$\mathbf K = S^n_+$ (nxn real symmetric positive semidefinite matrix). Show that the dual of $\mathbf K$
$\mathbf K^*=\{\mathbf Y | Tr(\mathbf X \mathbf Y) >0, \forall \mathbf X \ge \mathbf 0\}$
is self-dual, i.e., $\mathbf K^* = \mathbf K$.
Hint :Prove $\mathbf K^* \subseteq S^n_+$ and $S^n_+ \subseteq \mathbf K^*$
Here is parts of its solution
$1.$Prove $\mathbf K^* \subseteq S^n_+$
let $Y \in \mathbf K^*$,we need to prove $Y \in S^n_+ $
$1-1.$ Suppose $ Y \notin S^n_+$,there exists $q \in R^n$ so that $q^TYq=Tr(Yqq^T) \le 0$
$1-2.$ Let $X=qq^T \in S^n_+$,then $tr(XY) \lt 0$,it means that $Y \notin \mathbf K^*$ (Contradiction) ,so $Y \in S^n_+$
I have three problem in here
1.Why can we prove $\mathbf K^* = \mathbf K$ by proving $\mathbf K^* \subseteq S^n_+$ and $S^n_+ \subseteq \mathbf K^*$?
2.i don't understand the logic about the $1-1$ and $1-2$,why when $ Y \notin S^n_+$,then $q^TYq \le 0$?why when $X=qq^T \in S^n_+$,then $tr(XY) \lt 0$,it means that $Y \notin \mathbf K^*$?
3.Why is $X=qq^T \in S^n_+$,then $tr(XY) \lt 0$,shouldn't it be $tr(XY) \gt 0$ ?
4.Why can $X=qq^T \in S^n_+$ prove $Y \notin \mathbf K^*$?
5.Why can $Y \notin \mathbf K^*$ prove $\mathbf K^* \in S_+^n$ ?
Can anyone explain it to me?
In general, if two sets $A$ and $B$ satisfy $A \subseteq B$ and $B \subseteq A$, then $A = B$.
If this isn't clear, notice that $A \subseteq B$ means that every element in $A$ is also an element of $B$. Also, $B \subseteq A$ means that every element of $B$ is also an element of $A$. Hence, if we have both $A \subseteq B$ and $B \subseteq A$, then the sets $A$ and $B$ have the same elements, and thus, are equal.
Hence, $K^* \subseteq S^n_+$ and $S^n_+ \subseteq K^*$ means that $K^* = S^n_+$, i.e. $K^* = K$.
Using the property $\text{tr}(PQ) = \text{tr}(QP)$ for matrices of "compatible" sizes, we have $0 > \text{tr}(XY) = \text{tr}(qq^TY) = \text{tr}(q^TYq) = q^TYq$.
Since $q^TYq < 0$ for some vector $q$, we have that $Y$ is not a positive semidefinite matrix, i.e. $Y \not\in S^n_+ = K^*$.