Existence of a "basis" for the symmetric positive definite matrices Let $P_{\text{sym}}$ denote the convex cone of the symmetric positive definite real matrices (of size $n \times n$). 
Question: Is there a finite subset $B$ of matrices in $P_{\text{sym}}$ such that every matrix in $P_{\text{sym}}$ is a (non-zero) conic combination of elements of $B$?
Note: It is not possible for such an expression to be unique; Assuming otherwise, we have a finite set $\{P_1,...,P_k \} \subseteq P_{\text{sym}}$ such that any matrix $P \in P_{\text{sym}}$ can be expressed uniquely as: $P=\sum_{i=1}^k a_iP_i$ where all $a_i \ge 0$ and $(a_1,...,a_k)\neq \bar 0$.
This would imply $P_{\text{sym}}$ is homeomorphic to the set $D_k:= \{(a_1,...,a_k) \in \mathbb{R}^k | a_i \ge 0 \}$. (via the map $(a_1,...,a_k) \to \sum_{i=1}^k a_iP_i$).
However, it is known that $P_{\text{sym}}$ is homeomorphic to $\mathbb{R}^{n(n+1)/2}$, so $\mathbb{R}^{n(n+1)/2} \cong D_k$.
For $k \neq \frac{n(n+1)}{2}$ this is impossible, since this would imply an open subset of $\mathbb{R}^k$ is homeomorphic to an open subset of $\mathbb{R}^{n(n+1)/2}$, contradicting the invariance of domain. 
For $k=\frac{n(n+1)}{2}$, this is also impossible; Invariance of domain implies the only subsets of $\mathbb{R}^n$ which are homeomorphic to it are open, and $D_k$ is not open.
 A: Suppose such $P_i$ existed.  Take any nonzero vector $v$.  Let $t_i = \text{trace}(P_i) > 0$.
If $P = \sum_{i=1}^k a_i P_i$ with $a_i \ge 0$, then  $$ v^T P v = \sum_{i=1}^k a_i t_i v^T P_i v /t_i \ge \text{trace}(P) \min_i (v^T P_i v/t_i) $$
Thus $$ \dfrac{v^T P v}{\text{trace}(P)} \ge \min_i \dfrac{v^T P v}{\text{trace}(P_i)}$$
But there are positive definite matrices $P$ for which $v^T P v/\text{trace}(P)$ is arbitrarily close to $0$, so this is impossible.
A: Some pictures to help the intuition:
For $n=2$, any symmetric matrix can be written as $P=\begin{bmatrix}a+b & c \\ c & a-b\end{bmatrix}$. $P$ is positive definite if and only if $a>0$ and $a^2 > b^2 + c^2$. This is the interior of a right circular cone in $abc$-space.

For $n=3$, consider the affine subspace of symmetric matrices $P=\begin{bmatrix}1 & d & e \\ d & 1 & f \\ e & f & 1\end{bmatrix}$. $P$ is positive definite if and only if $d,e,f \in (-1,1)$ and $\det P = 1 - d^2 - e^2 - f^2 + 2def > 0$. This is the interior of the middle tetrahedron-like piece of Cayley's cubic surface.

Both these regions have curved boundaries, and so cannot be generated by conic combinations of a finite set of points.
