Eigenvalues for toral Anosov automorphisms It is well known that on every $d$-dimensional torus there exists linear Anosov automorphisms. 
My question is the following:
Given $k< d$ does there exists a linear Anosov automorphism of $\mathbb{T}^d$ with exactly $k$ eigenvalues smaller than $1$? If true (which I expect), does there exists an \emph{irreducible} linear Anosov automorphism of $\mathbb{T}^d$ with exactly $k$ eigenvalues smaller than $1$? 
This can be phrased in terms of matrices with integer coeficients (please add the corresponding relevant tags) as:
Given $k< d$ does there exists a matrix in $SL(d,\mathbb{Z})$ such that all eigenvalues have modulus different from $1$ and $k$ of them are of modulus smaller than $1$? What about if the characteristic polynomial is irreducible over $\mathbb{Q}$?. 
Some relevant related information can be found in this paper (http://arxiv.org/pdf/1009.2994v2.pdf) where some results of W. Duke, Z. Rudnick, P. Sarnak as well as of Nevo and Sarnak are refered to. 
 A: This is only a partial answer, which I shall delete if I find a better one. Every pair $(k,d)$ of the form
$$d=\frac12\phi(n),\qquad k={\rm card}(\frac{n}{6}\le j \le\frac{n}{2},j\wedge n=1)$$
is OK: take the cyclotomic polynomial $\Phi_n$ and form the irreducible polynomial $P_n\in{\mathbb Z}[X]$ defined by
$$\Phi_n(t)=t^{\frac{n}{2}}P_n\left(t+\frac1t\right).$$
The roots of $P_n$ are the numbers $2\cos\frac{2j\pi}{n}$ with $j\wedge n=1$, smaller than $1$ if and only if $\frac{n}{6}\le j \le\frac{n}{2}$.
If instead $k=d-1$, take any Pisot number. Edit (after Nikita's comment below): One may take the companion matrix of $X^d-X^{d-1}-\cdots-X-1$. Its only root of modulus greter than $1$ is a Pisot number, also called a multinacci number. If $d=2$, this is just the golden ratio, at the basis of the Fibonacci sequence, hence the `word' multinacci.
A: Given $k < d$, one can always construct a monic polynomial irreducible over $\mathbb Q$ with exactly $k$ roots less than 1 in modulus and $d-k$ roots greater than 1 in modulus. This follows from the general construction of algebraic units, namely, each group of units of an algebraic field contains a unit with a given $k$  -- see, e.g., [Borevich and Shafarevich]. 
Then you can simply take the companion matrix of such a polynomial. 
Or do you need an explicit construction?
