Whether the set of a symmetric copositive matrices $x^T A x \geq 0$ for $\forall x \in \mathbb{R}^n$ $x \geq 0$ is convex?

Question

Whether the set of a $n \times n$ symmetric copositive matrices $x^T A x \geq 0$ $\forall x \in \mathbb{R}^n$ and $x \geq 0$ is convex?

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Attempt:

If I show that the set of the positive semidefinite (psd) matrices is convex, then the intersection of the set of psd matrices and $x \geq 0$ is also convex. Is that correct?

Answer 1

No need to do that! Note that if $A,B$ are copositive, then for any $0<\lambda<1$ and $x\ge 0$:$$x^T\Bigg(\lambda A+(1-\lambda)B\Bigg)x=\lambda x^TAx+(1-\lambda )x^TBx\ge 0$$which implies that also $\lambda A+(1-\lambda)B$ is copositive and hence the set of copositive matrices is convex.

Answer 2

You can follow these steps.

Let $\mathcal{F}$ denote the family of all the set of a $n\times n$ symmetric copositive matrices.

Now prove the following two assertions: The first one is,

For any $\lambda>0$, $A\in\mathcal{F}$ implies $\lambda A\in\mathcal{A}$.

Which follows from the fact that $x^T (\lambda A) x = \lambda (x^T Ax)$.

The second is,

For any $A,B\in\mathcal{F}$, $A+B\in\mathcal{F}$.

Which you can do by calculation $x^T (A+B)x$.

Once done. You can combine those two to show that

For any $A,B\in\mathcal{F}$ and $\lambda\in[0,1]$ we have $\lambda A + (1-\lambda) B \in\mathcal{F}$

which means, by definition, that $\mathcal{F}$ is convex.