Let $A,B:E\to E$, where $E$ is a finite dimensional vector space, be two linear operators such that all of the eigenvalues are real numbers.
If $AB=BA$, prove that there exists a basis in which both the matrices of $A$ and $B$ are triangular.
2026-04-07 03:07:27.1775531247
If two operators with all eigenvalues real numbers commute then both are triangularizable!
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