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Show that a matrix's transpose has same eigenvalue.

  1. Nov 5, 2012 #1
    Show that a matrix and its transpose have the same eigenvalues.
    I must show that det(A-λI)=det(A^t-λI)
    Since det(A)=det(A^t)
    →det(A-λI)=det((A-λI)^t)=det(A^t-λI^t)=det(A^t-λI)
    Thus, A and A^t have the same eigenvalues.

    Is the above enough to prove that a matrix and its transpose have the same eigenvalues or am i missing something?
     
  2. jcsd
  3. Nov 5, 2012 #2

    Dick

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    No, I don't think you are missing anything. The eigenvalues of A and A^t satisfy the same equation.
     
  4. Nov 5, 2012 #3
    I don't think you've shown that they have the same eigenvalues.

    You want to show that a=b in the following equations; Ax=ax and A^T x=bx

    You know that Det[A-aI]=0 and Det[A^T-bI]=0 solves the eigenvalue problems.

    Right now I don't see a way to do this, but I'm sure if you look in a linear text you'll find some theorems in the determinant section.
     
  5. Nov 5, 2012 #4

    Dick

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    If you know a is an eigenvalue iff Det[A-aI]=0 and Det[A^T-bI]=Det[A-aI], aren't you kind of done?
     
  6. Nov 5, 2012 #5
    Does this mean that eigenvalue a necessarily equals eigenvalue b?
     
  7. Nov 6, 2012 #6

    haruspex

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    No, you don't need to do that, which is just as well because you can't.
    Each equation has multiple solutions. If you could show that an arbitrary pair, one from each equation, is the same then you would have shown all solutions are the same in each equation.
    It is only necessary to show that the set of solutions is the same for each equation, and that is clear since the set is determined by the roots of the polynomial det(A-λI).
     
  8. Nov 6, 2012 #7
    Thanks haruspex makes sense now.
     
  9. Nov 6, 2012 #8

    Dick

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    I meant to say Det[A^T-aI]=Det[A-aI]. I quoted the Det's from your comment and didn't notice the b.
     
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