I am currently studying for my linear algebra exam and I got quite confused when trying to find the Jordan normal form for some matrix. Let $A = \begin{pmatrix} 2 & 0 & 0 & 1\\ 0 & 0 & -1 & 0\\ -1 & 0 & 1 & -1\\ 0 & 1 & 1 & 1 \end{pmatrix}.$ The eigenvalue is $\lambda = 1$ with multiplicity 4. Then, $\ker(A-I)^2 = \langle \begin{pmatrix} 1\\ -1\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} 1\\ 0 \\ -1 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ 0 \\ -1 \end{pmatrix} \rangle$ and $\ker(A-i)^3 = \mathbb{R}^4$. Now I want to find $v \in \ker(A-I)^3$ such that $v \notin \ker(A-I)^2$. The solution now suggests me to take $v = \begin{pmatrix} 1\\ 0 \\ 0 \\ 0 \end{pmatrix}$ and I understand why this is a possible choice. However, if I am not mistaken, also $e_2, e_3$ and $e_4$ are a valid choice. How is this possible? Where is my mistake? Or are really all four vectors a valid choice?
EDIT: I think I misformulated the question - my mistake.
I know that $\ker(A-I)^2 \subset \ker(A-I)^3$ and I want to find $v \in \ker(A-I)^3$ such that $\langle v \rangle \oplus \ker(A-I)^2 = \ker(A-I)^3.$ Now if I think about this, $\ker(A-I)^2$ is a subset of $\ker(A-I)^3$ and I want to find a basis of the subset of $\ker(A-I)^3$ which - when added to $\ker(A-I)^2$ - gives $\ker(A-I)^3$. Then, if the basis vector $e_1$ spans this whole subset that means that the subset contains only vectors of the form $c \cdot e_1$. But if I can also take $e_2$, that means the subset contains only vectors of the form $c \cdot e_2$ which is impossible. Where is my mistake?
Sorry again for the incorrectly formulated original question.