TITLE: Is the absolute value of a P.D.S. matrix P.D.S.
QUESTION [6 upvotes]: Suppose that $A$ is a positive definite symmetric matrix (P.D.S.). Now consider the matrix $|A|$, the matrix arrived at by taking the absolute value of all the entries of $A$. Is $|A|$ also P.D.S.? I have been trying to construct a counter example, but I can't seem to get one. Can someone proved a proof or counterexample?

REPLY [2 votes]: One counterexample could be
$$
A=\left[\matrix{10 & 3 & -2 & 1\\3 & 10 & 0 & 9\\-2 & 0 & 10 & 4\\1 & 9 & 4 & 10}\right].
$$
$A$ is positive-definite, however $|A|$ is not positive-semidefinite.
Another one:
$$
A=\left[\matrix{1.5 & 1 & 0 & -1\\1 & 1.5 & 1 & 0\\0 & 1 & 1.5 & 1\\-1 & 0 & 1 & 1.5}\right].
$$