Find $v \in \ker(A-I)^3$ such that $\langle v \rangle \oplus \ker(A-I)^2 = \ker(A-I)^3$ 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.
 A: Every vector not in the hyperplane $H=\ker(A-I)^2$ is a solution. The analogue in two dimensions would be that if the kernel is $H=\text{Vect}(e_1-e_2)$, then every vector not in the line $H$ is a solution, for example both $e_1$ and $e_2$ are. One sees the two basis vectors are solutions.
Thus, the error (if any) is to think that for a given (strict) vector subspace $W$ of the vector space $V$ there is a unique subset $S$ of $V$ such that $\text{Vect}(S)\oplus W=V$. There are plenty. And there are even plenty of subspaces $U$ such that $U\oplus W=V$, as the 2D example I gave shows.
Maybe the origin of the trouble is to confuse the Euclidean space $E=\mathbb R^4$ and the (simple) vector space $V=\mathbb R^4$. The former structure is the latter plus the usual scalar product. The orthogonal supplement of a subspace $W$ of the Euclidean vector space $E=\mathbb R^4$ is indeed unique, in the case at hand it is $\text{Vect}(u)$ with $u=e_1+e_2+e_3+e_4$. But the supplements of subspace $W$ of the vector space $V=\mathbb R^4$ are many.
A: I see no mistake. Denote your three basis vectors of $\ker(A-I)^2$ as $v_1,v_2,v_3$, with $v_1=(1,-1,0,0)^T$. You can always extend $\{v_1,v_2,v_3\}$ to a basis of $\mathbb{R}^4$ by adding a fourth vector $u\notin{\rm span}\{v_1,v_2,v_3\}$, but then any fifth vector would be a linear combination of $v_1,v_2,v_3$ and $u$. For example, you may add $e_1$ or $e_2$ to $\{v_1,v_2,v_3\}$ to form a basis of $\mathbb{R}^4$, but you cannot add both, because five vectors in $\mathbb{R}^4$ would make a linearly dependent set. (In particular, $e_1-e_2=v_1$.)
