathrm {GL}_n(\mathbb Q)$ and $x\in\mathbb R^n$.

Let $A$ be a subspace of $\mathbb R^n$ of dimension $2$.
Does there exist $u,v\in A$ linearly independent such that $v$ is in the orbit of $u$ under the action of $\mathrm{GL}_n(\mathbb Q)$?

Remarks.

*

*We can reformulate the question this way: does there exist a rational transformation $\varphi\in\mathrm{GL}_n(\mathbb Q)$ such that $\varphi(u)=v$?


*I managed to prove this result for $n=3$ by constructing a rational rotation which sends $u$ to $v$.


*With a reasoning of cardinality, we can prove that this result is false if we fix $u\in A$, and we try to find $\varphi\in\mathrm{GL}_n(\mathbb Q)$ and $v\in A$ such that $v=\varphi(u)$.


*If $A$ contains a rational vector $x$, the symmetry with respect to $x$ will do the trick, so we can assume that $A$ does not contain any rational vector.


*Moreover, if $A$ intersect non-trivially a rational plane $B$, then we can consider the rotation of axis $B^\perp$ (of dimension $n-2$) and of angle $\pi/2$, and a little work shows this result. So we can assume now that for all rational planes $B$, $A\cap B=\{0\}$.


*I have no idea if this result is even true when $n\geqslant 4$.


*Any hints, references or proofs would be much appreciated.
 A: This is false for $n\geq 4$.  Consider the Grassmannian $\mathrm{Gr}(2,n)$ of all two-dimensional subspaces of $\mathbb{R}^n$, and recall that $\mathrm{Gr}(2,n)$ is a compact manifold of dimension $2n-4$.  For each $\varphi\in\mathrm{GL}_n(\mathbb{Q})$, let
$$
S_\varphi = \{A\in \mathrm{Gr}(2,n) \mid \varphi u=v \text{ for some linearly independent }u,v\in A\}.
$$
By the Baire category theorem, $\mathrm{Gr}(2,n)$ cannot be expressed as a union of countably many closed, nowhere dense sets.  Therefore, it suffices to prove that each $S_\varphi$ is closed and nowhere dense in $\mathrm{Gr}(2,n)$.
To that end, we decompose $S_\varphi$ as a disjoint union $T_\varphi \uplus U_\varphi$, where


*

*$T_\varphi$ is the set of all $A\in S_\varphi$ for which $\varphi(A) \ne A$, and

*$U_\varphi$ is the set of all $A\in S_\varphi$ for which $\varphi(A) = A$.
It suffices to prove that $T_\varphi$ and $U_\varphi$ are closed and nowhere dense in $\mathrm{Gr}(2,n)$.
Claim. $T_\varphi$ is either empty or is a submanifold of $\mathrm{Gr}(2,n)$ of dimension $n-1$.
Proof:  Suppose $T_\varphi$ is nonempty.  If $A\in T_\varphi$, then $A\cap\varphi^{-1}(A)$ is a one-dimensional subspace of $A$, and this contains exactly one pair $\{u,-u\}$ of unit vectors.  Such a $u$ has the property that $u,\varphi u\in A$ and $\{u,\varphi u,\varphi^2 u\}$ are linearly independent.  Let
$$
\widetilde{T_\varphi} = \{u\in \mathbb{R}^n : \|u\|=1\text{ and }u,\varphi u,\varphi^2 u\text{ are linearly independent}\}.
$$
Then $\widetilde{T_\varphi}$ is an open subset of the unit $(n-1)$-sphere in $\mathbb{R}^n$ and the map $p\colon \widetilde{T_\varphi}\to T_\varphi$ defined by $p(u) = \mathrm{Span}\{u,\varphi u\}$ is a degree two covering map, which proves the claim. $\square$
Since $n-1 < 2n-4$ for $n\geq 4$, this gives us the following.
Corollary. $T_\varphi$ is closed and nowhere dense in $\mathrm{Gr}(2,n)$ as long as $n\geq 4$.
Claim. $U_\varphi$ is a union of finitely many submanifolds of $\mathrm{Gr}(2,n)$, all of dimension at most $n-2$.
Proof: We separate the possible $A\in U_\varphi$ into three types, based on the eigenvalues of the restriction of $\varphi$ to $A$:


*

*The restriction of $\varphi$ to $A$ has two distinct real eigenvalues $\lambda,\mu$.

*The restriction of $\varphi$ to $A$ has one real eigenvalue $\lambda$ and is not diagonalizable.

*The restriction of $\varphi$ to $A$ has two complex eigenvalues $\lambda,\overline{\lambda}$.


In each case the eigenvalues of the restriction must also be eigenvalues of $\varphi$, of which there are only finitely many.  Our strategy is to analyze the set of all $A$ of a given type corresponding to a given eigenvalue or pair of eigenvalues.
For type (1), let $\lambda$ and $\mu$ be distinct real eigenvalues of $\varphi$, and let $E_\lambda$ and $E_\mu$ be the corresponding eigenspaces.  Then any $A$ corresponding to $\lambda$ and $\mu$ can be written uniquely as the sum of a one-dimensional subspace of $E_\lambda$ and a one-dimensional subspace of $E_\mu$.  If $\dim(E_\lambda) = d_\lambda$ and $\dim(E_\mu) = d_\mu$, then the set of all such $A$ is homeomorphic to $\mathrm{Gr}(1,d_\lambda) \times \mathrm{Gr}(1,d_\mu)$, which is a manifold of dimension $d_\lambda+d_\mu - 2$.  In particular, since $d_\lambda+d_\mu \leq n$, the set of all such $A$ for a given pair $\lambda,\mu$ is a submanifold of $\mathrm{Gr}(2,n)$ of dimension at most $n-2$.
For type (2), let $\lambda$ be a real eigenvalue of $\varphi$ with higher algebraic multiplicity than geometric multiplicity.  Let $E_\lambda$ be the eigenspace for $\lambda$ and let $E_\lambda'$ be the nullspace of $(\varphi-\lambda I)^2$.  Then any $A$ of type (2) corresponding to $\lambda$ has one-dimensional image in $E_\lambda'/E_\lambda$ and is entirely determined by this image. If $\dim(E_\lambda) = d_\lambda$ and $\dim(E_\lambda') = d_\lambda'$, then the set of all such $A$ is homeomorphic to $\mathrm{Gr}(1,d_\lambda'-d_\lambda)$, which is a manifold of dimension $d_\lambda'-d_\lambda - 1$. In particular, since $d_\lambda'-d_\lambda \leq n-1$, the set of all such $A$ for a given $\lambda$ is a submanifold of $\mathrm{Gr}(2,n)$ of dimension at most $n-2$.
For type (3), let $\lambda$ be a complex eigenvalue of $\varphi$, and let $E_\lambda$ be the eigenspace for $\lambda$ in $\mathbb{C}^n$.  Then any $A$ of type (3) corresponding to $\lambda$ is obtained by taking a subspace of $E_\lambda$ of complex dimension one and taking the real part of each vector.  If $\dim_{\mathbb{C}}(E_\lambda) = d_\lambda$, then the set of all such $A$ is homeomorphic to the complex Grassmannian $\mathrm{Gr}_{\mathbb{C}}(1,d_\lambda)$, which is a manifold of real dimension $2d_\lambda-2$. In particular, since $2d_\lambda \leq n$, the set of all such $A$ for a given $\lambda$ is a submanifold of $\mathrm{Gr}(2,n)$ of dimension at most $n-2$. $\square$
Corollary. $U_\varphi$ is closed and nowhere dense in $\mathrm{Gr}(2,n)$ for all $n\geq 3$.
Incidentally, what's going on here from an algebraic perspective should be roughly that each $S_\varphi$ is an algebraic subvariety of $\mathrm{Gr}(2,n)$ of dimension $n-1$, with $T_\varphi$ being the set of regular points of $S_\varphi$ and $U_\varphi$ being its set of singular points, but we don't need to know any of that to provide a topological proof that it's nowhere dense in $\mathrm{Gr}(2,n)$.
