should compatible operators have the same eigenvalues??
They have the same eigenvectors (more precise: same eigenspaces), their eigenvalues can be different. A trivial example for any operator A is the same operator with some prefactor: 2A, 3A, ...
No, they don't have to have identical eigenspaces, they have to be diagonalizable in the same basis. If you want to express that in terms of eigensubspaces it gets more complicated, because eigensubspaces can be a proper subspaces of an eigenspace of the other operator only, no identity required.
Ah, right, one operator can have different eigenvalues for the same eigenvalue of the other operator.
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