For given $n$, the following $n\times n$ real matrix $M=M^{T}$ is called *positive*, if 


$x^{T}M x\geq 0$


holds for all *non-negative* real $x\_1,x\_2,\cdots,x\_n$,
where $x=(x\_1,x\_2,\cdots,x\_n)^T$.


Notice here, the definition of positive is not the same as usual defination of positive semidefinite.


Now the goal is to characterize the set of all such $M$. In particular, I would like to show the following statement:


Such $M$ can always written as positive combination of some positive semidefinite matrix and some $E\_{ij}$s, where $E\_{ij}$ is the $n\times n$ matrix with the only non zero element 1 lies in the position $(i,j)$.



On the positivity of matrices
Such matrices are called [copositive](http://en.wikipedia.org/wiki/Copositive_matrix) in the literature. 


Moreover, the statement you want to show is known to be false. 
While I don't recall a counterexample right away,
an intuition for this is that it's NP-hard to check copositivity, but computationally easy to check the condition in your statement, via semidefinite programming, see e.g. this [text](http://www.mit.edu/~parrilo/pubs/files/Parrilo-Semidefinite%20programming%20based%20tests%20for%20matrix%20copositivity.pdf) by Pablo Parrilo.


EDIT: here is an explicit counterexample: the matrix
$$
M=\begin{pmatrix}
 1&-1& 1& 1&-1\\
-1& 1&-1& 1& 1\\
 1&-1& 1&-1& 1\\
 1& 1&-1& 1&-1\\
-1& 1& 1&-1& 1
\end{pmatrix}
$$
is copositive, but not equal to the sum of a positive semidefinite matrix $P$ and a nonnegative matrix $N$.
It is taken from an old paper by P.Diananda in 1962 ([Proc. Cambridge Phil.Soc., vol 58(1962)](http://journals.cambridge.org/article_S0305004100036185)), where it is also shown that $n=5$ is minimal size for which one has such a counterexample.


A quick way to see that $M\neq P+N$ is as follows. First of all notice that $N$ can be assumed to have 0s on the main diagonal. This the condition $M=P+N$, that we want to bring to a contradiction, is equivalent to the existence of a positive semidefinite $P$ with all 1s on the diagonal, and satisfying the conditions $M\_{ij}-P\_{ij}\geq 0$, for all $1\leq i<j\leq 5$. 


As $P$ is p.s.d., also for $i<j$ one has $|P\_{ij}|\leq 1$, as can be checked by computing
$x^\top P x$ for $x$ with $x\_i=x\_j=1$ and the remaining entries 0. Thus $P\_{ij}=-1$ for all $i,j$ s.t. $M\_{ij}=-1$. Further, all $P$'s satisfying these condition form a convex set $\Sigma$. 


Thus if $P\in\Sigma$ then $\Pi P\Pi^\top\in\Sigma$, for any permutation matrix $\Pi$ commuting with $M$, as $M=\Pi M\Pi^\top=P+N=\Pi (P+N)\Pi^\top$, i.e. $\Pi (P+N)\Pi^\top$ is another representation of $M$. Hence, by an averaging argument over the group $G$ of permutation matrices commuting with $M$, we can select $P$ to commute with all such $\Pi$'s. (More explicitly, take $\frac{1}{|G|}\sum\_{\Pi\in G}\Pi P\Pi^\top$.) It follows that there must exist $P$ of the form
$$
P=\begin{pmatrix}
 1&-1& p& p&-1\\
-1& 1&-1& p& p\\
 p&-1& 1&-1& p\\
 p& p&-1& 1&-1\\
-1& p& p&-1& 1
\end{pmatrix}, \quad -1\leq p\leq 1
$$
But none of such $P$'s are p.s.d., as its eigenvalues are
$$2 \, p - 1,
 -\frac{1}{2} \, p {\left(\sqrt{5} + 1\right)} - \frac{1}{2} \, \sqrt{5} + \frac{3}{2},
 \frac{1}{2} \, p {\left(\sqrt{5} - 1\right)} + \frac{1}{2} \, \sqrt{5} + \frac{3}{2}.
$$


