TITLE: Is the set of singular matrices ever a differentiable manifold?
QUESTION [17 upvotes]: I can see that invertible matrices are a differentiable manifold however I don't know how to show that something is not a differentiable manifold so easily.
Is it ever the case that singular matrices form a differentiable manifold?

REPLY [7 votes]: In the case $n=2$, consider $C = \{(x_{11},x_{12},x_{21},x_{22}) \in {\mathbb R}^4: x_{11} x_{22} - x_{12} x_{21} = 0\}$.  With the change of variables $y=x_{11}+x_{22},z=x_{11}-x_{22},v=x_{12}+x_{21},w=x_{12}-x_{21}$, this becomes
$\{(y,z,v,w) \in {\mathbb R}^4: y^2 - z^2 - v^2 + w^2 = 0\}$.  The intersection of this with the unit sphere is
$\{(y,z,v,w) \in {\mathbb R}^4: y^2 + w^2 = z^2 + v^2 = 1/2\}$, which is the Cartesian product of two circles, i.e. a $2$-torus.  But in ${\mathbb R}^3$ a region whose boundary is a $2$-torus is not simply connected.  So no neighbourhood of the origin in $C$ is homeomorphic to ${\mathbb R}^3$.