Hi Hamdi,

In this case you can set up a double for loop for variables a and b, check the feasibility of each point and store the result as a binary value (1 for feasible, 0 for infeasible).

Once you have done that, you can plot the binary matrix using spy function in matlab. Do it for each theorem you have.

good luck

Issam Chekakta thank you, but how to check the feasibility of each point? what is the method that the researchers use?

LMIs are a part of SDP (semidefinite programming) in convex optimization. There are several ways to solve these kinds of problems. First one is LMI toolbox of Matlab, you have also YALMIP and CVX for Matlab also you can download both of them from their official websites.

Issam Chekakta yes I have used Yalmip before to do that but I wanted to confirm if this is the used way for doing that, so after solving the problem I have to check the eigenvalues of the LMI if they are definite negative or not if they are definites negative so the problem is feasible, is this the used method or there is something missing?

Wail Hamdi Yes, of course. you can get the first decision on feasibility by the obj.problem instruction in Yalmip. If your LMIs are a bit complex and you need further checking, you can use the instruction of"[Primal, Dual]=check(your_problem)" to show primal and dual feasibility and see if the eigenvalues of each LMI are satisfied or not, and from that you can have your answer on feasibility.

Issam Chekakta yes I have done exactly like that but I faced two problems:

1 - the obj.problem instruction in Yalmip is not precise because barely almost always there is no problem, but if you applicate the LMI it will not work, and when you check the eigenvalue using check you will find that the LMI is not well solved.

so you have to use check instruction, not only obj.problem.

2 - it depends on the Solver, for example, Lmilab solver succeeds to solve the problem where Mosek solver did not, and sometimes the opposite, so there may be points feasible but the solver could not solve them, so this will give me imprecision in mapping.

what do you say about those two points?

Wail Hamdi Yes, as I said, obj.problem is used only with simple problems, so obviously you can not rely on it when it comes to complex LMIs.

2 - Regarding the feasibility with diferent solvers, since you are doing the feasibility for several theorems, there must be fair-play whatever the solver. So, basically, if you choose MOSEK, you have to stick with Mosek for all other tested theorems with the same conditions of feasibility (eigenvalues verification). In our work, we often used SeduMi or MOSEK.

Speaking of a solver's success and failure, this depends on the approach used in the solver and the way of handling the problem.

Thus, you can use an appropriate solver and give all tested theorems a fair-play (solver, feasibility criterion, number of iteration,...etc).

Issam Chekakta thank you very much sir issam,

this is my attempt, what do you think about it, do you work in the same way or there are any suggestions.

Wail Hamdi Yes, pretty much the same except for some minor variations (I use the spy function with a binary matrix of feasible points). But, overall, this is correct, and you can improve it further by adding (a and b steps, try to define you constraints using For loops(makes things easier when you have a lot of repetitive LMIs with different indexes) )....etc.

Issam Chekakta which one is better, Yalmip or LMI toolbox which one do you usually use?

I didn't use LMI toolbox before, is the method the same as in Yalmip, solving the LMI and then verifying the eigenvalues?

Wail Hamdi Your code is correct, and we usually use YALMIP (very handy).

Issam Chekakta thank you very much sir issam

You are most welcome Wail Hamdi

good luck

Usually, these problems are easily solved by YALMIP toolbox in MATLAB.

Here, I write the following pseudocode for your problem:

It can be solved by defining two 'for-loop' as follows:

for min(a)0

figure(1)

hold on

plot(a,b,'.k','MarkerSize',5)

end

end

end

sir Mohsen Farbood thank you very much

Good Luck :)

This approach will work out fine only if you have a very small number of parameters. You might instead want to switch to an optimisation problem of a single scalar, for example maximise k under Ai'*P+P*Ai < - k I with trace(P)=1. In this way the biggest the k, the "easiest" in principle the original LMI.