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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    $\begingroup$ 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 $B, C$. $\endgroup$
    – Sunni
    Jun 7 '10 at 16:37
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    $\begingroup$ 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. $\endgroup$ Jun 7 '10 at 16:53
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    $\begingroup$ 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 $n^{1/2} \|C\|$. $\endgroup$ Jun 8 '10 at 11:52
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    $\begingroup$ 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. $\endgroup$ Sep 28 '10 at 15:54
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    $\begingroup$ 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? $\endgroup$
    – Suvrit
    Jan 1 '11 at 13:20

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.


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

In a recent paper ([1]), Ravichandran and Srivastava (RS) study pavings for collections of matrices. Their main theorem claims to yield an improvement to the bound obtained by Johnson, Ozawa, and Schechtman (JOS). However, as noted by YCor in a comment, RS [1] cite the JOS work as satisfying a bound on $\max(\|B\|,\|C\|)$, instead of a bound on the product $\|B\| \|C\|$ as in Bill Johnson's answer above.

But as YCor notes, we can scale $B$ by $\|A\|$ (or both $B$ and $C$ by suitably, e.g., $\sqrt{\|A\|}$), to recover the inequality for the case noted in the OP and in the JOS paper.

In particular, Ravichandran and Srivastava's results imply the following:

Corollary (Corollary 3 in [1]). Every zero trace matrix $A \in M_n(\mathbb{C})$ may be written as $A=[B,C]$ such that $\|B\|$, $\|C\| \le K\log^2(n)\|A\|$ for some universal constant $K$.

(By suitable scaling, this translates into $\|B'\|\|C'\| \le K^2\log^4(n)\|A\|$, for $[B',C']=A$).

[1]. M. Ravichandran and N. Srivastava. Asymptotically Optimal Multi-Paving. arXiv. Jun 2017.

  • $\begingroup$ I believe existence of such an improvement was essentially prognosticated in the paper of Johnson, Ozawa, and Schechtman already. $\endgroup$
    – Suvrit
    Aug 11 '17 at 15:14
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    $\begingroup$ In the linked paper the BOS paper looks misquoted: the upper bound is on the product $\|B\|\cdot\|C\|$, not on $\max(\|B\|,\|C\|)$ (written $\|B\|,\|C\|\le\dots$ in your link and also in your answer). Or I miss something, could you clarify? $\endgroup$
    – YCor
    Aug 11 '17 at 16:02
  • $\begingroup$ @YCor -- indeed, you are right -- the paper that I cite, misquotes BOS, who as you say upper bound the product not the max. I am going to update my answer to reflect this. $\endgroup$
    – Suvrit
    Aug 11 '17 at 19:31
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    $\begingroup$ Anyway for $A\neq 0$ apply the result to $A/\|A\|$. It yields $B',C$ with $\|B'\|,\|C\|\le K\log(n)^2$ with $[B',C]=(1/\|A\|)A$. Then, for $B=\|A\|B'$, we have $[B,C]=A$, $\|B\|\le K\log(n)^2\|A\|$, $\|C\|\le K\log(n)^2$. Hence $\|B\|\cdot\|C\|\le K^2\log(n)^4\|A\|$ and of course the case $A=0$ works too. $\endgroup$
    – YCor
    Aug 11 '17 at 19:39
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    $\begingroup$ We have realized that that assertion in our paper ' Asymptotically Optimal Multi-Paving' that the existence of $(O(\epsilon^{-2}),\epsilon)$ pavings for zero diagonal matrices implies a polylogarithmic estimate for the quantitative commutator problem is incorrect. We have only been able to very slightly improve the Johnson-Ozawa-Schechtman result: We are able to show that given any zero diagonal $A \in M_m(\mathbb{C})$, there is a representation $A = [B,C]$ such that $||B|| ||C|| \leq C \operatorname{exp}(D\sqrt{\operatorname{log}(m)}) ||A||$, where $C, D$ are universal constants. $\endgroup$
    – mohanravi
    Aug 14 '17 at 21:21

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