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I General form of symmetric 3x3 matrix with only 2 eigenvalues

  1. May 10, 2016 #1
    I'm looking for the general form of a symmetric 3×3 matrix (or tensor) ##\textbf{A}## with only two different eigenvalues, i.e. of a matrix with the diagonalized form ##\textbf{D}=\begin{pmatrix}a& 0 & 0\\0 & b & 0\\0 & 0 & b\end{pmatrix} = \text{diag}(a,b,b)##.

    In general, such a matrix can be described by 4 parameters, e.g. the two eigenvalues ##a,b## and the direction of the eigenvector of ##a## defined by the angles ##\theta,\phi## (in spherical coordinates). The other eigenvectors are in the plane perpendicular to this direction (with arbitrary in-plane orientation).

    With these four parameters, ##(a,b,\theta,\phi)##, I can construct arbitrary matrices with the eigenvalues ##(a,b,b)## by multiplying ##\textbf{D}## with an appropriate rotation matrix ##\textbf{R}## (consisting basically of the eigenvectors): ##\textbf{A} = \textbf{R}^T \textbf{D} \textbf{R}##.

    Is there any other (well-known?) form or parametrization (with 4 independent parameters) of such matrices ##\textbf{A}##? Ideally, a parametrization without (spherical) angles, but closely related to the actual matrix entries in ##\textbf{A}##?

    The background of this question is that I want to find such a matrix ##\textbf{A}## by least-squares fitting to measured data that depend on ##\textbf{A}##, and up to now it seems more efficient to vary 6 independent parameters ##(s_1,\ldots,s_6)## defining a general symmetric matrix ##\textbf{S}=\begin{pmatrix}s_1&s_2&s_3\\s_2&s_4&s_5\\s_3&s_5&s_6\end{pmatrix}## than to vary e.g. the 4 parameters ##(a,b,\theta,\phi)## from above. (More efficient means that the fit converges faster - in spite of having more degrees of freedom - and typically does not run into wrong local minima which sometimes happens depending on the initial values of the angles ##\theta,\phi##.) It might help if I could express 2 of the 6 parameters ##(s_1,\ldots,s_6)## by the other 4 parameters and use these remaining 4 parameters for fitting, so I could perhaps rephrase my question: Are there two "simple" dependencies ##s_1=s_1(s_3,s_4,s_5,s_6)## and ##s_2=s_2(s_3,s_4,s_5,s_6)## (with ##s_1,\ldots,s_6## as in the form of the general symmetric matrix given above) to describe the matrix ##\textbf{A}##?

    (If this is a known problem, I'd be also grateful for pointing me to any text books or articles dealing with it - I wasn't able to find any.)
     
    Last edited: May 10, 2016
  2. jcsd
  3. May 10, 2016 #2

    Orodruin

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    You might want to rewrite your entire matrix as
    $$
    D = b I + (a-b) \begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}.
    $$
    You can then use ##b## and the components of a single column vector as the parameter space. You will also have to separately treat the case of ##a-b## being negative, which can also be done with the same type of parametrisation.
     
  4. May 10, 2016 #3
    Thanks for your suggestion! I think that the resulting parameters are very similar to another set of 4 numbers (that I had considered before, but didn't mention in my question above): the eigenvector of ##a## multiplied by ##a## and the eigenvalue ##b##?
    I tried least-squares fitting with these (latter) parameters as well, but this didn't work better than using ##(a,b,\theta,\phi)##. Somehow, fitting based on the (first) eigenvector (either in the form ##(\theta,\phi)## or in the form of a scaled 3-component eigenvector) is considerably worse than fitting with the (non-diagonalized) symmetric matrix ##\mathbf{S}## from above.
     
  5. May 11, 2016 #4
    Considering Orodruin's suggestion in more detail, I found that I can write the symmetric matrix ##\textbf{S}=\begin{pmatrix}s_1&s_2&s_3\\s_2&s_4&s_5\\s_3&s_5&s_6\end{pmatrix}## as ##v\textbf{1} + (u-v) (\textbf{v}_1\otimes\textbf{v}_1) = v\textbf{1} + (u-v) \begin{pmatrix}rr&rs&rt\\rs&ss&st\\rt&st&tt\end{pmatrix}## where the (first) eigenvector be ##\textbf{v}_1=(r,s,t)^T## with ##r^2+s^2+t^2=1##, so:
    $$\textbf{S} = v\textbf{1} + (u-v) \begin{pmatrix}r^2&rs&r\sqrt{1-r^2-s^2}\\rs&s^2&s\sqrt{1-r^2-s^2}\\r\sqrt{1-r^2-s^2}&s\sqrt{1-r^2-s^2}&1-r^2-s^2\end{pmatrix}$$
    or (all together):
    $$\begin{pmatrix}s_1&s_2&s_3\\s_2&s_4&s_5\\s_3&s_5&s_6\end{pmatrix}
    = \begin{pmatrix}
    v+(u-v)r^2 & (u-v)rs & (u-v)r\sqrt{1-r^2-s^2}\\
    (u-v)rs & v+(u-v)s^2 & (u-v)s\sqrt{1-r^2-s^2}\\
    (u-v)r\sqrt{1-r^2-s^2} & (u-v)s\sqrt{1-r^2-s^2} & v+(u-v)(1-r^2-s^2)\end{pmatrix}.$$
    Thus, I have expressed the symmetric matrix ##\textbf{S}## by a matrix parametrized by ##(u,v,r,s)##.
    Now, I would like to (partially) invert this and find some dependencies between the six parameters ##(s_1, \ldots, s_6)## based on this result. In theory, it must be possible to express e.g. ##s_3## and ##s_5## by the other matrix entries ##s_1, s_2, s_4, s_6##. I wonder if these dependencies can be found be staring long enough at these two matrices ...

    UPDATE (just to clarify): The last paragraph basically means that I would like to express e.g. ##s_3=(u-v)r\sqrt{1-r^2-s^2}## by an appropriate combination of the terms ##s_1, s_2, s_4, s_6##, i.e. by combining ##v+(u-v)r^2##, ##(u-v)rs##, ...
     
    Last edited: May 11, 2016
  6. May 12, 2016 #5
    Here's a solution (took some staring:smile:):
    $$s_1 = s_6 + s_3 (\frac{s_2}{s_5} - \frac{s_5}{s_2})$$
    and
    $$s_4 = s_6 + s_5 (\frac{s_2}{s_3} - \frac{s_3}{s_2}).$$
    With this solution, I can express ##s_1## and ##s_4## by the matrix entries ##s_2, s_3, s_5##, and ##s_6##.
    So, if the symmetric matrix ##\textbf{S}## has only two different eigenvalues ##(a,b,b)## and, thus, can be constructed as ##\textbf{S}=\textbf{R}^T \textrm{diag}(a,b,b)\textbf{R}##, then the elements of ##\textbf{S}## (indexed as defined in posting #1) satisfy the two given equations (of course, the off-diagonal elements of ##\textbf{S}## should not be zero ...)
     
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