Show eigenvalue does not depend on magnetic quantum number $m$ We have a scalar operator $A$, being invariant under rotations which commutes with the angular momentum, i.e.
$$[A,J_i]=0 \text{   where } i=x,y,z$$
$$[A,J^2]=0 $$
So eigenfunctions of $A$ can be chosen such that they are eigenfunctions of $J^2$ and $J_z$. I know the corresponding eigenvalues to these operators are $\hbar^2j(j+1)$ and $\hbar m$ respectively. 
I shall show, that the eigenvalues of $A$ do not depend on the magnetic quantum number $m$. I feel this should go along the lines of the commutator ones again but I don't get anywhere.
Something like this:
Suppose $\psi$ are eigenfunctions of all three operators and
$$A\psi = \lambda \psi$$
where $\lambda=\lambda(m)$ and lead this to a contradiction. 
Any help appreciated!
 A: First of all the question has to be stated into a more precise form.
The Hilbert space $H$ is an orthogonal direct sum of the eigenspaces $H_j$ of $J^2$:
$$H = \oplus_{j}H_j \tag{1}$$
where, obviously, $$J|_{H_j}= j(j+1)I\:.\tag{2}$$
Since $A$ and $J_k$ commute with $J^2$, every $H_j$ is invariant under the action of these operators:
$$A(H_j) \subset H_j\:,\quad J_k(H_j) \subset H_j\:.\tag{3}$$
It is cleat that (1) and (3) imply that $A$ is known provided it is known in each subspace $H_j$.
I therefore henceforth consider the restrictions $A|_{H_j}$ and $J_z|_{H_j}$  to a generic  $H_j$, considering $H_j$ as the Hilbert space of the theory, though I will use the simpler notation $A$ and $J_j$ in place of $A|_{H_j}$ and $J_z|_{H_j}$.
Since $A$ and $J_z$ commute, it may happen that,  $A=f(J_z)$ for some   non-constant function $f$. In other words, an eigenvector $|j,m\rangle$ of $J_z$ with eigenvalue $m$ is also an eigenvector  of $A$ with eigenvalue $f(m)$ for some non-constant function $f$. 
Let us prove that the function $f$ is actually constant. This is the thesis written into a more precise form.
In other words, $A$ restricted to $H_j$ is of the form $cI$.
As $L_k$ commutes with $A$, $A$ commutes with $J_\pm$ which are linear combinations of $J_x$ and $J_y$ and therefore, from
$$A|j,-j\rangle = f(-j)|j,-j\rangle$$
we have
$$J_+A|j,-j\rangle = f(-j)J_+|j,-j\rangle$$
that is
$$AJ_+|j,-j\rangle =C_j f(-j)|j,-j+1\rangle$$
namely
$$C_j A|j,-j+1\rangle = C_jf(-j)|j,-j+1\rangle\:.$$
For some non vanishing $C_j$, so that
$$A|j,-j+1\rangle = f(-j)|j,-j+1\rangle\:.$$
Repeating the operation (finding constants $C_m\neq 0$) we get
$$A|j,m\rangle = f(-j)|j,m\rangle \quad \mbox{if $m=-j,-j+1,\ldots, j$.}$$
In other words, explicitly writing the restriction to $H_j$, since the vectors $|j,m\rangle$ forms a basis of $H_j$, 
$$A|_{H_j} = \sum_{m}  f(-j) |j,m\rangle \langle j,m| = f(-j)I\:.$$ 
In every subspace $H_j$, $A$ is the constant $aI$ operator and this constant may depend on $j$. In the entire Hilbert space:
$$A = \sum_{j,m}  a_j |j,m\rangle \langle j,m| = \oplus_j a_jI_j\:.$$ 
As  a matter of fact we have established the most elementary version of Wigner-Eckart theorem.
