# Does the cone of copositive matrices include the cone of positive semidefinite matrices?

I'm trying to prove that the cone of copositive matrices is closed and in S. Boyd, L. Vandenberghe's Convex Optimization it says that:

K has nonempty interior, because it includes the cone of positive semidefinite matrices, which has nonempty interior.

I can't make sense of this. Since the definition of a copositive matrix is

$$x^TAx \geq0,\,\forall x \geq 0,$$

but for a positive semidefinite matrix all $x$ would be considered. It seems to me that positive semidefnite matrices are a more "general" concept and a less strict constraint than copositivity.

Am I misunderstanding the relationship between cones and sets?

• It is "easier" to be copositive than to be p.s.d. -- you have to satisfy the condition $x^TAx\ge0$ for fewer vectors. So any p.s.d. matrix is copositive, but a copositive matrix may not be p.s.d. (e.g. $A=\begin{bmatrix}1&2\\2&1\end{bmatrix}$). It's an easy mistake to make, I was confused for a minute there too. – Rahul Jan 5 '18 at 17:30
• @Rahul Thank you! The constraint on $x$ is something different from $x^TAx \geq 0$. Positive semidefiniteness must be satisfied for more vectors compared to what is required for copositivity! Will you supply an answer, then I'll accept it! – Henrik Hansen Jan 5 '18 at 17:49

1. Let $$C_1=\{A: \text{A is a real symmetric } n \times n \text{ matrix and } X'AX \ge 0 \text{ for all n vector } X \ge 0\}$$
$$C_2=\{A: \text{A is a real symmetric } n \times n \text{ matrix and } X'AX \ge 0 for all n vector X\}$$
1. For a copositive matrix A, the quadratic term X'AX must be nonnegative only for those vectors X satisfy the constraint $$X \ge 0$$, while for positive semidefiniteness the quadratic form must be satisfied the constraints $$X \ge 0$$, $$X \le 0$$ and all of vectors X which are neither nonnegative nor nonpositive. In other words, for positive semidefiniteness we need more constraints to be satisfied.