= Solution
\b[True.] All three <eigenvalues> are distinct in $\mathbb C$, so their nonzero <eigenvectors> are <linearly independent>. Here is a direct proof of that assertion. If $a_1v_1+a_2v_2+a_3v_3=0$, apply $(A-\lambda_2I)(A-\lambda_3I)$ to obtain
$$
a_1(\lambda_1-\lambda_2)(\lambda_1-\lambda_3)v_1=0.
$$
Distinctness makes both scalar factors nonzero, hence $a_1=0$. Applying the analogous products gives $a_2=a_3=0$. The three <eigenvectors> therefore form a <basis> of $\mathbb C^3$. Taking them as the columns of $P$ gives
$$
\boxed{P^{-1}AP=\operatorname{diag}\left(-1,\frac{1+i}{\sqrt2},\frac{1-i}{\sqrt2}\right)}.
$$
This proves that every such matrix is a <diagonalizable matrix> over $\mathbb C$, rather than merely showing it for the example.
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