Nilpotent matrices and idempotents of fixed rank

1. Nilpotent matrices. Done in Problem 3: $(tA)^m=t^mA^m=0$ when $A^m=0$, so $t\mapsto tA$ joins $A$ to $0$.

2. Idempotents of rank $k$.

*Step 1: conjugacy.* If $P^2=P$ then $\mathbb{R}^n=\operatorname{Im}P\oplus\ker P$ (every $x=Px+(x-Px)$, and $P$ is the identity on $\operatorname{Im}P$). In a basis adapted to this decomposition, $P$ becomes $E_k=\operatorname{diag}(I_k,0)$ with $k=\operatorname{rank}P$. So
$$\mathcal{P}_k=\{QE_kQ^{-1}:Q\in GL_n(\mathbb{R})\}.$$

*Step 2: we may take $\det Q>0$.* Let $D=\operatorname{diag}(-1,1,\dots,1)$. It is diagonal, so $DE_k=E_kD$. If $\det Q<0$, put $Q'=QD$. Then $\det Q'>0$ and
$$Q'E_kQ'^{-1}=QDE_kD^{-1}Q^{-1}=QE_kQ^{-1}.$$
So $\mathcal{P}_k=\{QE_kQ^{-1}:Q\in GL_n^+(\mathbb{R})\}$.

*Step 3.* The map $GL_n^+\to M_n$, $Q\mapsto QE_kQ^{-1}$ is continuous (the inverse is a rational function of the entries with non-vanishing denominator $\det Q$). Its image is $\mathcal{P}_k$, so $\mathcal{P}_k$ is the continuous image of a path connected set, hence path connected.

Remark. Why is $0<k<n$ needed? For $k=0$ and $k=n$ the set is a single point ($\{0\}$ or $\{I\}$), trivially connected. The role of $D$ in step 2 is the crucial subtlety: without a negative-determinant matrix in the centraliser of $E_k$, an orbit of $GL_n$ would split in two (see the case of $J$ in Problem 29).