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A Matrices commute & Eigenvectors question

  1. Feb 11, 2017 #1
    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?
     
  2. jcsd
  3. Feb 11, 2017 #2

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    ##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.

    EDIT: Take ##A = I_n## and ##B \in M_{n,n}(\mathbb{F})## where ##\mathbb{F}## is a field. Then you can see that this claim is obviously false.
     
  4. Feb 11, 2017 #3
    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?
     
  5. Feb 11, 2017 #4

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    Matrix A does have eigenvectors.. (but yes, A is not diagonalizable because we do not have enough linearly independent eigenvectors)

    ##det(A - \lambda I_n) = 0 \iff \lambda = 0##

    Thus ##0## is an eigenvalue. We deduce that the eigenvectors are ##V_0 = span(\{(1,0)\})##

    As for your question with the hermitian matrices, look at the edit in my reply to your first post.
     
    Last edited: Feb 11, 2017
  6. Feb 13, 2017 #5
    Thanks a lot. Is it possible to construct similar example for to Hermitian matrices?
     
  7. Feb 13, 2017 #6

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    Yes, look at the edit in my first post. (The identity matrix is hermitian)
     
  8. Feb 14, 2017 #7
    Yes. But the other one is not. Is there any example like this where both matrices ##A## and ##B## are hermitian.
     
  9. Feb 14, 2017 #8

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    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).
     
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