# 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 $$\begin{pmatrix}1&1&0\\0&1&0\\0&0&1\end{pmatrix}.$$

• I'd just start with a block diagonal matrix with Jordan or scalar blocks, and perform a similarity transformation with a unimodular matrix. – J. M. is a poor mathematician Sep 14 '11 at 12:01
• In particular, Pascal matrices are very handy. – user1551 Sep 14 '11 at 12:40
• Start with the matrix you want. Then pick the basis you want to be the Jordan canonical basis. Then perform the change-of-basis transformation to "disguise" the original matrix, and use that as your basic matrix. – Arturo Magidin Sep 14 '11 at 16:24
• Also: do not use only matrices with integer entries... The only effect on students is that the first time they need to deal with a $\sqrt2$ in a matrix or ---if the devils are having fun that day--- a $1+2i$, they have panic attacks. – Mariano Suárez-Álvarez Sep 14 '11 at 17:05
• Arturo: That usually results a non-integer matrix. For me it is important that the final matrix and Jordan's form are integer. The transformations are not that important. – Peter Patzt Sep 14 '11 at 18:33

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.
• The first example comes from the construction in my answer with $a=-3,b=8,c=-15,r=s=3,t=1$. The second one comes from $a=1,b=-2,c=5,r=1,s=3,t=1$. – Gerry Myerson Sep 15 '11 at 9:36
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, $$\pmatrix{\phantom{-}3&\phantom{-}6&\phantom{-}4\cr\phantom{-}4&\phantom{-}13&\phantom{-}8\cr-7&-21&-13\cr}$$