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

edited Nov 3 2010 at 15:33

asked Nov 3 2010 at 14:18

Denis Serre
18.7k●1●31●93

	What is $\mu(2)$? Nilpotent matrices there are easy to describe. – Mark Sapir Nov 3 2010 at 14:41
	@Mark. First of all, I edited my question, because obviously an inequality like $\\|B\\|\le \mu\\|A\\|$ cannot hold true. Because the best $B$ is homogeneous of degree zero as a function of $A$. – Denis Serre Nov 3 2010 at 14:55
	When $a_{ij}=0$ for every pair such that $j-i\ne1$, we may take $B$ diagonal, in such a way that $\\|B\\|_2=\frac{n-1}{2}$ and $\\|B\\|_F=\sqrt{n\frac{n^2-1}{12}}$. – Denis Serre Nov 3 2010 at 15:51
	@Denis: Thanks for the question. I did not know this characterization of nilpotent matrices. – Mark Sapir Nov 3 2010 at 18:42

Mark Sapir · Accepted Answer · 2010-11-03 19:55:22Z UTC

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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".

answered Nov 3 2010 at 19:55

Mark Sapir
27.5k●2●57●118

Right! One gets $b_{12}-b_{23}=2k$. Setting $A_k:=k^{-1}A$, which does not change $B$, this means that $B$ cannot be chosen continuously, at least when the rank of $A$ drops from $2$ to $1$. Amazing! Of course, this does not provide an answer to the older MO question, but it suggests that it is highly non trivial. – Denis Serre Nov 3 2010 at 20:53

I got $b_{12} - b_{23}=k$. Anyway, Mark's example works out. Very nice! – Andreas Thom Nov 3 2010 at 21:16

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Yeah, Denis; that question has made me regret leaving my nice commutative world. – Bill Johnson Nov 4 2010 at 0:53

Norm of commutators (bis)

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