 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 $T$ areLet $d=\dim \ker N$, which is the dimension of the only eigenspace. If $d=3$, then $N=0$ and $T=I$.

If $d=2$, rank-nullity theorem gives $\dim \mathrm{im}N=1$. The restriction of the nilpotent map $N$ to this invariant line is zero, so $N^2=0$. Choose $v_3$ with $v_2=N{v}_3\ne 0$, 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 $\mathrm{im}N$ has dimension two, so $N^2$ cannot vanish: otherwise $\mathrm{im}N\subseteq \ker N$. Choose $v_3$ with $N^2{v}_3\ne 0$ and set $v_2=N{v}_3$, $v_1=N^2{v}_3$. 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 exactly one of the three displayed forms.