Which matrix transforms my vector field $F(r,\theta,\phi)$ from cylindrical to spherical coordinates I am looking for the matrix that I have to apply my vector at the position $(r,\theta, z)$ to in order to get the appropriate vector in spherical coordinates.
I am totally okay, if you could give me a reference, thanks in advance
 A: Let us suppose, in Cartesian coordinates, $\vec{F} = A\hat{x}+B\hat{y}+C\hat{z}$ where $\hat{x} = \langle 1,0,0 \rangle$ and $\hat{x} = \langle 0,1,0 \rangle$ and $\hat{z} = \langle 0,0,1 \rangle$. These provide the global, constant, Cartesian frame.
Introduce spherical coordinates as follows:
$$ x = r\cos \phi \sin \theta, \ \ \ y = r\sin \phi \sin \theta, \ \ \ z = r\cos \theta $$
to derive the spherical frame a simple technique is to take gradients in the coordinate directions and normalize: $\nabla u / || \nabla u ||$ for $u=r, \theta, \phi$. Note: as $r^2=x^2+y^2+z^2$ we find:
$$ \nabla r = \langle \frac{x}{r},\frac{y}{r},\frac{z}{r} \rangle = \langle \cos \phi \sin \theta, \sin \phi \sin \theta, \cos \theta \rangle $$
which gives $\hat{r} = \langle \cos \phi \sin \theta, \sin \phi \sin \theta, \cos \theta \rangle$ as the gradient of $r$ is already unit-length. Some calculation further, we'll find: (using $\hat{x},\hat{y},\hat{z}$ to emphasize the connection to the Cartesian frame)
\begin{array}{l}
\hat{r} = \sin \phi \cos \theta \hat{x} + \sin \phi \sin \theta \hat{x} +\cos \phi \hat{z} \\
\hat{\phi} =  -\cos \phi \cos \theta\hat{x} - \cos \phi\sin \theta \hat{y} +\sin \phi \hat{z} \\
\hat{\theta} = -\sin \theta \hat{x}+\cos\theta \hat{y}.
\end{array}
Finally, to find components of a vector field given in Cartesian coordinates we need only calculate some dot-products as the frame above gives us an orthonormal basis at most points in $\mathbb{R}^3$ (ignoring once more the points where the angular coordinates fail to be coordinates, like the origin or the $z$-axis). Returning to $\vec{F} = A\hat{x}+B\hat{y}+C\hat{z}$ we calculate:
$$\vec{F} = (\vec{F} \cdot \hat{r}) \cdot \hat{r} + (\vec{F} \cdot \hat{\phi}) \cdot \hat{\phi}+(\vec{F} \cdot \hat{\theta}) \cdot \hat{\theta} $$
which yields:
$$ \vec{F} = (A\sin \phi \cos \theta + B\sin \phi \sin \theta  +C\cos \phi) \hat{r} +
( -A\cos \phi \cos \theta - B\cos \phi\sin \theta  +C\sin \phi  )\hat{\phi} +
( -A\sin \theta +B\cos\theta )\hat{\theta} $$
For example,  if $\vec{F} = \langle x,0,0 \rangle$ then
$$ \vec{F} = (x\sin \phi \cos \theta) \hat{r} +
( -x\cos \phi \cos \theta )\hat{\phi} +
( -x\sin \theta )\hat{\theta} $$
where $x$ should be replaced with $x=r\cos \phi \sin \theta$ to complete the reformulation into spherical coordinates. Let me know if you need further detail. My reference for this material is Griffith's Electrodynamics, but you can find in many advanced calculus texts of a certain bent. For example, Kaplan fifth ed. Chapter 3.
Here is an attempt at a visualizations of the spherical frame:

However, it's certainly easier to see one at a time: for example $\hat{r}$

or $\hat{\theta}$,

