If an $n$ by $n$ complex matrix $A$ has trace zero, then it is a commutator, which means that there are $n$ by $n$ matrices $B$ and $C$ so that $A= BC-CB$. What is the order of the best constant $\lambda=\lambda(n)$ so that you can always choose $B$ and $C$ to satisfy the inequality $\|B\|\cdot \|C\| \le \lambda \|A\|$?

Added June 10: Gideon Schechtman showed me that for normal $A$ you can take $B$ a permutation matrix and $\|C\|\le \|A\|$ s.t. $A=BC-CB$.