Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 339 / 3

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.

First suppose $A=P+N$, with $P$ a real positive semidefinite matrix and $N$ a symmetric nonnegative matrix. Factor $P=B^{T}B$, and write $v(z)=(z_1^2,\dots ,z_n^2{)}^T$. ThenEach term is a square multiplied by a nonnegative scalar, so $p$ is a sum of squares polynomial.

Conversely, suppose $p={\sum }_{\ell }{q}_{\ell }^2$. Because $p$ is a degree-four homogeneous polynomial, the homogeneous sum of squares representation allows every $q_{\ell }$ to be quadratic and homogeneous. Explicitly, higher-degree parts cannot cancel in a sum of squares; constant parts vanish because $p(0)=0$, and the degree-two part ${\sum }_{\ell }(q_{\ell }^{(1)}{)}^2$ forces all linear parts to vanish. WriteSince $p$ is invariant under every coordinate sign change, sign averaging of a sum of squares over independent Rademacher random variables givesThe cross terms vanish because their sign products contain an odd power of at least one independent sign. SetThen $P$ is a positive semidefinite matrix and $N$ is a symmetric nonnegative matrix. Comparing the coefficients of $z_i^4$ and $z_i^2{z}_j^2$ gives $A=P+N$. This proves the sum of squares criterion for a biquadratic form.

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.