Norm of commutators (bis) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T23:26:20Z http://mathoverflow.net/feeds/question/44680 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/44680/norm-of-commutators-bis Norm of commutators (bis) Denis Serre 2010-11-03T14:18:34Z 2010-11-03T19:55:22Z <p>This question is slightly related to a popular one with the same title (see <a href="http://mathoverflow.net/questions/27345" rel="nofollow">here</a>).</p> <p>Let $k$ be a field with characteristic zero. It is known (see <a href="http://umpa.ens-lyon.fr/~serre/DPF/exobis.pdf" rel="nofollow">Exercise 310</a>) 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.</p> <blockquote> <p>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 $\mu(n)$ ?</p> </blockquote> <p><strong>Edit</strong>. 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 $-\frac12,\frac12,0,\ldots,0$.</p> http://mathoverflow.net/questions/44680/norm-of-commutators-bis/44715#44715 Answer by Mark Sapir for Norm of commutators (bis) Mark Sapir 2010-11-03T19:55:22Z 2010-11-03T19:55:22Z <p>Consider the matrix $$\left[ \begin{array}{ccc} 0 &amp; 1 &amp; k \\ 0 &amp; 0 &amp; 1\\ 0 &amp; 0 &amp; 0\end{array}\right].$$ 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". </p>