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Norms of commutators of self-adjoint operators
J. Math. Anal. Appl. 342 (2008) 747–751 www.elsevier.com/locate/jmaa
Note
Norms of commutators of self-adjoint operators ✩ Yue-Qing Wang, Hong-Ke Du ∗ College of Mathematics and Information Science, Shaanxi Normal University, Xi’an 710062, PR China Received 28 September 2007 Available online 26 December 2007 Submitted by J.A. Ball
Abstract In this note, the estimate of norms of commutators of self-adjoint operators is established. © 2007 Elsevier Inc. All rights reserved. Keywords: Norm inequality; Commutator; Self-adjoint operator
Throughout this paper, H denotes a Hilbert space. The set of all bounded linear operators acting on H is denoted by B(H). An operator A is said to be self-adjoint if A = A∗ . An operator A is said to be positive if Ax, x 0 for each vector x ∈ H. An operator P ∈ B(H) is said to be an orthogonal projection if P = P ∗ = P 2 , where T ∗ denotes the adjoint of T . For two operators A, B ∈ B(H), the commutator of A and B is the operator AB − BA. As is well known, the research about norms of operator commutators has attracted much attention of many authors (see [1–3, 5–9]). In general, by the triangle inequality and the submultiplicativity of the usual operator norm that AB − BA 2AB.
(1)
For the case when A or B is positive, the inequality (1) has been recently improved by Kittaneh in [5] as follows: AB − BA AB.
(2)
The purpose of this paper is to establish inequalities for norms of commutators of self-adjoint operators by using operator spectral theory. We shall see that our inequalities improve the inequality in [5], and they seem natural and applicable to be widely useful. Firstly, we shall give some lemmas. Lemma 1. (See [9].) Let A and B ∈ B(H). Then AX − XB min A − λI + B − λI : λ ∈ C X for X ∈ B(H). ✩
This research was partially supported by the National Natural Science Foundation of China (10571113).
* Corresponding author.
E-mail addresses: [email protected] (Y.-Q. Wang), [email protected] (H.-K. Du). 0022-247X/$ – see front matter © 2007 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2007.12.005
748
Y.-Q. Wang, H.-K. Du / J. Math. Anal. Appl. 342 (2008) 747–751
Moreover, there exists a sequence {Xn } ⊆ B(H) with Xn = 1 such that lim AXn − Xn B = min A − λI + B − λI : λ ∈ C . n→∞
Lemma 2. (See [4].) If W and L are two closed subspaces of H and P and Q denote the orthogonal projections on W and L, respectively, then P and Q have the operator matrices P = I1 ⊕ I2 ⊕ 0I3 ⊕ 0I4 ⊕ I5 ⊕ 0I6 and
(3)
⎛ Q = I1 ⊕ 0I2 ⊕ I3 ⊕ 0I4 ⊕ ⎝
1
1
Q02 (I5 − Q0 ) 2 D
Q0
⎞ ⎠
(4) − Q0 ) − Q )D 5 0 6 with respect to the space decomposition H = i=1 Hi , respectively, where H1 = W ∩ L, H2 = W ∩ L⊥ , H3 = W ⊥ ∩ L, H4 = W ⊥ ∩ L⊥ , H5 = W (H1 ⊕ H2 ) and H6 = H ( 5j =1 Hj ), Q0 is a positive contraction on H5 , 0 and 1 are not eigenvalues of Q0 , and D is a unitary from H6 onto H5 . Ii is the identity on Hi , i = 1, . . . , 5. 1
D ∗ Q02 (I5
1 2
D ∗ (I
Theorem 3. Let A and B ∈ B(H) be self-adjoint and denote α2 = max λ: λ ∈ σ (A) α1 = min λ: λ ∈ σ (A) , and
β1 = min μ: μ ∈ σ (B) ,
β2 = max μ: μ ∈ σ (B) .
If 1 1 (α1 + α2 ) (β1 + β2 ), 2 2 then AX − XB (β2 − α1 )X. Proof. By Lemma 1, we only need to prove that min A − λI + B − λI : λ ∈ C = β2 − α1 . For convenience, the proof should be divided into four steps. Step 1. Suppose 12 (α1 + α2 ) λ 12 (β1 + β2 ). Observe that λ − α1 α2 − λ and λ − β1 β2 − λ, we have A − λI = λ − α1 and B − λI = β2 − λ. Hence A − λI + B − λI = (λ − α1 ) + (β2 − λ) = β2 − α1 . Step 2. Suppose λ < 12 (α1 + α2 ). Similarly, we get A − λI = α2 − λ and B − λI = β2 − λ. In this case, A − λI + B − λI = (α2 − λ) + (β2 − λ) = α2 + β2 − 2λ.
Y.-Q. Wang, H.-K. Du / J. Math. Anal. Appl. 342 (2008) 747–751
749
Noting that α2 + β2 − 2λ α2 + β2 − (α1 + α2 ) = β2 − α1 , then A − λI + B − λI β2 − α1 . Step 3. Suppose λ > 12 (β1 + β2 ). Similarly, we get A − λI = λ − α1 and B − λI = λ − β1 . In this case, A − λI + B − λI = (λ − α1 ) + (λ − β1 ) = 2λ − (α1 + β1 ). Noting that 2λ − (α1 + β1 ) (β1 + β2 ) − (α1 + β1 ) = β2 − α1 , then A − λI + B − λI β2 − α1 . Step 4. Suppose λ ∈ C \ R. If λ = a + ib, then A − λI + B − λI = A − aI − ibI + B − aI − ibI A − aI + B − aI β2 − α1 . Combining the four steps above, we have established min A − λI + B − λI : λ ∈ C = β2 − α1 .
2
Corollary 4. Let A ∈ B(H) be self-adjoint. If a = min{λ: λ ∈ σ (A) and b = max{λ: λ ∈ σ (A)}, then AX − XA (b − a)X for X ∈ B(H). Corollary 5. (See [5].) Let A ∈ B(H) be positive. Then AX − XA AX for X ∈ B(H). Proof. It is clear since A = max{λ: λ ∈ σ (A)} and 0 min{λ: λ ∈ σ (A)} if A is positive.
2
Corollary 6. Let A ∈ B(H) be positive and invertible. Then
−1 AX − XA A − A−1 X for X ∈ B(H). Proof. If A is positive and invertible, then A = max{λ: λ ∈ σ (A)} and A−1 −1 = min{λ: λ ∈ σ (A)}. By Corollary 4, we get
−1
2 AX − XA A − A−1 X. Corollary 7. Let A, B ∈ B(H) be self-adjoint. Then 1 AB − BA (α2 − α1 )(β2 − β1 ), 2 where α1 = min{λ: λ ∈ σ (A)}, α2 = max{λ: λ ∈ σ (A)} and β1 = min{μ: μ ∈ σ (B)}, β2 = max{μ: μ ∈ σ (B)}. Proof. For any pair λ and μ of complex numbers, we have AB − BA = (A − λI )(B − μI ) − (B − μI )(A − λI ). So
AB − BA = (A − λI )(B − μI ) − (B − μI )(A − λI ) 2 (A − λI )
(B − μI ) .
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Y.-Q. Wang, H.-K. Du / J. Math. Anal. Appl. 342 (2008) 747–751
Therefore,
AB − BA 2 min (A − λI ) : λ ∈ C min (B − μI ) : μ ∈ C .
Moreover, suppose A and B are self-adjoint, by Theorem 3, we have min{(A − λI ): λ ∈ C} = 12 (α2 − α1 ) and min{(B − μI ): λ ∈ C} = 12 (β2 − β1 ). Hence, 1 AB − BA (α2 − α1 )(β2 − β1 ). 2
2
Corollary 8. (See [6].) Let A, B ∈ B(H) be positive. Then 1 AB − BA AB. 2 2
Proof. By Corollary 7, it is clear.
Remarks. (1) The inequality of Corollary 8 is sharp. For example, if 2 × 2 matrices A and B are as follows:
1 1 1 0 2 2 A= , B= 1 1 ; 0 0 2
2
it is clear that A and B are positive, and A = B = 1. In this case, 1 0 2 AB − BA = . − 12 0 So 1 AB − BA = . 2 (2) By Corollary 8, if A, B are two positive contractions and AB − BA = 12 , then A = B = 1. (3) Let P and Q be orthogonal projections. Then P Q − QP = 12 if and only if 12 ∈ σ (P Q). In this case, by Lemma 2, ⎛ ⎞ 1 1 0 Q02 (I5 − Q0 ) 2 D ⎠, P Q − QP = 0I1 ⊕ 0I2 ⊕ 0I3 ⊕ 0I4 ⊕ ⎝ 1 1 2 ∗ 2 0 −D Q0 (I5 − Q0 ) so
1 1
P Q − QP = Q02 (I5 − Q0 ) 2 . 1
1
1
1
Since 0 Q0 1, Q02 (I5 − Q0 ) 2 12 . Moreover, Q02 (I5 − Q0 ) 2 = 12 if and only if 12 ∈ σ (Q0 ). We know that 1 1 1 1 2 ∈ σ (Q0 ) if and only if 2 ∈ σ (P Q). Hence, P Q − QP = 2 if and only if 2 ∈ σ (P Q). (4) In (2), note that 0 Q0 1. Then we get that 1 P Q − QP : P and Q are orthogonal projections in B(H) = 0, . 2 Corollary 9. (See [5].) Let A and B be positive contractions. Then AX − XB X. Proof. Because A and B are positive contractions, 0 α1 , α2 , β1 , β2 1, so |β2 − α1 | 1 and |α2 − β1 | 1. By Theorem 3, AX − XB X.
2
Y.-Q. Wang, H.-K. Du / J. Math. Anal. Appl. 342 (2008) 747–751
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Corollary 10. Let A and B ∈ B(H) be positive and invertible. Then
−1
−1 A − B max A − B −1 , B − A−1 . Proof. If 12 (A + A−1 −1 ) 12 (B + B −1 −1 ), then A − B A − B −1 −1 . Conversely, if 12 (B + B −1 −1 ) 12 (A + A−1 −1 ), then A − B B − A−1 −1 .
2
Corollary 11. If T = A + iB is the Cartesian decomposition of an operator T , then T ∗ T − T T ∗ 4 min A − λI : λ ∈ C min B − λI : λ ∈ C . Proof. Let α0 = min{α: α ∈ σ (A)} and β0 = min{β: β ∈ σ (B)} and denote λ0 = α0 + iβ0 . Since T ∗ T − T T ∗ = (T − λ0 )∗ (T − λ0 ) − (T − λ0 )(T − λ0 )∗ and A − α0 0 and B − β0 0, it follows that
T ∗ T − T T ∗ = (T − λ0 )∗ (T − λ0 ) − (T − λ0 )(T − λ0 )∗
= 2 (A − α0 I )(B − β0 I ) − (B − β0 I )(A − α0 I )
A − α0 B − β0 = 4 min A − λI : λ ∈ C min B − λI : λ ∈ C .
2
Corollary 12. (See [6].) Let T = A + iB be the Cartesian decomposition of an operator T . If A and B are positive, then T ∗ T − T T ∗ AB. Acknowledgment The authors would like to thank the referee for his/her suggestions and comments.
References [1] [2] [3] [4] [5] [6] [7] [8] [9]
R. Bhatia, F. Kittaneh, On the singular values of a product of operators, SIAM J. Matrix Anal. Appl. 11 (1990) 272–277. R. Bhatia, F. Kittaneh, Cartesian decomposition and Schatten norms, Linear Algebra Appl. 318 (2000) 109–116. R. Bhatia, X. Zhan, Norm inequalities for operators with positive real part, J. Operator Theory 50 (2003) 67–76. H.K. Du, X.Y. Yao, C.Y. Deng, Invertibility of linear combinations of two idempotents, Proc. Amer. Math. Soc. 134 (2005) 1451–1457. F. Kittaneh, Norm inequalities for commutators of positive operators and applications, Math. Z., in press. F. Kittaneh, Inequalities for commutators of positive operators, J. Funct. Anal. 250 (2007) 132–143. F. Kittaneh, On some operator inequalities, Linear Algebra Appl. 208 (1994) 19–28. F. Kittaneh, Commutator inequalities associated with the polar decomposition, Proc. Amer. Math. Soc. 130 (2002) 1279–1283. J.G. Stampfli, The norm of a derivation, Pacific J. Math. 33 (1970) 737–747.
Note
Norms of commutators of self-adjoint operators ✩ Yue-Qing Wang, Hong-Ke Du ∗ College of Mathematics and Information Science, Shaanxi Normal University, Xi’an 710062, PR China Received 28 September 2007 Available online 26 December 2007 Submitted by J.A. Ball
Abstract In this note, the estimate of norms of commutators of self-adjoint operators is established. © 2007 Elsevier Inc. All rights reserved. Keywords: Norm inequality; Commutator; Self-adjoint operator
Throughout this paper, H denotes a Hilbert space. The set of all bounded linear operators acting on H is denoted by B(H). An operator A is said to be self-adjoint if A = A∗ . An operator A is said to be positive if Ax, x 0 for each vector x ∈ H. An operator P ∈ B(H) is said to be an orthogonal projection if P = P ∗ = P 2 , where T ∗ denotes the adjoint of T . For two operators A, B ∈ B(H), the commutator of A and B is the operator AB − BA. As is well known, the research about norms of operator commutators has attracted much attention of many authors (see [1–3, 5–9]). In general, by the triangle inequality and the submultiplicativity of the usual operator norm that AB − BA 2AB.
(1)
For the case when A or B is positive, the inequality (1) has been recently improved by Kittaneh in [5] as follows: AB − BA AB.
(2)
The purpose of this paper is to establish inequalities for norms of commutators of self-adjoint operators by using operator spectral theory. We shall see that our inequalities improve the inequality in [5], and they seem natural and applicable to be widely useful. Firstly, we shall give some lemmas. Lemma 1. (See [9].) Let A and B ∈ B(H). Then AX − XB min A − λI + B − λI : λ ∈ C X for X ∈ B(H). ✩
This research was partially supported by the National Natural Science Foundation of China (10571113).
* Corresponding author.
E-mail addresses: [email protected] (Y.-Q. Wang), [email protected] (H.-K. Du). 0022-247X/$ – see front matter © 2007 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2007.12.005
748
Y.-Q. Wang, H.-K. Du / J. Math. Anal. Appl. 342 (2008) 747–751
Moreover, there exists a sequence {Xn } ⊆ B(H) with Xn = 1 such that lim AXn − Xn B = min A − λI + B − λI : λ ∈ C . n→∞
Lemma 2. (See [4].) If W and L are two closed subspaces of H and P and Q denote the orthogonal projections on W and L, respectively, then P and Q have the operator matrices P = I1 ⊕ I2 ⊕ 0I3 ⊕ 0I4 ⊕ I5 ⊕ 0I6 and
(3)
⎛ Q = I1 ⊕ 0I2 ⊕ I3 ⊕ 0I4 ⊕ ⎝
1
1
Q02 (I5 − Q0 ) 2 D
Q0
⎞ ⎠
(4) − Q0 ) − Q )D 5 0 6 with respect to the space decomposition H = i=1 Hi , respectively, where H1 = W ∩ L, H2 = W ∩ L⊥ , H3 = W ⊥ ∩ L, H4 = W ⊥ ∩ L⊥ , H5 = W (H1 ⊕ H2 ) and H6 = H ( 5j =1 Hj ), Q0 is a positive contraction on H5 , 0 and 1 are not eigenvalues of Q0 , and D is a unitary from H6 onto H5 . Ii is the identity on Hi , i = 1, . . . , 5. 1
D ∗ Q02 (I5
1 2
D ∗ (I
Theorem 3. Let A and B ∈ B(H) be self-adjoint and denote α2 = max λ: λ ∈ σ (A) α1 = min λ: λ ∈ σ (A) , and
β1 = min μ: μ ∈ σ (B) ,
β2 = max μ: μ ∈ σ (B) .
If 1 1 (α1 + α2 ) (β1 + β2 ), 2 2 then AX − XB (β2 − α1 )X. Proof. By Lemma 1, we only need to prove that min A − λI + B − λI : λ ∈ C = β2 − α1 . For convenience, the proof should be divided into four steps. Step 1. Suppose 12 (α1 + α2 ) λ 12 (β1 + β2 ). Observe that λ − α1 α2 − λ and λ − β1 β2 − λ, we have A − λI = λ − α1 and B − λI = β2 − λ. Hence A − λI + B − λI = (λ − α1 ) + (β2 − λ) = β2 − α1 . Step 2. Suppose λ < 12 (α1 + α2 ). Similarly, we get A − λI = α2 − λ and B − λI = β2 − λ. In this case, A − λI + B − λI = (α2 − λ) + (β2 − λ) = α2 + β2 − 2λ.
Y.-Q. Wang, H.-K. Du / J. Math. Anal. Appl. 342 (2008) 747–751
749
Noting that α2 + β2 − 2λ α2 + β2 − (α1 + α2 ) = β2 − α1 , then A − λI + B − λI β2 − α1 . Step 3. Suppose λ > 12 (β1 + β2 ). Similarly, we get A − λI = λ − α1 and B − λI = λ − β1 . In this case, A − λI + B − λI = (λ − α1 ) + (λ − β1 ) = 2λ − (α1 + β1 ). Noting that 2λ − (α1 + β1 ) (β1 + β2 ) − (α1 + β1 ) = β2 − α1 , then A − λI + B − λI β2 − α1 . Step 4. Suppose λ ∈ C \ R. If λ = a + ib, then A − λI + B − λI = A − aI − ibI + B − aI − ibI A − aI + B − aI β2 − α1 . Combining the four steps above, we have established min A − λI + B − λI : λ ∈ C = β2 − α1 .
2
Corollary 4. Let A ∈ B(H) be self-adjoint. If a = min{λ: λ ∈ σ (A) and b = max{λ: λ ∈ σ (A)}, then AX − XA (b − a)X for X ∈ B(H). Corollary 5. (See [5].) Let A ∈ B(H) be positive. Then AX − XA AX for X ∈ B(H). Proof. It is clear since A = max{λ: λ ∈ σ (A)} and 0 min{λ: λ ∈ σ (A)} if A is positive.
2
Corollary 6. Let A ∈ B(H) be positive and invertible. Then
−1 AX − XA A − A−1 X for X ∈ B(H). Proof. If A is positive and invertible, then A = max{λ: λ ∈ σ (A)} and A−1 −1 = min{λ: λ ∈ σ (A)}. By Corollary 4, we get
−1
2 AX − XA A − A−1 X. Corollary 7. Let A, B ∈ B(H) be self-adjoint. Then 1 AB − BA (α2 − α1 )(β2 − β1 ), 2 where α1 = min{λ: λ ∈ σ (A)}, α2 = max{λ: λ ∈ σ (A)} and β1 = min{μ: μ ∈ σ (B)}, β2 = max{μ: μ ∈ σ (B)}. Proof. For any pair λ and μ of complex numbers, we have AB − BA = (A − λI )(B − μI ) − (B − μI )(A − λI ). So
AB − BA = (A − λI )(B − μI ) − (B − μI )(A − λI ) 2 (A − λI )
(B − μI ) .
750
Y.-Q. Wang, H.-K. Du / J. Math. Anal. Appl. 342 (2008) 747–751
Therefore,
AB − BA 2 min (A − λI ) : λ ∈ C min (B − μI ) : μ ∈ C .
Moreover, suppose A and B are self-adjoint, by Theorem 3, we have min{(A − λI ): λ ∈ C} = 12 (α2 − α1 ) and min{(B − μI ): λ ∈ C} = 12 (β2 − β1 ). Hence, 1 AB − BA (α2 − α1 )(β2 − β1 ). 2
2
Corollary 8. (See [6].) Let A, B ∈ B(H) be positive. Then 1 AB − BA AB. 2 2
Proof. By Corollary 7, it is clear.
Remarks. (1) The inequality of Corollary 8 is sharp. For example, if 2 × 2 matrices A and B are as follows:
1 1 1 0 2 2 A= , B= 1 1 ; 0 0 2
2
it is clear that A and B are positive, and A = B = 1. In this case, 1 0 2 AB − BA = . − 12 0 So 1 AB − BA = . 2 (2) By Corollary 8, if A, B are two positive contractions and AB − BA = 12 , then A = B = 1. (3) Let P and Q be orthogonal projections. Then P Q − QP = 12 if and only if 12 ∈ σ (P Q). In this case, by Lemma 2, ⎛ ⎞ 1 1 0 Q02 (I5 − Q0 ) 2 D ⎠, P Q − QP = 0I1 ⊕ 0I2 ⊕ 0I3 ⊕ 0I4 ⊕ ⎝ 1 1 2 ∗ 2 0 −D Q0 (I5 − Q0 ) so
1 1
P Q − QP = Q02 (I5 − Q0 ) 2 . 1
1
1
1
Since 0 Q0 1, Q02 (I5 − Q0 ) 2 12 . Moreover, Q02 (I5 − Q0 ) 2 = 12 if and only if 12 ∈ σ (Q0 ). We know that 1 1 1 1 2 ∈ σ (Q0 ) if and only if 2 ∈ σ (P Q). Hence, P Q − QP = 2 if and only if 2 ∈ σ (P Q). (4) In (2), note that 0 Q0 1. Then we get that 1 P Q − QP : P and Q are orthogonal projections in B(H) = 0, . 2 Corollary 9. (See [5].) Let A and B be positive contractions. Then AX − XB X. Proof. Because A and B are positive contractions, 0 α1 , α2 , β1 , β2 1, so |β2 − α1 | 1 and |α2 − β1 | 1. By Theorem 3, AX − XB X.
2
Y.-Q. Wang, H.-K. Du / J. Math. Anal. Appl. 342 (2008) 747–751
751
Corollary 10. Let A and B ∈ B(H) be positive and invertible. Then
−1
−1 A − B max A − B −1 , B − A−1 . Proof. If 12 (A + A−1 −1 ) 12 (B + B −1 −1 ), then A − B A − B −1 −1 . Conversely, if 12 (B + B −1 −1 ) 12 (A + A−1 −1 ), then A − B B − A−1 −1 .
2
Corollary 11. If T = A + iB is the Cartesian decomposition of an operator T , then T ∗ T − T T ∗ 4 min A − λI : λ ∈ C min B − λI : λ ∈ C . Proof. Let α0 = min{α: α ∈ σ (A)} and β0 = min{β: β ∈ σ (B)} and denote λ0 = α0 + iβ0 . Since T ∗ T − T T ∗ = (T − λ0 )∗ (T − λ0 ) − (T − λ0 )(T − λ0 )∗ and A − α0 0 and B − β0 0, it follows that
T ∗ T − T T ∗ = (T − λ0 )∗ (T − λ0 ) − (T − λ0 )(T − λ0 )∗
= 2 (A − α0 I )(B − β0 I ) − (B − β0 I )(A − α0 I )
A − α0 B − β0 = 4 min A − λI : λ ∈ C min B − λI : λ ∈ C .
2
Corollary 12. (See [6].) Let T = A + iB be the Cartesian decomposition of an operator T . If A and B are positive, then T ∗ T − T T ∗ AB. Acknowledgment The authors would like to thank the referee for his/her suggestions and comments.
References [1] [2] [3] [4] [5] [6] [7] [8] [9]
R. Bhatia, F. Kittaneh, On the singular values of a product of operators, SIAM J. Matrix Anal. Appl. 11 (1990) 272–277. R. Bhatia, F. Kittaneh, Cartesian decomposition and Schatten norms, Linear Algebra Appl. 318 (2000) 109–116. R. Bhatia, X. Zhan, Norm inequalities for operators with positive real part, J. Operator Theory 50 (2003) 67–76. H.K. Du, X.Y. Yao, C.Y. Deng, Invertibility of linear combinations of two idempotents, Proc. Amer. Math. Soc. 134 (2005) 1451–1457. F. Kittaneh, Norm inequalities for commutators of positive operators and applications, Math. Z., in press. F. Kittaneh, Inequalities for commutators of positive operators, J. Funct. Anal. 250 (2007) 132–143. F. Kittaneh, On some operator inequalities, Linear Algebra Appl. 208 (1994) 19–28. F. Kittaneh, Commutator inequalities associated with the polar decomposition, Proc. Amer. Math. Soc. 130 (2002) 1279–1283. J.G. Stampfli, The norm of a derivation, Pacific J. Math. 33 (1970) 737–747.