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Is it possible to find matrices that commute but eigenvectors of one matrix are not the eigenvectors of the other one. Could you give me example for it?
##A = \pmatrix{0&1\\0&0}, B = \pmatrix{1&0\\0&1}## commute. ##B## has an eigenvector ##(0,1)##, which ##A## doesn't have. They do have a common eigenvector.Is it possible to find matrices that commute but eigenvectors of one matrix are not the eigenvectors of the other one. Could you give me example for it?
Matrix A does have eigenvectors.. (but yes, A is not diagonalizable because we do not have enough linearly independent eigenvectors)Matrix A practically do not have eigenvectors. Right? Because it is not diagonalizable.
What about two hermitian matrix. Is there any posiibility like this. Is it some easy way to construct this?
Yes. But the other one is not. Is there any example like this where both matrices ##A## and ##B## are hermitian.Yes, look at the edit in my first post. (The identity matrix is hermitian)
Take ##A = I_n## and ##B## such that ##B## is hermitian. You should do some effort yourself (almost any hermitian matrix for ##B## will give you what you search for).Yes. But the other one is not. Is there any example like this where both matrices ##A## and ##B## are hermitian.