Meaning of matlab coding(l1 minimization)...................

Question

Tankeswar Kumar 2013-2-22

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/64493-meaning-of-matlab-coding-l1-minimization

in the folloeing matlab coding:

%Solve
% min_x ||x||_1  s.t.  Ax = b
% Recast as linear program
% min_{x,u} sum(u)  s.t.  -u <= x <= u,  Ax=b
% Usage: xp = l1eq_pd(x0, A, At, b, pdtol, pdmaxiter, cgtol, cgmaxiter)
% x0 - Nx1 vector, initial point.
% A - Either a handle to a function that takes a N vector and returns a K 
%     vector , or a KxN matrix.  If A is a function handle, the algorithm
%     operates in "largescale" mode, solving the Newton systems via the
%     Conjugate Gradients algorithm.
% At - Handle to a function that takes a K vector and returns an N vector.
%      If A is a KxN matrix, At is ignored.
%
% b - Kx1 vector of observations.
% pdtol - Tolerance for primal-dual algorithm (algorithm terminates if
%     the duality gap is less than pdtol).  
%     Default = 1e-3.
%
% pdmaxiter - Maximum number of primal-dual iterations.  
%     Default = 50.
%
% cgtol - Tolerance for Conjugate Gradients; ignored if A is a matrix.
%     Default = 1e-8.
%
% cgmaxiter - Maximum number of iterations for Conjugate Gradients; ignored
%     if A is a matrix.
%     Default = 200.
% function programme...
function xp = l1eq_pd(x0, A, At, b, pdtol, pdmaxiter, cgtol, cgmaxiter)
largescale = isa(A,'function_handle');
if (nargin < 5), pdtol = 1e-3;  end
if (nargin < 6), pdmaxiter = 50;  end
if (nargin < 7), cgtol = 1e-8;  end
if (nargin < 8), cgmaxiter = 200;  end
N = length(x0);
alpha = 0.01;
beta = 0.5;
mu = 10;
gradf0 = [zeros(N,1); ones(N,1)];
% starting point --- make sure that it is feasible
if (largescale)
  if (norm(A(x0)-b)/norm(b) > cgtol)
disp('Starting point infeasible; using x0 = At*inv(AAt)*y.');
AAt = @(z) A(At(z));
[w, cgres, cgiter] = cgsolve(AAt, b, cgtol, cgmaxiter, 0);
if (cgres > 1/2)
disp('A*At is ill-conditioned: cannot find starting point');
xp = x0;
return;
end
x0 = At(w);
end
else
if (norm(A*x0-b)/norm(b) > cgtol)
disp('Starting point infeasible; using x0 = At*inv(AAt)*y.');
opts.POSDEF = true; opts.SYM = true;
[w, hcond] = linsolve(A*A', b, opts);
   if (hcond < 1e-14)
   disp('A*At is ill-conditioned: cannot find starting point');
xp = x0;
return;
end
   x0 = A'*w;
   end  
  end

Now my question is that A & At have different dimensione, that measns AAt have the same dimension with At. because of AAt = @(z) A(At(z)); is it write. What is the necessity of AAt and At....Please help...............