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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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    I think a rather simple one-dimensional counterexample is with $A = 2$ and $B = 1$ (so $c_0 = 0, c_1 = 1/2$).. Have I understood this correctly?2012-08-17
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    Regarding "It is well known that every matrix that commutes with A is a polynomial in A.": Matrix $C = \exp(A) = \sum_{k=0}^\infty \frac{1}{k!} A^k$ commutes with $A$ and is not a polynomial in $A$.2012-08-17
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    @Sasha An expression for $C$ which isn't a polynomial doesn't prove that there are no polynomial expressions. You can represent $C$ as a polynomial in $A$ of degree at most $n-1$, see the Cayley-Hamilton theorem.2012-08-17
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    Cocopuffs You Have not understood this correctly. if $A=2$ the minimum polynomila does not equal the characteristic polinimial. We assume that $n>1$.2012-08-17
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    Take the matrix $A$ first raw $(1,0)$ second raw $(0,0)$ and the matrx $B$ first raw $(2,0)$ and second raw $(0,0)$. Then they commute. The minimal is equal to the characteristic. and $A= \frac{1}{2} B$2012-08-17
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    @Cocopuffs I stand corrected, thanks for pointing it out.2012-08-17
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    @Zacarias The characteristic polynomial and minimal polynomial are both $t - 2$ in that case. We can take $n > 1$ but the same sort of counterexample will work, as clark pointed out.2012-08-17
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    Yes you are right Cocopuffs. Thanks2012-08-17
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    @Cocopuffs: you should post it as an answer. It is sort of trivial, but that saves the question from being unanswered...2012-08-17
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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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