Singular matrices and the nilpotent cone are star-shaped

Idea. Show that each set is *star-shaped about $0$*: if $A$ is in the set, so is $tA$ for $t\in[0,1]$.

Singular matrices. If $\det A=0$, then $\det(tA)=t^n\det A=0$ for every $t$. So $t\mapsto tA$, $t\in[0,1]$, is a continuous path inside the set joining $A$ (at $t=1$) to $0$ (at $t=0$).

Nilpotent matrices. If $A^k=0$, then $(tA)^k=t^kA^k=0$. The same path works.

Conclusion. Every point is joined to $0$ inside the set, so by concatenating two such paths (and reversing one) any two points are joined. Both sets are path connected.

Remark. The set of singular matrices is the hypersurface $\{\det=0\}$, which is *not* a manifold at $0$ (it is a cone with a singular vertex), but it is connected. By contrast, its complement $GL_n(\mathbb{R})$ is disconnected (Problem 2).