when will the surfficient large power of a rational matrix be a integer matrix? $A$ is a $n\times n$ matrix whose elements are all non-negative rational numbers and $Det(A)$ is a non-zero integer.Under what condition the following is true?(0) There exist a positive integer $M$ such that $M\cdot A^k$ is always a element in $\mathbb{Z}^{n\times n},\ \forall k\in\mathbb{N}.$ It's obviously that the following condition is sufficient. (1)$A\in\mathbb{Z}^{n\times n}$   
Is (0) equivalent to (1)? If not,can you give a equivalent condition to (0)?
 A: This is an elaboration on Tyler Lawson's comment: For $A \in \mathrm{Mat}_{n \times n}(\mathbb{Q})$, the following are equivalent:
(0) There is a nonzero integer $M$ so that $M \cdot A^r$ has integer entries for all $r \geq 0$.
(1) $A = g B g^{-1}$ for some $B \in \mathrm{Mat}_{n \times n}(\mathbb{Z})$ and $g \in GL_n(\mathbb{Q})$
(2) All the coefficients of the characteristic polynomial $\det(t \mathrm{Id} - A)$ are integers.
Proofs: $(0) \implies (1)$. Let $\Lambda$ be the subgroup of $\mathbb{Q}^n$ generated by $A^r \mathbb{Z}^n$ for all $r \geq 0$. Then $\frac{1}{M} \mathbb{Z}^n \subseteq \Lambda \subseteq \mathbb{Z}^n$ so $\Lambda$ is a free abelian group of rank $n$. Choosing a basis for $\Lambda$ places $A$ in the required form.
$(1) \implies (2)$ Obvious.
$(2) \implies (0)$ Let the characteristic polynomial be $t^n + b_{n-1} t^{n-1} + \cdots b_1 t + b_0$. Then, by the Cayley-Hamilton theorem, 
$$A^{r+n} = - \left( b_{n-1} A^{r+n-1} + \cdots + b_1 A^{r+1} + b_0 A^r \right).$$
So, for $s \geq n$, the matrix $A^s$ is in the integer span of $A^0$, $A^1$, ..., $A^{n-1}$. Choosing $M$ large enough to clear the denominators of these finitely many matrices proves the result.
Note that description (1) makes it easy to find many examples.
A: They are not equivalent.  Take a (15,8,8) design ( or the 15 by 15 0-1 matrix associated with the order 16 Sylvester-Hadamard matrix ). Multiply all elements by 1/2, and call the result A.  AA is an integer matrix equal to 2I + 2J, and so every larger power of A is also an integer matrix. But A is half (!?!) of an integer matrix with nonnegative entries and determinant of size 2.
Edit: As Gerry mentions, there are examples of smaller order that are not very hard to find.
(It helps to start with an integral matrix with determinant divisible by a power of an integer.)
