= Solution
{c}
Over the <complex numbers>, the <characteristic polynomial> splits into three distinct <roots of a polynomial>, $1,i,-i$. Direct multiplication gives the <eigenvectors>
$$
v_1=\begin{pmatrix}0\\0\\1\end{pmatrix},\qquad
v_i=\begin{pmatrix}1\\-i\\0\end{pmatrix},\qquad
v_{-i}=\begin{pmatrix}1\\i\\0\end{pmatrix},
\qquad Av_1=v_1,\quad Av_i=iv_i,\quad Av_{-i}=-iv_{-i}.
$$
They are <linearly independent>: the last two have two-dimensional <determinant> $2i\ne0$ in the first two coordinates, and the first has nonzero third coordinate. Thus
$$
\boxed{\mathcal B=(v_1,v_i,v_{-i}),\qquad[T']_{\mathcal B}=\operatorname{diag}(1,i,-i).}
$$
This is a <basis> of <eigenvectors> of $\mathbb C^3$, proving that $T'$ has a <diagonalizable matrix>.
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