= Horn copositive matrix
{c}
{title2=$H$}
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:
$$
H=\begin{pmatrix}
1&-1&1&1&-1\\
-1&1&-1&1&1\\
1&-1&1&-1&1\\
1&1&-1&1&-1\\
-1&1&1&-1&1
\end{pmatrix}.
$$
For $x\geq0$, a cyclic relabelling puts a smallest coordinate at $x_5$. The identity
$$
x^THx=(x_1-x_2+x_3-x_4+x_5)^2+4x_2x_5+4x_1(x_4-x_5)
$$
then 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 $w^TPw$ and $w^TNw$ must vanish. Positivity of the first three coordinates of $w$ forces every entry of the leading $3\times3$ 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 $Hw=(0,0,0,2,2)^T$.
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