I have asked this question in [math.se](https://math.stackexchange.com/questions/3365350/decomposing-a-entry-wise-positive-positive-semidefinite-matrix) 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 $k\leq n$.



Rank-one positive decomposition for a entry-wise positive positive definite matrix
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](https://www.sciencedirect.com/science/article/pii/S002437950900281X) (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](https://pdfs.semanticscholar.org/0daf/f8a13a4afebbfc4b39b7393e6c2b626d3a7f.pdf) (2015).


An alternative characterization is give in a 2011 [MO question:](https://mathoverflow.net/q/67641/11260) *A doubly nonnegative $n\times n$ matrix is completely positive if and only if the $n$
 vectors making up the [Gram matrix](https://en.wikipedia.org/wiki/Gramian_matrix) lie in the non-negative orthant of some space of dimension $>n$.*


