Can I write $\mathbb{S}_+^3$ as a norm cone? Let $\mathbb{S}^3_+$ be the set of $3 \times 3$ symmetric, positive semidefinite matrices. I wonder whether I can write $\mathbb{S}^3_+$ as a norm cone, i.e., 
$$\exists A\in \mathbb{R}^{m\times 9}, C, \|\cdot\|, s.t. X \in \mathbb{S}^n_+ \iff \|A\text{vec}(X)\|\leq \text{trace}(CX),$$
where $\text{vec}(X)$ means to put $X$ as a vector.
I  guess not as there could be so many polynomial inequalities for the positive semidefinite case (the determinant of all the principal minors). But I could not figure out why. I have tried to look at the dual cone (as $\mathbb{S}_+^3$ is self-dual) but it seems very hard to compute the dual cone of $\{X:\|A\text{vec}(X)\|\leq \text{trace}(CX)\}$.
I am thinking whether there is any property always preserved by linear transform but not sure there is really any.
 A: I'm going to assume that we may consider matrix norms. Then in fact, the answer is precisely analogous to the answer given for $\mathbb{R}^n_+$ in your other question.
Let $\|X\|_*$ be the nuclear norm of $X\in\mathbb{S}^n$; that is, $$\|X\|_*=\sum_{i=1}^n \sigma_i(X) = \sum_{i=1}^n|\lambda_i(X)|$$
Then we have
$$X\in\mathbb{S}^n_+ \quad\Longrightarrow\quad \|X\|_* \leq \mathop{\textrm{Tr}}(X)$$
After all, this inequality is equivalent to
$$\sum_{i=1}^n |\lambda_i(X)| \leq \sum_{i=1}^n \lambda_i(X)$$
and the only way for this to be true is if $\lambda_i(X)\geq 0$ for all $i$. Indeed, it holds with equality in that case.
One thing I should point out is that, in both cases, strict inequality cannot be satisfied; that is, there is no $X$ such that $\|X\|_*<\mathop{\textrm{Tr}}(X)$; or, in the vector case, no $x$ such that $\|x\|_1<\sum_i x_i$. This has some important practical consequences if, say, you intend to implement some sort of numerical method around this approach. Better to stick with the original conic form.
