quantum mechanics - Eigenspaces of angular momentum operator and its square (Casimir operator)

The casimir operator $\textbf{L}^2$ commutates with the elements $L_i$ of the angular momentum operator $\textbf{L}$:

$$ [\textbf{L}^2, L_i] = 0. $$

However, the $L_i$ do not commute among themselves:

$$ [L_i, L_j] = i\hbar\epsilon_{ijk}L_k. $$

This makes sense so far, but it leaves me wondering how their eigenspaces relate to each other. I remember some theorem that diagonalizable, commuting matrices share their eigenspaces. If those operators could be expressed as complex matrices (in the finite-dimensional case), they surely are diagonalizable. So it follows that $\textbf{L}^2$ has the same eigenspaces as the three $L_i$, but that would imply that they commute among themselves, which is not the case.

What am I missing? What is the relation between the eigenspaces of these operators?

Answer

OP is essentially pondering if commutativity is a transitive relation, ie. if three normal operators$^{1}$ $A$, $B$, and $C$ satisfies

$$ [A,B]~=~0\quad \wedge\quad [B,C]~=~0 \quad\stackrel{?}{\Rightarrow} \quad [A,C]~=~0 .\tag{T}$$

The answer is No, but OP argues via the existence of a common basis of eigenvectors for two commuting normal operators that eq. (T) should hold.

To most clearly expose the flaw in OP's argument, pick $B$ to be proportional to the identity. Then $B$ commutes with everything. Clearly we can then find two non-commuting normal operators $A$ and $C$, so that eq. (T) is violated! And a basis of eigenvectors for $A$ cannot be a basis of eigenvectors for $C$, and vice versa.

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$^{1}$We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer.