Let $A=(a\_{ij}) \in \mathbb{R}^{n \times n}$ be a matrix with $a\_{ij} = a\_{ji} \geq 0$ and
$B=(b\_{ij})$ with $b\_{ij} = \sqrt{a\_{ij}}$.


Is $B$ positive-definite whenever $A$ is?


In other words:


$\sum\_{1 \leq i,j \leq n} c\_i c\_j a\_{ij} > 0 \iff \sum\_{1 \leq i,j \leq n} c\_i c\_j \sqrt{a\_{ij}} > 0$
for every choice of $c\_i \in \mathbb{R}$?


A counterexample or any hint to a proof would be appreciated, thanks!



Is the componentwise square-root of a positive-definite matrix also pos.-def.?
Counterexample:


$B=\left(\begin{array}{ccc} 1 & 0 & 1 \\\\ 0 & 1 & 1 \\\\ 1 & 1 & \sqrt3 \end{array}\right)$.


This is for $n=3$, and easily extends to all $n\geq 3$.


For $n=2$ it holds, though.


