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?

  • 1
    $\begingroup$ 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. $\endgroup$ – Rahul Jan 5 '18 at 17:30
  • $\begingroup$ @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! $\endgroup$ – Henrik Hansen Jan 5 '18 at 17:49
  1. Let $$C_1=\{A: \text{A is a real symmetric } n \times n \text{ matrix and } X'AX \ge 0 \text{ for all n vector } X \ge 0\}$$


$$C_2=\{A: \text{A is a real symmetric } n \times n \text{ matrix and } X'AX \ge 0 for all n vector X\}$$

Obviously, if A belongs to C2, then A belongs to C1. Anyway, I think you are involved with this proposition: "more constraints cause smaller feasible space". Ok, we can see the proof from this angle:

  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.
  • $\begingroup$ Welcome on the site! I formatted a little bit your text, as you can see, we support Latex. I suggest to follow in your next posts. $\endgroup$ – peterh - Reinstate Monica May 10 '19 at 16:08

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