I'm reading baby Rudin, and I know that the set of invertible matrices forms an open set (under the topology induced by the metric induced by the norm $||A|| = \sup{|Ax|}$ where $x \in \mathbb{R}^n$). I also know that this set is a disconnected union of two connected components: matrices with positive determinant and matrices with negative determinant.
My topological intuition says the set of singular matrices is an $n$-manifold, and my guess is that $n = 2$. I am fairly confident that this is true for 2x2 matrices (because it can be identified with $\mathbb{R}^4$; I know they have different norms) but it quickly becomes difficult to picture this (nine dimensions are a bit too much for me). I would be very interested to see a proof, or a sketch of one, on this conjecture, but I unfortunately don't have a great deal of experience with manifolds or local homeomorphisms. Thanks!
Now, this criterion doesn't tell you if it is a topological manifold or not. For example, ${(x,y)\in\mathbb{R}^2:\ y^2-x^3=0}$ doesn't satisfy the Jacobian criterion at $(0,0)$, but it is a topological manifold anyways. You can parameterize it with the chart $t\mapsto (t^2,t^3)$.
– Aug 02 '18 at 13:45