Way to solve AX=XB

dim = 4;
X = sym('X',[dim dim]);
A = rand(dim);
B = A.';
[M, ~] = equationsToMatrix(A*X==X*B)
if rank(M) < size(A,1)^2
  N = null(M);
  for i = 1:size(N,2)
     S{i} = reshape(N(:,i),size(X));
     S{i}
     A*S{i}-S{i}*B
  end
end

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Answer 3

Matt J 2023-1-28

0
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1900510-way-to-solve-ax-xb#answer_1158480

编辑：Matt J 2023-1-28

在 MATLAB Online 中打开

[ma,na]=size(A);
[mb,nb]=size(B);
%size(X)=[na,mb]
X=null( kron(speye(mb),A) - kron(B.',speye(na))  );
X=reshape(X,na,mb,[]);

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the cyclist 2023-1-28

在 MATLAB Online 中打开

I couldn't get this method to work. Am I overlooking something dumb?

rng default
A = rand(5);
B = rand(5);
[ma,na]=size(A);
[mb,nb]=size(B);
X=null( kron(speye(mb),A) - kron(B.',speye(na))  );
Error using svd
SVD does not support sparse matrices. Use SVDS to compute a subset of the singular values and vectors of a sparse matrix.

Error in null (line 75)
    [V, s] = svd(A','vector');
X=reshape(X,na,mb,[]);

Bruno Luong 2023-1-28

编辑：Bruno Luong 2023-1-28

在 MATLAB Online 中打开

null can only work wth full matrix

rng default
A = rand(5);
XX = rand(5);
B = XX\(A*XX);
[ma,na]=size(A);
[mb,nb]=size(B);
K=null( kron(eye(mb),A) - kron(B.',eye(na)));
R = rand(size(K,2),1); % Any random vector with this size will do the job
X = reshape(K*R,[na,mb])
X = 5×5
    0.1100    0.1626    0.0772   -0.0458    0.0810
   -0.0536   -0.3884    0.0510    0.0552   -0.0602
    0.1937   -0.1993    0.3240    0.3508   -0.0352
   -0.1165    0.4388   -0.3413   -0.3089    0.1851
    0.0500    0.0277    0.0913    0.1389    0.0450
norm(A*X-X*B)
ans = 7.4501e-16