= Solution
For a <vector> in the <nonnegative orthant>, the PDF's <matrix> gives the <quadratic form>
$$
x^TAx=x_1^2+4x_1x_2+x_2^2\geq0.
$$
Thus it is a <copositive matrix>. On the other hand, with $u=(1,-1)^T$,
$$
\boxed{u^TAu=-2<0},
$$
so it is not a <positive semidefinite matrix>. Equivalently, the <eigenvalues> are $3$ and $-1$, with <eigenvectors> $(1,1)^T$ and $(1,-1)^T$.
\b[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.
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