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I am pretty sure that my problem has already discussed, but I didn't find. So, the question is how to prove that two commuting operators have a common eigenvector.

The first note is following: Let $A$ be the operator such that $Av = \lambda v$ (we are considering algebraically closed case, so such $v$ surely exists). Then $ABv = BAv = \lambda Bv$, so $Bv$ is eigenvector of $A$ with eigenvalue $\lambda$ as well. But I am not sure, what to do next.

Ben Grossmann
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Invincible
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    There is an answer to your question in the following post: http://math.stackexchange.com/questions/1227031/do-commuting-matrices-share-the-same-eigenvectors. In particular, see Algebraic Pavel's response. – Mark Yasuda Nov 22 '16 at 15:08
  • Remark: this can be used to prove that if a matrix $A$ commutes with $B$, then $A$ and $B$ are simultaneously triangularizable by orthogonal similarity transformation. This proof can be done following the proof of Schurs decomposition and applying induction. In particular, considering that normal upper triangular matrices are in fact diagonal, a pair of normal matrices $A,B$ commutes if and only if are simultaneously diagonalizable. – R. W. Prado Aug 31 '21 at 08:24

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Let $\lambda$ an eigenvalue of $A$, and $E=\ker(A-\lambda \operatorname{Id})$.

Then $B(E) \subset E$ as you have shown.

So $B|_{E}$ has en eigenvector $x$ and an eigenvalue $\mu$ such that $B(x)=\mu x$.

$x \in E$ so $A(x)=\lambda x$ too.

Ben Grossmann
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marco2013
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  • If image of $AB - BA$ has dimension equals 1, can I consider $AB - BA$ restricted to its kernel and use same arguments? – Invincible Nov 22 '16 at 14:53
  • Let $F= \ker(AB-BA)$ , I don't know if $A(F) \subset F$ and $B(F) \subset F$. – marco2013 Nov 22 '16 at 17:33
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    @Vladislav isn’t $AB-BA=0$ by hypothesis? – amd Nov 22 '16 at 18:27
  • But what if $B|_E$ has no eigenvector aka $\forall$ eigenvektors $x$ of $A: x \notin E$? – Roger Jul 19 '21 at 10:56
  • @Roger We need that the eigenspace $E$ is one dimensional, as explained in the other linked answer, in order to conclude that $B|_E$ has an eigenvector $x$ with an eigenvalue $\mu$ in $E$. – Apoorv Potnis Nov 24 '23 at 16:01