When I solve an optimization problem with an eigenvalue constraint, it gets two different results under the different position of function value().

[风清扬]

Dear Pro. Johan Lofberg

When I solve an optimization problem ,I meet a question.

The code of my optimization problem is:

%%%%%%%%%%%%%%

A = sdpvar(3,3,'full');

B = [1 0 0;0 1 1; 1 1 1];

cons = [max(abs(eig(A-B)))<=0.8];   %constrains

obj = trace((A - B)' * (A - B));  %objetive

solvesdp(cons,obj)            %solve

A_v = value(A);

type1 = eig(A_v-B)

type2 = value(eig(A-B))

%%%%%%%%%%%%%

the results of two variables are:

type1 =

   0.4158 + 2.2437i

   0.4158 - 2.2437i

   0.2080 + 0.0000i

type2 =

   -0.5604

    0.8000

    0.8000

%%%%%%%%%%%%%

But as far as I know， type1 should equal to type2. Why there are two different results? If it is wrong of my code, how to use yalmip to represent eigenvalue constraints?

Sincerely yours,

[Johan Löfberg]

don't use the eig operator. it is an experimental hack which really isn't supposed to be used (as it doesn't work). You will have to come up with a semidefinite representation (which looks impossible as you optimize over unsymmetric matrices)

[song yang]

It means that the eig operator is no use even though the solver gives a result. Is there only one solution to such an optimization problem which has eigenvalue constraint ?

If so, when (A-B) is symmetric in my code, how can I solve it? The norm-2 operator is ok?

thank you!

[Johan Löfberg]

For symmetric, you would simply write -0.8*I <= A-B <= 0.8*I

The quadratic objective is fine, although it is unnecessarily complicated for large problems to create the product (symbolic O(n^3) operation) and then take the trace. Either simply use norm(A-B,'fro')  or vectorize E = A-B;obj = E(:)'*E(:).

[song yang]

OK, thank you!