Interpret “distinct eigenvalues” as pairwise distinct eigenvalues in the chosen field. Select a nonzero eigenvector for each. A single nonzero vector is linearly independent. Suppose inductively that are linearly independent. If adding made the set linearly dependent, its nontrivial relation would have a nonzero coefficient of , since otherwise it would contradict the induction hypothesis. Solving for would place it in the linear span of the first eigenvectors.
Part (c) would then force for some , contradicting pairwise distinctness. Mathematical induction therefore proves that all selected eigenvectors are linearly independent, forming a basis. If the field is , the assertion requires real eigenvalues; a real matrix can instead have nonreal eigenvalues and a eigenvector basis over the complex numbers.
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