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In an $N$-dimensional real vector space, each member of a conic hull has a representation using at most $N$ generators. To prove it, take a finite positive-coefficient representation with more than $N$ terms. Its generators are linearly dependent, say ${\sum }_j{\mu }_j{s}_j=0$, with some ${\mu }_j>0$ after reversing the relation if necessary. Subtract $t{\mu }_j$ from each coefficient, taking $t={\min }_{{\mu }_j>0}{\lambda }_j/{\mu }_j$. All coefficients stay nonnegative and at least one vanishes. Iterate. The bound differs from the $N+1$ bound for a convex hull because the coefficient sum is unrestricted.

For a convex cone $D$ in an inner product space, use the nonnegative-pairing conventionHere the pairing is the Frobenius inner product on real symmetric matrices. Let $C_0$ be the conic hull of the nonnegative rank-one matrices $x{x}^T$, and let $C=\overline{C_0}$. For $A\in K$, $⟨A,x{x}^T{⟩}_F=x^{T}Ax\ge 0$. This extends to conic combinations, and by continuity to their limits. Hence $C\subseteq K^{\ast }$.

For the converse, if $B\notin C$, separation from a closed convex cone supplies a symmetric $H$ withIn particular $x^{T}Hx\ge 0$ for every $x\ge 0$, so $H\in K$. The negative pairing then excludes $B$ from $K^{\ast }$. ThereforeThe same generator test gives $C^{\ast }=K$. This is the duality of copositive and completely positive cones; the next argument removes the closure.

The real vector space of symmetric matrices has dimension $N=n(n+1)/2$. By the conic Carathéodory theorem, every member of $C_0$ is a conic combination of at most $N$ generators. Absorb each nonnegative coefficient into its vector through $\lambda x{x}^T=(\sqrt{\lambda }x)(\sqrt{\lambda }x{)}^T$.

If $B_{\ell }\in C_0$ converges to $B$, write, padding with zero vectors if necessary,The matrix trace satisfiesThe left side is bounded because $B_{\ell }$ converges. Thus the finite tuple $(x_{\ell 1},\dots ,x_{\ell N})$ is bounded. The Bolzano-Weierstrass theorem gives a subsequence on which every vector converges, say $x_{\ell j}\to x_j\ge 0$. Continuity of the outer product now gives $B={\sum }_{j=1}^N{x}_j{x}_j^T\in C_0$. HenceThis proves closedness of the completely positive cone. The uniform bound on the number of factors and the matrix trace bound are both essential: an arbitrary conic hull of a closed generating set need not be closed.