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

istrue for $n=2$ $\endgroup$ – Yemon Choi Jan 20 '11 at 21:501more comment