Let $\mathcal S \subset \mathbb R^n$ be a subspace in $\mathbb R^n$ with dimension $1 \le r < n$. For a matrix $A$, let us denote $A=(a_1, \dots, a_n)$ where $a_j$'s are columns of $A$. Let us define a set \begin{align*} \mathcal E = \{V \in GL_n(\mathbb R): v_1, \dots, v_r \text{ is a basis for } \mathcal S\}, \end{align*} where $v_1, \dots, v_r$ are the first $r$ columns of $V$. I am interested in determining whether $\mathcal E$ has two connected components (or more).
p.s. This is a modified question I asked here.
It has four connected components. Pick a basis $\{u_1,\ldots,u_r\}$ of $S$ and extend it to a full basis $\{u_1,\ldots,u_n\}$ of $\mathbb R^n$. Let $U$ be the augmented matrix of these $n$ basis vectors. Then every $V\in\mathcal E$ can be written in the form of $V=U\pmatrix{P&R\\ 0&S}$, where $P\in GL_r(\mathbb R),\ S\in GL_{n-r}(\mathbb R)$ and $R\in M_{r\times(n-r)}(\mathbb R)$. It follows that $\mathcal E$ has four connected components, which are respectively homeomorphic to \begin{cases} GL_r^+(\mathbb R)\times GL_{n-r}^+(\mathbb R)\times M_{r\times(n-r)}(\mathbb R),\\ GL_r^+(\mathbb R)\times GL_{n-r}^-(\mathbb R)\times M_{r\times(n-r)}(\mathbb R),\\ GL_r^-(\mathbb R)\times GL_{n-r}^+(\mathbb R)\times M_{r\times(n-r)}(\mathbb R),\\ GL_r^-(\mathbb R)\times GL_{n-r}^-(\mathbb R)\times M_{r\times(n-r)}(\mathbb R). \end{cases}