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If two operators $A$ and $B$ commute then any eigenvector of $A$ is an eigenvector of $B$? I know that if that happens there is a basis in which the eigenvectors of $A$ and $B$ are equal, but I don't know if that can be generalized to any basis.

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  • $\begingroup$ Minor language nitpick: it's probably more correct to say "commuting operators", and reserve the adjective "commutative" to refer to binary operations. $\endgroup$ – David H Sep 7 '14 at 19:16
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Not necessarily. Counterexample: Be $A$ the $2\times 2$ unit matrix, $B$ any $2\times 2$ diagonal matrix with two different entries in the diagonal, and $v=\begin{pmatrix}1\\1\end{pmatrix}$.

Then $v$ is an eigenvector of $A$ (since every vector is an eigenvector of the identity matrix), but $v$ is not an eigenvector of $B$.

However, if $v$ is an eigenvector of $A$ with non-degenerate eigenvalue $\alpha$, then $v$ is also an eigenvalue of $B$.

Proof: We have $$ABv = BAv = B\alpha v = \alpha Bv.$$ Therefore $Bv$ is also an eigenvector of $A$ with the same eigenvalue. But since we assumed that the eigenvalue is non-degenerate, $Bv$ must be a multiple of $v$, that is, $v$ is an eigenvector of $B$.

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Take $J=\begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}$. Then $I,J$ commute.

Any basis forms a basis of eigenvectors for $I$, but at most one element of any basis can be an eigenvector for $J$.

(See https://math.stackexchange.com/a/6261/27978 for relevant comments.)

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