TITLE: Characterising semi-definite positiveness on vectors with non-negative entries
QUESTION [8 upvotes]: My problem is to characterise (or find useful information on) the cone $C$ of $N\times N$ matrices $M$ ($N\geq 1$) such that $$V^t M V\geq 0$$ for every vector $V $ with non-negative entries. Is this cone of matrices familiar to anyone?
Remark 1: $C$ clearly contains the convex cone $S$ of semi-definite positive matrices, and that of matrices $P$ with non-negative entries. Even though it looks too easy, I could not prove that $C$ is not the convex hull of $S$ and $P$.
Remark 2: My original problem is to characterise the dual cone of $C$, containing the covariances of random vectors of $\mathbb{R}^N$ with a.s. non-negative entries.

REPLY [6 votes]: (After Noah Stein's answer) By definition, the dual cone $C^\star$ is spanned by matrices $v\otimes v$ with $v\ge0$. The following counter-example is due to Hall. The $5\times5$ symmetric matrix
$$S=\begin{pmatrix}  4 & 0 & 0 & 2 & 2 \\\\ 0 & 4 & 3 & 0 & 2 \\\\ 0 & 3 & 4 & 2 & 0 \\\\ 2 & 0 & 2 & 4 & 0 \\\\ 2 & 2 & 0 & 0 & 4 \end{pmatrix}$$
has non-negative entries and is positive semi-definite. Therefore, it belongs to $(S\cup P)^\star$. Yet, it cannot be written as the sum of $v_j\otimes v_j$ where the vectors $v_j$ are non-negative. Therefore $A\not\in C^\star$. By duality, this proves that $C$ is not the convex hull of $S\cup P$.
The example is analyzed in details in Exercise 347 of my list.