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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. $A=BC-CB$.

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I guess it is Frobenius norm first. For Frobenius norm it is true that $\sqrt{2} \|B\|\cdot \|C\| \ge \|A\|$ for all complex matrices $B, C$. –  Sunni Jun 7 '10 at 16:37
I mean the operator norm, $\|A\|= \max \{\|Ax\|: \|x\|=1\}$ with $\|x\|$ the Euclidean norm. However, I do not know the answer if you use the Frobenius (Hilbert-Schmidt) norm. @Gil: I do not understand your comment. –  Bill Johnson Jun 7 '10 at 16:53
Schechtman showed that $\lambda(n) \le n$. WLOG (conjugate with an appropriate unitary) $C$ has zero diagonal and choose $A$ diagonal so that the magnitude of the difference of any two diagonal entries is at least one and the magnitude of every diagonal entry is less than $(n/2)^{1/2}$ (or a bit larger if $n$ is not a square). When you solve $AB-BA =C$ you see that the norm of $B$ is at most $n^{1/2} \|C\|$. –  Bill Johnson Jun 8 '10 at 11:52
Off-line more has happened--the latest upper bound is a power of $\log n$ (sixth power, I think), resulting from combined efforts with N. Ozawa and G. Schechtman. I thought this thread had died and so did not post. The proofs are a bit beyond what should go on MO, but eventually we'll write what we can do and I'll then post a link here. –  Bill Johnson Sep 28 '10 at 15:54
I wanted to know if you are already aware of the result that for the Frobenius norm, the ratio $\|BC-CA\|/ (\|B\|\|C\|)$ for randomly chosen $B$ and $C$, tightly concentrates around a number that goes to zero as $n\to \infty$. Thus, it suggests that $\lambda(n) \to \infty$ as $n \to \infty$, right? –  Suvrit Jan 1 '11 at 13:20
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2 Answers

Ozawa, Schechtman, and I finally wrote up what we know on this question. The estimate is that for every $\epsilon > 0$ there is a constant $C_\epsilon$ so that for every $n$, $\lambda(n)\le C_\epsilon n^{\epsilon}$. The paper can be downloaded from the arXiv.

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Almost the references cited below discuss upper bounds (i.e., norm(commutator) $\le$ something). One of the most relevant results is in reference #3 that I alluded to in my comment above.

  1. A short note on the Frobenius norm of the commutator
  2. The Frobenius norm and the commutator
  3. How big can the commutator of two matrices be and how big is it typically?
  4. Commutators, Pinching, and Spectral variation (Bhatia and Kittaneh)
  5. Norm inequalities for commutators of normal operators

If you chase the citations to these papers in google scholar, you will find several more very interesting and relevant papers---though, I have not been able to (yet) find a paper that discusses lower-bounds like yours.

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Thanks, Suvrit. Unfortunately, these do not address the problem I raised. BTW: The link for (3) does not work. –  Bill Johnson Jan 17 '11 at 13:08
Thanks for the update Bill, and also for posing such a nice question (I fixed the link though) –  Suvrit Jan 17 '11 at 13:28
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