most recent 30 from http://mathoverflow.net 2013-06-19T10:11:50Z http://mathoverflow.net/feeds/question/116781 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116781/when-are-cones-of-matrices-generated-by-vectors Felix Goldberg 2012-12-19T11:58:27Z 2012-12-19T12:16:55Z

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

http://mathoverflow.net/questions/116781/when-are-cones-of-matrices-generated-by-vectors/116782#116782 Denis Serre 2012-12-19T12:16:55Z 2012-12-19T12:16:55Z

<p>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 $K^0$. </p> <p>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 $xx^T$.</p> <p>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.</p>