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The real symmetric matrices that are copositive matrices form a closed convex cone:Each fixed $x$ gives a closed linear inequality in $A$. Their intersection is therefore closed and convex, and it is preserved by nonnegative scaling.

Let $\alpha =\min \{x^{T}Qx:x\ge 0,\Vert x{\Vert }_2=1\}$ for a real symmetric matrix $Q$. Compactness gives attainment. Homogeneity shows $Q-\lambda I$ is a copositive matrix exactly when $\lambda \le \alpha$, provingIts conic program dual is the trace-normalized completely positive optimization problem. A minimizing vector gives $X=x{x}^T$, which certifies equality and dual attainment directly. In general $\alpha$ differs from the unrestricted smallest eigenvalue.

Under the nonnegative-pairing convention for the dual cone, ${\mathrm{CP}}_n^{\ast }={\mathrm{COP}}_n$ follows from $⟨A,x{x}^T{⟩}_F=x^{T}Ax$. Conversely, if $B\notin {\mathrm{CP}}_n$, separation from a closed convex cone gives a separating $A$ nonnegative on all generators and negative on $B$. That $A$ is a copositive matrix, so $B\notin {\mathrm{COP}}_n^{\ast }$. This proves the displayed equality using closedness of the completely positive cone.

The five-dimensional Horn copositive matrix has diagonal entries $1$, entries $-1$ on the edges of the five-cycle, and entries $1$ on the remaining pairs:For $x\ge 0$, a cyclic relabelling puts a smallest coordinate at $x_5$. The identitythen proves that $H$ is a copositive matrix.

However, $H$ is outside the positive-semidefinite-plus-nonnegative cone. Set $w=(1,2,1,0,0{)}^T$. Its quadratic form is zero. If $H=P+N$ with $P$ a positive semidefinite matrix and $N$ a symmetric nonnegative matrix, both $w^{T}Pw$ and $w^{T}Nw$ must vanish. Positivity of the first three coordinates of $w$ forces every entry of the leading $3\times 3$ block of $N$ to vanish. Applying the argument to all cyclic shifts of $w$ forces every entry of $N$ to vanish, since each pair of indices lies in a cyclic interval of length three. This would make $H$ a positive semidefinite matrix, but zero quadratic form of a positive semidefinite matrix would then give $Hw=0$, whereas $Hw=(0,0,0,2,2{)}^T$.

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.

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 Horn copositive matrix is invariant under cyclic permutation of its five coordinates: its negative entries correspond precisely to neighboring indices on the five-cycle. For any $x\ge 0$, cyclically relabel the coordinates so that $x_5$ is a smallest coordinate. In particular, $x_4-x_5\ge 0$.

Using the stated identity in this coordinate order,The square is nonnegative, and both remaining terms are nonnegative since $x_i\ge 0$ and $x_4\ge x_5$. Cyclic invariance means the relabeling has not changed the quadratic form. Thus $H$ is a copositive matrix for every original ordering of $x$.

Take the free squared mass $m^2\ge 0$, as for the ordinary massive free theory. The quartic part of the scalar potential can be writtenIt is nonnegative for every real field value exactly whenNecessity follows from the axis ${\varphi }_2=0$ and the diagonal ${\varphi }_1={\varphi }_2$: a negative coefficient on either ray makes $V$ tend to minus infinity under rescaling. The displayed sum proves sufficiency, since $m^2({\varphi }_1^2+{\varphi }_2^2)/2$ is also nonnegative. This is a copositive matrix condition on the quadratic form in the nonnegative variables $({\varphi }_1^2,{\varphi }_2^2)$, rather than positive definiteness on all independent signed variables.

The strict inequalities $\lambda >0$, $\kappa >-\lambda$ make the quartic part grow in every nonzero direction. They are not necessary for a stable vacuum when $m^2>0$. For example, $\lambda =\kappa =0$ gives two stable free massive fields; $\lambda >0$, $\kappa =-\lambda$ has a flat quartic diagonal but a confining quadratic term. If $m^2=0$, boundary cases have flat vacuum directions and need not have a unique isolated minimum. If one instead allows negative squared mass, a flat quartic ray makes the full potential unbounded below; the stability conditions must then be strengthened. This distinguishes boundedness from strict quartic coercivity.

Under the generator of a, the variation isIt vanishes identically if and only if $\kappa =\lambda$. In that case $V_4=\lambda ({\varphi }_1^2+{\varphi }_2^2{)}^2$ preserves the entire global U(1) symmetry. More explicitly, the interacting Euler-Lagrange equations giveHenceA special solution lying on an axis does not establish an operator conservation law. Among the stable couplings, the symmetry-preserving line is $\kappa =\lambda \ge 0$.