= Solution
For $T_1$, choose the third <standard basis> <vector> $v=e_3$. Its first three orbit <vectors> are
$$
v=e_3,\qquad T_1v=e_2+e_3,\qquad T_1^2v=e_1+2e_2+e_3.
$$
They are <linearly independent>: their <matrix> of columns has <determinant> $-1$. They therefore span $\mathbb R^3$, proving \b[$T_1$ has a <cyclic vector>, namely $e_3$].
For $T_2$, write $T_2=I+N$ where $N^2=0$. The <binomial theorem> gives $T_2^jv=v+jNv$ for every integer $j\geq0$. The entire orbit of any $v$ lies in $\operatorname{span}\{v,Nv\}$, which has <dimension of a vector space> at most two. Therefore \b[$T_2$ has no <cyclic vector>] in the three-dimensional space. Equivalently, its <minimal polynomial> is $(t-1)^2$, while its <characteristic polynomial> is $(t-1)^3$.
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