Determine rotational ellipsoid from main orientation and Eigenvalues for my application I end up at each pixel with a 3D vector $\vec{v}$ and an error for every vector component $\sigma_{v_i}$ representing its uncertainty. I want to visualize this with help of rotational ellipsoids so that the main axis of the ellipsoid is given by the vector and its semiaxes by the orientation errors. Visualization programs of the Diffusion MRI community require the ellipsoid to be given in Matrix form.
First I had looked at Determine a matrix knowing its eigenvalues and eigenvectors
So I thought I have the Eigenvalues $\lambda_i=\sigma_{v_i}$ and Eigenvectors  $\vec{e}_i$ (essentially unit vectors) in the laboratory system. Now it seems I just have to rotate the unit vectors into the ellipsoid frame and sum the product of Eigenvalues and Eigenvectors to get the ellipsoid matrix $\Lambda_{rot}$: 
$$
\Lambda_{rot}=\sum\lambda_i\cdot\vec{e}_{i,rot}\vec{e}_{i,rot}^T/|\vec{e}_{i,rot}|^2
$$
I used the rotations:
$$
\begin{align}
R & =R^z(\varphi)R^y(\alpha)R^z(-\varphi)\\
& = \begin{pmatrix}
\cos(\varphi) & \sin(\varphi) & 0 \\ -\sin(\varphi) & \cos(\varphi) & 0 \\ 0 & 0 & 1
\end{pmatrix}
\cdot\begin{pmatrix}
\cos(\alpha) & 0 & \sin(\alpha) \\ 0 & 1 & 0 \\ -\sin(\alpha) & 0 & \cos(\alpha)
\end{pmatrix}
\cdot\begin{pmatrix}
\cos(\varphi) & -\sin(\varphi) & 0 \\ \sin(\varphi) & \cos(\varphi) & 0 \\ 0 & 0 & 1
\end{pmatrix}
\end{align}
$$
But when I computed the first Eigenvector of the $\Lambda_{rot}$ it didnt't agree with the main orientation anymore. This left me confused. Any idea what went wrong? Especially with the coordinate transformation maybe?
I also read Change of Eigenvalues of Ellipsoid Tensor during Rotation but I think it still has to work with a coordinate transformation.
Thanks in advance!
 A: You wrote: "...the main axis is given by the vector and its semiaxes by the orientation errors".
Suppose for a moment that the vector is $(3, 0, 0)$, and the orientation errors are $2$ and $1$. What are the semiaxes going to be? $(0,2,0)$ and $(0, 0, 1)$? Maybe $(0,0,2)$ and $(0, 1, 0)$? Maybe 
$$
2(0, s, c)\\
1(0, -c, s)
$$
where $s$ and $c$ are the sine and cosine of some arbitrary angle? 
What I'm getting at is that the problem here is that the thing you're trying to do is under-specified. Now you might say "just choose the vector for the larger error to always be orthogonal to, say, $(0, 1, 0)$". That's not a bad idea if your vector $v$ is never $(0, 1, 0)$, but if it IS....then the solution is underspecified. 
So you really need, for any nonzero vector $v$, a perpendicular vector $w$ that varies continuously as a function of $v$, so that when your rotational ellipsoid axis is $v$, you can put the larger (or the first, or whatever) error-value in the $w$ direction. 
Such a choice, restricted to the case of $v$ lying in the unit sphere, gives you a function 
$$
w : S^2 \to \S^2 : v \mapsto w(v)
$$
with the property that for every $v$, the dot product $v \cdot w(v) = 0$. You can visualize $w(v)$, which is orthogonal to $v$, as being placed as a vector with its base at the tip of $v$, in which case it looks like a tangent-vector to $S^2$. And then the function $w$ gives you an everywhere nonzero vector field on the unit sphere in 3-space...which is impossible. In short: there doesn't seem to be any continuous way of doing what you're asking. 
Gratuitous observation: one reason people use ellipsoid visualization of (symmetric) tensors is that a tensor contains a lot of information in a not-very-easily disentangled form. You, on the other hand, have data in a very nice form: a vector, with the error components along each axis already given. I don't see a good reason for transforming these error-values into measurements in some alternative coordinate system in which the direction in which the error is displayed is no longer correlated with the direction in which the error is measured. 
In short, this sounds as if it's not only impossible, because of the vector fields on spheres theorem, but also a not very good way to look at the data. Rather than trying to cast your data into a form that it doesn't really fit (but for which there are already viz tools), I'd suggest that you spend a bit more time thinking about what you'd like to be able to understand from looking at your visualization, and make sure that the process you follow leads to that result. 
