= Solution
Interpret “distinct <eigenvalues>” as \b[$n$ 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 $z_1,\ldots,z_k$ 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 $i\leq k$, contradicting pairwise distinctness. <Mathematical induction> therefore proves that \b[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>.
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