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A real symmetric matrix $A$ is copositive if its quadratic form satisfies $x^{T}Ax\ge 0$ for every $x$ in the nonnegative orthant. Every positive semidefinite matrix and every symmetric nonnegative matrix is copositive, as is their sum. The converse fails for the Horn copositive matrix.

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.

For a vector in the nonnegative orthant, the PDF's matrix gives the quadratic formThus it is a copositive matrix. On the other hand, with $u=(1,-1{)}^T$,so it is not a positive semidefinite matrix. Equivalently, the eigenvalues are $3$ and $-1$, with eigenvectors $(1,1{)}^T$ and $(1,-1{)}^T$.

The local TeX has corrupted the matrix entries. The PDF has diagonal entries one and off-diagonal entries two. The all-twos matrix in the TeX would instead have quadratic form $2(x_1+x_2{)}^2$ and would be a positive semidefinite matrix; it cannot demonstrate the requested distinction.

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 pointed cone has no nonzero line through the origin: equivalently $K\cap (-K)=\{0\}$. If $A\in K\cap (-K)$, its quadratic form vanishes on the nonnegative orthant. Testing the coordinate vectors gives $A_{ii}=0$. Testing $e_i+e_j$ then givesSince $A$ is a symmetric matrix, every entry is zero. Thus $\boxed{K\cap (-K)=\{0\}}$.

The identity matrix gives an interior point. For a symmetric perturbation $E$ with operator norm $\Vert E{\Vert }_2<1$, the Cauchy-Schwarz inequality givesThis open norm ball around $I$ lies even in the positive semidefinite cone, hence in $K$. Consequently $\boxed{I\in \mathrm{int}K}$ and the interior is nonempty.

More generally, the interior of the copositive cone consists exactly of strictly copositive matrices. Positivity on the compact nonnegative unit sphere has a positive minimum and persists under small perturbations; a zero there is destroyed by an arbitrarily small negative multiple of $I$.

Let $U=\{x\ge 0:\Vert x{\Vert }_2=1\}$. This set is nonempty and compact for $n\ge 1$, and the quadratic form is continuous. Its minimum $\alpha$ is therefore attained.

For every nonzero $y$ in the nonnegative orthant, normalizing $y$ givesThe zero vector imposes no further condition. It follows that $Q-\lambda I$ is a copositive matrix exactly when $\lambda \le \alpha$. Thus the feasible scalar set is $(-\infty ,\alpha ]$, andBoth optima are attained. This copositive reformulation of an orthant Rayleigh minimum follows directly from homogeneity; it does not require invoking a duality theorem for the original nonconvex constraint.

The restriction to $x\ge 0$ matters. For the matrix in part (a), this minimum is $1$ whereas the unrestricted smallest eigenvalue is $-1$.

For $z\in {\mathbb{R}}^n$, define $x_i=z_i^2$. Then $x$ lies in the nonnegative orthant andIf $A$ is a copositive matrix, this is nonnegative for every $z$, so $p$ is a globally nonnegative polynomial. Conversely, every $x\ge 0$ has the form $x_i=z_i^2$ by taking $z_i=\sqrt{x_i}$. If $p$ is globally nonnegative, then $x^{T}Ax=p(z)\ge 0$ for every such $x$, proving that $A$ is a copositive matrix. The map $z\mapsto (z_i^2{)}_i$ covers the entire nonnegative orthant, which is the key to both directions.

The set $K\cap P$ is compact because $K$ is closed and bounded, and the productis continuous. By the extreme value theorem, $G$ has a maximizer. Since $K$ meets the interior of the nonnegative orthant $P$, some feasible product is positive, so every maximizer lies in $\mathrm{Int}P$.

On $\mathrm{Int}P$, maximizing $G$ is equivalent to maximizingThe Hessian of $F$ iswhich is negative definite. Thus $F$ is a strictly concave function, and it has at most one maximizer on the convex set $K\cap P$. Denote this unique point by $x^{\ast }$.

For any $x\in K\cap P$, the line segmentlies in $K\cap \mathrm{Int}P$. Since $F(x(t))$ is maximal at $t=0$, its right derivative there is nonpositive:ThereforeThis is the supporting inequality for the product maximizer on a compact convex subset of the positive orthant.

Suppose now that $K$ is invariant under cyclic coordinate permutation. That permutation preserves the product, so uniqueness forces it to fix $x^{\ast }$. Hencefor some $r>0$. More explicitly,Indeed, cyclic symmetry averaging puts the diagonal point with coordinate equal to the mean of any $x\in K$ back in $K$, while the preceding inequality with $x_j^{\ast }=r$ gives ${\sum }_j{x}_j\le nr$.

Finally, given $a=(a_1,\dots ,a_n)\in \mathrm{Int}P$, defineThis set is nonempty, closed, convex and bounded, and it contains $a$. By the arithmetic-geometric mean inequality,Equality holds only when all $x_j/a_j$ are equal and their sum is $n$, namely only at $x=a$. Thus this $K$ has

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.

A self-dual cone equals its dual cone under the specified inner product. The nonnegative orthant, second-order cone and positive semidefinite cone are standard examples. Self-duality of a set alone does not establish primal or dual feasibility.

A real symmetric matrix whose quadratic form is positive at every nonzero vector in the nonnegative orthant. By compactness of the nonnegative unit sphere, this is equivalent to a positive uniform lower bound $x^{T}Ax\ge \delta \Vert x{\Vert }_2^2$ on that orthant. It need not be a positive-definite matrix.