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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 $\det A$.

Attempt.

(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?

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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 $A$.

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

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