I have asked this question in math.se without any success.
Let $\mathbf{A}$ be a symmetric $n\times n$ positive semi-definite matrix and also such that each of its entries is positive. Does $\mathbf{A}$ have a decomposition of the form \begin{align} \mathbf{A} \,=\,\sum_{i=1}^{k}\mathbf{y}_i\mathbf{y}_i^T \end{align} where each vector $\mathbf{y}$ is also entry-wise positive and $k\leq n$.
No, there exist doubly nonnegative matrices which are not completely positive, see for example The difference between 5 x 5 doubly nonnegative and completely positive matrices (2009).
A graph-based characterization of doubly nonnegative matrices which are completely positive is: Every doubly nonnegative matrix realization of a graph $G$ is completely positive if and only if $G$ does not contain an odd cycle of length at least 5, see Open problems in the theory of completely positive and copositive matrices (2015).
An alternative characterization is give in a 2011 MO question: A doubly nonnegative $n\times n$ matrix is completely positive if and only if the $n$ vectors making up the Gram matrix lie in the non-negative orthant of some space of dimension $>n$.
Not necessarily.
If this would hold, all components of the $y_i$ would be bounded in terms of $A$; so, by compactness argument, the same would hold for matrices and columns with non-negative entries. We will show that this is not the case.
Let $I$ and $J$ be the identty and the all-ones matrices of order $3\times3$. Set $$ A=\left[\matrix{ 100I&J\\J&100I}\right]. $$ Then each $y_i$ may contain at most one non-zero among the first three entries, the same for the last three. On the other hand, for each $a=1,2,3$ and $b=4,5,6$ there should be a $y_i$ with non-zero $a$th and $b$th entries. Thus $k\geq9$.
Remark. This argument shows that, in general, $k\geq [n^2/4]$. It cannot show more, since the edges of an arbitrary graph on $n$ vertices may be split into at most $n^2/4$ cliques. It may happen that this estimate is sharp.
