$$C = conv\{xx^T | x\succeq0\}$$
As far as I know, $x\succeq0$ means x is a symmetric, positive semidefinite matrix, therefore the $xx^T$ here is also symmetric and positive semidefinite.
As the "Positive Semedefinite Cone" is defined as $\mathbb{S}^{n}_{+} = \{\mathbf{X}\in\mathbb{S}^{n}: \mathbf{X}\succeq\mathbf{0}\}$, then $\{xx^T | x\succeq0\}$ should've already been a convex cone here.
How to understand this $C$, does it over-define a convex set?
Thank you!
I don't know what "over-define" means. Certainly each non-extreme element of any convex set $C$ has more than one representation as a convex combination of elements of $C$, so it is over-defined in one sense. I also don't completely understand your definition of $C$.
The set $D=\operatorname{conv}\{XX^T: X\in \mathbb S_+^n\}$ is equal to the set $\mathbb S_+^n$ as follows. Among the elements of $\mathbb S_+^n$ are the rank $1$ matrices of form $X=vv^T$ for $v\in\mathbb R^n$. But the convex hull of such matrices is the set of psd matrices, so $D \subseteq \mathbb S_+^n$. Since elements of $\mathbb S_+^n$ posessess psd square roots, each element $Y\in\mathbb S_+^n$ can be written in the form $Y=XX=XX^T$ for some $X\in\mathbb S_+^n$. Hence $\mathbb S_+^n\subseteq D$. Hence $D=\mathbb S_+^n$.
If your $C$ is my $D$ then it true that your definition of $C$ is another equivalent characterization of the psd matrices.
