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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 \([1,\infty[\).

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