Invertible matrices acting on set of subspaces of the same dimension

Question

I'm reading a paper and have stumbled upon the following phrases (everything over an arbitrary field $\mathbb{K}$):

"As the natural action of $\operatorname{GL}_n(\mathbb{K})$ on the set of linear hyperplanes of $\mathbb{K}^n$ is transitive..."

"Let $H$ be a subspace of $\mathbb{K}^n$ of codimension $2$. Since $\operatorname{GL}_n(\mathbb{K})$ acts transitively on the set of $(n − 2)$-dimensional linear subspaces of $\mathbb{K}^n$, we may also assume that $H$ is the subspace defined by the following system of two (independent) linear equations... "

What is this action exactly? The first thing I could think of was that it was simply permutation matrices permuting the basis vectors of each subspace in the set, and my initial reaction was to choose (e.g. for the hyperplane case) the orbit of the vectors with last coordinate $0$, whence (by permutation), we obtain the subspaces with $0$ each of the $n$ coordinates. However, I stumbled upon the following obstacles:

1. I think (I may be wrong on this) that, for this to be valid, we first need to choose a basis for each subspace, and I don't know if this is permitted.

2. Even after choosing a basis, I can see how to include subspaces, whose basis is a subset of the standard basis, in the orbit, but there are infinitely many hyperplanes and permutting a subset of the standard basis to get a not-so-nice looking basis for another subspace is not clear to me.

3. As I'm writing 2), I realise that I have probably made a choice of basis implicitly when I'm speaking of permutation matrices: They contain $0$'s and $1$'s, so they are supposed to permute the vectors of the standard basis, so now I'm asking myself how would we go about permuting vectors of a non-standard basis. Off the top of my head, I can think of Gram-Schmidt, but this does not feel like it solves the problem completely.

I'm sorry about the mumbo-jumbo, but I'm really trying to understand this intuitively, so I may be overthinking quite a bit. Is there any merit in my thoughts, or the action does not have anything to do with permutation matrices whatsoever?

Answer 1

$\text{GL}_n(\mathbb{K})$ acts on all points of an $n$-dimensional vector space by left multiplication, since an element of $\text{GL}_n(\mathbb{K})$ is a matrix.

This implies though that all points of a hyperplane will be mapped onto the points of a (possibly different) hyperplane. And analogously for other dimensional subspaces. That is what they mean by $\text{GL}_n(\mathbb{K})$ acting on hyperplanes/$(n-2)$-spaces.

The fact that those actions are transitive comes from the fact that one can map every basis to any other basis using elements of $\text{GL}_n(\mathbb{K})$. Since if you have bases $B_1$ and $B_2$ and $Q_1$ and $Q_2$ are the matrices with as columns the elements of respectively $B_1$ and $B_2$, then $Q_2^{-1}\cdot Q_1$ maps the first basis onto the second and is an element of $\text{GL}_n(\mathbb{K})$ $(Q_1$ and $Q_2$ are elements of $\text{GL}_n(\mathbb{K})$ since their columns are linearly independant, so their product is also element of $\text{GL}_n(\mathbb{K}))$.

Any linearly independant set of $k$ vectors that span a subspace can be expanded to a basis and then mapped onto another basis where the first $k$ vectors are a linearly independant set that span another subspace (of the same dimension) of your choice.