= Solution
Write $N=T-I$. <Cayley-Hamilton theorem> gives $N^3=0$. Its <minimal polynomial> is therefore $X^k$ for exactly one $k\in\{1,2,3\}$, so the possibilities for that of $T$ are
$$
\boxed{X-1,\quad(X-1)^2,\quad(X-1)^3.}
$$
Let $d=\dim\ker N$, which is the <dimension> of the only <eigenspace>. If $d=3$, then $N=0$ and $T=I$.
If $d=2$, <rank-nullity theorem> gives $\dim\operatorname{im}N=1$. The restriction of the nilpotent map $N$ to this invariant line is zero, so $N^2=0$. Choose $v_3$ with $v_2=Nv_3\ne0$, and choose $v_1\in\ker N$ independent of $v_2$. Then $v_3\notin\ker N$, so these three vectors form a <basis>; its <matrix> for $T$ is the printed $J_1(1)\oplus J_2(1)$.
If $d=1$, then $\operatorname{im}N$ has <dimension> two, so $N^2$ cannot vanish: otherwise $\operatorname{im}N\subseteq\ker N$. Choose $v_3$ with $N^2v_3\ne0$ and set $v_2=Nv_3$, $v_1=N^2v_3$. Applying $N^2$ and then $N$ to a putative linear relation proves these three vectors independent. The <matrix> is the printed $J_3(1)$.
These cases exhaust the possible <eigenspace> dimensions. They are pairwise dissimilar because similarity preserves the <eigenspace> <dimension>, equivalently the three different minimal <polynomials>. Hence every <matrix> is similar to \b[exactly one] of the three displayed forms.
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