1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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.

  2. jcsd
  3. Aug 24, 2012 #2
    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

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

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

    User Avatar
    Science Advisor
    Homework Helper

    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]

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook