max_det_psd_completion.m

来自「斯坦福大学Grant和Boyd教授等开发的凸优化matlab工具箱」· M 代码 · 共 43 行

% Exercise 4.47: Maximum determinant PSD matrix completion
% Boyd & Vandenberghe "Convex Optimization"
% Almir Mutapcic - Jan 2006
%
% Given a symmetric matrix A in R^(n-by-n) with some entries unspecified
% we find its completion such that A is positive semidefinite and
% it has a maximum determinant out of all possible completions.
% This problem can be formulated as a log det (and det_rootn) problem.
%
% This is a numerical instance of the specified book exercise.

% problem size
n = 4;

% create and solve the problem
cvx_begin sdp
  % A is a PSD symmetric matrix (n-by-n)
  variable A(n,n) symmetric;
  A >= 0;

  % constrained matrix entries.
  A(1,1) == 3;
  A(2,2) == 2;
  A(3,3) == 1;
  A(4,4) == 5;
  % Note that because A is symmetric, these off-diagonal
  % constraints affect the corresponding element on the
  % opposite side of the diagonal.
  A(1,2) == .5;
  A(1,4) == .25;
  A(2,3) == .75;

  % find the solution to the problem
  maximize( log_det( A ) )
  % maximize( det_rootn( A ) )
cvx_end

% display solution
disp(['Matrix A with maximum determinant (' num2str(det(A)) ') is:'])
A
disp(['Its eigenvalues are:'])
eigs = eig(A)