# Relations between eigenvalues and determinant of an integer matrix

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?

## 2 Answers

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