Let $A$ be an $n \times n$ matrix with positive real entries such that rank($A$) > 1. Is it possible to find an $n \times 1$ vector $v$ such that
$$ vv^T = A $$
where $v^T$ is the transpose of $v$? The obvious technique is to make a series of equations with unknowns and and attempt to use substitution to solve, however are any other techniques available to try even if a solution is not guaranteed to exist or to be unique?
The solution for the case where rank($A$) = $0$ or rank($A$) = $1$ is given in the following link: Express a matrix as a vector multiplied by its transpose . I am inquiring about the case where rank($A$) > 1 .
Any matrix of the form $vv^\top$ necessarily has rank equal to $1$ (except when $v=0$). One way to prove this is to note that the columns are scalar multiples of each other, so the column space is one-dimensional.