The space of positive semidefinite $n \times n$ matrices is a cone The positive semedefinite cone is
$$\mathbb{S}^{n}_{+} = \{\mathbf{X}\in\mathbb{S}^{n}: \mathbf{X}\succeq\mathbf{0}\} .$$
To my knowledge, this representation gives two restrictions on ${\bf X}$: $\mathbf{X}$ is a positive semidefinite matrix, and $\mathbf{X}$ is a symmetric matrix. But it is shown that $\mathbb{S}_+^n$ is also a convex cone.
I do not know how to prove it, can anyone help me, please? Also, what is the relation between a symmetric matrix and a cone?
 A: Here, $\mathbb{S}^n$ is a vector space, and we're asked to check that the subset $\mathbb{S}^n_+ \subset \mathbb{S}^n$ has some special properties:
(1) $\mathbb{S}^n_+$ is a cone, that is, for every $\textbf{X} \in \mathbb{S}^n_+ - \{\mathbf{0}\}$, the entire ray with vertex $\bf 0$ passing through $\bf X$ is also contained in $\mathbb{S}^n_+$. Algebraically, any matrix in this ray can be written as $\lambda \bf X$ for some $\lambda \geq 0$.
(2) $\mathbb{S}^n_+$ is convex, that is, for semidefinite matrices $\textbf{A}, \textbf{B} \in \mathbb{S}^n_+$, the line segment in $\mathbb{S}^n$ with endpoints $\textbf{A}, \textbf{B}$ is contained in $\mathbb{S}^n_+$. Algebraically, any matrix on this line segment can be written as a unique convex combination of $\bf A$ and $\bf B$, that is, as $t \textbf{A} + (1 - t) \textbf{B}$ for a unique $t \in [0, 1]$.
We can show readily that $\mathbb{S}^n_+$ has both of these properties by appealing to the definition of positive semidefinite matrix, namely that a given matrix $\bf X$ satisfy
$${}^t \textbf{zXz} \geq 0$$ for all $\mathbf{z} \in \mathbb{R}^n$.
