Simple example: consider in C^3, the unity operator, 1, and the projector on the first coordinate Px (Px(x,y,z) = (x,0,0) )
Clearly they commute: [1,Px] = 0 (as the unity operator commutes with all).
now, consider the vector (1,2,0). This is an eigenstate of 1 (all vectors are eigenstates of 1). However, it is not an eigenstate of Px. Only vectors of the kind (x,0,0) are eigenstates of Px with eigenvalue 1, and vectors of the kind (0,y,z) are eigenstates of Px with eigenvalue 0. But (1,2,0) is neither.
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