$Q=(P+\frac12I)$ is invertible where $P$ is a square matrix with integral entries Let $P $ be $n×n$ matrix with integral entries and $Q=P+\frac12I$  where I denote the $n×n$ identity matrix . Then  $Q$ is invertible. 
My attempt to prove:
$Q$ is invertible iff $0$ is not an eigen value.
If possible, let $0$ be eigen value. Then $\exists$ $v\in R^n$ s.t $Qv=0$ 
Then $Pv=-\frac12v$. So $-\frac12$ is an eigen value of $P$ with eigen vector $v$ .
So $(2x+1)$ is a divisor of the characteristics polynomail $p(x)$ (say) of $P$ which is of course a monic polynomial with integer coefficients.
So $\exists g(x)$ s.t $(2x+1)g(x)=p(x)$ . 
I am not sure how to proceed from here because $g(x)$ may not itself be monic but the product is monic. . This may be  elementary question but please guide me if I am wrong or give a better proof.A lot of thanks for your time!!
 A: It suffices to prove that $2Q$ is invertible. But $2Q$ is an integer matrix with all even entries except for odd entries on the diagonal. In other words, $2Q\equiv I\pmod2$, and so $\det(2Q)\equiv \det(I)=1\pmod2$. In particular, $\det(2Q)$ is an odd integer and hence nonzero, so $2Q$ is invertible. (And $\det Q$ is an odd integer divided by $2^n$, which could probably be proved directly if desired.)
A: $Q$ is invertible iff $2Q$ is invertible. In $2Q$, the diagonal entries are all odd and the off-diagonal entries are all even. Now if we apply the Leibniz formula for $\det2Q$
$$\det 2Q=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^n(2Q)_{i,\sigma(i)}$$
we see that the identity permutation corresponds to a selection of exactly the diagonal entries, which multiply to an odd number, while all other permutations correspond to a selection containing at least one even entry, so their contribution to the sum is even. $\det2Q$ is thus the sum of one odd and several even numbers, which must be odd, hence necessarily non-zero. So $2Q$ is invertible, implying $Q$ is invertible as well.
A: As you mentioned, the characteristic polynomial of $P$ is a polynomial with integer coefficients. The leading and last coefficients are
$$ x^n + \dots + (-1)^n \det P$$
By the rational root theorem, all rational eigenvalues of $P$ must be integer divisors of $\det P$. In particular,  $-1/2$ cannot be an eigenvalue.
