Since the action of a unitary matrix $U$ on the Bloch sphere is defined by
$$\sigma_j\mapsto U\,\sigma_j\,U^\dagger = U\,\sigma_j\,U^{-1}\tag{1}$$
with $\sigma_j$ standing for the Pauli matrices, any scalar phase factor $e^{i\,\phi}$ is quotiented away by the map (which is the big A Adjoint representation of the unitary group). Unitary matrices always act on the Bloch sphere by rotation, never reflexion. This is because $U(2)$, as opposed to $O(3)$, is a topologically connected group and a reflexion matrix, say $\left(\begin{array}{cc}-1&0\\0&1\end{array}\right)$, is simply the $SU(2)$ member $i\,\sigma_z=\left(\begin{array}{cc}i&0\\0&-i\end{array}\right)$ multipied by the phase factor $i$. So the action of the two are the same rotation.
Given 1, the Bloch vector $S =\sum\limits_j s_j\,\sigma_j$ becomes $S^\prime = \sum\limits_j s_j^\prime\,\sigma_j$ where:
$$s_j^\prime = \frac{1}{2}\,\mathrm{tr}(\sigma_j\,\sum\limits_k\,(s_k\,U\,\sigma_k\,U^\dagger))\tag{2}$$
So the matrix of your Bloch sphere rotator is the $3\times 3$ matrix $\left[\frac{1}{2}\,\mathrm{tr}(\sigma_j\,U\,\sigma_k\,U^\dagger))\right]$.
You can find the axis of rotation by finding the eigenvector of the rotation matrix that corresponds to the eigenvalue 1, which is usually the only real eigenvector (unless the rotation is trivial).
Alternatively you should first divide out any phase factor to set your matrix's determinant to 1, so that you're dealing with a member of $SU(2)$. This is not essential for the method above (because the phase is cancelled by the Adjoint representation) but it is here. Now calculate the components of the Pauli matrices in your operator:
$$u_j = -\frac{i}{2}\,\mathrm{tr}(\sigma_j\,U)$$
The vector with the $u_j$ as components is the axis of rotation, where the $j^{th}$ co-ordinate axis is the axis of rotation of the Pauli matrix $\sigma_j$. The magnitude $\sqrt{\sum_j\,u_j^2}$ is the sine of half the rotation angle.
Question from OP
Anyway, why is it possible to find a rotation axis for any unitary matrix?It is possible to decompose a unitary matrix in the product of the three Operators x,y,z , so there shoul be 3 axes? –
Answer: (1) The image of the group $U(2)$ under the Adjoint representation is the group $SO(3)$; then (2) Euler's rotation theorem shows that all members of the latter group are rotations.
I tried to find the rotation axis of the matrix .... This gives me the $e_x$
as rotation axis which is not correct.
This is what I get for your matrix. Applying my second method, the matrix is, modulo phase, equivalent to the unimodular (member of $SU(2)$) matrix $\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1&i\\i&1\end{array}\right)$ (multiply everything by $-i$). This, by inspection, is the matrix $\exp\left(i\,\frac{\pi}{4}\,\sigma_x\right) = \cos\left(\frac{\pi}{4}\right)\,\mathrm{id}+ i\,\sin\left(\frac{\pi}{4}\right)\,\sigma_x$. So, depending on how you define your rotation axes, it's a rotation of $2\times \frac{\pi}{4}$, i.e. a right angled turn about the axis corresponding to $\sigma_x$. So, unless I have done something dreadfully wrong, I agree with your calculations and there must be a typo somewhere if the answer is meant to be a rotation about $(1,\,1,\,1)$.
For a matrix equivalent (modulo a phase) to $\frac{1}{\sqrt{2}} \begin{pmatrix} 1&-i\\1 &i\end{pmatrix}$, we multiply by $\exp(-\pi\,i/4)$ to get an $SU(2)$ member. Then my second method entails the following calculations (I checked the matrix was unitary first):
So indeed the axis is along $(1,\,1,\,1)$. Alternatively, we can simply calculate the matrix of the the transformation and work out which eigenvector corresponds to the eigenvalue 1; the Frobenius (trace norm) of the matrix logarithm gives us the angle of rotation. So its a rotation of through an angle of $2\pi/3$ about the axis $(1,1,1)$. These are the relevant calculations:
Best Answer
Two somewhat trivial observations:
Topologically, the non-entangled states must be two spheres, ie $S^2 \times S^2$. This is because the product state form of your middle equation is specified by two separate and completely independent pure states, which each topologically comprise an ordinary Bloch sphere.
These two Bloch spheres $S^2 \times S^2$ comprise a four-dimensional manifold; meanwhile, the total Bloch space is six (seven?) dimensional, as you observe. Thus, the non-entangled states cannot possibly cut the total Bloch space into two regions, in the same way a one-dimensional manifold cannot cut a three-dimensional manifold into two regions.