Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 339 / 1 / a / Solution

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.