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Some context :

(Notations : For the rest of the post, I will identify $\mathrm{M}_n(\mathbb{C})$ with $\mathbb{C}^{n^2}$, and I will note $\mathcal{S}^{n^2-1}$ the unit sphere of $\mathbb{C}^{n^2}$)

I was solving this small exercise : "Show that any hyperplane of $\mathrm{M}_n(\mathbb{C})$ contains an invertible matrix".

From there I was able to prove that any circle $\mathcal{S}^{n^2-1}\cap H$ (btw is this $\mathcal{S}^{n^2-2}$ ? or homeomorphic to it ?) contains an invertible matrix.

What I would like to prove (if it is true) is that any spherical sector of a circle contains an invertible matrix.

The problem is that I don't know how to generalize a circular arc to higher dimension. What I would like is the equivalent of saying "the circular arc from $0$ to $\pi/2$" when working with $\mathcal{S}^{1}$.

Any help would be great !

Edit :

I tried to define a spherical sector of angle $\alpha$ by saying it is the intersection between the sphere and the following convex cone around the matrix $P$ : $$ \mathcal{C}_\alpha(P) = \{ M \in \mathrm{M}_n(\mathbb{C}) : \frac{\langle M\,,P\rangle}{\|M\|\|P\|} \geq \cos(\alpha) \} $$ Is this a good way to define it ? Does it truely define what I think it does ?

My conjecture would then be :

$$ \forall \, H, \quad \forall \, P \in H\setminus \{0\}, \quad \forall \, \alpha \in ]0, 2\pi], \quad \quad \mathcal{C}_\alpha(P) \cap \mathcal{S}^{n^2-1}\cap H \cap \mathrm{GL}_n(\mathbb{C}) \neq \emptyset $$

$H$ being an hyperplane.

Timothe Schmidt
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1 Answers1

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See https://math.stackexchange.com/a/2026205 for a proof that $\mathrm{GL}_n(\mathbb{C})$ is dense in $\mathbb{C}^{n^2}$. Now, whatever your definition of a spherical sector $S$ is, it should contain a non-empty open set of $S^{2n^2 - 1}$. Switching to polar coordinates, $$\{x \in \mathbb{C}^{n^2} \mid ||x|| \in (1-\epsilon, 1 + \epsilon), x/||x|| \in S\} \cong S \times (1-\epsilon, 1 + \epsilon) $$ is open in $\mathbb{C}^{n^2} - 0 \subset \mathbb{C}^{n^2}$ and therefore contains an invertible matrix $x$. But then $x/||x|| \in S$ is also invertible.

We can now consider the situation where we restrict everything to a linear subspace $L$, ie replace $\mathrm{GL}_n(\mathbb{C})$ by $\mathrm{GL}_n(\mathbb{C}) \cap L$ (which is still the complement of the zero set of $\det$, restricted to $L$), $S$ by $S \cap L$ (which is now open in $S^{2n^2 - 1} \cap L$) etc. As long as $L$ contains some non-singular matrix, $\det$ restricted to $L$ must be a non-zero polynomial, so $\mathrm{GL}_n(\mathbb{C}) \cap L$ will be dense in $L$, and the same argument goes through. This is guaranteed if $\dim L > n^2 - n$, by Max dimension of a subspace of singular $n\times n$ matrices, in particular if $L$ is a hyperplane.

ronno
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  • Thank you very much for your answer ! The key insight I was missing is that $\mathrm{GL}_n(\mathbb{C}) \cap H$ is still dense in $H$ – Timothe Schmidt May 14 '24 at 01:35
  • Now is there a limit to the number of dimensions we can substract and still have invertible matrices ? Like what if we intersect again with a new hyperplane (not parallel to the previous one) ? and again ? Is there a point where this does not work ? I guess my question is kind of "How dense are the invertible matrices ?" Are there more powerful results that makes my questions trivial ? – Timothe Schmidt May 14 '24 at 01:41
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    @TimotheSchmidt ${\det \ne 0}$ is dense as soon as it is non-empty, so $\dim H > n^2 + n$ works: see https://math.stackexchange.com/questions/66877/max-dimension-of-a-subspace-of-singular-n-times-n-matrices – ronno May 14 '24 at 06:46
  • Thanks ! Btw didn't you mean $\dim H > n^2 - n$ ? Would you agree then that the best I can do with this approach is this result : $$ \forall , P \in \mathcal{S}^{2n^2-n}, \quad \forall , \alpha \in ]0, 2\pi], \quad \quad \mathcal{C}_\alpha(P) \cap \mathcal{S}^{2n^2-n}\cap \mathrm{GL}_n(\mathbb{C}) \neq \emptyset $$ $ \mathcal{S}^{2n^2-n}$ being the intersection between the $(2n^2-1)$-unit sphere in $\mathbb{R}^{2n^2}$ and a subspace of dimension $n^2 -(n-1)$ – Timothe Schmidt May 15 '24 at 00:45
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    Yes, $n^2 - n$, and I would write that intersection sphere as $S^{2n^2 - 2n + 1}$ since it's the unit sphere in $\mathbb{C}^{n^2 -n + 1} = \mathbb{R}^{2n^2 - 2n + 2}$. – ronno May 15 '24 at 06:09
  • Thanks again ! btw this is my first time asking a question, should I delete my comments on your answer now that you have edited it ? – Timothe Schmidt May 15 '24 at 07:08
  • I try to incorporate relevant comments into my answers since they are in principle ephemeral. It's reasonable to delete the comments that are completely covered by the (edited) answer since they do little but add noise. But you should keep any that you think adds to the post. – ronno May 15 '24 at 07:15