direct sum of $T$-invariant subspaces

Question

Let $T:V \rightarrow V $ be a linear transformation. For $v \in V$ I construct the $I(v)=\mathrm{span} \{v,Tv,...T^{k-1}v\}$.

If $\mathrm{dim} (I(v)) < \mathrm{dim}V$, I can choose a $u$ which does not belong to $I(v)$ and similarly I take $I(u)$. Keep going like that i want to prove that at the end I will have created a basis for $V$.

IS it true that $I(u)$ won't have any common element with $I(v)$?

Answer 1

It is not true. Consider $$ A=\left(\begin{array}{ccc} 0&0&0\\1&0&1\\0&0&0\end{array}\right). $$ For a standard basis $\{e_1,e_2,e_3\}$, it holds $I(e_1)=\text{span}\{e_1,e_2\}$ and $I(e_3)=\text{span}\{e_2,e_3\}$. Hence $I(e_1)\cap I(e_3)\neq\varnothing.$