 Solution

For $T_1$, choose the third standard basis vector $v=e_3$. Its first three orbit vectors areThey are linearly independent: their matrix of columns has determinant $-1$. They therefore span ${\mathbb{R}}^3$, proving $T_1$ has a cyclic vector, namely $e_3$.

For $T_2$, write $T_2=I+N$ where $N^2=0$. The binomial theorem gives $T_2^{j}v=v+jNv$ for every integer $j\ge 0$. The entire orbit of any $v$ lies in $\mathrm{span}\{v,Nv\}$, which has dimension of a vector space at most two. Therefore $T_2$ has no cyclic vector in the three-dimensional space. Equivalently, its minimal polynomial is $(t-1{)}^2$, while its characteristic polynomial is $(t-1{)}^3$.