Matrices of rank at least k are path connected

Idea. The two components of $GL_n$ are connected to each other *through* the rank $n-1$ matrices.

Step 1: the three pieces. $\mathcal{R}=GL_n^+\cup GL_n^-\cup\mathcal{M}_{n-1}$ with $\mathcal{M}_{n-1}$ the set of rank $n-1$ matrices. We know:
- $GL_n^+$ is path connected (Problem 8).
- $GL_n^-=D\cdot GL_n^+$ with $D=\operatorname{diag}(-1,1,\dots,1)$ is path connected too (image of $GL_n^+$ under the continuous map $A\mapsto DA$).
- $\mathcal{M}_{n-1}$ is path connected (Problem 21, case $r=n-1<n$, whose proof does not use the present problem).

Step 2: joining them. Consider $E(s)=\operatorname{diag}(1,\dots,1,s)$, $s\in[-1,1]$. For $s>0$, $E(s)\in GL_n^+$; for $s<0$, $E(s)\in GL_n^-$; and $E(0)$ has rank $n-1$. So the path $s\mapsto E(s)$ runs inside $\mathcal{R}$ from a point of $GL_n^-$ ($s=-1$) through a point of $\mathcal{M}_{n-1}$ ($s=0$) to a point of $GL_n^+$ ($s=1$).

Step 3: union of connected sets. Let $X_1=GL_n^+$, $X_2=\mathcal{M}_{n-1}$, $X_3=GL_n^-$. Each is path connected, $X_1\cup X_2$ is path connected because the path $E(s)$, $s\in[0,1]$, links a point of $X_2$ to a point of $X_1$, and similarly $X_2\cup X_3$. Hence $\mathcal{R}=X_1\cup X_2\cup X_3$ is path connected.

Remark. The condition rank $\ge n-1$ is sharp: the set of matrices of rank $\ge n-2$ is also connected, but the argument for rank $\ge n-1$ is what allows the determinant to change sign continuously. In particular, a continuous path from $GL_n^+$ to $GL_n^-$ in $M_n(\mathbb{R})$ *must* pass through singular matrices (intermediate value theorem), but need not pass through a matrix of rank $<n-1$.