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Diagonalization of 8x8 matrix with Euler angles

  1. May 18, 2014 #1
    I am trying to diagonalize the following matrix:
    [itex] M =
    \left( \begin{array}{cccc}
    0 & 0 & 0 & a \\
    0 & 0 & -a & 0 \\
    0 & -a & 0 & -A \\
    a & 0 & -A & 0
    \end{array} \right) [/itex]
    a and A are themselves 2x2 symmetric matrices: [itex] a = \left( \begin{array}{cc} a_{11} & a_{12}\\ a_{12} & a_{22} \end{array} \right) [/itex] and [itex] A = \left( \begin{array}{cc} A_{11} & A_{12} \\ A_{12} & A_{22} \end{array} \right) [/itex].

    Step 1: I diagonalize M treating a and A as numbers. Let U be an orthogonal matrix, then: [itex] D = U^{-1} M U =
    \left( \begin{array}{cccc}
    -\frac{A}{2}-\frac{1}{2} \sqrt{A^2+4a^2} & 0 & 0 & 0 \\
    0 & -\frac{A}{2}+\frac{1}{2 }\sqrt{A^2+4a^2} & 0 & 0 \\
    0 & 0 & \frac{A}{2}-\frac{1}{2} \sqrt{A^2+4a^2} & 0 \\
    0 & 0 & 0 & \frac{A}{2}+\frac{1}{2} \sqrt{A^2+4a^2}
    \end{array} \right) [/itex]
    Step 2: Under the assumption: [itex] A>>a [/itex] the square root becomes: [itex] \sqrt{A^2+4a^2} ≈ A + 2\frac{a^2}{A}[/itex] and consequently the four eigenvalues: [itex] \pm A \pm \frac{a^2}{A}[/itex] and [itex]\pm \frac{a^2}{A}[/itex]. We can return now to our original a and A matrices and the diagonal matrix D becomes:
    [itex] D =
    \left( \begin{array}{cccc}
    -A-aA^{-1}a^{T} & 0 & 0 & 0 \\
    0 & aA^{-1}a^{T} & 0 & 0 \\
    0 & 0 & -aA^{-1}a^{T} & 0 \\
    0 & 0 & 0 & A+aA^{-1}a^{T}
    \end{array} \right) [/itex]
    Step 3: If we now find an 8x8 matrix: [itex]\left( \begin{array}{cccc}
    R_1 & 0 & 0 & 0 \\
    0 & R_2 & 0 & 0 \\
    0 & 0 & R_3 & 0 \\
    0 & 0 & 0 & R_4
    \end{array} \right) [/itex], where [itex]R_i [/itex] are the orthogonal matrices, which diagonalize the 2x2 eigenvalues of D, then we finally arrive at the matrix:
    [itex]D' = R^{-1} D R = diag(\lambda_i) [/itex] with [itex]\lambda_i [/itex] the final eigenvalues with over 40 terms combining the various components of a and A.

    Now I understand that probably nobody is gonna go through all this mess, but if some hero would do me this favour I have the following questions to ask:
    • I want U to be a 4x4 orthogonal matrix parametrized with the four-dimensional Euler angles. Does somebody know how to find such a matrix? I know 4D rotations can be expressed with two quaternions and each quaternion corresponds to 3 angle parameteres. U would all in all be parametrized by 6 angles!
    • [itex]R_i [/itex] on the other hand can be easily represented by a 2D rotation matrix. The difficult part is to find a closed expression for the one angle parameter as a function of the components of a and A. On condition [itex] A>>a [/itex], what further simplifications would you propose?

    I bet this thread will remain barren, but what the heck :D
     
  2. jcsd
  3. Jul 2, 2014 #2
    I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?
     
  4. Jul 2, 2014 #3
    Well, I needed this math in order to study a special scenario of neutrino oscillations with extra sterile particles. It is impossible to find an exact solution, but using certain approximations, which are plausible given the nature of the problem, it is possible to reduce the diagonalization task first to a 4x4 and then to a 2x2 case.
    The study is still on-going, so I unfortunately I cannot go into any details yet.
     
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