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.

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    $\begingroup$ It suffices to note that within the subspace of symmetric matrices, the positive definite matrices form an open set. $\endgroup$ – Omnomnomnom Jan 4 '17 at 15:34
  • $\begingroup$ @Omnomnomnom Thanks! I add a constraint that the trace of the matrices are all $1$. What about the new case? $\endgroup$ – Eden Harder Jan 4 '17 at 15:50
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    $\begingroup$ Same idea. Now, we're considering an open subset of the affine space of symmetric matrices with trace $1$. $\endgroup$ – Omnomnomnom Jan 4 '17 at 16:07
  • $\begingroup$ @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. $\endgroup$ – Eden Harder Jan 4 '17 at 16:55
  • $\begingroup$ @Rahul I'm sorry. We can just append the third diagonal elements $0$ for $A,B$, then they will be positive semidefinite. $\endgroup$ – Eden Harder Jan 4 '17 at 17:12

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