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Pisot numbers, primitive matrices and beta-conjugates. (Nombres de Pisots, matrices primitives et bêta-conjugués.) (French. English summary) Zbl 1285.11131
A subset $$\mathbb H$$ of $$\mathbb N$$ is relatively dense if there exists an integer $$m$$ such that, for all $$n\in\mathbb N$$, one at least of the numbers $$n,n+1,\dots,n+m$$ belongs to $$\mathbb H$$. A square matrix with all coefficient $$\geq 0$$ is said positive. A positive matrix is said primitive if there exists an integer $$k$$ such that all the coefficients of $$B^k$$ are $$>0$$. Perron has proved that a primitive matrix has a positive eigenvalue strictly greater than the module of all its other eigenvalues (called the dominant eigenvalue of the matrix).
A Perron number is an algebraic integer strictly greater than the module of all its conjugates. D. A. Lind [Ergodic Theory Dyn. Syst. 4, 283–300 (1984; Zbl 0546.58035)] has proved that: Every Perron number is the dominant eigenvalue of a matrix $$B$$ with coefficients in $$\mathbb N$$. A Pisot number is a real algebraic integer $$>1$$ whose all conjugates are of module $$<1$$. The author proves:
Let $$\beta$$ be Pisot number of degree $$d$$. There exists an integer $$n_0$$ such that for every $$n>n_0$$, we can find a primitive square matrix $$B_n$$ of order $$d$$, with coefficients in $$\mathbb N$$, whose $$\beta^n$$ is an eigenvalue.
The only eigenvalues of $$B_n$$ are then $$\beta^n$$ and its conjugates and they are simple.
If the irreducible polynomial of $$\beta^n$$ has coefficients $$\geq 0$$, then we can chose for $$B_n$$ its companion matrix. If not, we can find a matrix of form $\left( \begin{matrix} 1 & 0 & 0 & \dots &\cdot & 0& c_d\\ 1 & 0 & \dots & \dots&\cdot &0 & b_{d-1}\\ 0 & 1 & \dots& \dots &\cdot&\cdot \cdot& b_{d-2}\\ \cdot & 0 & 1 & \dots &\cdot& \cdot&\cdot\\ \cdot & \cdot & \dots & \dots&\cdot &\cdot & \cdot\\ 0 & \cdot & \cdot & \dots&\cdot &\cdot& b_2\\ 0 & 0 & 0 & \dots &0 & 1 &b_1 \end{matrix} \right)$ Now, let $$\beta>1$$ be a real number $$>1$$ and $$\beta=a_1/\beta+a_2/\beta^2+\dots+a_n/\beta^n+\dots$$ be the $$\beta$$-expansion of $$\beta$$. If the sequence $$(a_n)_{n\geq 1}$$ is ultimately periodic of period $$k$$ after the rank $$n_0$$ ( i.e. for any $$n\geq n_0$$, we have $$a_{n+k}=a_n$$), then $$\beta$$ is called a Parry number; $$\beta$$ is then an algebraic integer, strictly dominant root of the Parry polynomial, $X^{n_0+k}-(a_1X^{n_0+k-1}+\dots+a_{n_0+k})-(X^{n_0}-(a_1X^{n_0}+\dots+a_0)),$ where $$n_0$$ and $$k$$ are supposed minimum. There are two cases: either the degree of $$\beta$$ is equal to the degree of the Parry polynomial which is then the minimal polynomial of $$\beta$$ with $$d=n_0+k$$, or it is strictly smaller and the Parry polynomial has other roots than the conjugates of $$\beta$$. These other roots are called $$\beta$$-conjugates or pirat conjugates of $$\beta$$. It is known that Pisot numbers are Parry numbers (see [A. Bertrand, C. R. Acad. Sci., Paris, Sér. A 285, 419–421 (1977; Zbl 0362.10040); ibid. 289, 1–4 (1979; Zbl 0418.10049)]).
The author proves: Let $$\beta$$ be a Pisot number of degree $$d$$. Let $$\mathcal F_\beta$$ be the set of integers $$n$$ such that $$\beta^n$$ have no $$\beta$$-conjugate (i.e. the minimal polynomial of $$\beta^n$$ and its Parry polynomial are the same). Then $$\mathcal F_\beta$$ is relatively dense in $$[1,\infty[$$.

##### MSC:
 11R06 PV-numbers and generalizations; other special algebraic numbers; Mahler measure 11B85 Automata sequences 15A18 Eigenvalues, singular values, and eigenvectors
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##### References:
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