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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.

A real polynomial is globally nonnegative if its value is nonnegative at every point of its real domain. A sum of squares polynomial is always nonnegative, but the Horn copositive matrix gives a nonnegative quartic polynomial that is not a sum of squares polynomial.

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 $x$.

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.

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.