A spectrahedron is a set defined by a linear matrix inequality (https://en.wikipedia.org/wiki/Spectrahedron).

I was wondering if the boundary of such a set is almost always smooth. By "smooth" I mean that it admits a tangent hyperplane at any point of the surface.

  I say "almost" because for the special case of the polytope, the boundary has many "edges", so the answer in this special case would be no.

A spectrahedron is a convex set defined by a linear matrix inequality (LMI).

Is the boundary of such a set almost always smooth?

By "smooth" I mean that it admits a tangent hyperplane at any point of the surface. I say "almost" because, for the special case of the polytope, the boundary has many "edges". In this special case, the answer would be no.

A spectrahedron is a set defined by linear matrix inequality (https://en.wikipedia.org/wiki/Spectrahedron)

I was wondering if the boundary of such set is almost always smooth? By smooth I mean that it admits a tangent hyperplane at any point of the surface.

I say almost because for the special case of the polytope, the boundary has many "edges", so the answer in this special case would be no.

A spectrahedron is a set defined by a linear matrix inequality (https://en.wikipedia.org/wiki/Spectrahedron).

I was wondering if the boundary of such a set is almost always smooth. By "smooth" I mean that it admits a tangent hyperplane at any point of the surface.

I say "almost" because for the special case of the polytope, the boundary has many "edges", so the answer in this special case would be no.