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The cone $P_{n}$ of positive semidefinite matrices of order $n$ can be represented in this form: $P_{n}=\{A|\forall x\geq 0: \langle A,xx^{T}>0 \rangle \}$ with $x$ running over $\mathbf{R}^{n}-0$.

Question: can every convex cone of matrices be represented in this way, i.e. if $K$ is a cone of (say, real symmetric) matrices, does there exist a cone $k \in \mathbf{R}^{n}$ so that $K=\{A|\forall x \in k-0: \langle A,xx^{T}>0 \rangle \}$?

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Let $K$ be a closed convex cone in ${\bf Sym}_n({\mathbb R})$. I assume a generic cone: non void interior, strictly convex. Let $$K^0=\{ S\in{\bf Sym}_n({\mathbb R})\quad|\quad{\rm Tr}(SH)\ge0,\quad\forall H\in K\}.$$ be its dual. Then $K=(K^0)^0$. If $K=Z^0$ for some conical set $Z$ (that is, $tZ=Z$ for $t>0$), it is necessary that $Z\subset K^0$ and $Z$ contains the extremal lines of $K^0$.

Therefore the cone $K$ has the property that you request if, and only if, the extremal lines of its dual $K^0$ are spanned by matrices of the form $xx^T$.

This happens to be true if $K=K^0={\bf Sym}_n^+$, but it fails in general. Just take any finite set of lines in an open half-space, not all of them being spanned by a rank-one symmetric matrix, take $C$ their convex hull, and choose $K=C^0$. Then $K^0=C$ has an extremal line not spanned by an $xx^T$. Such a $K$ does not share the expected property.

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