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Eigenvalues of the power of a matrix

  1. Aug 24, 2012 #1
    1. The problem statement, all variables and given/known data

    If [itex]\lambda_i[/itex] are the eigenvalues of a matrix [itex]A^2[/itex], and [itex]A[/itex] is symmetric, then what can you say about the eigenvalues of A?

    2. Relevant equations



    3. The attempt at a solution

    I know how to prove that if [itex]\sqrt(\lambda_i)[/itex] is an eigenvalue of A, then [itex]\lambda_i [/itex] is an eigenvalue of [itex]A^2[/itex].

    I don't know how to prove the converse though.

    Thanks!
     
  2. jcsd
  3. Aug 24, 2012 #2

    micromass

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    Let [itex]\lambda[/itex] be an eigenvalue of [itex]A^2[/itex]. What can you say about [itex]A^2-\lambda I[/itex]?

    Now, use that

    [tex]A^2-\lambda I=(A-\sqrt{\lambda}I)(A+\sqrt{\lambda} I)[/tex]
     
  4. Aug 24, 2012 #3
    Then:

    [tex]det(A-\sqrt{\lambda}I)*det(A+\sqrt{\lambda} I) = 0[/tex]

    Not sure how to proceed...
     
  5. Aug 24, 2012 #4

    micromass

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    If xy=0, then what can you say about x or y?
     
  6. Aug 24, 2012 #5
    Oh so all we can say is that either [itex]\sqrt{\lambda}[/itex] or [itex]-\sqrt{\lambda}[/itex] is an eigenvalue of A, but not both?
     
  7. Aug 24, 2012 #6

    micromass

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    Right.
     
  8. Aug 24, 2012 #7
    Thank you.
     
  9. Aug 24, 2012 #8
    One more question, how do we know that all the eigenvalues of [itex]A^2[/itex] are positive? Since A is symmetric, if [itex]\lambda < 0[/itex] were an eigenevalue of [itex]A^2[/itex] we'd run into a problem...
     
  10. Aug 24, 2012 #9

    micromass

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    We can do the same with [itex]\sqrt{\lambda}[/itex] complex numbers, no?
     
  11. Aug 24, 2012 #10
    But since A is symmetric, it must have real eigenvalues, no?
     
  12. Aug 24, 2012 #11

    micromass

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    Yes, sure. So what we do is we argue the same argument where [itex]\lambda[/itex] might be negative and where [itex]\sqrt{\lambda}[/itex] might be complex. Then we can deduce that either [itex]\sqrt{\lambda}[/itex] or [itex]-\sqrt{\lambda}[/itex] are eigenvalues of A. Since A is symmetric, we have that [itex]\sqrt{\lambda}[/itex] must be real, so [itex]\lambda[/itex] was positive to begin with.
     
  13. Aug 24, 2012 #12
    Ah ok, thanks again!
     
  14. Aug 24, 2012 #13

    Ray Vickson

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    No, you can have both. The matrix
    [tex] A = \pmatrix{2 & 0 \\0 & -2}[/tex] has
    [tex] B = A^2 = \pmatrix{4 & 0 \\0 & 4}, [/tex]
    and so both [itex] \sqrt{4}[/itex] and [itex] -\sqrt{4}[/itex] are eigenvalues of [itex]A.[/itex]

    RGV
     
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