# 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.

• There is a similar statement about any integer roots of a monic polynomial (with all integer coefficients) dividing the constant term... – Joppy Apr 28 '18 at 17:49
• Eigenvalues are exactly the roots of some well-known polynomial... – Gabriel Romon Apr 28 '18 at 17:52
• I'm assuming the matrix in concern is invertible right? – chhro Apr 28 '18 at 18:13
• Have you heard of the rational root theorem? – Jyrki Lahtonen Apr 28 '18 at 19:41

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 $\det M$.

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.