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Finding a matrix, given one eigenvalue

  1. Mar 26, 2013 #1
    1. The problem statement, all variables and given/known data
    Suppose B is a real 2x2 matrix with the following eigenvalue:

    [itex]\frac{√3}{2} + \frac{3i}{2}. [/itex]

    Find B[itex]^3[/itex].


    2. Relevant equations
    One of the hints is to consider diagonalization over C together with the fact
    that [itex](\frac{1}{2} + \frac{√3}{2}i)^3 = -1[/itex].


    3. The attempt at a solution
    I found B by going backwards through the quadratic formula, and found B =[itex]
    \begin{pmatrix} 0 & 3\\-1 & √3\end{pmatrix}
    [/itex]

    How can I do this without the method I used? How can I apply the hint given above?
     
  2. jcsd
  3. Mar 26, 2013 #2

    Mark44

    Staff: Mentor

    What your hint is telling you is that 1/2 + √3i/2 is a cube root of -1. Obviously, one cube root is -1. The third cube root is the complex conjugate of the one you have. Use it as another eigenvalue to get a second eigenvector.
     
  4. Mar 26, 2013 #3

    HallsofIvy

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    I am confused by fact that the given eigenvalue is [itex]\sqrt{3}/2+ 3i/2[/itex], NOT [itex]1/2+ \sqrt{3}i/2[/itex].
     
  5. Mar 26, 2013 #4

    Mark44

    Staff: Mentor

    My guess is that there's a typo, and the the given eigenvalue should be 1/2 + √3i/2, which actually is a cube root of -1.
     
  6. Mar 26, 2013 #5

    Dick

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    Homework Helper

    I would guess it means what it says, so one eigenvalue is sqrt(3) times a cube root of -1. They have a cube root of -1 correct in the hint. Since you know what the other eigenvalue must be it's pretty easy to find B^3 when you consider the diagonalization.
     
  7. Mar 26, 2013 #6
    Ok, thanks for the help everyone. Does this mean then that B^3 has eigenvalues that are equal to -3√3? I just cubed the eigenvalues for B after factoring out a √3.
     
  8. Mar 26, 2013 #7

    Dick

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    B^3 for sure has one eigenvalue that -3√3. But that's not the question. The question is what is the matrix of B^3. There's a little bit more to do.
     
  9. Mar 26, 2013 #8
    The other eigenvalue is the same isn't it? So there's only one distinct eigenvalue? For the matrix B^3, I put the eigenvalues on the diagonal and then 0's in the other two spots.
     
  10. Mar 26, 2013 #9

    Dick

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    Sure. But a matrix can have only one eigenvalue and not be diagonalizable. That's not the case here. You just have to say why. That's all.
     
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