Jordan normal form with generalized eigenvector for initial value problem For one of my subjects I have gotten this question:  
Determine the solution of the initial value problem
$y(n+1)= \begin{pmatrix} 1 & 0 & 2\\
0 & 1 & 3\\
0 & 0 & 1
\end{pmatrix} \ y(n), \ \ y(0)=\begin{pmatrix} 1 \\ -1 \\ 1\end{pmatrix}$  
My solution thus far:
The characteristic polynomial is given as $(\lambda-1)^3=0$ and hence, the eigenvalues are $\lambda_1=\lambda_2=\lambda_3=1$ with algebraic multiplicity 3.
Two eigenvectors, which are linearly independent, are $\mathbf{u}=\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} 0 \\ 1\\ 0 \end{pmatrix}$. And consequently one may find a generalized eigenvector $\mathbf{w}=\begin{pmatrix} 0 \\ 0 \\ 1\end{pmatrix}$. So far, no problems at all. I am fairly certain that these are correct. The important part of this (for my question), is that we have a geometric multiplicity of 2. My issue is actually in the  "J" itself. I know that there are merely ones on the diagonal. So we have:
$J=\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1 
\end{pmatrix}$ or 
$J=\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 1\\
0 & 0 & 1 
\end{pmatrix}$
or even 
$J=\begin{pmatrix}
1 & 1 & 1\\
0 & 1 & 1\\
0 & 0 & 1 
\end{pmatrix}$
What I do not understand, is what and when there should be numbers above the diagonal entries. I am sorry, because I am not sure how to really phrase the question. Perhaps this is due to the face that I am no entirely certain about this concept. Is there anyone that could give me a little push in the right direction?
Thanks a lot. 
 A: Well, I don't know if the information is sufficient, but I'll make a start.  You have calculated the geometric multiplicity of the eigenvalue $\lambda = 1,$ which is equal to $\gamma = 2.$ 


*

*Since the geometric multiplicity of the eigenvalue $\lambda = 1$ does not equal its algebraic multiplicity, the Jordan normal form cannot be diagonal, which practically means there will be a number of  $1$'s on the superdiagonal.


The Jordan normal form consists of Jordan blocks corresponding to the eigenvalues (here we have only one eigenvalue), which are of the following form. A Jordan block is a square matrix that has the eigenvalue on the main diagonal and $1$'s on the superdiagonal.
The number of Jordan blocks we are going to construct is equal to the number of the geometric multiplicity. In our example, we are going to construct exactly $2$ Jordan blocks, say $J_1$ and $J_2.$ But what about the size of each Jordan block? This has to do with the Jordan chains.
In our specific example, we create 2 Jordan chains. The first chain is $\{\mathbf v_1, \mathbf v_2\},$ where $\mathbf v_1= \begin{bmatrix} 2\\3\\0\end{bmatrix}$ is one ordinary eigenvector of rank 1 and $\mathbf v_2 = \begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}$ is a generalised eigenvector of rank 2 (which is related to $\mathbf v_1$) and the second chain contains only the other ordinary eigenvector, say, $\{\mathbf u_1\},$ with $\mathbf u_1 = \begin{bmatrix} 0 \\ 1 \\ 0\end{bmatrix}.$
So, the Jordan blocks are going to be:
$$J_1 = \begin{bmatrix} 
1 & 1\\
0 & 1
\end{bmatrix}, \quad J_2 = \begin{bmatrix}  1 \end{bmatrix}.$$
Thus, the Jordan normal form $\mathbf J$ is going to be of the form (up to rearrangements of the Jordan blocks):
$$\mathbf J = \begin{bmatrix} J_1 \\ &  J_2 \end{bmatrix} = \begin{bmatrix}
\color{blue}{1} & \color{blue}{1} & 0 \\
\color{blue}{0} & \color{blue}{1} & 0 \\
0 & 0 & \color{red}{1}
\end{bmatrix}.$$
You are going to observe that if 
$$ Q = \Big[ \underbrace{ \mathbf v_1 \quad \mathbf v_2}_{\text{corr.  to $J_1$}} \quad \underbrace{\mathbf u_1}_{\text{corr. to $J_2$}} \Big],$$
then:
$$ Q \cdot \mathbf J \cdot Q^{-1} = \begin{bmatrix} 1 & 0 & 2\\
0 & 1 & 3\\
0 & 0 & 1
\end{bmatrix}.$$

Remark:
I skipped the calculation of the eigenvectors (ordinary and generalised), but you may find this answer helpful along with the wikipedia site.
