Norm of commutators (bis)

2010-11-03T14:18:34Z

This question is slightly related to a popular one with the same title (see here).

Let $k$ be a field with characteristic zero. It is known (see Exercise 310) 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.

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 $\mu(n)$ ?

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 $-\frac12,\frac12,0,\ldots,0$.

Answer by Mark Sapir for Norm of commutators (bis)

2010-11-03T19:55:22Z

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".

Norm of commutators (bis) - MathOverflow

Norm of commutators (bis)

Answer by Mark Sapir for Norm of commutators (bis)