D. 18 Eigen­func­tions of com­mut­ing op­er­a­tors Any two op­er­a­tors and that com­mute, , have a com­mon set of eigen­func­tions, pro­vided only that each has a com­plete set of eigen­func­tions. Thus the restrictions of the operators repre- COMMON EIGENVECTORS 15 sented by A and B to the subspace Xare commutative and hence they have a common eigenvector, which is, of course, a common eigenvector of A and B. Let be T,S two self adjoint linear Operator on a Hilbert Space $mathcalH$ with pure point spectrum. This does not imply that that every A and B have the same exact set of eigenstates Now note that A and B commute on .N ([ A, B ] x = 0 for every x E X). I am struggling to find a precise definition of this line from my quantum mechanics textbook: If $[A,B] = 0$, then the operators commute, and "commuting operators share common eigenstates". Uni­tary, anti-Her­mit­ian, etcetera, op­er­a­tors all qual­ify.) 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. We prove if x is a common eigenvector of two matrices A and B, then it is an eigenvector of A+B and AB as well. Thus the restrictions of the operators repre- COMMON EIGENVECTORS 15 sented by A and B to the subspace .Ware commutative and hence they have a common eigenvector, which is, of course, a common eigenvector of A and B. If they have a common Hilbert basis of eigenvectors (where now the $\lambda_n$ and $\mu_n$ can be complex) everything works. Regarding (2) it seems to me that only the fact that A and B are closed operators matters. The Ohio State University Linear Algebra exam. COMMON EIGENVECTORS 13 The other direction is obvious. In order to finish the proof the invariance of Xshould be proved. (In other words, the op­er­a­tors do not nec­es­sar­ily have to be Her­mit­ian. $\endgroup$ – Valter Moretti Dec 9 '13 at 22:21 If two matrices commute: AB=BA, then prove that they share at least one common eigenvector: there exists a vector which is both an eigenvector of A and B. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share …