Why do the eigenvalues of an operator sit exactly on the diagonal of any upper-triangular matrix it has?
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Prove the product kills , then , then each in turn, using and commutativity of the factors.
Each diagonal entry is an eigenvalue
Show that maps into , so it is not injective there and yields an eigenvector.
No eigenvalues off the diagonal
Show is a multiple of the minimal polynomial, then read off the eigenvalues of the example.