For a matrix with all integer called entries, any integer eigenvalue will divide the determinant. I have to show that for a matrix with all integer valued entries, any integer eigenvalue will divide the determinant.
I know that the product of the eigenvalues is the determinant, but there can be non integral eigenvalues too. Also I know the sum of the eigenvalues, which is the trace of the matrix, is also integral. 
Any hint will be good. Thank you.
 A: Let
$A \in M_n(\Bbb Z) \tag 1$
be such a square matrix with all integer entries.  The eigenvalues of $A$, whether integral or not, all satisfy the characteristic polynomial of $A$, 
$\chi_A(x) = \det(xI - A); \tag 2$
it is easy to see that
$\chi_A(x) \in \Bbb Z[x], \tag 3$
that is, that $\chi_A(x)$ has all integer coefficients; it is clearly monic, and the constant term of $\chi_A(x)$ is $(-1)^n \det A \in \Bbb Z$. If we write
$\chi_A(x) = \displaystyle \sum_0^n c_i x^i, \; c_i \in \Bbb Z, \tag 4$
then
$c_0 = (-1)^n \det A \tag 5$
as we have said; if $\lambda$ is any eigenvalue of $A$ then (4) yields
$\displaystyle \sum_0^n c_i \lambda^i = \chi_A(\lambda) = 0, \tag 6$
which indeed may be cast as
$\lambda \displaystyle \sum_1^n c_i \lambda^{i - 1} = \sum_1^n c_i \lambda^i = (-1)^{n + 1} \det A; \tag 7$
if now $\lambda \in \Bbb Z$, we also have
$\displaystyle \sum_1^n c_i \lambda^{i - 1} \in \Bbb Z; \tag 8$
this shows that
$\lambda \mid (-1)^{n + 1} \det A, \tag 9$
that is,
$\lambda \mid \det A. \tag{10}$
Note Added in Edit, Saturday 28 April 2018, 12:09 PM PST:  I suppose a few words about the definition of "divides" might be in order.  I taken the definition to be, for $a, b \in \Bbb Z$, 
$a \mid b \Longleftrightarrow \exists \; c \in \Bbb Z, \; ac = b; \tag{11}$
with this definition, we have $a \mid 0$ for every $a \in \Bbb Z$, since $a0 = 0$; we also have
$0 \mid b \Longrightarrow b = 0, \tag{12}$
since then
$\exists \; a \in \Bbb Z, \; b = a \cdot 0 = 0; \tag{13}$
of course there is a slight possibility of confusion with the definition of "zero divisors" relevant in certain rings; this definition, as I recall, generally declares that $a$ and $b$ are zero divisors provided
$a \ne 0 \ne b \; \text{and} \; ab = 0; \tag{14}$
clearly such $a$ and $b$ do not exist in $\Bbb Z$, which is an integral domain. In the present answer I have used (11) and its implications (12) and (13) without further qualification; this approach appears to be self-consistent, though I am not sure how other authors might address the issue.  End of Note.
A: Let $p(x) = x^n + c_1 x^{n-1} + \cdots + c_{n-1} x + c_{n}$ be the characteristic polynomial of the matrix $M$, so $c_n = \pm \det M$. If $M$ has all integer entries, then every coefficient $c_i$ is an integer. Now suppose that $M$ has an integer eigenvalue $\lambda$, meaning an integer such that $p(\lambda) = 0$.
Subbing $\lambda$ into $p(x)$ and rearranging, we get
$$ -c_n = \lambda(\lambda^{n-1} + c_1 \lambda^{n-2} + \cdots + c_{n-1})$$
and so we have found that $\pm \det M$ may be written as a product of two integers, $\lambda$ and $(\lambda^{n-1} + c_1 \lambda^{n-2} + \cdots + c_{n-1})$. In particular, $\lambda$ is a factor of $\det M$.
