Notations.
Let $n\ge 4$ be an integer, $d,e\in\{1,\ldots,n-1\}$ and $j\in\{1,\ldots,\min(d,e)\}$.
We say that a subspace of $\mathbb R^n$ is rational if it admits a rational basis. We denote by $\mathfrak R_n(e)$ the set of rational subspaces of dimension $e$.
We denote by $\mathfrak I_n(d,e)_j$ the set of subspaces $A$ of $\mathbb R^n$ of dimension $d$ such that
$$\forall B\in\mathfrak R_n(e),\quad \dim(A\cap B)<j.$$
The question.
We want to show that if $d+e\leqslant n$ and $j\in\{1,\ldots,\min(d,e)\}$, then
$$A\in\mathfrak I_n(d,e)_j\implies A^\perp\in\mathfrak I_n(n-d,n-e)_j.$$
What I did.
I don't know if this result is true, but I can't find a counterexample.
If we assume that there exists $B\in\mathfrak R_n(n-e)$ such that $\dim(A^\perp\cap B)\geqslant j$, then there exists $C$ a subspace of $\mathbb R^n$ of dimension $k\geqslant j$ such that
$$A^\perp\cap B=C,$$
then
$$(A^\perp\cap B)^\perp=C^\perp$$
so
$$A+B^\perp=C^\perp,$$
so
$$d+e\leqslant n-j$$
but it does not help that much...
Any help or reference to prove or disprove the result would be of great help!