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Diagonalization and Unitary Matrices

  1. Feb 29, 2016 #1
    \1. The problem statement, all variables and given/known data
    Let B = ##
    \left[ \begin{array}{ccc} -1 & i & 1 \\ -i & 0 & 0 \\ 1 & 0 & 0 \end{array} \right]
    ##. Find a Unitary transformation to diagonalize B.

    2. Relevant equations
    N/A

    3. The attempt at a solution
    I have found both the Eigenvalues (0, 2, -1) and the Eigenvectors, which are ##<0,i,1>,\ \ <2,-i,1>,## and ##<-1,-i,1>##. Vectors 1 and 3 are both already normalized, but I normalized vector two to be ##<1,\frac{-i}{2},\frac{1}{2}>##.

    They are now orthonormal, but the matrix formed from them,

    $$U =
    \left[ \begin{array}{ccc} 0 & 1 & -1\\ i & \frac{-i}{2} & -i \\ 1 & \frac{1}{2} & 1 \end{array} \right]$$

    is not a unitary matrix. It will still succesfully diagonalize B, but I don't know what I've done wrong. Thank you!
     
  2. jcsd
  3. Mar 1, 2016 #2

    Mark44

    Staff: Mentor

    The first and third eigenvectors aren't normalized, nor is the last one you show. For example, the magnitude of the first vector is ##\sqrt{2}##. The magnitude of a normalized vector is 1.
     
  4. Mar 1, 2016 #3
    Oh shoot, yeah. My "norm" function in maple was spitting out ones, but I realized that I was using the wrong one.

    It looks like two out of my three vectors are fine and normalized (## V_1 = <-1, -i, 1>, V_2 = <1, \frac{-i}{2}. \frac{1}{2}>##), but the third one, ##V_3 = < 0, -i, 1> ## has a magnitude of zero, so I don't know how to normalize it.... I'm not sure where to go from here.
     
  5. Mar 1, 2016 #4

    Mark44

    Staff: Mentor

    No, they are not normalized. For a complex vector v, ##||v|| = \sqrt{v \cdot \bar{v}}##. ##||v_1|| = \sqrt{3}##, so it's clearly not normalized.
    No. The only vector with a magnitude of zero is the zero vector. If Maple is giving you the wrong values, do them by hand.
     
  6. Mar 1, 2016 #5

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Are you sure 2 and -1 are eigenvalues?
     
  7. Mar 1, 2016 #6

    Mark44

    Staff: Mentor

    That's a very good question.
     
  8. Mar 1, 2016 #7
    Yep, about that I'm certain.

    $$(Q - \lambda I) =
    \left[ \begin{array}{ccc} 1-\lambda & i & 1\\ -i & -\lambda & 0 \\ 1 & 0 & -\lambda \end{array} \right]$$
    Therefore ##det(Q - \lambda I) = 0## is
    $$(1-\lambda)(-\lambda)(-\lambda) + \lambda + \lambda = 0$$
    which is simplified to
    $$ \lambda(\lambda - 2)(\lambda + 1) = 0$$
    Therefore, ##\lambda = 0, 2, -1##.
     
  9. Mar 1, 2016 #8
    Mark44, thank you so much for your help! I'd forgotten that when finding the norm of a complex vector, you need the conjugate. I'm getting the right answer now.

    Thank you!
     
  10. Mar 1, 2016 #9

    Samy_A

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    Science Advisor
    Homework Helper

    In your matrix, the term in column 1, row 1 is -1, not 1 as you assume here.
     
  11. Mar 1, 2016 #10
    Oh wow. That's a typo in my original post, the number in the question is 1.
     
  12. Mar 1, 2016 #11

    Ray Vickson

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    Homework Helper

    Maple gives the correct answer if the command is used correctly.
     
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