Let $P$ be a $n\times n$ matrix with integral entries and $Q=P+\frac{1}{2}I$, where $I$ denotes the $n\times n$ identity matrix. Then what can you say about $Q$.
Is it Idempotent?
Is it Invertible?
Is it Nilpotent?
Is it Unipotent?
According to me It is not necessary that it is Idempotent, Nilpotent, Unipotent.
For a counter example take the diagonal matrix $B=\left( b_{ij}\right)$ with $b_{ii}=i$.
I claim that it is invertible.
Consider the matrix $2Q=2P+I$. Reduce each entry modulo 2. We get $I$ whose determinant is $1$ hence determinant of $2Q$ must be odd. Hence $2Q$ is invertible and hence $Q$ is invertible.
Is this method correct?
Do you have any other better method?
Best Answer
Another way to get invertibility of $Q$ is as follows:
Suppose $\lambda$ is an eigenvalue of $P$, then $\lambda+\frac{1}{2}$ will be an eigenvalue of $Q$. We will show that $\lambda \neq -\frac{1}{2}$.
Consider the characteristic polynomial of $P$ (namely $\det(P-\lambda I)$). This is a monic polynomial with integer coefficients, so by the rational root theorem, all it's rational roots must be integers. Thus $\lambda$ cannot be $-1/2$, hence $Q$ does not have $0$ as an eigenvalue. This means $Q$ is invertible.