Suppose we are given the following:
An $n \times n$ matrix $A$ (with rational entries).
$\operatorname{trace}(A)=0$
$B=A+I$ where $I$ is the $n \times n$ identity matrix.
Question. What conditions on $A$ would guarantee the following to be true:
$\det(B) \leq 1$ and,
$\det(B)=1$ if and only if $A$ is nilpotent.
Example Condition: The eigenvalues of $A$ are real and at least $-1$.
To see why, let $\lambda_1,\dotsc,\lambda_n$ be the eigenvalues of $A$. The trace condition on $A$ gives us $\sum_{i=1}^n \lambda_i=0$. Now the eigenvalues of $B$ are $1+\lambda_1,\dotsc,1+\lambda_n$ which are all nonnegative, so applying AM–GM and using the trace condition on $A$, we get
$$\det(B) = \prod_i (1+\lambda_i) \leq \left(\frac{n+\sum_i \lambda_i}{n} \right)^n=1.$$
Equality holds if and only if $1+\lambda_1=\dotsb=1+\lambda_n$, which implies all the $\lambda_i$'s are equal, and subsequently by the trace condition on $A$ that all the $\lambda_i$'s are $0$. In other words, equality holds if and only if $A$ is nilpotent.
