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 $K=\{A|\forall x \in k-0: \langle A,xx^{T}>0 \rangle \}$?


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.