Almost commuting matrices with respect to normalized Hilbert-Schmidt norm

Abstract

Almost-commuting matrices with respect to the normalized Hilbert-Schmidt norm are considered. Normal almost commuting matrices are proved to be near commuting. Comment: 11 pages

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arXiv:1002.3082v1 [math.AG] 16 Feb 2010
Almost commuting matrices with respect to normalized
Hilbert-Schmidt norm.
Lev Glebsky
February 16, 2010
Abstract
Almost-commuting matrices with respect to the normalized Hilbert-Schmidt norm are consid-
ered. Normal almost commuting matrices are proved to be near commuting.
1 Introduction
Let Cn×n be the set of complex n × n-matrices. Let Hn,Un,Nn ⊂ Cn×n be the sets of self-adjoin
(Hermitian), unitary and normal matrices, correspondingly. For X,Y ∈ Cn×nlet [X,Y ] = XY −Y X.
The following problem is classic
Problem 1. (Must almost commuting matrices be nearly commuting?)
Let Sn= Cn×n,Hn,Unor Nn.
For each ǫ > 0, is there a δ = δ(ǫ) > 0 such that for each positive integer n, if A,B ∈ Sn with
?A?,?B? ≤ 1 and ?AB−BA? ≤ δ, then there exist˜A,˜B ∈ Snwith [˜A,˜B] = 0 and ?A−˜A?,?B−˜B? ≤
ǫ?
Here δ = δ(ǫ) is independent of n; the non-uniform version of the problem (δ = δ(n,ǫ)) has
affirmative answers [1, 7, 9]. In spite of the equivalence of norms for finite dimensional spaces, the
answer on the uniform problem depends on the norms ? · ?n: Cn×n→ R. Indeed, the equivalence
may be non-uniform with respect to n. In the series of works [2, 4, 5, 6, 11] the complete answer on
Problem 1 have been found for ? · ? = ? · ?op. Where
?A?op= sup{?Ax? : ?x? = 1}.1
The answer on Problem 1 is affirmative for Sn= Hnand negative for all other cases (?·? = ?·?op). We
didn’t know any results for other norms. In the present paper we consider Problem 1 for ?·? = ?·?tr,
the normalized Hilbert-Schmidt norm:
?A?tr=
?
?
?
?1
n
n
?
j,k=1
|Aj,k|2.
Our interest in normalized Hilbert-Schmidt norm arises from its use in factor II von Neumann algebras
and hyperlinear groups, [8, 10]. It turns out that the normalized Hilbert-Schmidt norm is more friendly
for Problem 1. We manage to prove that the answer is affirmative for Sn= Hn,Un,Nn, even if we speak
about several almost-commuting matrices. The reason why it is true is the following. Transform the
matrix A into its diagonal form. In this basis one can approximate ?·?tr-almost commuting matrices
A,B by block diagonal matrices˜A,˜B such that all blocks of˜A are multiples of unit matrices.
1We consider Cnas a Hilbert space with scalar product (x,y) =
?x? =
?(x,x).
?x∗
iyi. It defines the Hilbert norm on Cn,
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The technique of the paper is elementary. We systematically use that the squire of the normalized
Hilbert-Schmidt norm of a block diagonal matrix is a convex combination of the squires of the norms
of its blocks and concavity of some estimates, see Section 3 for details.
All estimates in the theorems are given in the form “? · ? < Cǫα”, where C is an integer. We
do not try to optimize the values of C, we just have decided that using some proper numbers is less
awkward than the use of expression of the type “there exists C > 0 such that...”.
2 Notations and inequalities
We consider Cnas a Hilbert space with the scalar product (x,y) =?x∗
include C ⊂ Cn×nby constant diagonal matrices. So, 1 ∈ Cn×n, some times 1n∈ Cn×ndenotes the
unit matrix. For A = {Ai,j} ∈ Cn×nwe define the normalized trace
iyi. It defines the Hilbert
norm on Cn, ?x? =
?(x,x). Set Cn×nof complex n×n matrices naturally acts on Cn. As usual, we
tr(A) = 1/n
n
?
i=1
Ai,i.
It defines a scalar product on Cn×n:
?A,B? = tr(A∗B) =1
n
?
i,j
A∗
ijBij
and the normalized trace norm (normalized Hilbert-Schmidt norm)
?A?tr=
?
?A,A? =
??
i,j
|Aij|2
We also need the uniform operator norm
?A?op= sup{?Ax? : ?x? = 1}
We list some useful well-known inequalities, see [8], in the following:
Lemma 1.1. |?A,B?| ≤ ?A?tr?B?tr (the Cauchy-Schwarz inequality); substituting 1 → B gives
|tr(A)| ≤ ?A?tr;
2. ?A + B?tr≤ ?A?tr+ ?B?tr
3. ?AB?tr≤ ?A?op?B?trand ?BA?tr≤ ?A?op?B?tr
4. ?A?tr≤ ?A?op≤√n?A?tr
5. If P is an orthogonal projector on k-dimensional subspace, then ?P?tr=
6. If a matrix A is of rank k, then there exists an orthogonal projector P of rank k such that
?
7. ?A?2
Remark 1. ?·?opis an algebraic norm: ?AB?op≤ ?A?op?B?op, but normalized trace norm is not a
good algebraic norm. We have only ?AB?tr≤√n?A?tr?B?tr(it follows from 3,4 of Lemma 1).
Remark 2. It is well known and easy to check that for unitary matrices U,V ?UXV ?i= ?X?i, where
i = op or tr. So, the norms ? · ?idefine norm on the set of linear operators from a Hilbert space to
another Hilbert space.
√k
√n.
A = PA and, by items 3,5 ?A?tr≤
k
n?A?op.
op= ?A∗A?op; ?A?2
tr= ?A,A? = ?1,A∗A? ≤ ?A∗A?tr
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The following lemma says that if ?A?tris small then there is an orthogonal projector P of a large
rank, such that ?PA?opis small. Precisely,
Lemma 2. For any A ∈ Cn×nthere exists an orthogonal projector P such that
• ?E − P?tr<??A?tr,
• If A is normal, then, in addition, AP = PA.
Proof. Observe that ?A?op=??A∗A?op=??AA∗?opand AA∗is positive. Let 0 ≤ λ1≤ λ2≤ ... ≤
?PA?op<
??A?tr
λnbe the eigenvalues of AA∗. So,
?A?2
tr= tr(AA∗) =1
n
?
i
λi≥δ2
n|{i | λi≥ δ2}|.
So, we have |{i | λi≥ δ2}| ≤ n?A?2
all eigenvectors of AA∗with λi< δ2. Then ?PA?op=??PAA∗P?op< δ and ?(E−P)?tr≤ ?A?tr/δ.
of P.
tr/δ2. Let Pδbe the orthogonal projector on the space spanned by
Putting δ =??A?trproves the first part of the lemma. The second part easily follows by construction
3 The concave estimate principle
We will need the following
Claim 1. Let φ : R+→ R+be a concave increasing function with φ(0) = 0. Then x → φ2(√x) is a
concave increasing function.
Proof. On the set of functions R+→ R+we define an operation T: (Tφ)(x) = φ2(√x). It is clear
that if φ1≤ φ2then Tφ1≤ Tφ2. Here φ1≤ φ2if φ1(x) ≤ φ2(x) for all x ∈ R+. For all x0∈ R+there
exists α,β ≥ 0 such that φ(x0) = αx0+ β and φ(x) ≤ αx + β. (We have used here that φ(0) = 0.)
Observe that T(αx + β) = α2x + β2+ 2αβ√x is concave. So, for any x ∈ R+there exists a concave
fx, such that Tφ(x) = fx(x) and Tφ ≤ fx. We deduce that Tφ is concave.
Let P ∈ C[x1,x2,...,xk,x∗
preted as complex conjugate to xi).
1,x∗
2,...,x∗
k] (polynomial with complex coefficients, where x∗
iis inter-
Definition 1. 1. We say that matrices A1,..,Akare an ǫ-solution of P (ǫ-satisfy P) if
?P(A1,...,Ak,A∗
1,...,A∗
k)?tr≤ ǫ
2. Let δ : R+→ R+be such that lim
A1,A2,...,Ak∈ Cn×nof P there exists an exact solution˜A1,˜A2,...,˜Ak∈ Cn×nof P δǫ-close to
A1,...Ak, that is ?Ai−˜Ai?tr< δǫfor i = 1,...,k. Polynomial P is called stable if it is δǫ-stable
for some δǫ, lim
ǫ→0δǫ= 0. Polynomial P is called δǫ-stable if for any ǫ-solution
ǫ→0δǫ= 0. (Note that δǫis independent of n.)
Let I = ?I1,I2,...,Ir? be an ordered partition of {1,...,n}. A matrix A is said to be I-block
diagonal (or just I-diagonal) if nonzero elements of A appear on Ij×Ijplaces only. (It is clear that
this is a usual block-diagonal matrix, after conjugation with a permutation.) Similarly, we call matrix
A cyclically I-three-diagonal if nonzero elements of A appear on Ij⊕1× Ij, Ij× Ijand Ij× Ij⊕1
places only. Here ⊕ is the sum
Solution (ǫ-solution) A1,...,Ak of a polynomial P is called I-diagonal solution (I-diagonal ǫ-
solution), if all matrices A1,...,Akare I-diagonal. Under our assumptions, P(A1,...,A∗
if all Aiare.
mod r.
k) is I-diagonal
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Lemma 3. Suppose, that a polynomial P is δ(ǫ)-stable with a concave δ(ǫ). Then for any I-diagonal
ǫ-solution of P there exist an I-diagonal solution of P that is δ(ǫ)-close to this ǫ-solution.
Proof. The proof uses the following facts:
1.
?
d1,...,dr, correspondingly. Then
jαjδ(xj) ≤ δ(?
2. Let A be an I-diagonal matrix and A1,A2,··· ,Arits diagonal components of dimensions
jαjxj) for xj,αj≥ 0,?
jαj= 1 and concave δ.
?A?2
tr=
?
j
dj
n?Aj?2
dj,
where n = d1+ d2+ ··· + drand ? · ?dis the normalized Hilbert-Schmidt norm on Cd×d.
Let A1,...,Akbe an I-diagonal solution of P with diagonal components Al
We have ǫ2≥?dl
?˜Al
Solution˜Ajis constructed by blocks˜Al
j. Let ?P(Al
1,...,Al
k)? = ǫl.
nǫ2
l. There exists a solution˜Al
1,...,˜Al
kof P with
√
ǫ2).
j− Al
j?2
tr≤ δ2(
j. Now
?˜Aj− Aj?2
tr=
?
l
dl
n?˜Al
j− Al
j?2≤
?
l
dl
nδ2(
?
ǫ2
l) ≤ δ2
???
l
dl
nǫ2
l
?
≤ δ2(ǫ).
Here we use concavity of δ2(√x) by Claim 1.
4Almost unitary matrices are near unitary
Lemma 4. Let B : L1→ L2 be an unitary operator from a Hilbert space L1 to a Hilbert space L2
such that ?B∗B − 1L1?op≤ ǫ ≤ 1/3. Then there exists an unitary operator V : L1→ L2 such that
?B − V ?op< 2ǫ.
Proof. Just take V = B(B∗B)−1/2, where (B∗B)−1/2= (1L1− X)−1/2=
denoted X = B∗B − 1.) Make the following estimates
∞
?
j=0
(2j)!Xj
22j(j!)2. (We have
?(B∗B)−1/2− 1?op≤
∞
?
j=1
(2j)!ǫj
22j(j!)2<
∞
?
j=1
ǫj=
ǫ
1 − ǫ.
Now,
?V − B?op≤ ?B?op?(B∗B)−1/2− 1?op<ǫ(ǫ + 1)
1 − ǫ
≤ 2ǫ,
for ǫ ≤ 1/3. (We were using the fact that ?B?op =
1?op+ 1.)
Lemma 5. Let ?A∗A − 1?tr ≤ ǫ ≤ 1/3 for a matrix A. Then there exists a unitary U, such that
?A − U?tr≤ 5ǫ
Proof. By Lemma 2 there exists orthogonal projector P, ?1−P?tr≤√ǫ such that ?PA∗AP −P?op≤
√ǫ. Let X = ImP. Consider the restriction B = A|X: X → Y = A(X), then B∗: Y → X. Observe
??B∗B?op ≤
??B∗B − 1?op+ 1 ≤ ?B∗B −
1
4 and ?A − U?tr≤ (3 + ?A?op)√ǫ.
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that B∗= PA∗|Y. So, ?B∗B −1X?op≤√ǫ and, by Lemma 4, there exists a unitary V : X → Y with
?V − B?op< 2√ǫ. Let˜V be any unitary operator from X⊥to Y⊥. Take U = V ⊕˜V . We estimate:
?A − V ⊕˜V ?tr≤ ?PYAPX− PYV PX?tr+ ?AP⊥
?PYAPX− PYV PX?op+ ?˜V ?op?P⊥
Now, the first inequality of the lemma follows from
X?tr+ ?P⊥
X?tr≤ 3√ǫ + ?AP⊥
Y˜V P⊥
X?tr≤
X?tr+ ?AP⊥
X?tr.
?AP⊥
X?2
tr≤ ?P⊥
XA∗AP⊥
√ǫ + ?PX?2
X?tr≤ ?P⊥
op?A∗A − 1?tr≤√ǫ + ǫ.
X?tr+ ?P⊥
XA∗AP⊥
X− P⊥
X?tr≤
The second inequality of the lemma follows from
?AP⊥
X?tr≤ ?A?op?P⊥
X?tr≤ ?A?op
√ǫ.
We will work with I-diagonal matrices, so we need a global concave estimate.
Corollary 1. Let ?A?op ≤ 3.
6??A∗A − 1?tr.
Proof. For ?A∗A−1?tr≤ 1/3 it is Lemma 5. Further, ?A−V?tr≤ ?A?tr+1 and ?A?2
?A∗A−1?tr+1. So, ?A−V ?tr≤??A∗A − 1?tr+ 1+1. It remains to check that 6√x >√x + 1+1
Then there exists a unitary matrix V such that ?A − V ?tr ≤
tr≤ ?A∗A?tr≤
for x > 1/3.
5 Almost commuting unitary matrices are near commuting
Theorem 1. Let U1and U2be unitary matrices. Then there exists unitary matrices A1,A2, [A1,A2] =
0 such that ?U1− A1?tr≤ 30(?[U1,U2]?tr)1/9and ?U2− A2?tr≤ 30(?[U1,U2]?tr)1/9. In addition,
[A1,U1] = 0.
Before the proof of the theorem we consider an example where U2is a cyclic permutation and U1
its diagonal form: U1= diag(w,w2,...,wn= 1) with w = exp(2πi
n) and U2= Pnwith
Pj,k=
?
1
0
if
overwise
j = k + 1 mod n
.
This is a counterexample to Problem 1 for ? · ? = ? · ?opand Sn= Unfound by Voiculescu, [6] (he
proves that it is indeed a counterexample). One has ?[U1,U2]?op= ?[U1,U2]?tr= |1 − w| → 0 when
n → ∞. Suppose, for simplicity, that n = md for large m and d. Then we can take as A1 and
A2the following block-diagonal matrices: A1= diag( ˜ w1d, ˜ w21d,..., ˜ wm1d), where ˜ w = exp(2πi
A2= diag(Pd,Pd,...,Pd).
m) and
Proof. After transformation U1→ V−1U1V , U2→ V−1U2V , assume that U1= diag(α1,α2,....,αn).
The main idea of the proof is the following. Change some elements of U2by 0 and approximate U1
by a diagonal matrix with spectrum exp(2πij
m) for proper m in such a way that U1and U2become
block diagonal matrices with all blocks of U1 being multiples of unit matrices. New U1and U2are
commuting, but now U2is not unitary. Approximate U2by an unitary matrix, conserving its block
structure. It can be done using Corollary 1 and Lemma 3. Let us describe this procedure in details.
Let ?[U1,U2]?tr= ǫ. Take a positive integer t ≥ 6 that will be optimized latter. Let w = exp(2πi
Let |1 − w| = ∆. One has
6/t ≤ ∆ ≤ 2π/t < 7/t.
t).
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1. Let U2= {ujk}. Define˜U2= {˜ ujk} by the following rule:
˜ ujk=
?
ujk
0
if
if
|αj− αk| < ∆
|αj− αk| ≥ ∆
One has
?˜U2− U2?tr≤ ǫ/∆
Indeed,
nǫ2≥ n?[U1,U2]?2=
n,n
?
j=1,k=1
|αk− αj|2|ujk|2≥ ∆2
?
|αk−αj|≥∆
|ujk|2= ∆2n?˜U2− U2?2
tr
One can check that ?[U1,˜U2]? ≤ ?[U1,U2]? ≤ ǫ.
2. Approximate U1by a diagonal matrix˜U1with spectrum {wj: j = 0,1,...,t − 1}. Precisely,
let I = {I0,I1,...,It−1} with Ij= {l : αl∈ (wj−1/2,wj+1/2]}, where (x,y] is a semiopen arc
of the unit circle in C. Then˜U1is an I-diagonal matrix with j block Uj
?˜U1− U1?tr≤ ∆
and ?[˜U1,˜U2]?tr≤ ǫ + 2∆. Observe that˜U2is a cyclically I-three-diagonal matrix.
3. Fix another parameter a ∈ N that will be optimized latter. One can find S = {s0,s2,...,sc−1} ⊂
{0,1,2,...,t − 1} such that
• a ≤ |sr⊕1− sr| ≤ 3a, for any r = 0,1,...,c − 1. Here ⊕ is the sum
• |Ij| ≤n
• |S| = c ≤ t/a
4. In order to construct A1we make a more rough partition˜I = {˜I0,˜I2,...,˜Ic−1}. Where˜Ij =
Isj∪Isj+1∪···∪Isj⊕1−1, where ± is
matrix with the blocks Aj
1= w
1= wj. One has that
mod c.
afor any j ∈ S.
mod t and ⊕ is
2(sj+sj⊕−1). By the first item of 3) we have
mod c. Now, A1is a cyclically˜I-diagonal
1
?˜U1− A1?tr≤ |1 − w
3
2a| ≤3
2a∆
and ?[A1,˜U2]?tr≤ ǫ + 2∆ + 3a∆.
5. We give our construction of A2in two steps. Recall that˜U2is I-three-diagonal. We construct
B by removing from˜U2the blocks Ij−1×Ijand Ij×Ij−1for each j ∈ S. The resulting matrix
B is˜I-diagonal and, consequently, [A1,B] = 0. We estimate:
?˜U2− B?tr=
?
j∈S
(?Uj−1,j
2
?tr+ ?Uj,j−1
2
?tr) ≤ 2|S|
√a≤ 2
t
a3/2.
For the first inequality we use ?Uj−1,j
?Uj−1,j
the matrix). The same inequalities are valid for ?Uj,j−1
of S.
2
?tr ≤
?
|Ij|
n?Uj−1,j
2
?op (the item 6. of Lemma 1) and
2
?op≤ ?U2?op= 1 (the operator norm of a submatrix is less than the operator norm of
2
?. The second inequality is a property
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6. The matrix B,A1 are˜I-diagonal and each block of A1 is a multiple of the unit matrix, so
[A1,B] = 0. The problem is that B is not unitary. ?U2−B?tr≤
and ?B?op ≤ 3 (B is I-three-diagonal with the operator norm of each block less than 1 as
submatrices of a unitary matrix.) It follows that
ǫ
∆+2
t
a3/2≤ ǫt/6+2ta−3/2= γ.
?B∗B − 1?tr= ?B∗B − U∗
≤ ?B∗?op?B − U2?tr+ ?U2?op?B∗− U∗
The matrix B is an˜I-diagonal matrix. By Lemma 3 and Corollary 1 there exists a unitary
˜I-diagonal matrix A2with
2U2?tr≤ ?B∗B − B∗U2?tr+ ?B∗U2− U∗
2U2?tr
2?tr≤ 4γ.
?B − A2?tr≤ 12√γ = 12
?ǫt
6+ 2ta−3/2
It is clear that [A1,A2] = 0.
7. We only need to choose a, t and estimate ?Ui−Ai?tr. Suppose, for a moment2, that ǫ ≤ 6−9/7,
choose a,t ∈ N such that ǫ−7/9≤ t ≤ 2ǫ−7/9and ǫ−2/3≤ a ≤ 2ǫ−2/3. We have:
?U1− A1?tr≤ ∆ +3
2a∆ ≤7
t+21a
2t
≤ 7ǫ7/9+ 21ǫ1/9< 30ǫ1/9.
Further,
?U2− A2?tr≤1
6ǫt + 2ta−3/2+ 12
?1
6ǫt + 2ta−3/2≤13
3ǫ2/9+ 12
?13
3ǫ1/9< 30ǫ1/9.
For ǫ ≥ 6−9/7we have
?Ai− Ui?tr≤ 2 ≤ 30 · 6−1/7≤ 30ǫ1/9
The pair A1,A2satisfies the statement of the theorem.
We need the following
Claim 2. Let ?A?op,?B?op,?˜A?op,?˜B?op≤ 1. Then
?[˜A,˜B]?tr≤ ?[A,B]?tr+ 2(?A −˜A?tr+ ?B −˜B?tr).
Proof. ?AB −˜A˜B?tr = ?AB − A˜B + A˜B −˜A˜B?tr ≤ ?A?op?B −˜B?tr+ ?A −˜A?tr?˜B?op ≤ ?B −
˜B?tr+ ?A −˜A?tr. Combining it with the same estimate for ?BA −˜B˜A?trwe get the claim.
Theorem 2. There exists δ(ǫ,k), δ(ǫ,k) → 0 when ǫ → 0 for any k ∈ N, such that if ?[Ui,Uj]?tr≤ ǫ
for unitary U1,U2,...,Uk, then there exist pairwise commuting unitary matrices A1,...,Aksuch that
?Uj− Aj?tr≤ δ(ǫ,k).
Proof. Let ψ(x) = 30x1/9and φj(·) be defined by the relation:
φ0(x) = x,φj+1(x) = 4ψ(φj(x)) + x.
For r = 1,...k − 1 we prove by induction the following statement
There exist a unitary matrix V , a partition Irof {1,...,n}, and Ir-diagonal matrices˜U1,˜U2,...,˜Uk,
such that
• All blocks of˜U1,...˜Urare multiples of the unit matrix.
2The condition on ǫ is to guarantee t ≥ 6.
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• ?˜Uj− V−1UjV ?tr≤ ψ (φj−1(ǫ)), for j ≤ r and ?˜Uj− V−1UjV ?tr≤ ψ (φr−1(ǫ)), for j > r.
The theorem follows from the Statement for r = k − 1 and δ(ǫ,k) = ψ (φk−1(ǫ)). Let us proof the
Statement.
r = 1. In the proof of Theorem 1 matrix A1and partitions I and˜I is independent of U2. The
construction of A2depends on partitions I and˜I only. So, we may construct˜I-diagonal˜U1,˜U2,...,˜Uk
satisfying the Statement for r = 1.
r → r + 1. Let˜U1,...,˜Ukbe as in the Statement. Then ?[˜Ui,˜Uj]? = 0 for i < r and, by Claim 2,
?[˜Ui,˜Uj]? ≤ φr(ǫ) for i,j ≥ r. Let Il∈ Ir. Work with˜Ul
apply Lemma 3.
r,...,˜Ul
kas in the proof for r = 1. Then
6 Self-adjoint matrices.
For every almost-commuting self-adjoint matrices A,B we construct commuting self-adjoint matrices
with the same operator norm and close to A,B by the normalized Hilbert-Schmidt norm. In order to
preserve the operator norm we need
Lemma 6. Let A, B, C = A + B be self adjoint matrices. Let D(B) and D(C) be the decreasing
diagonal form of B and C, correspondingly. Then ?D(C) − D(B)?op≤ ?A?op
Proof. Let α1≥ α2≥ ··· ≥ αn, β1≥ ··· ≥ βnand γ1≥ ··· ≥ γnbe (ordered) eigenvalues of A,B and
C, correspondingly. The H.Weyl inequality [3, 12] states:
γj+k−1≤ αj+ βk.
Writing −C = −A − B and reordering the eigenvalues we get:
γk−j+1≥ αn−j+1+ βk.
Putting j = 1 in the both inequalities and using the fact that α1,−αn≤ ?A?opwe get
−?A?op+ βk≤ γk≤ ?A?op+ βk.
Corollary 2. Let A,C be self-adjoint and C be I-diagonal. Then there exists I-diagonal self-adjoint
matrix˜C such that ?˜C?op≤ ?A?opand ?˜C − C?tr≤ ?C − A?tr
Proof. We can choose˜C such that D(˜C) = D(C) − D(C − A).
Theorem 3. Let H1 and H2 be self-adjoint matrices, such that ?Hi?op ≤ 1, i = 1,2. Then there
exists self-adjoint matrices A1,A2, [A1,A2] = 0 such that ?H1− A1?tr ≤ 12(?[H1,H2]?tr)1/6and
?H2− A2?tr≤ 12(?[H1,H2]?tr)1/6, ?Ai?op≤ 1. In addition, [A1,H1] = 0.
Proof. We follow the same routine as in the proof of Theorem 1. Instead of Lemma 5 we use Corollary 2
to keep the operator norm.
Let ?[H1,H2]?tr= ǫ We suppose that H1= diag(α1,α2,....,αn), −1 ≤ α1≤ α2≤ ··· ≤ αn≤ 1.
Take a positive integer t that will be optimized latter.
1. Let H2= {hjk}. Define˜H2= {˜hjk} by the following rule:
?
As in the proof of Theorem 1 one has
˜hjk=
hjk
0
if
if
|αj− αk| < 1/t
|αj− αk| ≥ 1/t
?˜H2− H2?tr≤ ǫt.
Clearly,˜H2is self-adjoin.
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2. Approximate H1 by a diagonal matrix˜H1 with spectrum {j
I = {I−t,I−t+1,...,It} with Ij= {l : αl∈ (2j−1
Then˜H1is an I-diagonal matrix with the j-th block Hj
t
: j = −t,...,t}. Precisely, let
2t]}, where (x,y] is a semiopen interval.
1=j
2t,2j+1
t. One has that
?˜H1− H1?tr≤1
t
and that˜H2is an I-three-diagonal matrix (not cyclically I-three-diagonal).
3. Fix another parameter a ∈ N that will be optimized latter. One can find S = {s0,s2,...,sc−1},
−t ≤ s1< s1< ...sc−1≤ t such that
• a ≤ |sr+1− sr| ≤ 2a, for any r = 0,1,...,c − 1.
• |Ij| ≤n
• |S| = c ≤ (2t + 1)/a
4. In order to construct A1 we make more rough partition˜I = {˜I0,˜I2,...,˜Ic−1}. Where˜Ij =
Isj∪ Isj+1∪ ··· ∪ Isj+1−1. Now, A1is cyclically˜I-diagonal matrix with block Aj
By the first item of 3) we have
?˜H1− A1?tr≤a
afor any j ∈ S.
1=
sj+sj+1−1
2t
.
t
5. We give construction of A2in two steps. Recall that˜H2I-three-diagonal. We construct B by
removing from˜H2 blocks Ij−1× Ij and Ij× Ij−1 for each j ∈ S. The resulting matrix B is
˜I-diagonal and, consequently, [A1,B] = 0. We estimate:
?˜ H2− B?tr=
?
j∈S
(?Hj−1,j
2
?tr+ ?Hj,j−1
2
?tr) ≤ 2|S|
√a≤ 22t + 1
a3/2.
For the first inequality we use ?Hj−1,j
?Hj−1,j
the matrix). The same inequalities are valid for ?Hj,j−1
of S.
2
?tr ≤
?
|Ij|
n?Hj−1,j
2
?op (the item 6. of Lemma 1) and
2
?op≤ ?H2?op≤ 1 (the operator norm of a submatrix is less than the operator norm of
2
?. The second inequality is a property
6. The matrix B is˜I-diagonal, self-adjoint, and
?H2− B?tr≤ ǫt + 22t + 1
a3/2.
So by Corollary 2 there exists˜I-diagonal self-adjoint A2with
?H2− A2?tr≤ 2ǫt + 42t + 1
a3/2.
It is clear that [A1,A2] = 0.
7. We only need to choose a, t and estimate ?Hi− Ai?tr. Suppose, for a moment, that ǫ ≤ 4−1,
choose a,t ∈ N such that1
?H1− A1?tr≤1
2ǫ−5/6≤ t ≤ ǫ−5/6and ǫ−2/3≤ a ≤ 2ǫ−2/3. We have:
t+a
t≤ 2ǫ5/6+ 4ǫ1/6< 12ǫ1/6
Further,
?H2− A2?tr≤ 2ǫ1/6+ 8ǫ1/6+ 4ǫ < 12ǫ1/6
For ǫ ≥ 4−1we have
?Ai− Hi?tr≤ ?Ai− Hi?op≤ 2 < 12(4−1/6) < 12ǫ1/6
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The pair A1,A2satisfies the statement of the theorem.
Theorem 4. There exists δ(ǫ,k), δ(ǫ,k) → 0 when ǫ → 0 for any k ∈ N, such that if ?[Hi,Hj]?tr≤ ǫ
for self-adjoint matrices H1,H2,...,Hk with ?Hi?op≤ 1, then there exist pairwise commuting self-
adjoint matrices A1,...,Aksuch that ?Uj− Aj?tr≤ δ(ǫ,k) and ?Ai? ≤ 1.
Proof. The same as for Theorem 2.
7 Normal matrices
Observe that Theorem 4 implies the existence of commuting normal matrices close to almost com-
muting ones. Observe also, that Theorem 3 implies the existence of a normal matrix N close to an
? · ?tr-almost normal matrix M. Could it be done in a way that ?N?op≤ ?M?op? In the section we
give the affirmative answer to this question (Corollary 3)
Theorem 5. Let U and H be unitary and positive matrices, correspondingly. Let ?H?op≤ 1. Then
there exists unitary and positive matrices V,A such that [V,A] = [H,A] = 0, ?V −U?tr≤ 30?[U,H]?1/9
and ?H − A?tr≤ 30?[U,H]?1/9
Proof. Let H = diag(h1,...,hn). Make partition I and˜I as in the proof of Theorem 3. Construct A
as A1in Theorem 3 and V as U2in Theorem1.
tr
tr, ?A?op≤ 1.
Lemma 7. Let A,B be positive commuting matrices. Then ?A − B?tr≤
Proof. Without loss of generality we may assume that A = diag(a1,a2,...,an) and B = diag(b1,b2,...,bn).
Now,
tr=1
n
j=1
??A2− B2?tr.
?A − B?2
n
?
(aj− bj)2≤
(a)
1
n
n
?
j=1
|a2
j− b2
j| =
1
n
n
?
j=1
?
(a2
j− b2
j)2≤
(b)
?
?
?
?1
n
n
?
j=1
(a2
j− b2
j)2= ?A2− B2?tr
The inequality (a) is due to (a −b)2≤ |a2− b2| for a,b ≥ 0; the inequality (b) is due to concavity of
√·.
Theorem 5 with Lemma 7 implies
Corollary 3. Let M be a matrix with ?MM∗− M∗M?tr≤ ǫ and ?M?op≤ 1. Then there exists a
normal matrix N such that ?M − N?tr≤ 36ǫ1/18and ?N?op≤ 1.
Proof. Let M = UH with unitary U and positive H.
Theorem 5 we can find positive A and unitary V such that ?H2−A?tr≤ 30ǫ1/9, ?U −V ?tr≤ 30ǫ1/9
and [H2,A] = [V,A] = 0. By Lemma 7 we have ?H − A1/2? ≤ 6ǫ1/18and N = V A1/2satisfies the
Corollary.
We have ?UH2− H2U?tr ≤ ǫ. So, by
8Concluding remarks
We see that the normalized Hilbert-Schmidt norm is more friendly for almost-near questions for the
commutator. We think that it is interesting to consider other relations. For example, if almost
solutions of
Uk= V−1UV
are near solutions?
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Acknowledgement Some questions answered in the paper arise during my talks at “Nonstandard
Analysis” seminar at Urbana-Champaign. I am thankful to E. Gordon, P. Leob and W. Henson who
was listening my messy talks and gave useful suggestions. I think that I have had much more benefits
from the talks then they had.
The “design” of the introduction is almost copied from [2].
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