Is the componentwise square-root of a positive-definite matrix also pos.-def.? 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!
 A: 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.
A: In the tradition (local to this question) of pointing out special cases or related results as answers, there is the amazing theorem of I. J. Schoenberg, of which I state the simplest form: Let $a_1, \dotsc, a_{n(n-1)/2}$ be the edge lengths of a simplex in $\mathbb{R}^n.$ Then, so are $a_1^\alpha, \dotsc, a_{n(n-1)/2}^\alpha,$ for any $0\leq \alpha \leq 1.$ This theorem is really about positive-semidefinite functions in the sense of Bochner (but has a direct spectral interpretation if you use the Cayley-Menger matrix).
A: As a slightly longish comment, however, I would like to point out that there is a rich class of matrices, for which component-wise square root remains positive definite. The simplest example is the "unit" of this class of matrices, namely the matrix of all ones.
In general, there are several wonderful matrices, for which not only componentwise squareroot, but any root, continues remaining positive definite (in other words, the matrix $a_{ij}^\epsilon$ is positive definite too). 
Simple, but instructive examples include:


*

*$a_{ij} = \frac{1}{\lambda_i + \lambda_j}$, where the $\lambda$s are nonnegative.

*$a_{ij} = \min(i,j)$

*$a_{ij} = \mbox{gcd}(i,j)$


These matrices are called infinitely divisible matrices. Please see the linked PDF for a very nice introduction to such matrices. Additional references can also be found in my related answer to an old question
A: Whenever $A$ is negative semi-definite on the subspace of vectors $(a_1,\dots,a_n)$ with $\sum_i a_i=0$, then the same will be true for $B$. This is a result of Schoenberg.
