1
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ $\{(a,b,c,d)\in\mathbb{R}^4:\ ad-bc=0\}$ 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
4
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.