- #1
Trifis
- 167
- 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 going to go through all this mess, but if some hero would do me this favour I have the following questions to ask:
I bet this thread will remain barren, but what the heck :D
[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 going to 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