Matrix with even integers entries doesn't have odd eigenvalue

We would like to show, in other words, that $A - kI$ is invertible for any odd integer $k$. To see that this is the case, we compute $\det(A - k I)$ modulo $2$. Notably, the determinant is a polynomial on the entries of a matrix, which means that the determinant defines a function over $M_n(\Bbb F_2)$. That is, it holds that $$ A \equiv B \pmod 2 \implies \det(A) \equiv \det(B) \pmod 2. $$ Because the entries of $A$ are even, we have $A - k I \equiv -kI \equiv I \pmod 2$. Thus, $$ \det(A - kI) \equiv \det(I) \equiv 1 \pmod 2. $$ That is, $\det(A - kI)$ is odd and thus non-zero. So, $(A - k I)$ has a non-zero determinant and is thus invertible.

---

Another proof: let $B = A/2$. We note that by Gauss's lemma, any eigenvalue of $B$ is either an integer or irrational.

Now if $\lambda$ is an eigenvalue of $A$, then $\lambda = 2 \mu$ for some eigenvalue $\mu$ of $B$. If $\mu$ is irrational, then $\lambda = 2 \mu$ is also irrational and hence not an integer. If $\mu$ is an integer, then $\lambda = 2\mu$ is even. In either case, $\lambda$ cannot be an odd integer.

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.