Integer matrix with particular Jordan's form For teaching purposes I would like to find integer matrices with a particular Jordan's form. Is there some kind of technique to find nice examples? For example for $$\begin{pmatrix}1&1&0\\0&1&0\\0&0&1\end{pmatrix}.$$
 A: As J.M. pointed out in his comment, for any Jordan form $J$ and unimodular matrix $U$, the matrix product $UJU^{-1}$ will do.  In particular, you can get sufficient varieties by building $P$ from a Pascal matrix. For example, the upper triangular order-3 Pascal matrix is
$$
P=\begin{pmatrix}1&1&1\\0&1&2\\0&0&1\end{pmatrix}.
$$
You can take $U$ as any product of $P,\ P^T,\ P^{-1},\ (P^T)^{-1}$, diagonal matrices with diagonal entries $=\pm1$ and permutation matrices. To illustrate, let $J$ be the Jordan form in your example.  Then
$$
\begin{eqnarray}
&&U=PP^TP \Rightarrow UJU^{-1} =\begin{pmatrix}-8&24&-45\\-9&25&-45\\-3&8&-14\end{pmatrix},\\
&&U=P^2\begin{pmatrix}0&-1&0\\0&0&1\\-1&0&0\end{pmatrix}(P^T)^{-1} \Rightarrow UJU^{-1} =\begin{pmatrix}2&-2&5\\3&-5&15\\1&-2&6\end{pmatrix}.
\end{eqnarray}
$$
You can generate a lot of integer matrices with identical Jordan form but very different appearances using this method.
A: Let's instead go for $$\pmatrix{0&1&0\cr0&0&0\cr0&0&0\cr}$$ Then we can just add the identity matrix and get what you asked for. 
So, we're looking for an integer matrix of rank 1 with zero as a triple eigenvalue. Rank 1 means $$\pmatrix{ar&br&cr\cr as&bs&cs\cr at&bt&ct\cr}$$ as each row must be a multiple of each other row. Add the condition $ar+bs+ct=0$ and I think we are there. 
For example, taking $a=1,r=2,b=3,s=4,c=2,d=-7$, and remembering to add in the identity at the end, we get the example, $$\pmatrix{\phantom{-}3&\phantom{-}6&\phantom{-}4\cr\phantom{-}4&\phantom{-}13&\phantom{-}8\cr-7&-21&-13\cr}$$
