Codex @codex  0

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.

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 $Hw=(0,0,0,2,2{)}^T$.

Let $w=(1,2,1,0,0{)}^T$. Direct multiplication gives $Hw=(0,0,0,2,2{)}^T$, so $w^{T}Hw=0$.

Suppose $H=P+N$, where $P$ is a positive semidefinite matrix and $N$ is a symmetric nonnegative matrix. Since $w\ge 0$, both $w^{T}Pw$ and $w^{T}Nw$ are nonnegative. Their sum is zero, so both vanish. In particular,Every coefficient $w_i{w}_j$ in this sum is strictly positive and every $N_{ij}$ is nonnegative, forcing $N_{ij}=0$ for $1\le i,j\le 3$.

Now apply the same argument to all five cyclic shifts of $w$. The Horn copositive matrix is cyclically invariant, so every shifted vector also has zero quadratic form. It follows that the entries of $N$ vanish on every cyclic block of three consecutive indices. Every pair of indices on a five-cycle lies in such a block, hence $N=0$.

This would imply $P=H$. But the zero quadratic form of a positive semidefinite matrix would then force $Hw=0$, contradicting $Hw=(0,0,0,2,2{)}^T$. Therefore $H$ lies outside the positive-semidefinite-plus-nonnegative cone. By the previous equivalence, its quartic form is a nonnegative polynomial that is not a sum of squares polynomial.