Good news first. Given a 2-dimensional subspace of (or ) such that (), we can always find a coordinate subspace such that and are in generic position. (Recall that generic position means , which is feasible only when the dimension of and is half the dimension of the ambient space.)
Indeed, suppose that none of the subspaces , , are in generic position with . By the pigeonhole principle, either or meets at least two of these subspaces. We may assume meets and . Since is two-dimensional, it follows that . Hence , a contradiction.
The story is different in and higher dimensions. Consider the space
The condition holds. Yet, for any 3-dimensional coordinate space , either or its complement contains at least two of vectors , and therefore intersect . Damn perfidious pigeons.
So it’s not enough to demand that does not contain any basis vectors. Let’s ask it to stay away from the basis vectors, as far as possible. By the Pythagorean theorem, the maximal possible distance of to is , attained when is equidistant from and . Let’s call an equidistant subspace if this holds for all . There are at least two other natural ways to express this property:
- In the basis , the projection onto is a matrix with 1/2 on the diagonal
- Reflection across sends into a vector orthogonal to . As a matrix, this reflection has 0s on the diagonal.
In two dimensions, the only equidistant subspaces are and . In higher dimensions they form a connected subset of the Grassmannian (self-promotion).
Is every equidistant subspace in generic position with some coordinate subspace?
We already saw that the answer is yes when . It is also affirmative when (G. Weiss and V. Zarikian, “Paving small matrices and the Kadison-Singer extension problem”, 2010). I am 75% sure that the answer is yes in general.