Let $V=\mathbb{R}^{3}$ and $T(x,y,z)=(-3x-4y,2x+3y.-z)$. Let $\alpha=(e_{1},e_{2},e_{3})$ the usual basis for $V$. We will denote the characteristic polynomial of $T$ by $c_{T}(t)$ and the minimal polynomial by $m_{T}(t)$. The matrix of $T$ with respect to the basis $\alpha$ is
$$ T_{\alpha\leftarrow\alpha}=\begin{bmatrix}-3 & -4 & 0\\2 & 3 & 0\\ 0 & 0 & -1\end{bmatrix} $$ We have $c_{T}(t)=(t+1)^{2}(t-1)$ and $m_{T}(t)=(t+1)(t-1)$.
We will denote $V_{t+1}=\textrm{Ker}(A+I)$ and $V_{t-1}=\textrm{Ker}(A-I)$. We can find $V_{t-1}=\textrm{Span}\{(1,-1,0)\}$ and $V_{t+1}=\textrm{Span}\{(-2,1,0),(0,0,1)\}$. I need to find a cyclic basis for $V$.
We will denote the space T-cyclic generated by $v$ by $C_{T}(v)$.
I know that $V_{t-1}=C_{T}(1,-1,0)$, but I'm not getting a cyclic basis for $V_{t+1}$, because for all $v\in V_{t+1}$ that I take, $T(v)$ is linearly dependent with $v$. What can I do?
Not every subspace of $V$ will be $T$-cyclic. You are guaranteed that (over an algebraically complete field such as $\mathbb{C}$) you can decompose $V$ into a direct sum of subspaces such that the restriction of $T$ to each will have only one eigenvalue.
But that does not mean that each eigenspace will have a $T$-cyclic basis. In particular, if the restriction of $T$ to $W$ is diagonalizable and $\dim(W) \geq 2$, then $W$ will not have a $T$-cyclic basis. This applies to your space $V_{t+1}$, since it has a basis of eigenvectors for $T$.