A $n\times n$ symmetric matrix $A$ is said to be copositive if $\mathbf{x}^{\top}A\mathbf{x} \geq 0$ for all $\mathbf{x}\in\mathbb{R}^{n}$ and $\mathbf{x} \geq 0$. And I want to show that the set of $n \times n$ non-copositive matrices is nonconvex unless $n=1$.
I can prove that the set of $n \times n$ copositive matrices is convex, which directly follows the definition of the convex set. But for the non-copositive part, if I pick two matrices $A$, $B$ which are not copositive, we have $\mathbf{x}_1^{\top}A\mathbf{x}_1 < 0$ and $\mathbf{x}_2^{\top}B\mathbf{x}_2 < 0$ for some $\mathbf{x}_1,\mathbf{x}_2\geq 0$. How can I prove that $\theta A+(1-\theta)B$ is copositive for some $\theta$?
Besides, I have a question about the definition of the copositive matrix: is the copositive matrix always positive semidefinite or elementwise nonnegative? Could anyone give me an example?
Thanks.
Best Answer
Let $$A = \begin{bmatrix} 100 & 0 \\0 & -1\end{bmatrix}, B = \begin{bmatrix}-1 & 0 \\0 & 100\end{bmatrix}.$$ They are both non-copositive, but the average of them $$\frac{A+B}{2} = \begin{bmatrix}49.5 & 0 \\ 0 & 49.5\end{bmatrix}$$ is copositive. Since the average is their convex combination, we get that the set of non-copositive matrices isn't convex.