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So any square matrix $A$ can be decomposed into $A = S J S^{-1}$ where $J$ has a normal Jordan form, moreover $A$ and $J$ are similar matrices.

My question is quite straightforward. Given arbitrary normal Jordan matrix $J$, can I always find $S$ such that the resulting matrix $A$ has real integer entries?

Edit: So we have some necessary conditions now namely that the diagonal elements of $J$ must be real or come in complex conjugate pairs.

It is still unclear though what are other conditions and most importantly what is a reasonable approach in order to actually calculate some $A$ given $J$.


Besides general approach I'm also considering a particular (not overly complicated I believe) case $$ S \begin{pmatrix} i & 1 & 0 \\ 0 & i & 0 \\ 0 & 0 & i \end{pmatrix} S^{-1} = A \in \mathbb{Z}^{3 \times 3}, $$ but I'm not sure how to proceed. (Proved impossible, see edit.)

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It seems the following.

The obvious necessary condition is

(C) the characteristic polynomial $\det (\lambda I-J)$ of the matrix $J$ has integer coefficients.

If all roots of the polynomial $\det (\lambda I-J)$ are mutually different, then Condition C is also sufficient. Indeed, in this case the matrix $J$ is diagonal, and each matrix $A$ with $\det (\lambda I-A)=\det (\lambda I-J)$ has Jordan normal form $J$. So it suffices to find a matrix $A$ with integer entries such that $\det (\lambda I-A)=\det (\lambda I-J)$, which is easy to do. Indeed, if $$\det (\lambda I-J)=a_0+a_1\lambda+\dots+a_{n-1}\lambda^{n-1}+\lambda^n$$ then we can put

$$A=\begin{pmatrix} 0 & 0 & \dots & 0 & -a_0 \\ 1 & 0 & \dots & 0 & -a_2 \\ 0 & 1 & \dots & 0 & -a_3 \\ \dots & \dots & \dots & \dots & \dots \\ 0 & 0 & \dots & 0 & -a_{n-2} \\ 0 & 0 & \dots & 1 & -a_{n-1} \end{pmatrix}.$$

Another sufficient condition is when all roots of the polynomial $\det (\lambda I-J)$ are already integers.

Consider the case when order $n$ of the matrix $J$ equals $2$. If its characteristic polynomial $\det (\lambda I-J)$ has different roots, then Condition C is both necessary and sufficient. Assume that the polynomial $\det (\lambda I-J)$ has only one root $\lambda_0$. Then $\det (\lambda I-J)=(\lambda-\lambda_0)^2$. If Condition C is satisfied, then both $2\lambda_0$ and $\lambda_0^2$ are integers. So $\lambda_0$ is a integer too, and thus the matrix $J$ has integer entries. So in this case Condition C is both necessary and sufficient too.

Similarly, if the matrix $J$ has a unique root $\lambda_0$ and Condition C is satisfied, then both $n\lambda_0$ and $\lambda_0^n$ are integers, so $\lambda_0$ is an integer too, and thus the matrix $J$ has integer entries. So in this case Condition C is both necessary and sufficient too.

The further steps in this investigation are to consider case $n=3$ and to find a counterexample of the matrix $J$ satisfying Condition $C$, but which is not similar to a matrix $A$ with integer entries.

PS. To be continued...

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  • $\begingroup$ Great answer. The companion matrix...But why does $J$ has to be diagonal? $\endgroup$ – Troy Woo May 17 '15 at 9:15
  • $\begingroup$ @TroyWoo Thanks. $J$ has not to be diagonal. But if all roots of the polynomial $\det(\lambda I−J)$ are mutually different then $J$ is diagonal. $\endgroup$ – Alex Ravsky May 17 '15 at 11:39
  • $\begingroup$ Yes. My point is that companion matrix always exists. $\endgroup$ – Troy Woo May 17 '15 at 12:49
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The answer depends. The diagonal elements of $J$ are eigenvalues of $A$. So if $A$ is real, its characteristic polynomial also has real coefficients, and therefore the eigenvalues come in complex conjugate pairs. the $3\times 3$ case you give obviously does not satisfy.

The whole problem can be seen as deciding whether the orbit of $J$ under similarity transformation of $S\in GL(n,\mathbb C)$ has intersection with $\mathbb Z^{n\times n}$. This is simply beyond my knowledge. Wiki seems to show little information.

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  • $\begingroup$ No doubt your reasoning here is correct and my random example shall be refuted. So we have some necessary conditions for diagonal elements of $J$. $\endgroup$ – Pranasas May 16 '15 at 20:59
  • $\begingroup$ @Pranasas I admit the whole question is not easy (for me) to answer. Since $\det A=\det J$, you should also expect $J$ to have integer determinant (i.e. the product of all diagonal elements). And this might not be the sufficient condition. $\endgroup$ – Troy Woo May 16 '15 at 21:08

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