1) Let us for simplicity assume that the Hilbert space is finite-dimensional.
Let $\hat{A}:H\to H$ and $\hat{B}:H\to H$ be two Hermitian operators (a.k.a. quantum observables). Their two spectra must then be finite sets of eigenvalues
$${\rm Spec}(A)~=~\{a_1, \ldots, a_n\} \qquad {\rm and} \qquad
{\rm Spec}(B)~=~\{b_1, \ldots, b_m\}.
$$
2) In the question(v1) OP specifically asks what happens if the spectra are degenerate? We can decompose the Hilbert space
$$H~=~\oplus_{i=1}^n E^{(\hat{A})}_i$$
in orthogonal eigenspaces
$$E^{(\hat{A})}_i~:=~ {\rm ker}(\hat{A}-a_i{\bf 1}) ~\subseteq~H$$
for $\hat{A}$. Let us define corresponding projection operators $\hat{P}^{(\hat{A})}_i:H\to E^{(\hat{A})}_i$.
Then we may decompose
$$\hat{A}~=~\sum_{i=1}^n a_i\hat{P}^{(\hat{A})}_i. $$
3) Let us for simplicity assume that the initial state is a pure state given by a ket $\mid\psi\rangle$. (For the case of a mixed state, see this answer.) A measurement of the observable $\hat{A}$, with the outcome $a_i$, collapses the initial state $\mid\psi\rangle$ into a new state
$$ \mid\psi^{\prime}\rangle~=~\frac{\hat{P}^{(\hat{A})}_i\mid\psi\rangle}{\sqrt{\langle\psi\mid\hat{P}^{(\hat{A})}_i\mid\psi\rangle}}.$$
4) Similarly, we can do the same thing with the other operator,
$$H~=~\oplus_{j=1}^m E^{(\hat{B})}_j,$$
$$E^{(\hat{B})}_j~:=~ {\rm ker}(\hat{B}-b_j{\bf 1}) ~\subseteq~H,$$
$$\hat{B}~=~\sum_{j=1}^m b_j\hat{P}^{(\hat{B})}_j. $$
5) Since the two operators $\hat{A}$ and $\hat{B}$ commute $[\hat{A},\hat{B}]=0$, i.e., are compatible, there exists a common set of orthogonal eigenspaces
$$E_{ij} ~:=~ E^{(\hat{A})}_i \cap E^{(\hat{B})}_j~\subseteq~H, \qquad\qquad H~=~\oplus_{i=1}^n\oplus_{j=1}^m E_{ij}.$$
Note that some of the subspaces $E_{ij}$ could be trivial (zero-dimensional). All the projection operators $\hat{P}^{(\hat{A})}_i$ and $\hat{P}^{(\hat{B})}_j$ will also commute.
6) If we next perform a measurement of the observable $\hat{B}$, with result $b_j$, on the state $\mid\psi^{\prime}\rangle$ from section 3, the state collapses into
$$ \mid\psi^{\prime\prime}\rangle
~=~\frac{\hat{P}^{(\hat{B})}_j\mid\psi^{\prime}\rangle}{\sqrt{\langle\psi^{\prime}\mid\hat{P}^{(\hat{B})}_j\mid\psi^{\prime}\rangle}},$$
which after some straightforward algebra reduces to
$$ \mid\psi^{\prime\prime}\rangle
~=~\frac{\hat{P}^{(\hat{B})}_j\hat{P}^{(\hat{A})}_i\mid\psi\rangle}{\sqrt{\langle\psi\mid\hat{P}^{(\hat{B})}_j\hat{P}^{(\hat{A})}_i\mid\psi\rangle}}. $$
7) The latter expression is symmetric in $\hat{A} \leftrightarrow \hat{B}$,
and therefore performing the measurements in opposite order, i.e., measuring first $\hat{B}$, with outcome $b_j$, and then $\hat{A}$, with result $a_i$ would produce the same final state $\mid\psi^{\prime\prime}\rangle.$
8) So to answer the question, in the degenerate case there may still be a collapse associated with the second measurement. However, if the spectrum of the first observable is non-degenerated, then there is no second collapse.
- Do $A$ and $B$ have the same amount of eigenvalues?
They don't have to, $S_z$ and the identity commute. The former has two eigenvalues. The latter has one. Other examples exist too, such as the operator, $H=p^2/2m+e/r$ which has an infinite number of eigenvalues but can commute with operators with a finite number of eigenvalues such as both operators listed above. However, if you look at the dimension of each eigenspace and add up all those dimensions for each eigenvalue then you get the same number for any hermitian operator (regardless of whether they commute) because since they are hermitian that sum just gives you dimension of the whole space. And that sum is just the dimension of the whole space and it happened regardless of whether they commuted, so it is really besides the point. The answer is that there is no real relationship in general for the number of eigenvalues of operators that commute.
- Do $A$ and $B$ have the same amount of eigenvectors?
They don't have to, $S_z$ and the identity commute. If the operators act on a spin space the former has two eigenvectors (modulo scalars) and the latter has uncountably many (a whole two dimensional eigenspace of eigenvectors with uncountably many directions). Again you could count the dimension of each eigenspace and add them up, but you'll just get the dimension of the whole space again. There will still literally be two directions that are eigen to one operator and uncountably many directions that are eigen to the other operator. And adding the dimensionality of the eigenspaces for each operator will again just give you the dimension of whole space. And again it happenes regardless of whether they commuted, so it is really besides the point. The answer is that there is no real relationship in general for the number of eigenvector directions of operators that commute.
- What do the previous answers mean for the cardinality of the set?
Nothing, the sets could be finite or countably infinite or for eigenvectors there can be uncountably many. Even $AB$ could have uncountably many different directions of eigenvectors. If you have not just two but have a collection of mutually commuting observables where you can't add more observables that also commute with those (a CSCO) then the vectors that are common eigenvectors to all the operators will have fixed directions so now there is a fixed set of directions. And the cardinality of those directions will be the dimension of the whole space but now it is actually related to the number of eigenvector directions that are eigen to all the operators, the eigenvectors that are common to all the operators give you real directions. And you can count them, but counting them will just give you the dimensionality of the whole space.
Best Answer
It is possible for two non commuting operators to have an eigenvector in common.
Consider two operators $A$ and $B$, that in some basis can be written as
$$ A = \begin{pmatrix} 1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 3 \end{pmatrix}$$
$$ B = \begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{pmatrix}$$
$$ AB-BA = \begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & -1\\ 0 & 1 & 0 \end{pmatrix} \neq 0$$
Yet, the vector $$\textbf{v} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$$ is a common eigenvector.
But the inequality is still satisfied, because on the RHS there is the expected value of $[A,B]$ over the state. $\textbf{v}$ is an eigenstate of $[A,B]$ with eigenvalue 0, therefore the expected value is 0 and the inequality holds.
This is a general result. If two operators share a common eigenvector, it will be also an eigenvector of their commutators, with eigenvalue 0.
Proof: If $A\textbf{v} = a\textbf{v}$ and $B\textbf{v} = b\textbf{v}$, then $$(AB-BA)\textbf{v} = ab\textbf{v}-ba\textbf{v} = 0$$