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I can prove that the set of symmetric positive-definite matrices with trace $1$ in the space of all symmetric matrices is convex, because $a A + (1-a) B$ is also symmetric positive-definite with trace $1$ when $A, B$ are so and $a\in [0,1]$.

But I have no idea about whether this set is strictly convex and how to prove that.

Eden Harder
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    It suffices to note that within the subspace of symmetric matrices, the positive definite matrices form an open set. – Ben Grossmann Jan 04 '17 at 15:34
  • @Omnomnomnom Thanks! I add a constraint that the trace of the matrices are all $1$. What about the new case? – Eden Harder Jan 04 '17 at 15:50
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    Same idea. Now, we're considering an open subset of the affine space of symmetric matrices with trace $1$. – Ben Grossmann Jan 04 '17 at 16:07
  • @Rahul $A=\begin{bmatrix}0.4&0\0&0.6\end{bmatrix}, B=\begin{bmatrix}0.6&0\0&0.4\end{bmatrix}$ while their average is half of the identity matrix. – Eden Harder Jan 04 '17 at 16:55
  • @Rahul I'm sorry. We can just append the third diagonal elements $0$ for $A,B$, then they will be positive semidefinite. – Eden Harder Jan 04 '17 at 17:12
  • @Rahul it is not at all clear what your comment has to do with the question – Ben Grossmann Jan 05 '17 at 00:12
  • @Omnomnomnom: Oops, I was thinking of strict convexity of the set of symmetric positive semidefinite matrices (which is false). –  Jan 05 '17 at 00:56
  • What does it even mean for a set to be strictly convex? Strict convexity is something you talk about with functions. – Michael Grant Jan 06 '17 at 04:54
  • @MichaelGrant "Furthermore, a set (in the vector space) is strictly convex if every point on the line segment connecting x and y other than the endpoints is inside the interior of the set." from Convex set. – Eden Harder Jan 06 '17 at 10:54
  • Hmm. So that means it's non-polyhedral (and has no hyperplane faces at all), and has a non-empty interior. I've not personally seen that definition used in practice. I wonder who does. – Michael Grant Jan 06 '17 at 12:37
  • So I looked at the Wikipedia history. The term "strictly convex" was added by an anonymous author, that's all they added, and they provided no reference to support the use of the term. I suppose it's definition is reasonably sensible, but still, I've not seen it used, and I've never taught it. I question how useful it is. Do you have a particular reason you need to prove strict convexity? – Michael Grant Jan 06 '17 at 12:43
  • Related (but not a duplicate): http://math.stackexchange.com/questions/984963/strictly-convex-set http://math.stackexchange.com/questions/408947/strictly-convex-sets – Michael Grant Jan 06 '17 at 12:47
  • The definitions I am seeing involve the boundary points, which are semidefinite. You can't limit your view to the interior points. If that is the case, then the set of positive definite matrices is not strictly convex. – Michael Grant Jan 06 '17 at 13:34

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