Way to solve AX=XB

Is there any implementation of Tsai and lenz's (Or any other) method for solving AX=XB for hand- Eye Calibration?

Answer 1

[the cyclist]

This is a special case of the Sylvester equation.

Looks like the sylvester function will be helpful for you.

You might also be interested in this blog post on the topic by Cleve Moler.

[SATISH SONWANE]

Sorry. That didn't help.

Answer 2

[Torsten]

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

Answer 3

[Matt J]

[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,[]);

[the cyclist]

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]

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