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Let $A\in M_n(\mathbb C)$ be a square matrix of order $n$. Suppose that the characteristic polynomial of $A$ equals its minimum polynomial. It is well known that every matrix that commutes with $A$ is a polynomial in $A$.

Suppose that $A$ has integral elements (that is $a_{ij}\in\mathbb Z$) and $B$ is a matrix also with integral elements that commutes with $A$. So, $B$ is of the form

$B=c_0+c_1A+\cdots +c_kA^k.$

Can we conclude that $c_i\in\mathbb Z$ for $i=1,\ldots ,k$?

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    @sacha, in fact exp(A) is a polynomial of A. Easy to prove and to intuite when you know that$A$has a polynomial such as: P(A) = 0, and that a subspace with a finite dimension is always closed2014-10-22

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No. For example, let $A$ be a diagonal matrix with entries $a_{kk} = 2k$ and $B$ a diagonal matrix with entries $b_{kk} = k$. Then $A$ and $B$ commute and there is the representation $B = \frac{1}{2}A$. $A$ has the same minimal and characteristic polynomials as all eigenvalues are distinct and nonzero.