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

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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. 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-Alvarez 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

## 2 Answers

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.

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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
For more variety with unimodulars, one could also consider unit (upper or lower) triangular (i.e., all 1's on the diagonal) matrices with integer entries, possibly varying the signs of the 1's on the diagonal. – J. M. Sep 15 '11 at 9:36
@J.M.: Yes, that is certainly a more general method. – user1551 Sep 15 '11 at 10:23
@Gerry Myerson: Yes, but considering a general Jordan form might be more interesting. – user1551 Sep 15 '11 at 10:25

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, $$\pmatrix{\phantom{-}3&\phantom{-}6&\phantom{-}4\cr\phantom{-}4&\phantom{-}13&\phantom{-}8\cr-7&-21&-13\cr}$$

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