Upper bound for determinant of square matrix

Question

Let $A \in \Bbb C^{n \times n}$ be a non-Hermitian matrix whose entries are denoted by $a_{i,j}$.

1. What is the best upper bound that we have for $\det (A)$ in term of $\mbox{trace}(A)$?

2. Does the following inequality hold for any complex matrix $A$?

$$\det(A) \leq \bigg(\frac{\mbox{trace}}{n}\bigg)^n$$

Answer 1

The best upper bound is $+\infty$. Consider the companion matrix $$ C=\pmatrix{0&&&&d\\ 1&\ddots&&&0\\ &\ddots&\ddots&&\vdots\\ &&\ddots&0&0\\ &&&1&t}. $$ Then $\operatorname{tr}(C)=t$ and $\det(C)=(-1)^{n-1}d$. Since $d$ and $t$ do not depend on each other, $|\det(C)|$ can assume any nonnegative real value, given $\operatorname{tr}(C)$.