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Show that A and its transpose have the same eigenvalues

  1. Dec 9, 2016 #1
    1. The problem statement, all variables and given/known data
    Show that ##A## and ##A^T## have the same eigenvalues.

    2. Relevant equations


    3. The attempt at a solution
    If they have the same eigenvalues, then ##Ax = \lambda x## iff ##A^T x = \lambda x##. In other words, we have to show that ##|A - \lambda I| = 0## iff ##|A^T - \lambda I| = 0##. From the properties of transpose, we see that ##(A - \lambda I)^T = A^T - \lambda I##. It's a property of transposes that ##A^T## is invertible iff ##A## is also invertible. So we have shown that ##A - \lambda I## is invertible iff ##A^T - \lambda I## is also invertible. So this shows that they have the same eigenvalues.

    I think that this is the correct solution, but I am a little confused about the beginning part of the proof. Does this imply that A and its transpose also have the same eigenvectors? Or do they have the same eigenvlues but different eigenvectors?
     
  2. jcsd
  3. Dec 9, 2016 #2

    fresh_42

    Staff: Mentor

    You don't need to consider invertibility here. And you have already said everything you need. Just clean it up a little.
    What are the eigenvalues of ##A##?
    What are the eigenvalues of ##A^T##?
    And how are ##\det B## and ##\det B^T## related for any matrix ##B##? (Esp. for ##B=A-\lambda I##.)
     
  4. Dec 9, 2016 #3
    ##det(A) = det(A^T)## for all square ##A##. So we have that ##det(A - \lambda I) = det((A - \lambda I)^T) = det(A^T - \lambda I)##
     
  5. Dec 9, 2016 #4
    So as a last note do ##A## and ##A^T## also have the same eigenvectors?
     
  6. Dec 10, 2016 #5

    fresh_42

    Staff: Mentor

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