A linear transformation $T$ such that $(v_1, \dots, v_k, Tv_1, \dots, Tv_k)$ is a basis for $\mathbb F^{2k}$

Question

Let $T: \mathbb R^{2k} \to \mathbb R^{2k}$ be a bijective linear transformation. We define a set \begin{align*} S = \{ (v_1, \dots, v_k) \subset \mathbb R^{2k}: (v_1, \dots, v_k, Tv_1, \dots, Tv_k) \text{ is a basis for $\mathbb R^{2k}$ } \}. \end{align*} Is it possible to characterize the set $S$? In particular, the topological property, such as connectedness?

Answer 1

A bit too long for a comment:

Per this construction, we can identify the $k$ dimensional subspaces of $\Bbb R^k$ with the corresponding orthogonal projections, i.e. the symmetric $2k\times 2k$ matrices satisfying $X^2 = X$ and $\operatorname{tr}(X) = k$. A set $B = (v_1,\dots,v_k)$ will be contained in $S$ if and only if the subspace $U$ spanned by $B$ satisfies $$ U \cap T(U) = \{0\} $$ This is in turn equivalent to the statement that the projection $X$ onto $U$ satisfies $$ XTX = 0 $$ So, if you'd like to characterize the space of $k$-dimensional subspaces of $V$ satisfying $V = U \oplus T(U)$, then you could identify this set with the set of projection matrices $X$ given by $$ \{X \in \Bbb R^{2k \times 2k} : X = X^T, X = X^2, \operatorname{tr}(X) = k, XTX = 0\} $$