Let $A \in M_{n}(\mathbb{Z})$.

(1) Prove that if $k \in \mathbb{Z}$ is an eigenvalue of $A$, then $k$ divides $\det A$.

(2) Let $j \in \mathbb{Z}$ such that the sum of all entries in each line of $A$ is equal to $j$. Prove that $j$ divides $\det A$.


(1) Let

$$A = \left(\begin{array}{ccc} a_{11} & \cdots & a_{1n}\\ a_{21} & \cdots & a_{2n}\\ \vdots & \ddots & \vdots\\ a_{n1} & \cdots & a_{nn} \end{array}\right).$$

We can write

$$A = \left(\begin{array}{ccc} \alpha_1 & \cdots & \times\\ 0 & \alpha_2 & \times\\ \vdots & \ddots & \vdots\\ 0 & \cdots & \alpha_n \end{array}\right).$$

Thus, $\det A = \alpha_1 \cdots \alpha_n$. If $k$ is an eigenvalue, so $k$ is a root of $\det(A - xI) = \pm(\alpha_1 - x)\cdots(\alpha_n - x)$. Therefore, $k = \alpha_i$ for some $\alpha_i$.

(2) Let

$$A^T = \left(\begin{array}{ccc} a_{11} & \cdots & a_{1n}\\ a_{21} & \cdots & a_{2n}\\ \vdots & \ddots & \vdots\\ a_{n1} & \cdots & a_{nn} \end{array}\right).$$

Note that

$$A^T e_i = (a_{1i},a_{2i},...,a_{ni})^T.$$

So, $$A^T h = A^T (e_1 + \cdots + e_n) = (a_{11},a_{21},...,a_{n1})^T + (a_{1n},a_{2n},...,a_{nn})^T = (a_{11} + \cdots + a_{1n},...,a_{n1} + ... + a_{nn})^T = j(1,...,1)^T = j(e_1 + \cdots + e_n) = jh.$$

Therefore, $j$ is an eigenvalue of $A^T$. Since $A$ and $A^T$ has the same eigenvalues, the result follows by the previous item.

Is this correct?


Your proof for part (2) is great!

Your proof for part (1) is incorrect; in particular, note that if your proof were correct, then it would also apply to matrices that do not have integer entries. Note that $$ A = \pmatrix{k&0\\0&1/k} $$ has $k$ as an eigenvalue, but has determinant $1$, which is not divisible by $k$ (for an integer $k \geq 2$). I think that that the simplest approach is to apply the rational root theorem to the characteristic polynomial of $A$.


Modulo $k$ we have $\ \det A \equiv \det(A-k I_n)=0$


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.