Finding dual cone for a set of copositive matrices This is a question from the textbook Convex Optimization by Stephen Boyd and Lieven Vandenberghe (2.35). I did read through the solution manual but I couldn't figure out why it is written the way it is and I wish if I could get some helps from this community.


Given a matrix $\mathbf{X} \in \mathbb{S}^n$ is copositive, i.e. $\mathbf{z^TXz} \geq 0,  \mathbf{z} \succeq0$, also given the fact that the set of copositive matrices is a proper cone. Find the dual cone.


My solution is as follows:
Suppose $K$ is the set of copositive matrices in $\mathbb{S}^n$, $K = \{\mathbf{X} \in \mathbb{S}^n : \mathbf{z^TXz} \geq 0, \forall \mathbf{z} \succeq 0\}$. $K^*$ is a set of normal vectors of all halfspaces containing K, defining as $K^* = \{\mathbf{Y}:\mathrm{tr}\mathbf{(Y^TX)}, \forall \mathbf{X} \in K \}$.
We know the equality that $\mathbf{z^TXz} = \mathrm{tr}(\mathbf{X^Tzz^T})=\langle \mathbf{X,zz^T} \rangle \geq 0$, so let $\mathbf{Y = zz^T}$.
\begin{align*}
K^* &= \{\mathbf{Y}: \mathbf{\langle \mathbf{X,Y}  \rangle}\geq 0, \forall \mathbf{X} \in K\} \\
&=\{ {\mathbf{zz^T}}: \mathbf{\langle \mathbf{X,zz^T}\rangle}\geq 0\ ,\forall \mathbf{X} \in K\} \quad \text{(Need to get rid of $\mathbf{X}$ from this expression)}\\
&=\{{\mathbf{zz^T}}:\mathbf{z} \succeq 0\} \quad \text{(as $K$ is pointed and non-empty, and $\mathbf{X} \in K$)}
\end{align*}
But according to the solution manual, it is the actually convex hull of the set that I just wrote above, which is $K^* = \mathbb{conv}\{ \mathbf{zz^T}:\mathbf{z} \succeq 0\}$. Why this extra $\mathbb{conv}$ there, is it because that the set $\{ \mathbf{zz^T}:\mathbf{z} \succeq 0\}$ is not convex and $K^*$ must be closed and convex? I suspect that I did some very cheeky maths here, and I deeply appreciate any guidance.
 A: One "sanity check" in computing dual cones is that if your new cone is smaller, then your dual cone is bigger. In your case, a copositive cone is bigger than a semidefinite cone, and the dual of a semidefinite cone is the semidefinite cone, so we should expect the dual of the copositive cone to be smaller than the semidefinite cone.
With that in mind, I'm going to make some guesses. Denote $K$ and $K^*$ as the copositive and dual of copositive cone, respectively.

*

*If $z\geq 0$, then for any $X\in K$, $z^TXz \geq 0$. Then any matrix
$$
Y = \sum_{i=1}^m a_ia_i^T
$$
where $a_i\geq 0$ is in $K^*$, since $$\mathbf{tr}(XY) = \sum_{i} a_i^TXa_i \geq 0.$$
Note that the answer doesn't matter if $a_i$ is strictly negative, since the sign cancels out in the product. Therefore
$$
K^* \supseteq \{ \sum_{i} a_ia_i^T : a_i \geq 0 \text{ or } a_i \leq 0\}.
$$


*In fact, this is the entire set for $K^*$. To see this, suppose that $Y = zz^T$ where $z\ngeq 0$ and $z\nleq 0$. Then we can find a copositive $X$ where $z^TXz < 0$. Then $\mathbf{tr}(XY) < 0$.
This is clearly a set smaller than the set of positive semidefinite matrices (which is the proposed set but with no constraints in $a_i$) and thus passes our sanity check.
