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Linear question

  1. Dec 15, 2003 #1
    I am trying to show that a matrix A and A^t have the same eigenvalues. Well im sure its because det(a) = det(A^t), the characteristic polynomial is det(A-xI). But this question im trying to solve also asks for an example of a 2x2 matrix A where A and A^t have different eigenvectors. Im kinda lost of this one. Wouldn't they have the same eigenvectors because they have the same eigenvalues?
     
  2. jcsd
  3. Dec 16, 2003 #2
    no. consider the eivenvalue equation:

    [tex]A\mathbf{v}=\lambda\mathbf{v}[/tex]
    then take the transpose of that equation:

    [tex]\mathbf{v}^TA^T=\lambda\mathbf{v}^T[/tex]

    so the eigenvector of A becomes something you might call a "left eigenvector" of A^T, but there is no reason to think that it should also be a regular eigenvector.

    the eigenvalues must be the same, however, for the reason you stated.

    i imagine, that to find an example of a matrix whose transpose has different eigenvectors, you should just grab any old generic nonsymmetric matrix, it will probably do.
     
    Last edited: Dec 16, 2003
  4. Dec 16, 2003 #3
    thanks lethe,

    I feel kinda stupid that i overlooked that equation. Its really fundemental stuff[b(]
     
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