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.

  • 7
    $\begingroup$ There is a similar statement about any integer roots of a monic polynomial (with all integer coefficients) dividing the constant term... $\endgroup$ – Joppy Apr 28 '18 at 17:49
  • $\begingroup$ Eigenvalues are exactly the roots of some well-known polynomial... $\endgroup$ – Gabriel Romon Apr 28 '18 at 17:52
  • $\begingroup$ I'm assuming the matrix in concern is invertible right? $\endgroup$ – chhro Apr 28 '18 at 18:13
  • 1
    $\begingroup$ Have you heard of the rational root theorem? $\endgroup$ – 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 $\det M$.



$A \in M_n(\Bbb Z) \tag 1$

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

$\chi_A(x) = \displaystyle \sum_0^n c_i x^i, \; c_i \in \Bbb Z, \tag 4$


$c_0 = (-1)^n \det A \tag 5$

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.