Least Squares with constraint on absolute value

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,

Answer 1

[Torsten]

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

[L]

Thanks!

Answer 2

[Matt J]

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.

[L]

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

[Matt J]

I meant that you should follow the example here,

https://www.mathworks.com/help/optim/ug/fit-model-to-complex-data.html#d124e72611