Least Squares with constraint on absolute value

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Hi , I need to solve a least squares values of the form
, where x is a 32x1 vector and B is a 32x32 matrix.
Howerver, x is complex and I need to constraint the solutions to make each element of vector x to have absolute value of 1.
Is that possible?
Best,

采纳的回答

Torsten
Torsten 2023-6-16
编辑:Torsten 2023-6-16
rng("default")
n = 32;
y = rand(n,1) + 1i*rand(n,1);
B = rand(n) + 1i*rand(n);
x0 = rand(n,1) + 1i*rand(n,1);
x0 = [real(x0);imag(x0)];
x0 = x0./[sqrt(x0(1:n).^2+x0(n+1:2*n).^2);sqrt(x0(1:n).^2+x0(n+1:2*n).^2)];
fun = @(x)(B*(x(1:n)+1i*x(n+1:2*n))-y)'*(B*(x(1:n)+1i*x(n+1:2*n))-y);
fun(x0)
ans = 1.2661e+04
nonlcon = @(x)deal([],x(1:n).^2+x(n+1:2*n).^2-ones(n,1));
sol = fmincon(fun,x0,[],[],[],[],[],[],nonlcon,optimset('MaxFunEvals',10000,'TolFun',1e-12,'TolX',1e-12))
Local minimum possible. Constraints satisfied. fmincon stopped because the size of the current step is less than the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance.
sol = 64×1
0.6418 -0.9517 -0.9328 -0.8359 -0.9372 0.4603 0.7496 0.8249 0.9990 -0.2242
fun(sol)
ans = 10.3184
sol(1:n).^2+sol(n+1:2*n).^2-ones(n,1)
ans = 32×1
1.0e-15 * 0 0 -0.2220 0 0 0 0 0 0 0

更多回答(1 个)

Matt J
Matt J 2023-6-15
编辑:Matt J 2023-6-15
You'll need to write the problem in terms of the real-valued components xi and xr of x,
x=xr+1i*xi
Once you do that, your absolute value constraints become quadratic,
xr^2+xi^2=1
and you can solve with fmincon.
  2 个评论
L
L 2023-6-16
编辑:L 2023-6-16
Than
ks for your answer.
Is this correct?
n = @(x) vecnorm( y - B*x);
A = [];
b = [];
Aeq = [];
beq = [];
lb = [];
ub = [];
nonlcon = @unity;
x0 = zeros(32,1);
x = fmincon(fun,x0,A,b,Aeq,beq,lb,ub,nonlcon)
function [c,ceq] = unity(x)
c = real(x)^2 + 1*iimg(x)^2 - 1;
ceq = [];
end

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