Suppose I have an $n \times n$ matrix $A$ with positive real entries. Is it possible to find an $n \times 1$ vector $v$ such that
$$ vv^T = A $$
where $v^T$ is the transpose of $v$?
Is it possible to use singular value decomposition or any technique to achieve this?
Best Answer
The rank of $vv^T$ is 1 or 0, so only square matrices of rank 1 or 0 can have such a representation.
If $A=0$ then $v=\vec 0$.
In general if $v=(a_1,...,a_n)^T$ then $vv^T$ has rows $a_1v^T,..., a_nv^T$, so you can find $v$ by looking at the diagonal entries of $A$.