Question regarding Jordan canonical forms of linear transformation. Suppose that $V$ is a $5$ dimensional complex vector space over $\mathbb{C}$. Moreover, let $T : V \rightarrow V$ be such that $\exists v \in V$ with $\{T^{i}(v)\}_{i=0}^{4}$ a spanning set of $V$, and $T$ has eigenvalues $\{1,2\}$. Determine the possible jordan canonical forms. I think if I can determine the characteristic polynomial of $T$, then I should be able to determine the possible Jordan Canonical Forms. I know that $\exists a_{0} , \dots , a_{5} \in \mathbb{C} : \sum_{i=0}^{5} a_{i} T^{i}(v) = 0$, but I don't know how to deduce anything about the characteristic polynomial from this.
 A: Hint: There exists a $v$ such that $\{v,Tv,\dots,T^4v\}$ are linearly independent.
It follows that the matrices $I,T,\dots,T^4$ are linearly independent.  Consider the minimal polynomial of $T$.
A: The matrix could be 
$\begin{bmatrix}1 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 2\end{bmatrix}$
if the eigen-space corresponding to eigenvalue 1 has dimension 3 and the eigen-space corresponding to eigenvalue 2 has dimension 2.
The matrix could be 
$\begin{bmatrix}1 & 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 2\end{bmatrix}$
if the eigen-space corresponding to eigenvalue 1 has dimension 2 and the eigen-space corresponding to eigenvalue 2 has dimension 2.
The matrix could be 
$\begin{bmatrix}1 & 1 & 0 & 0 & 0\\ 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 2\end{bmatrix}$
if the eigen-space corresponding to eigenvalue 1 has dimension 1 and the eigen-space corresponding to eigenvalue 2 has dimension 2.
The matrix could be 
$\begin{bmatrix}1 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 0 & 2\end{bmatrix}$
if the eigen-space corresponding to eigenvalue 1 has dimension 3 and the eigen-space corresponding to eigenvalue 2 has dimension 1.
The matrix could be 
$\begin{bmatrix}1 & 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 0 & 2\end{bmatrix}$
if the eigen-space corresponding to eigenvalue 1 has dimension 2 and the eigen-space corresponding to eigenvalue 2 has dimension 1.
The matrix could be 
$\begin{bmatrix}1 & 1 & 0 & 0 & 0\\ 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 0 & 2\end{bmatrix}$
if the eigen-space corresponding to eigenvalue 1 has dimension 2 and the eigen-space corresponding to eigenvalue 2 has dimension 1.
