This question is slightly related to a popular one with the same title (see [here](https://mathoverflow.net/questions/27345)).


Let $k$ be a field with characteristic zero. It is known (see [Exercise 310](http://umpa.ens-lyon.fr/~serre/DPF/exobis.pdf)) that a matrix $A\in M\_n(k)$ is nilpotent if and only if it is a commutator of its own: there exists a $B$ such that $A=AB-BA$. Of course, $B$ is not unique.



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> Consider the complex case ($k=\mathbb C$). Endow $M\_n(\mathbb C)$ with your beloved norm, preferably either the operator norm $\|\cdot\|\_2$ or the Schur--Frobenius--Hilbert--Schmidt norm $\|\cdot\|\_F$. If $A$ is nilpotent, what is the smallest value of $\|B\|$, where $B$ is a factor in $A=AB-BA$ ? What is the smallest constant $\mu(n)$ such that for every $n\times n$ nilpotent $A$, there exists such a $B$ with $\|B\|\le\mu(n)$ ? Actually, is there such a finite $\mu(n)$ ?
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**Edit**. When ${\rm rk}A=1$, that is $A=xy^\*$ with $y^\*x=0$, one can always take $B$ such that $\|B\|\_2=\frac12$ or $\|B\|\_F=\sqrt2/2$. Just take $B$ diagonal in a unitary basis $\{\frac{x}{\|x\|\_2},\frac{y}{\|y\|\_2},\ldots\}$, with eigenvalues $-\frac12,\frac12,0,\ldots,0$.



Norm of commutators (bis)
Consider the matrix $$\left\[ \begin{array}{ccc} 0 & 1 & k \\\ 0 & 0 & 1\\\ 0 & 0 & 0\end{array}\right\].$$ Then the norm of $B$ depends on $k$ (just solve the system of linear equations $AB-BA=A$). So the answer to your question seems to be "no". 


