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?
Best Answer
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.