Set of all positive definite matrices with off diagonal elements negative Please, help me to find the set of all positive definite matrices (PDM) of which off diagonal elements are negative.
Considering the case with n=2, the symmetric mat $A=[a_1, a_2;a_2, a_3]$ needs to be a PD. i.e.  $x'Ax>0$,  $a_2\neq 0$, $x\neq 0$ where $x=[x_1,x_2]'$.
Always we have $a_1>0$ and $a_3>0$ as they are principal minors.
I guess such matrices exists as $x_1^2a_1+2x_1x_2a_2+x_2^2a_3$ can be positive with either of $x_1$ or $x_2$ negative and $a_2$ negative. 
Also, in case which $x_1>0$, $x_2>0$ and $a_2<0$, the condition $x'Ax>0$ is met if the absolute value of $x_1^2a_1+x_2^2a_3$ is greater than absolute value of $2x_1x_2a_2$.
How can I genearlize this to denote and derive set of all possible positive definite matrices? I mean can you give some examples of A which is positive definite with offdiagonal elements negative?
or suggest  method to get them? please?
 A: If you want to find all possible values of off-diagonal elements of a matrix A such that A is positive definite and all the off-diagonal elements are negative:
First assume that the diagonal elements are given, $a_{11}, a_{22},a_{33},a_{44}$, which must be positive.  I will presume that by positive definite, you mean symmetric positive definite.  Let the 6 upper triangular elements be variables, a_{12},a_{13},a_{14},a_{23},a_{24},a_{34}. The lower triangle is determined by symmetry.
Using MATLAB notation, you need all principal minors to be positive:
det(A(1:2,1:2)) > 0 , det(A(1:3,1:3)) > 0, det(A) > 0  (note that det(A(1:1,1:1)) = $a_{11}$ which is > 0 by assumption). You can expand these out symbolically, which results in a set of 3 strict inequalities in 6 variables. Additionally. there are negativity constraints on all 6 variables.  The set of all solutions to these 9 inequalities is the solution to your problem.  However, from a practical perspective, this is not a good way to compute things, and really just serves as a theoretical exercise.
As for the question in your comment "can you please suggest a way to show that all such matrices, that is those with off diagonal elements negative but still positive definite form a convex set?":  You should already know that the set of all positive definite matrices is convex (a convex cone).  Any convex combination of negative numbers will be negative, therefore, a convex combination of matrices having negative off-diagonal elements will have negative off-diagonal elements. Hence the set of all matrices having negative off-diagonal elements is convex.  Hence, the set of all positive definite matrices having negative off-diagonal elements is the intersection of two convex sets, hence is convex.
