It is well-known that a complex m m matrix A is a commutator

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1 A quantitative version of the commutator theorem for zero trace matrices William B. Johnson a, Narutaka Ozawa b, and Gideon Schechtman c, a Deartment of Mathematics, Texas A&M University, College Station, TX 77845; b Research Institute for Mathematical Sciences, Kyoto University, Kyoto -85, Jaan; and c Deartment of Mathematics, Weizmann Institute of Science, Rehovot 7, Israel Edited by* Robion C. Kirby, University of California, Berkeley, Berkeley, CA, and aroved July 3, (received for review February 5, ) Let A be an m m comlex matrix with zero trace and let ε >. Then there are m m matrices B and C such that A ¼½B,CŠ and B C K ε m ε A where K ε deends only on ε. Moreover, the matrix B can be taken to be normal. It is well-known that a comlex m m matrix A is a commutator (i.e., there are matrices B and C of the same dimensions as A such that A ¼½B; CŠ ¼BC CB) if and only if A has zero trace. In such a situation clearly A B C where D denotes the norm of D as an oerator from l m to itself. Is it true that the converse holds? That is, if A has zero trace are there m m matrices B and C such that A ¼½B; CŠ and B C K A for some absolute constant K? Here we rovide a weaker estimate. Theorem. For every ε >othere is a constant K ε such that if an m m matrix A has zero trace then there are m m matrices B and C such that A ¼½B; CŠ and B C K ε m ε A. Moreover, the matrix B can be taken to be normal. The roof will be resented in the next section. It is self-contained excet for two facts. The first is a relatively easy result of Rosenblum () which gives a solution for X of the matrix equation A ¼ SX XT where all matrices are square and S and T have searated sectra in the sense that there is a domain D, whose boundary is a simle curve, which contains the sectrum of S and is disjoint from the sectrum of T. The solution then is, as follows: X ¼ Z ðzi SÞ AðzI TÞ dz: πı D which is a translate of ε Λ n and for each two of them their rojection on either the real or imaginary axis is ε searated. As we shall discuss below, every square matrix with zero trace is unitarily equivalent to a matrix with zero diagonal. It is thus enough to consider such matrices A. In our main result the matrix B can then be chosen to be a diagonal matrix. This fact is the reason for the definitions of μ and λ below. We do not know if one can get better results with more general B. Given a 4 n 4 n matrix A with zero diagonal denote by μðaþ the smallest number μ such that there is a diagonal matrix B with diagonal elements exactly the oints of Λ n and a 4 n 4 n matrix C such that A ¼½B; CŠ ¼BC CB and C μ. Note that because A has zero diagonal, for each diagonal matrix B with distinct diagonal entries fb i g such a matrix C exists and its nondiagonal entries are uniquely defined by c ij ¼ a ij ðb i b j Þ. Put also μð4 n Þ¼max μðaþ where the max ranges over all zero diagonal 4 n 4 n matrices of norm one. Similarly, for m not necessarily of the form 4 n, we denote by λðaþ the smallest number λ such that there is a diagonal matrix B with diagonal elements in ½ ; Š ½ ı; ıš and an m m matrix C such that A ¼½B; CŠ ¼BC CB and C λ. Put λðmþ ¼ max λðaþ where the max ranges over all m m matrices of zero diagonal and norm one. Given an m m, m ¼ 4 n, matrix A write it as a 4 4 block matrix with blocks of size 4 n 4 n A A A 3 A 4 A A A 3 A 4 B A 3 A 3 A 33 A 34 A : A 4 A 4 A 43 A 44 The second fact is a heavy theorem of Bourgain and Tzafriri () related to restricted invertibility of matrices and to the Kadison Singer conjecture. It is stated as Theorem in the sequel. We also remark in Claim 3 that if the statement of our main theorem, Theorem 3, which is a somewhat stronger version of the theorem above, holds with an absolute constant relacing K ε m ε, then the Kadison Singer conjecture holds. After two of us were led to this roblem while considering classification roblems for commutators in saces of oerators on Banach saces, one of us raised the roblem discussed here on MathOverFlow (htt://mathoverflow.net/questions/7345) Although the MathOverFlow discussion did not roduce a solution to the roblem, it did ut the authors in contact with one another and the discussion itself contains some useful tidbits. The Main Result Given < ε <, define a sequence of sets Λ n inductively: Λ is the set of 4 oints f ıg and Λ n ¼ Λ n ε ı ε : Note that Λ n is a subset of the square ½ ; Š ½ ı; ıš of cardinality 4 n and that it consists of a disjoint union of four sets each of Claim. In articular Also, and μðaþ max i4 μða iiþ A ε : μð4 n Þ μð4n Þ ε : λðaþ max i4 λða iiþ A ε λð4 n Þ λð4n Þ ε : [] Author contributions: W.B.J., N.O., and G.S. erformed research. The authors declare no conflict of interest. *This Direct Submission article had a rearranged editor. To whom corresondence should be addressed. gideon@weizmann.ac.il. PNAS November, 3 vol. no

2 Proof: Let B ii be diagonal matrices with diagonal entries in Λ n and C ii 4 n 4 n matrices with A ii ¼½B ii ;C ii Š and C ii ¼ μða ii Þ.Let fb ii g4 i¼ ¼ Ba b;a b a ε ıb ε I 4 n a;b¼ (the order doesn t matter), and, for i j, letcij be defined (uniquely) by A ij ¼ Bii C ij C ij B jj : Then by the result mentioned in the Introduction (see refs. or 3), Cij ¼ Z ðzi Bii πı Þ A i;j ðzi Bjj Þ dz; D ij where D ij is the boundary curve of any domain containing the sectrum of Bii and disjoint from the sectrum of Bjj. Because we can easily find such a curve of distance at least ε from the sectra of Bii and Bjj and of length 4 4ε < 8 we get that Cij < A ε ij. Let Cii ¼ ε C ii and set B B B ¼ B B33 A B44 and Then C ¼ðC ij Þ i;j¼;;3;4: C max μða ii Þ i;i ε A : Here the first summand on the right-hand side comes from the estimate of the diagonal art of the decomosition of C and the second from the rest (the constant is not imortant here but is easy to achieve by decomosing the off diagonal art of C into three matrices each of which is a ermutation of a diagonal block matrix). This estimate gives the claim for μ and the roof for λ is almost identical. Note that to estimate λð4 n Þ in terms of λð4 n Þ we only used the fact that A ii A. If we could get a better estimate on A ii we would get a better estimate on λð4 n Þ. This observation will be used in the roof of the main theorem below. In the roof of the main theorem we shall only use the arameter λ. The reason we also included μ here is that the matrices B in the roof for the roerty of μ deend only on ε and not on the matrices A. Otimizing over ε we get Corollary. (i) For each m there is an m m diagonal matrix B with sectrum in the square ½ ; Š ½ ı; ıš such that for each m m matrix A with diagonal zero there is an m m matrix C with norm at most Oððlog mþ 3 ffi m Þ A such that A ¼½B; CŠ. (ii) For each m ¼ 4 n there is a subset Λ m of ½ ; Š ½ ı; ıš such that for any trace zero m m matrix A there is a normal matrix B with sectrum Λ m and a matrix C with norm at most Oððlog mþ 3 ffi m Þ A such that A ¼½B; CŠ. Proof: For each < ε <, m of the form 4 n, and an m m matrix A with norm and zero diagonal, Claim gives, as long as μðm 4Þ, that ε ε ε μðmþ ε μðm 4Þ: Let k be the largest natural number smaller than log 4 m such that ε ε ε μðm 4k Þ. (If no such k exists take k ¼ log 4 m and change the argument below a bit, getting a better estimate.) Then μðmþ For ε ¼ k we get ε kμðm4 k Þ ε k μðm4 ðkþ Þ ε k ε 3 ε ε 3 ε k: ε μðmþ k 3 k 3 k k: Because k is at most log 4 m we get (i). To get (ii) use the fact (see e.g. refs. 4 or 5) that any trace zero matrix is unitarily equivalent to a matrix with zero diagonal. Remark : The ower of m in the first art of Corollary cannot be lowered. Indeed, if B is any m m diagonal matrix with sectrum in ½ ; Š ½ ı; ıš then there are i j in f; ; ;mg with ji jj 8 m.ifais the m m matrix with in the i; j lace and zero elsewhere and A ¼½B; CŠ, thenitiseasyto see that the absolute value of the i; j entry of C is at least m 8. Note that the constant ε in Eq. is what leads to the ower of m in the Corollary above. If we could relace it with ε we could eliminate the ower of m altogether and be left with only a log factor. The way we aroach this strengthening was hinted at above we make sure that the norms of the A ii jj are substantially smaller than that of A. Unfortunately, we can assure that only for half of the i-s (and we shall say more about it before Theorem ). The next Claim hints at how to deal with the full matrix once we take good care of half of it. We lace it here rather than just before it is used because its roof is similar to that of Claim. Claim shows that if a zero diagonal m m matrix A has its two m m central submatrices having substantially different λ values and the smaller one is substantially larger than the norm of the matrix, then λðaþ is, u to a multilicative constant close to, basically the same as the larger of these two values. This reasoning will be used in the roof of the main theorem. Claim. Let A ¼ A A A A be a m m matrix with zero diagonal where the A ij are all m m matrices. Assume also that λða ii Þ c i where c c < 4. Then λðaþ ð Kððc c Þ A c ÞÞc for some absolute constant K>. Proof: Write A ii ¼ B ii C ii C ii B ii, i ¼ ; where the B ii are diagonal matrices with sectrum in ½ ; Š ½ ı; ıš and C ii ¼ λða ii Þ c i. For any > δ c c ut 95 Johnson et al.

3 and B ¼ð δþi δb ;B ¼ δi ð δþb C ¼ δ C ;C ¼ð δþ C : Then A ii ¼ Bii C ii C ii B ii and the Bii -s are diagonal matrices with sectrum in ½ ; Š ½ ı; ıš. Moreover, the sectrum of B lies to the left of the vertical line Rz ¼ δ and that of B to the right of the vertical line Rz ¼ 4δ. Also max i¼; C ii maxfδ c ; ð δþ c g¼ c δ : Define Cij, i j ;, by A ij ¼ B ii C ij C ij B jj then, by the same argument as in the roof of Claim, using Rosenblum s result, C ij K A δ for some universal K. Define B ¼ B B and C ¼ C C C C then A ¼ B C C B, B is a diagonal matrix with sectrum in ½ ; Š ½ ı; ıš and C c δ K A δ : Taking δ ¼ðc c Þ we get that λðaþ ð Kððc c Þ A c ÞÞc for some absolute constant K (which, as a careful examination of the roof shows, can be taken to be 4 π). Claim together with the remark following the roof of Claim hints that to evaluate λðmþ in terms of λðmþ it is enough, given a m m matrix A with zero diagonal, to find an m m central submatrix and a 4 4 block decomosition of it so that the 4 central submatrices (corresonding to A ii in Claim ) have relatively small norms. To achieve this decomosition and similar ones we use a theorem of Bourgain and Tzafriri (): Theorem. For some absolute constant K>, ifa is an m m matrix with zero diagonal then for all ε > there is a central (i.e., whose diagonal is a subset of the diagonal of A) submatrix A of dimension ε m ε m whose norm is at most Kε A. Consequently, if A is a norm one 4 n 4 n matrix with zero diagonal then for all l n there are 4 l disjoint subsets σ i of ; ; ; 4 n each of size 4 n l such that all the submatrices corresonding to the entries in σ i σ i have norm at most K l. The last aragrah in the theorem above is a simle and known consequence of the first one choose σ, remove it from the set of rows and columns obtaining a somewhat smaller matrix [of size ð 4 n 4 n l Þ ð 4 n 4 n l ÞŠ. Choose σ from the set of rows (and columns) of this new matrix, remove it and continue in that manner 4 l times. This choice is ossible because all the submatrices from which we are choosing 4 n l 4 n l central submatrices are of size at least 4 n 4 n. We are now ready for the statement and roof of our main theorem, which easily imlies Theorem. Theorem 3. (i) For each ε > there is a constant K ε such that for all m λðmþ K ε m ε : (ii) For each ε > there is a constant K ε such that for all m and every m m zero trace matrix A there is a normal matrix B with sectrum in ½ ; Š ½ ı; ıš and a matrix C with norm at most K ε m ε A such that A ¼½B; CŠ. Proof: Let A be a 4 n 4 n matrix with zero diagonal and norm one. Let l n and let A be the 4 n 4 n submatrix corresonding to the entries in 4 l i¼ σ i 4 l i¼ σ i where σ i are given by Theorem. Let Aii l denote the submatrix corresonding to the entries in σ i σ i, i ¼ ; ; ; 4 l. Divide ; ; ; 4 l into 4 l disjoint sets each a union of 4 σ i -s and let Aii l, i ¼ ; ; ; 4 l, denote the 4 n l 4 n l submatrices corresonding to the entries corresonding to these sets. Continue in this manner to define Aii s, i ¼ ; ; ; 4s for each s ¼ ; ; ; ;l where for s A s ii is a 4n s 4 n s submatrix of one of the A s jj that A ¼ A. Now, By Claim for each ε >, λða Þ max i4 λða ii Þ ε max i λða ii Þ ε max λða l i4 l ii Þ l ε lλð4 n l ÞK l ε ;. Note where the last ste is the lace we use Theorem. Now use Corollary to get that for some absolute constants K (not necessarily the same in each row) lλð4 λða Þ K n l Þl K For ε ¼ l we get and taking l ¼ n gives lðn lþ 3 n l l λða Þ Kððn lþ 3 n l l 3 l Þ ε l ε : [] λða Þ Kn 3 n ¼ Kðlog mþ 3 m 4 : [3] We managed to reduce the ower of m in the bound on λðaþ from m to m 4 but only for a large submatrix. Next we are going to utilize Claim to get a similar bound for the whole matrix. Let σ c ¼f; ; ; 4 n g \ 4 l i¼ σ i and let A be the submatrix of A with entries in σ c σ c. Put c ¼ Kðlog mþ 3 m 4 and c ¼ maxfkðlog mþ 7 m 4 ; λða Þg. Then A, A ¼ A and A ¼ A satisfy the assumtions of Claim with c ;c. Consequently, Johnson et al. PNAS November, 3 vol. no

4 λðaþ ð Kðlog mþ Þ maxfkðlog mþ 7 m 4 ; λða Þg; where we continue to use K to denote a universal constant, ossibly different in different occurrences, and for m ¼ 4 n, n, λðmþ ð Kðlog mþ Þ maxfkðlog mþ 7 m 4 ; λðmþg: Reeating the argument again reducing from matrices of size 4 n 4 n to ones of size 4 n 4 n and combining with the above we get, for m ¼ 4 n, λð4mþ ð Kðlog mþ Þ maxfkðlog mþ 7 m 4 ; λðmþg: Let k m be the largest ower of 4 such that λðkþ Kðlog 4 kþ 7 k 4. Then λð4mþ log Y4 m s¼log 4 k ð Ks Þ Kðlog kþ 7 k 4 : For some other absolute constant K this last quantity is at most Kðlog mþ 7 m 4. We thus imroved the revious bound on λðmþ (for m ¼ 4 n )to λðmþ Kðlog mþ 7 m 4 for some absolute K. Reeating the argument one can imrove the bound further: Go back to Eq. and lug this new bound to get lðn λða Þ K lþ 7 ðn lþ l ε : For ε ¼ l we get and taking l ¼ n 3 gives λða Þ Kððn lþ 7 ðn lþ l 3 l Þ λða Þ Kn 7 n 3 ¼ Kðlog mþ 7 m : relacing Eq. 3 with this new estimate and following the rest of the argument above leads to λðmþ Kðlog mþ m : Iterating, this argument leads to a bounds of the form: λðmþ K k ðlog mþ 4k m k [4] for every m ¼ 4 n and every ositive integer k, where K k deends only on k. This inequality gives the statement of the theorem for m being a ower of 4. For a general m m zero diagonal matrix A, comlete it to a 4 n 4 n matrix A where 4 n < m 4 n by adding zero entries and keeing A suorted on f; ; ;mg f; ; ;mg. Aly the theorem to A and note that the fact that B is diagonal imlies that we can assume that C has nonzero entries only in f; ; ;mg f; ; ;mg. We thus roved the first art of the theorem. The second follows from the fact that any trace zero matrix is unitarily equivalent to a zero diagonal matrix. Concluding Remarks. Recall that the aving conjecture states that for every ε > there is a ositive integer nðεþ such that any norm one zero diagonal matrix has a aving of length at most nðεþ and norm at most ε. By a aving of A we mean a block diagonal submatrix of A whose diagonal is the same as that of A. The length of a aving is the number of blocks. Anderson () showed that this conjecture is equivalent to the Kadison Singer conjecture (7) on the extension of ure states. For a recent exository aer on these conjectures see ref. 8. It is clear from the roof above that if the aving conjecture holds with the right arameters then the roof can be simlified and the main result strengthened to get a olylog estimate on λðmþ. We next show that the reverse holds in a very strong sense. In articular if λðmþ is bounded indeendently of m then the aving conjecture holds. Claim 3. Assume A ¼½B; CŠ with B an m m diagonal matrix with sectrum in ½ ; Š ½ ı; ıš and C an m m matrix. Then for every < ε < A has a aving of length ε and norm ε C. Proof: Partition ½ ; Š into ε disjoint intervals I i of length at most ε each. Let Bði; jþ be the central (diagonal) submatrix of B whose diagonal entries are in I i ıi j, let Aði; jþ and Cði; jþ be the central submatrices of A and C; resectively, with the same suort as Bði; jþ. Aði; jþ, i; j ¼ ; ; ; ε, is a aving of A and it is enough to rove that Aði; jþ ε C. Clearly Aði; jþ ¼½Bði; jþ;cði; jþš. Pick i; j, let b be the center of the square I i ıi j and note that bi Bði; jþ [with I the identity matrix of the same dimensions as Bði; jþ] is a diagonal matrix with entries of absolute value at most ε. Therefore Aði; jþ ¼ ðbði; jþ biþcði; jþ Cði; jþðbði; jþ biþ ε C :. A more careful examination of the roof of Theorem 3 shows that the constant we get in Eq. 4 is λðmþ K k ðlog mþ 4k m k for some absolute constant K. Otimizing over k gives Kðlog log m log mþ λðmþ m for some absolute K. 3. It is quite easy to see that is also the best constant for K in the second aragrah of the introduction (assume A has zero diagonal and take B to be diagonal with diagonal elements and ). It also follows that λðþ ¼. We did not try to comute the best constants for other small values of the dimension. 4. Although the roblem we discuss seems basic enough not to need further motivation, we would like to indicate one. If any trace zero matrix A could be written as A ¼½B; CŠ with B C K A for a universal K, then we would get a simle characterization of the commutators in an imortant class of II factors, the Wright factors; an element there would be a commutator if and only if it has zero trace. See ref. 9 for this and related matters. 5. One can ask similar questions to the one addressed here for norms other than the oerator norm. In articular, what is the (order of) the best constant in B C HS K A HS where A ranges over all trace zero m m matrices, A ¼ BC CB and HS denotes the Hilbert Schmidt norm? We checked that a roof with a similar idea but much simler gives K ¼ Oððlog mþ 3 Þ. In this case we can also rove a lower bound for K of order ðlog mþ. This roof uses a quantitative version of an argument from ref.. Details will aear elsewhere Johnson et al.

5 ACKNOWLEDGMENTS. W.B. Johnson was suorted in art by National Science Foundation Grant DMS-3 and US-Israel Binational Science Foundation. N. Ozawa was suorted in art by Jaan Society for the Promotion of Science.. Rosenblum M (95) On the oerator equation BX XA ¼ Q. Duke Math J 3:3 9.. Bourgain J, Tzafriri L (99) On a roblem of Kadison and Singer. J Reine Angew Math 4: Marcoux L-W () Sums of small number of commutators. J Oerat Theor 5: Fillmore P-A (99) On similarity and the diagonal of a matrix. Am Math Mon 7: Halmos P-R (98) A Hilbert sace roblem book. Graduate Texts in Mathematics (Sringer, New York), nd Ed. G. Schechtman was suorted in art by US-Israel Binational Science Foundation. Ozawa and Schechtman were articiants of the National Science Foundation Worksho in Analysis and Probability, Texas A&M University.. Anderson J (979) Extensions, restrictions and reresentations of states on C -algebras. Trans Am Math Soc 49: Kadison R, Singer I (959) Extensions of ure states. Am J Math 8: Casazza P-G, Edidin D (7) Equivalents of the Kadison-Singer roblem. Function saces. Contem Math, : Dykema K, Skrika A () On single commutators in II -factors. Proc AMS 4: Dykema K, Figiel T, Weiss G, Wodzicki M (4) Commutator structure of oerator ideals. Adv Math 85: 79. Johnson et al. PNAS November, 3 vol. no