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.
$\blacksquare$
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 $\lambda_1,\ldots, \lambda_m$ are the eigenvalues of $A$, and $E_{\lambda_i}$ is the eigenspace for $\lambda_i$.
(Here $m \leq n$, 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 $A (B v) = (AB)v = (BA)v = B(Av) = B(\lambda_i v) = \lambda_i Bv.$
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 $B$.
$\blacksquare$
Proposition
If two operators commute, they share a common eigenvector.
$\blacksquare$
Proof
Check it here for matrices.
See this video for operators.
$\blacksquare$
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Author of the notes: Antonio J. Pan-Collantes
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