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!

  • $\begingroup$ $\{(a,b,c,d)\in\mathbb{R}^4:\ ad-bc=0\}$ is singular at the origin $a=b=c=d=0$. $\endgroup$ – user580373 Aug 2 '18 at 3:42
  • $\begingroup$ One can test if $S=\{(a,b,c,d)\in\mathbb{R}^4:\ ad-bc=0\}$ is a submanifold using the Jacobian criterion. The Jacobian of $ad-bc$ is $(d,-c,-b,a)$, which has rank $4-(4-1)=1$ at all points of $S$, except for the origin $a=b=c=d=0$. Therefore, $S$ is not a smooth submanifold of $\mathbb{R}^4$. 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 $t\mapsto (t^2,t^3)$. $\endgroup$ – user580373 Aug 2 '18 at 13:45
  • $\begingroup$ The question if it is a topological manifold has been asked here before but no one has answered. $\endgroup$ – user580373 Aug 2 '18 at 13:46

No, it's not true. A square matrix is singular iff its determinant is $0$. That means the singular matrices form a variety of dimension $n^2-1$ in the $n^2$-dimensional space of $n \times n$ matrices. But it's not a manifold. Thus in the case $n=2$, the cross-section $d=0$ of the set of singular matrices $\pmatrix{a & b\cr c & d\cr}$ is the union of the two planes $b=0$ and $c=0$ intersecting at right angles.

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    $\begingroup$ A cross section of a manifold can be two planes meeting at right angles, for example $xy - z = 0$ in R^4. $\endgroup$ – Lorenzo Aug 2 '18 at 5:37
  • $\begingroup$ (I like your answer I just think that point is potentially misleading.) $\endgroup$ – Lorenzo Aug 2 '18 at 5:43

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