Common eigenvalues

Diagonalizable matrices case

Proposition
Two matrices which commute and are also diagonalizable are simultaneously diagonalizable, i.e there exists a common basis of eigenvectors, not necessarily with the same eigenvalues.
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The proof is particularly simple if at least one of the two matrices has distinct eigenvalues. The proof of both cases is in this video.

Proof
(Here)
Suppose that $A$ and $B$ are $n\times n$ matrices, with complex entries say, that commute.
Then we decompose ${\mathbb{C}}^n$ as a direct sum of eigenspaces of $A$, say
${\mathbb{C}}^n=E_{{\lambda }_1}\oplus \cdots \oplus E_{{\lambda }_m}$, where ${\lambda }_1,\dots ,{\lambda }_m$ are the eigenvalues of $A$, and $E_{{\lambda }_i}$ is the eigenspace for ${\lambda }_i$.
(Here $m\le n$, but some eigenspaces could be of dimension bigger than one, so we need not have $m=n$.)

Now one sees that since $B$ commutes with $A$, $B$ preserves each of the $E_{{\lambda }_i}$:
If $A v = \lambda_i v, $ then $A(Bv)=(AB)v=(BA)v=B(Av)=B({\lambda }_{i}v)={\lambda }_{i}Bv.$

Now we consider $B$ restricted to each $E_{{\lambda }_i}$ separately, and decompose
each $E_{{\lambda }_i}$ into a sum of eigenspaces for $B$. Putting all these decompositions together, we get a decomposition of ${\mathbb{C}}^n$ into a direct sum of spaces, each of which is a simultaneous eigenspace for $A$ and $B$.
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General case, including operators

Proposition
If two operators commute, they share a common eigenvector.
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Proof
Check it here for matrices.
See this video for operators.
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