Finding a generalized eigenvector I'm asked to find a Jordan normal form and a Jordan basis for a matrix A such that
$$A=
\begin{bmatrix}
    1   & 0 & 0 & 2 \\
    2   & -1 & 0 & 2\\
    2   & 0 & -1 & 2 \\
    0   & 0 & 0 & 1
\end{bmatrix}$$
I was able to find the eigenvalues $\lambda_{1}=1$ and $\lambda_{2}=-1$ just fine, and both have algebraic multiplicities equal to 2. For the first eigenvalue, I also got a single eigenvector, whereas for $\lambda_{2}$ I found two. The problem starts when I try to complete the Jordan basis. I assume I need to find a generalized eigenvector from $\lambda_{1}$, since it's geometric multiplicity is just 1. From other similar questions, I gathered that:
$$Ker(A-\lambda_{1}{I})^k$$
Is the generalized eigenspace which also contains the eigenvector I need to fill the basis. Then, it seems I'd just need to work with the following matrix:
$$(A-1{I})=
\begin{bmatrix}
    0   & 0 & 0 & 2 \\
    2   & -2 & 0 & 2\\
    2   & 0 & -2 & 2 \\
    0   & 0 & 0 & 0
\end{bmatrix}$$
But the problem is that the matrix $(A-1I)$ isn't nilpotent, so I can't never seem to get to a point where $(A-1I)^k$ is the zero matrix, and thus I can't find a "final" matrix for the sequence $Ker(A-1I) \subset Ker(A-1I)^2 \subset \dots \subset Ker(A-1I)^k \subset \dots$
According to the method I've used to find a Jordan basis, I need to find generalized eigenvectors from $Ker(A-\lambda{I})^k$ that aren't eigenvectors from $Ker(A-\lambda{I})^{k-1}$, assuming $k$ is the index of nilpotence of the matrix. However, since the sequence is infinite in this case, I don't know how to find the missing eigenvector. Could someone please point out the flaw in my process?
 A: so I think the quastion for the Eigenvalue $\lambda_1 = -1$ is clear and as you already said this is the Eigenvalue with geometric multiplicity 2. So you can pick up two vectors from the eigen space namely the standard basis vectors: 
\begin{align}
\{e_2,e_3 \}
\end{align}
Now to the eigenvalue $\lambda_2 = 1$. We can find a eigenvector from the eigenspace, which is of dimension one, this is :
\begin{align}
b := e_1+e_2+e_3
\end{align}
and choose any vector $v \in \Re^4$, which is linear independent to $\{e_1,e_2,b\}$ and obtain that this vector must be in the space: 
\begin{align}
Ker(A-I)^2 \setminus Ker(A-I)
\end{align}
So your Jordan basis is now given by the construction by:
\begin{align}
\{ v,(A-I)v,e_2,e_3 \}
\end{align}
Note that the Matrix $A-I$ does not have to nilpotent on $\Re^4$, but only on the restriction to the generlized eigenspace to $\lambda_2 = 1$. You can check it by the choice of the eigenspace spanned by $\{e_2,e_3\}$. The restriction of $(A-I)^k$ is not the zero Mapping for all integers $k$. If you namely assume that there is an integer $k_0$ with:
\begin{align}
(A-I)^{k_0} = 0
\end{align}
then $(A-I)^{k_0-1}$ is an eigenvector to the eigenvalue $\lambda_2 = 1$, which is a linear combination of $\{e_2,e_3\}$, which can not happen as the eigenvectors to different eigenvalues are linear independent.  
A: If $A-\lambda I$ is nilpotent, that means some power is zero, that means that the generalised eigenspace is the whole space ${\Bbb F}^n$, for your example ${\Bbb C}^4$ (or ${\Bbb R}^4$).  But in this case you already know that that is not true, because you have eigenvectors and generalised eigenvectors associated with two different eigenvalues.
In fact, the generalised eigenspace for $\lambda$ will always be $\ker(A-\lambda I)^a$, where $a$ is the algebraic multiplicity.  Sometimes this will also be true for some exponent less than $a$, as in your example with $\lambda=-1$: the generalised eigenspace is $\ker(A+I)$, but it is also $\ker(A+I)^2$, which turns out to be the same (check it!!).
For $\lambda=1$ the generalised eigenspace is $\ker(A-I)^2$, and in this case it isn't equal to $\ker(A-I)$.  The matrices
$$(A-I)^2\ ,\quad (A-I)^3\ ,\quad (A-I)^4\ ,\ldots$$
may not all be the same, but their kernels will be the same.
So to sum up: the generalised eigenspace you want is $\ker(A-I)^2$.
