Why can a generalized eigenvector belong to only one eigenvalue, and how does that make lists for distinct eigenvalues independent?
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Each generalized eigenvector has exactly one eigenvalue
State the result and write out its proof: take minimal with , expand binomially, apply , and conclude .
Generalized eigenvectors for distinct eigenvalues are independent
State the theorem and reproduce the minimal-counterexample proof: apply to and show each is a nonzero generalized eigenvector for .
Comparison with the eigenvector independence theorem
Write the earlier theorem for ordinary eigenvectors beside this one, and record which steps of the two minimal-counterexample proofs correspond, including where commuting powers and the uniqueness result enter.