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The convex cone of completely positive matrices:It is a closed convex cone and the dual cone of the copositive cone under the Frobenius inner product. The finite-sum definition imposes no closure by fiat; closedness of the completely positive cone supplies that fact.

A convex cone is a subset $C$ of a real vector space closed under nonnegative linear combinations: $ax+by\in C$ whenever $x,y\in C$ and $a,b\ge 0$. The positive semidefinite cone and the copositive cone are examples, ordered by inclusion through the positive-semidefinite-plus-nonnegative cone.

Conic optimization using the copositive cone. The constraint $Q-\lambda I\in {\mathrm{COP}}_n$ supplies an exact reformulation of a Rayleigh quotient minimum on the nonnegative orthant. Exact conic formulation does not itself provide an efficient membership algorithm.

The interior consists exactly of strictly copositive matrices. If $x^{T}Ax$ has positive minimum $\delta$ on the compact nonnegative unit sphere, perturbations of operator norm less than $\delta$ preserve this positive lower bound there. Conversely, if a unit nonnegative vector has zero quadratic form, $A-\epsilon I$ leaves the copositive cone for every $\epsilon >0$. Thus such a matrix is not interior. The identity matrix is a convenient interior point.

For $x$ in the nonnegative orthant, the positive semidefinite matrix $P$ satisfies $x^{T}Px\ge 0$. The nonnegative matrix $N$ satisfiessince every summand is nonnegative. Adding these inequalities provesConsequently $A=P+N$ is a copositive matrix. The symmetry of $N$ follows from that of $A$ and $P$; entrywise nonnegativity alone need not imply positive semidefiniteness.

This gives the inclusion of the positive-semidefinite-plus-nonnegative cone in the copositive cone. It is only a sufficient construction; the argument does not claim that every copositive matrix has this decomposition in arbitrary dimension.

For each fixed $x$ in the nonnegative orthant, the map $A\mapsto x^{T}Ax$ is a continuous linear function on the vector space of real symmetric matrices. Thusis an intersection of closed half-spaces, using the Frobenius inner product $⟨A,B{⟩}_F=\mathrm{tr}(AB)$ on this space. Arbitrary intersections of closed sets are closed, so $K$ is closed.

If $A,B\in K$ and $s,t\ge 0$, then $x^T(sA+tB)x=s{x}^{T}Ax+t{x}^{T}Bx\ge 0$ for every $x\ge 0$. Therefore $sA+tB\in K$, including the zero matrix when $s=t=0$. HenceThe copositive cone is defined in the real space of symmetric matrices; symmetry is needed later to recover every matrix entry from these quadratic forms.

A convex cone containing no nonzero line through the origin, equivalently $K\cap (-K)=\{0\}$. The copositive cone is pointed: a matrix in both it and its negative has zero quadratic form on the nonnegative orthant; testing $e_i$ and $e_i+e_j$ kills all entries.

The sums $P+N$ of a real positive semidefinite matrix $P$ and a symmetric nonnegative matrix $N$ form a convex cone inside the copositive cone. Both terms have nonnegative quadratic forms on the nonnegative orthant. The Horn copositive matrix shows that the inclusion is strict in dimension five; the sum of squares criterion for a biquadratic form explains this cone's relation to semidefinite programming.