Understanding about $S^n_+=\cap_{z \in R^n}S_z$,so it is convex.

Question

A set $S^{+}$ of positive semi-definite matrices (PSD) is defined as

$S^+=\{\mathbf X \in S^n |\mathbf z^T \mathbf X \mathbf z \ge 0,\forall z \in R^n,\mathbf X^T=X \}$

Use the property of which intersection of the halfspaces is also convex to prove $S^+$ is convex set.

And the solution is as below

Define $S_z=\{\mathbf X \in S^n |\mathbf z^T \mathbf X \mathbf z \ge 0\}=\{\mathbf X \in S^n |Tr(\mathbf X \mathbf z\mathbf z^T )\ge 0\}=\{\mathbf X \in S^n |Tr(\mathbf X \mathbf Z)\ge 0\}$,so $S_z$ is a halfspace,so $S^n_+=\{\mathbf X \in S^n |\mathbf z^T \mathbf X \mathbf z \ge 0,\forall z \in R^n \}=\cap_{z \in R^n}S_z$,so it is convex.

I can't understand the last part of the solution,i mean the relation between

1.$S_z$ is a halfspace

2.$S^n_+=\{\mathbf X \in S^n |\mathbf z^T \mathbf X \mathbf z \ge 0,\forall z \in R^n \}=\cap_{z \in R^n}S_z$

3.so it is convex.

By the way,why is $S^n_+=\{\mathbf X \in S^n |\mathbf z^T \mathbf X \mathbf z \ge 0,\forall z \in R^n \}=\cap_{z \in R^n}S_z$,i can't understand this either,can anyone explain them to me?

Answer 1

The following steps are equivalent:

1. $X$ is positive-semidefinite.
2. By definition, $\forall z\in{\Bbb R}^n$ we have $z^TXz\ge 0$ (that is, $X\in S_+^n$).
3. $\forall z\in{\Bbb R}^n$ we have $z^TXz\ge 0$ can be interpreted as $\forall z\in{\Bbb R}^n$ it holds $X\in S_z$.
4. By definition of intersection, the latter gives $X\in \cap_{z\in{\Bbb R}^n}S_z$.

In particular, equivalence of $2$ and $4$ means that $S_+^n=\cap_{z\in{\Bbb R}^n}S_z$. $S_z$ is a half-space as it is given by inequality "the inner product with a fixed vector is non-negative". A half-space is convex, hence, $S_+^n$ is convex as an intersection of convex sets.