= Solution
{c}
Over the <complex numbers>, the <eigenvalues>
$$
1,\qquad \lambda_+=\frac{1+\sqrt5}{2},\qquad
\lambda_-=\frac{1-\sqrt5}{2}
$$
are distinct. Thus <distinct eigenvalues imply diagonalizability>, and \b[$A$ is diagonalizable over $\mathbb C$], in fact already over $\mathbb R$.
An explicit <eigenbasis> is
$$
e_3=\begin{pmatrix}0\\0\\1\end{pmatrix},\qquad
v_+=\begin{pmatrix}1\\\lambda_+\\\lambda_++1\end{pmatrix},\qquad
v_-=\begin{pmatrix}1\\\lambda_-\\\lambda_-+1\end{pmatrix}.
$$
For either quadratic <root> $\lambda$, the identity $\lambda^2=\lambda+1$ verifies $Av_\lambda=\lambda v_\lambda$. Taking these three columns as $P$ gives $P^{-1}AP=\operatorname{diag}(1,\lambda_+,\lambda_-)$. Their distinct <eigenvalues> ensure <linear independence>.
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