Commutator with subspaces belonging to the same eigenvalue I have two operators, which commute with each other: $[A,B]=0$.
The operator $A$ is Hermitian and has a degenerate spectrum.
Does it follow from $[A,B]=0$ that the operator $B$ commutes with every projection on a subspace belonging to the same eigenvalue, i.e.
$$\forall_m [\Pi_m,B]=0$$
with $A \Pi_m = a_m \Pi_m$? That is, $\forall_i A |v_i\rangle = a_m |v_i\rangle \Rightarrow \Pi_m|v_i\rangle\langle v_i|=|v_i\rangle\langle v_i|$?
 A: If I understand you correctly, the answer is no. The reason is that even though the subspace in question might be degenerate for $A$, it might not be for $B$. In other words, if $v_1$ and $v_2$ are both eigenvalues of $A$ with eigenvalue $\lambda$, the eigenvectors of $B$ can be some linear combinations of $v_{1,2}$ and can have different eigenvalues.
As a simple counterexample, consider 
$$A=\begin{pmatrix}1&0\\0&1\end{pmatrix}\,,$$
which clearly is Hermitean, has a degenerate spectrum, and commutes with all operators $B$. Two projectors $\Pi_m$ according to your question could be 
$$
\Pi_1=\begin{pmatrix}1&0\\0&0\end{pmatrix}\,,\qquad \Pi_2=\begin{pmatrix}0&0\\0&1\end{pmatrix}\,.$$
Clearly, not all operators $B$ commute with $\Pi_1$ or $\Pi_2$ -- consider e.g.
$$B=\begin{pmatrix}0&1\\1&0\end{pmatrix}\,.$$
The eigenvectors of $B$ are $(1,1)$ and $(1,-1)$ with eigenvalues $\pm1$.
UPDATE
Apparently I misunderstood the question: The projectors are supposed to project onto the full subspace. In that  case, the answer is yes: If $A$ and $B$ commute, $B$ does not mix the eigenspaces of $A$, and hence it commutes with the projectors. To see this, assume an eigenvector $v$ with eigenvalue $\lambda$, $Av=\lambda v$. If $B v=w$ would not be in the eigenspace corresponding to $\lambda$, we would have 
$$A B v=A w\neq \lambda w=B\lambda v=B A v \,.$$
Hence $B$ maps the eigenspace onto itself, and the projector is a multiple of the unit on the eigenspace, so they commute.
