butools.reptrans.SimilarityMatrix¶

butools.reptrans.SimilarityMatrix()¶

Matlab:	B = SimilarityMatrix(A1, A2)
Mathematica:	B = SimilarityMatrix[A1, A2]
Python/Numpy:	B = SimilarityMatrix(A1, A2)

Returns the matrix that transforms A1 to A2.

Parameters:	A1 : matrix, shape (N,N) The smaller matrix A2 : matrix, shape (M,M) The larger matrix (M>=N)
Returns:	B : matrix, shape (N,M) The matrix satisfying \(A_1\,B = B\,A_2\)

Notes

For the existence of a (unique) solution the larger matrix has to inherit the eigenvalues of the smaller one.

Examples

For Matlab:

>>> A1m = [0.2, 0.8, 0.; 1.2, -0.4, 0.1; -0.2, 0.7, 0.5];
>>> T = [1., 2., -4., 6.; 0., 8., -9., 7.; -3., 7., 8., -2.];
>>> A2m = pinv(T)*A1m*T;
>>> B = SimilarityMatrix(A1m, A2m);
>>> err = norm(A1m*B-B*A2m);
>>> disp(err);
   1.5605e-15

For Mathematica:

>>> A1m = {{0.2, 0.8, 0.},{1.2, -0.4, 0.1},{-0.2, 0.7, 0.5}};
>>> T = {{1., 2., -4., 6.},{0., 8., -9., 7.},{-3., 7., 8., -2.}};
>>> A2m = PseudoInverse[T].A1m.T;
>>> B = SimilarityMatrix[A1m, A2m];
>>> err = Norm[A1m.B-B.A2m];
>>> Print[err];
1.0553656492422553*^-15

For Python/Numpy:

>>> A1m = ml.matrix([[0.2, 0.8, 0.],[1.2, -0.4, 0.1],[-0.2, 0.7, 0.5]])
>>> T = ml.matrix([[1., 2., -4., 6.],[0., 8., -9., 7.],[-3., 7., 8., -2.]])
>>> A2m = la.pinv(T)*A1m*T
>>> B = SimilarityMatrix(A1m, A2m)
>>> err = la.norm(A1m*B-B*A2m)
>>> print(err)
9.4758466024e-16