= Solution
{c}
A real number $\lambda$ is an <eigenvalue> of $A$ if there is a nonzero <vector> $v$ with $Av=\lambda v$; such a $v$ is an <eigenvector> for $\lambda$. The nonzero condition is essential.
For distinct <eigenvalues> $\lambda_1,\ldots,\lambda_r$, choose corresponding nonzero <eigenvectors> $v_1,\ldots,v_r$. Prove their <linear independence> by induction on $r$. The claim for one vector is immediate. Given a relation $\sum_{i=1}^r a_iv_i=0$, apply $A-\lambda_rI$ to get
$$
\sum_{i=1}^{r-1}a_i(\lambda_i-\lambda_r)v_i=0.
$$
By the induction hypothesis, every coefficient here is zero. Distinctness gives $a_i=0$ for $i<r$, and the original relation then gives $a_r=0$ since $v_r\ne0$. Thus the <eigenvectors> are <linearly independent>.
With $n$ distinct real <eigenvalues>, the $n$ real <eigenvectors> form a <basis> of $\mathbb R^n$. Put them into the columns of $P$. Then $P$ is invertible and
$$
AP=P\operatorname{diag}(\lambda_1,\ldots,\lambda_n),
\qquad
\boxed{P^{-1}AP=\operatorname{diag}(\lambda_1,\ldots,\lambda_n).}
$$
This proves <diagonalizability> directly.
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