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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.

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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.