Let $A$ be real square matrix.


Let $\mathcal{F}(A)$ be the set of real matrices $A'$ of the same size such that $A'\_{ii}=A\_{ii}$ for all $i$, and for all $i,j$, $A\_{ij}=0\Rightarrow A'\_{ij}=0 \land A\_{i,j} \ne 0 \implies A'\_{i,j} \ne 0$: in other words, obtained from $A$ by modifying nonzero nondiagonal entries.



> 
> Q1 What are sufficient conditions for $A$ s.t. exists 
>  a positive definite $A'$ in $\mathcal{F}(A)$?
> 
> 
> 


Since there are several definitions of positive definite,
we don't require neither $A$ or $A'$ to be symmetric.


In practice this is possible for some $A$.


Partial answers are welcome.
The definition of positive definite from [mathworld](http://mathworld.wolfram.com/PositiveDefiniteMatrix.html)


$M$ is positive definite if for all nonzero real vectors $x$, $x^T M x >0$.



Making a real matrix positive definite by replacing nonzero and nondiagonal entries with arbitrary nonzero reals
The answer is quite boring... Necessary and sufficient conditions are $A\_{ii}>0$ for each $i$. In this case, let $B\_\varepsilon$ the matrix with entries
$$
(B\_\varepsilon)\_{ij} = \begin{cases}
A\_{ij} & i=j,\\
\varepsilon & i\neq j, A\_{ij}\neq 0,\\
0 & i\neq j, A\_{ij}=0.
\end{cases}
$$
The matrix $B\_0$ is diagonal and posdef, so for any sufficiently small value of $\varepsilon$ the matrix $B\_\varepsilon$ (which belongs to $\mathcal{F}(A)$) is posdef, too, since positive definiteness is an open condition.


On the other hand, if $A\_{ii}\leq 0$ for some $i$ there is no hope, since for each $A'\in\mathcal{F}(A)$ we have $x^\top A'x \leq 0$ when $x$ is the vector $e\_i$ of the canonical basis.


