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This topic has been discussed in various other posts but I haven't been able to find specifically what I'm looking for. I've seen the proof in the case of an infinite number of commuting operators, that of course applies also to two of them. But it seems to me, given how my professor stated the fact, that there should be an easy proof regarding just two operators. All the ones I've been able to find on math.stackexchange, like

Commuting Operators Have the Same Eigenvectors, but not Eigenvalues.

or

common eigenvectors of commuting operators

Suppose that the eigenvector is associated to a non-degenerate eigenvalue, that is that two eigenvectors corresponding to this same eigenvalue cannot be linearly independent. Stated in another way, that the dimension of the corresponding eigenspace is 1.

So, given that if $|a〉$ is an eigenvector of $A$ with eigenvalue $a$ we have

$$A|a〉 = a|a〉 ⇒ A(B|a〉) = BA|a〉 = B a|a〉 = a(B|a〉) ⇒ B|a〉 ∈ Sa$$

that is, knowing that also $B|a〉$ is an eigenvector of $A$ associated to the same eigenvalue, is it possible to quickly prove that $|a〉$ is a common eigenvector without adding that it is non-degenerate? Or one needs the complete and more complicated induction proof usually used to prove it for a numerable infinity of operators? Hope it is clear enough.

Ricky
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GN00Fu
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    It is not true that if $A$ and $B$ are commuting operators, then every eigenvector for $A$ is also an eigenvector for $B$. For example, if $A=I$, the identity operator, then every vector is an eigenvector and every operator $B$ commutes with $A$, yet not every vector must be an eigenvector of $B$. – MaoWao Sep 14 '23 at 09:22
  • Ok that's very important. However, is it true that if two operators commute than they have at least one eigenvector in common? Without the need to ask for non degenerate associated eigenvalues? – GN00Fu Sep 14 '23 at 10:03

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Yes, it is still true that commuting operators on a finite-dimensional complex vector space (or over any other algebraically closed field) have an eigenvector in common. Instead of arguing in terms of eigenvectors we argue in terms of eigenspaces: if $A$ and $B$ commute then $B$ preserves each of the eigenspaces $\text{ker}(A - \lambda I)$ of $A$, and then $B$ must itself have an eigenvector acting on each of these eigenspaces, so this eigenvector is a simultaneous eigenvector for both $A$ and $B$.

Qiaochu Yuan
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  • The fact is that if B preserves the eigenspace, this implies that Bx where x is an eigenvector of the eigenspace, only is an eigenvector itself is the dimension of the eigenspace is 1. Otherwise, this is not generally true. As stated in a comment by MaoWao what I was getting wrong is that commuting operators don't have all eigenvectors in common, and that is actually obvious but to my defense I was mislead in believing so by the hastiness of the notes I'm using. Thanks for the answer. – GN00Fu Sep 15 '23 at 12:30
  • @GN00Fu: no, it is still true with no assumptions on the eigenspace, and your calculation still shows it. $Bx$ is still an eigenvector of $A$, it just isn't necessarily an eigenvector of $B$. But because $B$ preserves the eigenspace, $B$ still has an eigenvector on it. – Qiaochu Yuan Sep 15 '23 at 17:44