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Problem about an Eigenvector

  1. Sep 25, 2016 #1
    1. The problem statement, all variables and given/known data
    If ##AA^T=A^TA##, then prove that ##A## and ##A^T## have the same eigenvectors.

    2. Relevant equations


    3. The attempt at a solution
    ##Ax=\lambda x##
    ##A^TAx=\lambda A^Tx##
    ##AA^Tx=\lambda A^Tx##
    ##A(A^Tx)=\lambda (A^Tx)##
    So, ##A^Tx## is also an eigenvector of ##A##.
    What should be the next steps?
     
  2. jcsd
  3. Sep 25, 2016 #2

    Mark44

    Staff: Mentor

    What you have to show is that if x is an eigenvector of A, then x, not ATx, is also an eigenvector of AT.
     
  4. Sep 25, 2016 #3

    Mark44

    Staff: Mentor

    A fact that might be pertinent here is that for square matrices A and B, if AB = BA, then each one is the inverse of the other.

    Edit: As pointed out by Ray, this is not true. Mea culpa.
     
    Last edited: Sep 25, 2016
  5. Sep 25, 2016 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    If
    $$A = \pmatrix{0&1\\1&0} \; \text{and} \; B = \pmatrix{1&0\\0&1}$$
    then ##AB = BA = A##, but ##B \neq A^{-1}## and ##A \neq B^{-1}##.

    However, if you mean that ##AB = (BA)^{-1}##, etc., then that seems OK, in this example at least.
     
    Last edited: Sep 25, 2016
  6. Sep 25, 2016 #5

    Mark44

    Staff: Mentor

    No, I meant the first, so my "fact" was incorrect. I guess I had that confused with if AB = I, then A and B are inverses.
     
  7. Sep 26, 2016 #6

    pasmith

    User Avatar
    Homework Helper

    I don't think we are entitled to assume that 0 is not an eigenvalue of [itex]A[/itex].
     
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