Commuting operators have a common set of eigenvectors ↩

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created: 2022-01-06 02:05:34
modified: 2022-01-10 04:13:04

Let

and

.
Statement:

and

have a common set of eigenvectors.

Proof

Let be a nondegenerate eigenvector of :

Since and commute:

Since is the only eigenvector of with eigenvalue , must be proportional to , so:

Which means that is an eigenvector of as well.