Triangularization of some integer matrices?

Question

Let $n\geq 2$, $a_1,\ldots, a_n \in \mathbb Z$, and $A\in M_n(\mathbb Z)$. If the characteristic polynomial of $A$ splits over $\mathbb Q$ with roots $a_i$, does there necessarily exist $P\in \mathrm{GL}_n(\mathbb Z)$ and an upper triangular matrix , $T$, with coefficients in $\mathbb Z$, such that $A=PTP^{-1}$? I think I can prove this result for $n=2$, but not beyond.

Answer 1

It is true. We use a recurrence reasoning with respect to $n$. Let $v=(v_1,\cdots,v_n)\in \mathbb{Q}^n$ s.t. $Av=a_1v$ ; we may assume that the $(v_i)_i$ are in $\mathbb{Z}$ and that $gcd(v_1,\cdots,v_n)=1$. Then there is $P\in GL_n(\mathbb{Z})$ s.t. $v$ is the first column of $P$. Consequently $P^{-1}AP=\begin{pmatrix}a_1&R\\0&A_1\end{pmatrix}\in M_n(\mathbb{Z})$, where $A_1\in M_{n-1}(\mathbb{Z})$ and $spectrum(A_1)=\{a_2,\cdots,a_n\}$. Finally, we apply to $A_1$ the recurrence hypothesis.

EDIT 1. We can generalize the previous result as follows: Assume that the characteristic polynomial of $A\in M_n(\mathbb{Z})$ is $f(x)=f_1(x)\cdots f_k(x)$ where the $f_i$ are irreducible. Then $A$ is integrally similar to a block-triangular matrix $T$ with diagonal elements $T_i$ that have $f_i$ as characteristic polynomial.

EDIT 2. On the other hand, even if the $(a_i)_i$ are distinct, we cannot, in general, diagonalize $A$ over $\mathbb{Z}$. Indeed, in such a case, the number of similarity classes over $\mathbb{Z}$ of matrices $M$ s.t. $spectrum(M)=(a_i)_i$ is finite but may be $>1$. Example : $\begin{pmatrix}1&1\\0&-1\end{pmatrix}$ is not diagonalizable over $\mathbb{Z}$ (work over $\mathbb{Z}/2\mathbb{Z}$).