On copositive matrix criteria For a while, I have been reading on copositive matrices and the required criteria for a $2 \times 2$ matrix to be copositive. Nevertheless, I can't yet understand why the requirement of it to be copositive simply reduces to having
$$ a_{11} \geq 0, \qquad a_{22} \geq 0, \qquad a_{12} + \sqrt{a_{11} a_{22}} \geq 0 $$
 A: Let us for simplicity rename $a:=a_{11}$, $b:=a_{12}=a_{21}$ and $c:=a_{22}$.
In other words the matrix$^1$ is $A~:=~\begin{pmatrix} a & b \cr b & c \end{pmatrix}$. Call the cone of copositive $2\times 2$ matrices for ${\cal C}_2$.
Necessary condition: Assume $A\in{\cal C}_2$. Then $x^TAx\geq 0$ for $x\geq 0$. It is easy to see that $a,c\geq 0$ by picking $x$ to be the 2 unit-vectors. Next pick $x=\begin{pmatrix} t \cr 1 \end{pmatrix}$, where $t\geq 0$. If $b\geq 0$ then nothing to prove. Assume therefore $b<0$.

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*Case $a=0$.  By taking the limit $t\to\infty$ we see that $x^TAx$ becomes negative. Contradiction. $\Box$


*Case $a>0$. $x^TAx$ is a parabola with positive minimum point. The discriminant $D=4(b^2-ac)\leq 0$ must be non-positive. Then $-b \leq \sqrt{ac}$. $\Box$
Sufficient condition: Assume $a,c\geq 0$. Define $A_0~:=~\begin{pmatrix} a & -\sqrt{ac} \cr -\sqrt{ac} & c \end{pmatrix}$. It is easy to see that $x^TA_0x=(\sqrt{a}x_1-\sqrt{c}x_2)^2\geq 0$, so $ A\geq A_0 \in {\cal C}_2\Rightarrow A\in {\cal C}_2$. $\Box$
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$^1$ We will assume that a copositive matrix is symmetric, although the definition in Wikipedia (July 2020) doesn't mention this.
