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

• I

## Main Question or Discussion Point

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.)

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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.

• odietrich
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.

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 $v+(u-v)r^2$, $(u-v)rs$, ...

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Here's a solution (took some staring ):
$$s_1 = s_6 + s_3 (\frac{s_2}{s_5} - \frac{s_5}{s_2})$$
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 ...)