Matrix with even integers entries doesn't have odd eigenvalue Let $A \in M_n(\mathbb{Z})$ with even entries. Prove that $A$ doesn't have odd eigenvalue.

Suppose $A$ has an odd eigenvalue $k$. Then the characteristic polynomial $\det(A-\lambda I)$ can be written as:
$$\lambda^n+a_{n-1}\lambda^{n-1}+\ldots+a_0=(\lambda-k)(b_{n-1}\lambda^{n-1}+b_{n-2}\lambda^{n-2}+\ldots+b_1\lambda+b_0)$$
where each $a_i$ is even $(i=0, \ldots, n-1)$.
The right hand side evaluates to:
$$b_{n-1}\lambda^{n}+(b_{n-2}-kb_{n-1})\lambda^{n-1}+(b_{n-3}-kb_{n-2})\lambda^{n-2}+\ldots+(b_0-kb_1)\lambda-kb_0$$
Each of the coefficients must match the even $a_i$, so:
$$b_{n-1}=1\text{ which is odd},\\
a_{n-1}=b_{n-2}-kb_{n-1}\text{ even} \implies b_{n-2}\text{ odd},\\
a_{n-2}=b_{n-3}-kb_{n-2}\text{ even} \implies b_{n-3}\text{ odd},\\
\cdots \\
a_{1}=b_0-kb_1\text{ even} \implies b_0 \text{ odd}, \\
a_{0}=-kb_0\text{ even} \implies k\text{ even}.
$$
This is a contradiction.
Qed.
