Matrices of fixed rank

Idea. The singular value decomposition parametrises $\mathcal{M}_r$ by rotations and positive numbers, all of which are connected.

Step 1: SVD. Every $A$ of rank $r$ can be written $A=P\Sigma Q^T$ with $P,Q\in O(n)$ and
$$\Sigma=\operatorname{diag}(\sigma_1,\dots,\sigma_r,0,\dots,0),\quad\sigma_i>0.$$

Step 2: for $r<n$, we may assume $P,Q\in SO(n)$. Let $D=\operatorname{diag}(1,\dots,1,-1)$. Since $r<n$, the last diagonal entry of $\Sigma$ is $0$, so $D\Sigma D=\Sigma$. Therefore
$$(PD)\,\Sigma\,(QD)^T=PD\Sigma DQ^T=P\Sigma Q^T=A.$$
If $\det P=-1$ replace $P$ by $PD\in SO(n)$; if $\det Q=-1$ replace $Q$ by $QD\in SO(n)$ (independently).

Step 3: the path ($r<n$). Choose paths $P_t,Q_t$ in $SO(n)$ from $P,Q$ to $I$ (Problem 16) and let $\Sigma_t=\operatorname{diag}((1-t)\sigma_1+t,\dots,(1-t)\sigma_r+t,0,\dots,0)$. Then
$$A_t=P_t\Sigma_tQ_t^T$$
is continuous, has rank exactly $r$ for all $t$ (the first $r$ diagonal entries of $\Sigma_t$ stay positive, and rank is preserved by multiplication by invertible matrices), and goes from $A$ to $E_r=\operatorname{diag}(I_r,0)$. Every matrix of rank $r$ is joined to $E_r$ inside $\mathcal{M}_r$, so $\mathcal{M}_r$ is path connected.

Step 4: the case $r=n$. $\mathcal{M}_n=GL_n(\mathbb{R})$ has exactly two components $GL_n^+$ and $GL_n^-$ (Problems 2 and 8).

Remark. The argument fails for $r=n$ precisely because $D\Sigma D\ne\Sigma$ is not a symmetry any more: the sign of $\det P\det Q$ cannot be changed without changing $A$. Topologically, $\mathcal{M}_r$ is a connected manifold of dimension $r(2n-r)$.