Why is the boundary of spectrahedra “more pointy” at matrices of lower rank?

Asked Sep 03 '21 at 16:58

Active Sep 03 '21 at 21:15

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In the following expository article about spectrahedra, it is established informally that the boundary of spectrahedra is “more pointy” at matrices of lower rank.

Cynthia Vinzant, What is a... Spectrahedron?, Notices of the AMS, Volume 61, Number 5, May 2014.

Is there any formal statement related to this property? Is there in general, a relationship between the geometric dimension in $\mathbb R^n$ and the rank of the corresponding matrix?

edited Sep 03 '21 at 21:15

Rodrigo de Azevedo

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asked Sep 03 '21 at 16:58

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Have you taken a look at Blekherman & Parrilo & Thomas's Semidefinite Optimization and Convex Algebraic Geometry? – Rodrigo de Azevedo Sep 03 '21 at 21:20
I guess by "more pointy" you mean that the normal cone at that point is larger? $\mathbb S_n^+$ has larger normal cone at low rank matrices, since a spectrahedron is the intersection of an affine manifold with the space of PSD matrices, then we expect a spectrahedron to be more pointy at these points as well. – dmh Feb 01 '22 at 15:37

Why is the boundary of spectrahedra “more pointy” at matrices of lower rank?

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