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PGRE | Eigenvalue Problem

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  1. Nov 3, 2011 #1
    1. The problem statement, all variables and given/known data
    The problem amounts to finding the eigenvalues of the matrix

    |0 1 0|
    |0 0 1|
    |1 0 0|
    (I have no idea how to set up a matrix in the latex format, if anyone can tell me that'd be great)

    2. Relevant equations
    The characteristic equation for this matrix is

    [itex]\lambda^{3}=1[/itex]

    3. The attempt at a solution
    The solution to this problem can be found on grephysics.net.
    The characteristic equation can be solved by noting that
    1=e[itex]^{2\pi i}[/itex]

    Using this fact, the eigenvalues as noted in the solution are
    [itex]\lambda_{n}=e^{\frac{2\pi i n}{3}}[/itex], (n=1,2,3)

    What I don't understand, is how one goes from

    [itex]\lambda^{3}=e^{2\pi i}[/itex]

    to

    [itex]\lambda_{n}=e^{\frac{2\pi i n}{3}}[/itex]


    If [itex]\lambda^{3}=e^{2\pi i}[/itex] then we can take both sides to the power of [itex]\frac{1}{3}[/itex] to get [itex]\lambda=e^{\frac{2\pi i}{3}}[/itex]. But how can you just throw the n in the exponent and call these (n=1,2,3) the 3 eigenvalues?
     
  2. jcsd
  3. Nov 3, 2011 #2

    vela

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    Check out the brand new LaTeX FAQ!

    https://www.physicsforums.com/showthread.php?t=546968
    It's because [itex]e^{2\pi n i}=1[/itex] for all integer n. When you take the cube root of that, you find you get three distinct solutions, and the rest are repeats.
     
  4. Nov 3, 2011 #3
    Ah yes, after using euler's formula it's much clearer now. Thanks Vela
     
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