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I working with an open, simply-connected cone in $\mathbb{R}^{n \times n}$ with vertex at the origin, which is the set of all real matrices which have eigenvalues with negative real part (paper on it). This set is obviously invariant under similarity transforms (as these preserve eigenvalues). What does this say about the geometry of the cone? Specifically:

  1. Is it spherical (by which is meant that the intersection of the cone and any hyperplane in $\mathbb{R}^{n\times n}$ that is normal to the axis of the cone is an $(n^2-1)$-ball)?
  2. Can something be said about the direction of it's axis?
  3. Can something be said about its aperture?

If none of the above can be concluded based on this symmetry, then what can?

If you can spot any other symmetries, please do let me know.

EDIT: The first example on this website may apply, if the cone I'm working with is indeed spherical. On the linked site, it is said that

"The automorphism group of [the (bounded, which mine isnt', red.) spherical cone] is the direct product of a subgroup of index 2 of the Lorentz group $O_{n,1}(\mathbb{R})$ (isomorphic to the group of motions of the $n$-dimensional Lobachevskii space) and the group $\mathbb{R}^+$ of homotheties with positive coefficients,"

however, I have barely any idea what this means (haven't studied abstract algebra yet), but I thought it might be relevant.

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    The set of Hurwitz matrices is not a convex set. – 2017-01-04
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    That is, the first sentence is false for $n>1$. – 2017-01-04
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    @copper.hat What is your basis for saying this? But it might very well be true, I think I might have misread that in the article! :S – 2017-01-04
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    The matrices $A=\begin{bmatrix} -1 & 1000 \\ 0 & -1 \end{bmatrix}$ and $A^T$ are both Hurwitz, but their average has an eigenvalue at $499$ so is not Hurwitz. – 2017-01-04
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    @copper.hat Thanks for pointing that out! Does this mean that the geometry of the cone is not so simple anymore, or is it just the higher-dimensional version of a regular 2D or 3D cone with an aperture larger than $\pi$? – 2017-01-04
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    I'm not sure what you mean by simple, but the space is not convex. – 2017-01-04
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    @copper.hat Yeah, sorry, I know it is vague. Do you think that it being non-convex *and* simply-connected implies that the cone is a single structure (and not, say, the union of two or more convex cones), which would necessarily be the higher-dimensional analogue of a 2D or 3D cone with aperture larger than $\pi$? – 2017-01-04

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