 Solution

Interpret “distinct eigenvalues” as $n$ pairwise distinct eigenvalues in the chosen field. Select a nonzero eigenvector $z_i$ for each. A single nonzero vector is linearly independent. Suppose inductively that $z_1,\dots ,z_k$ are linearly independent. If adding $z_{k+1}$ made the set linearly dependent, its nontrivial relation would have a nonzero coefficient of $z_{k+1}$, since otherwise it would contradict the induction hypothesis. Solving for $z_{k+1}$ would place it in the linear span of the first $k$ eigenvectors.

Part (c) would then force ${\lambda }_{k+1}={\lambda }_i$ for some $i\le k$, contradicting pairwise distinctness. Mathematical induction therefore proves that all $n$ selected eigenvectors are linearly independent, forming a basis. If the field is $\mathbb{R}$, the assertion requires $n$ real eigenvalues; a real matrix can instead have nonreal eigenvalues and a eigenvector basis over the complex numbers.