We prove several singular value inequalities for sum and product of compact
operators in Hilbert space. Some of our results generalize the previous
inequalities for operators. Also, applications of some inequalities are given.Comment: 1

Darvish, V

Dragomir, Sever S

Nazari, HM

Taghavi, A

arXiv

Dragomir, Silvestru

Name not available

Some results on singular value inequalities of compact operators in Hilbert space

Victoria University Institutional Repository (VUIR)

Darvish, Vahid

Victoria University Eprints Repository

arXiv:1510.00114v1  [math.FA]  1 Oct 2015SOME RESULTS ON SINGULAR VALUE INEQUALITIES OFCOMPACT OPERATORS IN HILBERT SPACEA. TAGHAVI, V. DARVISH, H. M. NAZARI, S. S. DRAGOMIRAbstract. We prove several singular value inequalities for sum and productof compact operators in Hilbert space. Some of our results generalize theprevious inequalities for operators. Also, applications of some inequalities aregiven.1. IntroductionLet B(H) stand for the C∗-algebra of all bounded linear operators ona complex separable Hilbert space H with inner product 〈·, ·〉 and letK(H) denote the two-sided ideal of compact operators in B(H). ForA ∈ B(H), let ‖A‖ = sup{‖Ax‖ : ‖x‖ = 1} denote the usual operatornorm of A and |A| = (A∗A)1/2 be the absolute value of A.An operator A ∈ B(H) is positive and write A ≥ 0 if 〈Ax, x〉 ≥ 0 forall x ∈ H . We say A ≤ B whenever B −A ≥ 0.We consider the wide class of unitarily invariant norms ||| · |||. Eachof these norms is defined on an ideal in B(H) and it will be implicitlyunderstood that when we talk of |||T |||, then the operator T belongsto the norm ideal associated with ||| · |||. Each unitarily invariant norm||| · ||| is characterized by the invariance property |||UTV ||| = |||T |||for all operators T in the norm ideal associated with ||| · ||| and for allunitary operators U and V in B(H). For 1 ≤ p < ∞, the Schatten p-norm of a compact operator A is defined by ‖A‖p = (tr |A|p)1/p, wheretr is the usual trace functional. Note that for A ∈ K(H) we have,‖A‖ = s1(A), and if A is a Hilbert-Schmidt operator, then ‖A‖2 =(∑∞j=1 s2j(A))1/2. These norms are special examples of the more generalclass of the Schatten p-norms, which are unitarily invariant [2].2010 Mathematics Subject Classification. 15A60, 47A30, 47B05, 47B10.Key words and phrases. Singular value, compact operator, normal operator, unitarily invariantnorm , Schatten p-norm.12 A. TAGHAVI, V. DARVISH, H. M. NAZARI, S. S. DRAGOMIRThe direct sum A⊕B denotes the block diagonal matrix[A 00 B]defined on H ⊕H , see [1, 9]. It is easy to see that(1.1) ‖A⊕B‖ = max(‖A‖, ‖B‖),and(1.2) ‖A⊕B‖p = (‖A‖pp + ‖B‖pp)1/p.We denote the singular values of an operator A ∈ K(H) as s1(A) ≥s2(A) ≥ . . . are the eigenvalues of the positive operator |A| = (A∗A)1/2and eigenvalues of the self-adjoint operator A denote as λ1 ≥ λ2 ≥ . . .which repeated accordingly to multiplicity.There is a one-to-one correspondence between symmetric gauge func-tions defined on sequences of real numbers and unitarily invariantnorms defined on norm ideals of operators. More precisely, if ||| · ||| isunitarily invariant norm, then there exists a unique symmetric gaugefunction Φ such that|||A||| = Φ(s1(A), s2(A), . . .),for every operator A ∈ K(H). Let A ∈ K(H), and if U, V ∈ B(H) areunitarily operators, thensj(UAV ) = sj(A),for j = 1, 2, . . . and so unitarily invariant norms satisfies the invarianceproperty|||UAV ||| = |||A|||.In this paper, we obtain some inequalities for sum and product ofoperators. Some of our results generalize the previous inequalities foroperators.2. Some singular value inequalities for sum and product ofoperatorsIn this section we give inequalities for singular value of operators.Also, some norm inequalities are obtained as an application.SOME RESULTS ON SINGULAR VALUE INEQUALITIES 3First we should remind the following inequalities. We apply inequalities(2.1) and (2.3) in our proofs.The following inequality due to Tao [8] asserts that if A,B,C ∈K(H) such that[A BB∗ C]≥ 0, then(2.1) 2sj(B) ≤ sj[A BB∗ C],for j = 1, 2, . . ..Here, we give another proof for above inequality.Let[A BB∗ C]≥ 0 then[A −B−B∗ C]≥ 0 and have the same sin-gular values (see[1, Theorem 2.1]). So, we can write[0 2B2B∗ 0]≤[A BB∗ C],and [0 −2B−2B∗ 0]≤[A −B−B∗ C].On the other hand, we know that for every self-adjoint compact oper-ator X we have sj(X) ≤ λj(X ⊕−X), for all j = 1, 2, . . .. By using ofthis fact we obtainsj([0 2B2B∗ 0])= λj([0 2B2B∗ 0]⊕[0 −2B−2B∗ 0])≤ λj([A BB∗ C]⊕[A −B−B∗ C])= sj([A BB∗ C]⊕[A −B−B∗ C]).So, we obtainsj([0 2B2B∗ 0])≤ sj([A BB∗ C]⊕[A −B−B∗ C]).4 A. TAGHAVI, V. DARVISH, H. M. NAZARI, S. S. DRAGOMIREquivalently,2sj(B ⊕B∗) ≤([A BB∗ C]⊕[A −B−B∗ C]).Since sj(B) = sj(B∗) and sj([A BB∗ C])= sj([A −B−B∗ C]),we have2sj(B) ≤ sj([A BB∗ C]).In [1, Remark 2.2], Audeh and Kittaneh proved that for every A,B,C ∈K(H) such that[A BB∗ C]≥ 0, then(2.2) sj([A BB∗ C])≤ 2sj(A⊕ C),for j = 1, 2, . . .. Therefore, by inequality (2.1) we have the followinginequality(2.3) sj(B) ≤ sj(A⊕ C),for j = 1, 2, . . .. Since every unitarily invariant norm is a monotonefunction of the singular values of an operator, we can write(2.4)∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣[A BB∗ C]∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣ ≤ 2|||A⊕ C|||.We can obtain the reverse of inequality (2.4) for arbitrary operatorsX, Y ∈ B(H) by pointing out the following inequality holds because ofnorm property|||X + Y ||| ≤ |||X|||+ |||Y |||.Replace X and Y by X − Y and X + Y , respectively. We have2|||X||| ≤ |||X − Y |||+ |||X + Y |||,SOME RESULTS ON SINGULAR VALUE INEQUALITIES 5for all X, Y ∈ B(H).Let X =[A 00 C]and Y =[0 BB∗ 0]in above inequality. So,2∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣[A 00 C]∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣ ≤∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣[A BB∗ C]∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣+∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣[A −B−B∗ C]∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣= 2∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣[A BB∗ C]∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣ .Hence,|||A⊕ C||| ≤∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣[A BB∗ C]∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣ ,for all A,B,C ∈ B(H).[A 00 C]is called a pinching of[A BB∗ C].For operator norm we havemax{‖A‖, ‖C‖} ≤∥∥∥∥∥[A BB∗ C]∥∥∥∥∥ .Here we give a generalization of the inequality which has been provedby Bhatia and Kittaneh in [5]. They have shown that if A and B aretwo n× n matrices, thensj(A+B) ≤ sj ((|A|+ |B|)⊕ (|A∗|+ |B∗|)) ,for 1 ≤ j ≤ n.For giving a generalization of above inequality, we need the followinglemmas.In the rest of this section, we always assume that f and g are non-negative functions on [0,∞) which are continuous and satisfying therelation f(t)g(t) = t for all t ∈ [0,∞).The following lemma is due to Kittaneh [7].6 A. TAGHAVI, V. DARVISH, H. M. NAZARI, S. S. DRAGOMIRLemma 2.1. Let A,B, and C be operators in B(H) such that A andB are positive and BC = CA. If[A C∗C B]is positive in B(H ⊕H),then[f(A)2 C∗C g(B)2]is also positive.Let T be an operator in B(H). We know that[|T | T ∗T |T ∗|]≥ 0, ifT is normal then we have[|T | T ∗T |T |]≥ 0, ( see [4]).Lemma 2.2. Let A be an operator in B(H). Then we have(2.5)[|A|2α A∗A |A∗|2(1−α)]≥ 0,where 0 ≤ α ≤ 1.Proof. It is easy to check that A|A|2 = |A∗|2A, then we have A|A| =|A∗|A for A ∈ B(H). Now by making use of Lemma 2.1, for f(t) = tαand g(t) = t1−α, 0 ≤ α ≤ 1, and positivity of[|A| A∗A |A∗|], we obtainthe result. Theorem 2.3. Let A and B be two operators in K(H). Then we havesj(A+B) ≤ sj((|A|2α + |B|2α)⊕ (|A∗|2(1−α) + |B∗|2(1−α))) ,for j = 1, 2, . . . where 0 ≤ α ≤ 1.Proof. Since sum of two positive operator is positive, Lemma 2.2 im-plies that [|A|2α + |B|2α A∗ +B∗A+B |A∗|2(1−α) + |B∗|2(1−α)]≥ 0,By inequality (2.3) we have the result. Corollary 2.4. Let A and B be two operators in K(H). Then we havesj(A+B) ≤ sj ((|A|+ |B|)⊕ (|A∗|+ |B∗|)) ,for j = 1, 2, . . ..SOME RESULTS ON SINGULAR VALUE INEQUALITIES 7Proof. Let α = 12in Theorem 2.3. It is easy to see that if A and B are normal operator in K(H), thenwe havesj(A +B) ≤ sj ((|A|+ |B|)⊕ (|A|+ |B|)) ,for j = 1, 2, . . ..On the other hand, for α = 1 in Theorem 2.3, we havesj(A +B) ≤ sj(|A|2 + |B|2 ⊕ 2I)= sj(|A|2 + |B|2) ∪ sj(2I)= sj(A∗A+B∗B) ∪ sj(2I),for j = 1, 2, . . ..Theorem 2.5. Let A,B and X be operators in B(H) such that X iscompact. Then we have the followingsj (AXB∗) ≤ sj(A∗f(|X|)2A⊕ B∗g(|X∗|)2B) ,for j = 1, 2, . . ..Proof. Since[|X| X∗X |X∗|]≥ 0, by Lemma 2.1 we haveY =[f(|X|)2 X∗X g(|X∗|)2]≥ 0.Let Z =[A 00 B]. Since Y is positive, we haveZ∗Y Z =[A∗f(|X|)2A A∗X∗BB∗XA B∗g(|X∗|)2B]≥ 0.Hence, by inequality (2.3), we have the desired result. In above theorem, let X be a normal operator. Then we havesj (AXB∗) ≤ sj(A∗f(|X|)2A⊕ B∗g(|X|)2B) ,for j = 1, 2, . . ..8 A. TAGHAVI, V. DARVISH, H. M. NAZARI, S. S. DRAGOMIRCorollary 2.6. Let A, B and X be operators in B(H) such that X iscompact. Then we havesj (AXB∗) ≤ sj (A∗|X|A⊕ B∗|X∗|B) ,for j = 1, 2, . . ..Proof. Let f(t) = t12 and g(t) = t12 in Theorem 2.5. Here, we apply above corollary to show that singular values of AXB∗are dominated by singular values of ‖X‖(A⊕B). For our proof we needthe following lemma.Lemma 2.7. [2, p. 75] Let A,B ∈ B(H) such that B is compact.Thensj(AB) ≤ ‖A‖sj(B),for j = 1, 2, . . ..Theorem 2.8. Let A,B,X ∈ B(H) such that A and B are arbitrarycompact. Then, we havesj(AXB∗) ≤ ‖X‖s2j(A⊕ B),for j = 1, 2, . . ..SOME RESULTS ON SINGULAR VALUE INEQUALITIES 9Proof. From Corollary 2.6 we havesj(AXB∗) ≤ sj(A∗|X|A⊕ B∗|X∗|B)= sj([A 00 B]∗[|X| 00 |X∗|][A 00 B])= sj([A 00 B]∗[|X| 12 00 |X∗| 12]∗[|X| 12 00 |X∗| 12][A 00 B])= sj(([|X| 12 00 |X∗| 12][A 00 B])∗([|X| 12 00 |X∗| 12][A 00 B]))= sj∣∣∣∣∣[|X| 12 00 |X∗| 12][A 00 B]∣∣∣∣∣2= s2j([|X| 12 00 |X∗| 12][A 00 B])≤∥∥∥∥∥[|X| 12 00 |X∗| 12]∥∥∥∥∥2s2j(A⊕ B)= ‖X‖s2j(A⊕B),for j = 1, 2, . . .. The last inequality follows by Lemma 2.7. In Theorem 2.8, let A and B be positive operators in K(H). Thenwe have(2.6) sj(A12XB12 ) ≤ ‖X‖sj(A⊕B),for j = 1, 2, . . ..Corollary 2.9. Let A and B be two operators in K(H). Then we have(2.7) sj(AB∗) ≤ sj (A∗A⊕ B∗B) ,for j = 1, 2, . . ..Proof. Let X = I in Corollary 2.6. 10 A. TAGHAVI, V. DARVISH, H. M. NAZARI, S. S. DRAGOMIRMoreover, we can write inequality (2.7) in the following formsj(AB∗) ≤ sj(|A|2 ⊕ |B|2)= s2j(|A| ⊕ |B|) = s2j(A⊕B),for j = 1, 2, . . ..We should note here that inequality (2.7) can be obtained by Theorem1 in [3] and Corollary 2.2 in [6].Here, we give two results of Corollary 2.9. As the first application, letA =[X Y0 0]and B =[Y −X0 0], such that X, Y ∈ K(H) thenby easy computations we havesj(XY∗ − Y X∗) ≤ sj((XX∗ + Y Y ∗)⊕ (XX∗ + Y Y ∗)),for j = 1, 2, . . ..For obtaining second application, replace A and B in (2.7) by AXαand BX(1−α) respectively, where X is a compact positive operator andα ∈ R. So, we havesj(AXB∗) ≤ sj(XαA∗AXα ⊕X(1−α)B∗BX(1−α))= sj([XαA∗AXα 00 X(1−α)B∗BX(1−α)])= sj([XαA∗ 00 X(1−α)B∗][AXα 00 BX(1−α)])= sj([AXα 00 BX(1−α)][XαA∗ 00 X(1−α)B∗])= sj([AX2αA∗ 00 BX2(1−α)B∗])= sj(AX2αA∗ ⊕ BX2(1−α)B∗),for all j = 1, 2, . . ..Finally, we have(2.8) sj(AXB∗) ≤ sj(AX2αA∗ ⊕ BX2(1−α)B∗),SOME RESULTS ON SINGULAR VALUE INEQUALITIES 11for all j = 1, 2, . . ..By a similar proof of Theorem 2.8 to inequality (2.8), we obtainsj(AXB∗) ≤ max{‖X2α‖, ‖X2(1−α)‖}s2j(A⊕ B),for all j = 1, 2, . . ..In above inequality, for positive operators A and B in K(H) we havesj(A12XB12 ) ≤ max{‖X2α‖, ‖X2(1−α)‖}sj(A⊕ B)for all j = 1, 2, . . ..3. Some singular value inequalities for normal operatorsHere we give some results for compact normal operators. For everyoperator A, the Cartesian decomposition is to write A = ℜ(A)+iℑ(A),where ℜ(A) = A+A∗2and ℑ(A) = A−A∗2i. If A is normal operator thenℜ(A) and ℑ(A) commute together and vice versa.Theorem 3.1. Let A1, A2, . . . , An be normal operators in K(H). Thenwe have1√2sj (⊕ni=1(ℜ(Ai) + ℑ(Ai))) ≤ sj(⊕ni=1Ai)≤ sj (⊕ni=1(|ℜ(Ai)|+ |ℑ(Ai)|)) ,for j = 1, 2, . . ..Proof. Let A1, A2, . . . , An be normal operators, then⊕ni=1Ai =A1 0A2. . .0 Anis normal, so we have(⊕ni=1ℜ(Ai))(⊕ni=1ℑ(Ai)) = (⊕ni=1ℑ(Ai))(⊕ni=1ℜ(Ai))).By above equation, we obtain the following√(⊕ni=1Ai)∗(⊕ni=1Ai) =√(⊕ni=1ℜ(Ai))2 + (⊕ni=1ℑ(Ai))2.12 A. TAGHAVI, V. DARVISH, H. M. NAZARI, S. S. DRAGOMIRSosj(⊕ni=1Ai) = sj(| ⊕ni=1 Ai|)= sj(√(⊕ni=1Ai)∗(⊕ni=1Ai))= sj(√(⊕ni=1ℜ(Ai))2 + (⊕ni=1ℑ(Ai))2),for j = 1, 2, . . ..By using Weyl’s monotonicity principle [2] and the inequality√(⊕ni=1ℜ(Ai))2 + (⊕ni=1ℑ(Ai))2 ≤ | ⊕ni=1 ℜ(Ai)|+ | ⊕ni=1 ℑ(Ai)|,we have the followingsj(√(⊕ni=1ℜ(Ai))2 + (⊕ni=1ℑ(Ai))2)≤ sj(|⊕ni=1ℜ(Ai)|+|⊕ni=1ℑ(Ai)|),for j = 1, 2, . . .. Now for proving left side inequality, we recall thefollowing inequality0 ≤ (ℜ(Ai) + ℑ(Ai)∗(ℜ(Ai) + ℑ(Ai) ≤ 2(ℜ(Ai)2 + ℑ(A2)2).Therefore, by using the Weyl’s monotonicity principle we can writesj(√((⊕ni=1ℜ(Ai)) + (⊕ni=1ℑ(Ai)))∗ ((⊕ni=1ℜ(Ai)) + (⊕ni=1ℑ(Ai)))),which is less than√2sj(√(⊕ni=1ℜ(Ai))2 + (⊕ni=1ℑ(Ai))2).for j = 1, 2, . . .. Therefore,sj((⊕ni=1ℜ(Ai)) + (⊕ni=1ℑ(Ai))) = sj(|(⊕ni=1ℜ(Ai)) + (⊕ni=1ℑ(Ai))|)≤√2sj(√(⊕ni=1ℜ(Ai))2 + (⊕ni=1ℑ(Ai))2).for j = 1, 2, . . .. The following example shows that normal condition is necessary.Example 3.2. Let A =[−1 + i 1i 1 + 2i], then a calculation showss2(ℜ(A) + iℑ(A)) ≈ 1.34 > s2(|ℜ(A)|+ |ℑ(A)|) ≈ 1.27.SOME RESULTS ON SINGULAR VALUE INEQUALITIES 13Corollary 3.3. Let A be a normal operator in K(H). Then we have(1/√2)sj(ℜ(A) + ℑ(A)) ≤ sj(A) ≤ sj(|ℜ(A)|+ |ℑ(A)|),for j = 1, 2, . . ..For each complex number x = a+ib, we know the following inequalityholds(3.1)1√2|a+ b| ≤ |x| ≤ |a|+ |b|.Now, by applying Corollary 3.3, we can obtain operator version of in-equality (3.1).Here, we determine the upper and lower bound for A+ iA∗.Theorem 3.4. Let A1, A2, . . . , An be in K(H). Then√2sj (⊕ni=1(ℜ(Ai) + ℑ(Ai))) ≤ sj(⊕ni=1(Ai + iA∗i ))≤ 2sj(⊕ni=1(ℜ(Ai) + ℑ(Ai))),for j = 1, 2, . . ..Proof. Note that Ai + iA∗i is normal operator for i = 1, . . . , n, so T =⊕ni=1(Ai + iA∗i ) is normal. On the other hand, we can write T =ℜ(T ) + iℑ(T ) whereℜ(T ) = (⊕ni=1(Ai + A∗i ) + i⊕ni=1 (A∗i − Ai))/2,ℑ(T ) = (⊕ni=1(Ai −A∗i ) + i⊕ni=1 (A∗i + Ai))/2i.It is enough to compare ℜ(T ) and ℑ(T ) to see ℜ(T ) = ℑ(T ). So(3.2) ℜ(T ) + ℑ(T ) = ⊕ni=1(Ai + A∗i ) + i⊕ni=1 (A∗i − Ai).Now apply Theorem 3.1, we have(1/√2)sj(ℜ(T ) + ℑ(T )) ≤ sj(ℜ(T ) + iℑ(T ))≤ sj(|ℜ(T )|+ |ℑ(T )|),(3.3)for j = 1, 2, . . .. Put (3.2), ℜ(T ) + iℑ(T ) = ⊕ni=1(Ai + iA∗i ) and ℜ(T )in (3.3) to obtain(3.4) (1/√2)sj(⊕ni=1(Ai+A∗i )+i⊕ni=1 (A∗i−Ai)) ≤ sj(⊕ni=1(Ai+iA∗i )),14 A. TAGHAVI, V. DARVISH, H. M. NAZARI, S. S. DRAGOMIRandsj(⊕ni=1(Ai + iA∗i )) ≤ 2sj(⊕ni=1(Ai + A∗i )/2 + i⊕ni=1 (A∗i − Ai)/2)= sj(⊕ni=1(Ai + A∗i ) + i⊕ni=1 (A∗i −Ai)),for j = 1, 2, . . .. By writing ℜ(⊕ni=1Ai) = ⊕ni=1(Ai+A∗i )/2 and ℑ(⊕ni=1Ai) =⊕ni=1(Ai −A∗i )/2i we have(1/√2)sj(2ℜ(⊕ni=1Ai) + 2ℑ(⊕ni=1Ai)) ≤ sj(⊕ni=1(Ai + iA∗i ))≤ sj(2ℜ(⊕ni=1Ai) + 2ℑ(⊕ni=1Ai)),for j = 1, 2, . . .. Finally√2sj (⊕ni=1(ℜ(Ai) + ℑ(Ai))) ≤ sj(⊕ni=1(Ai + iA∗i ))≤ 2sj(⊕ni=1(ℜ(Ai) + ℑ(Ai))),for j = 1, 2, . . .. References[1] W. Audeh and F. Kittaneh, Singular value inequalities for compact operators, Linear AlgebraAppl. 437 (2012) 2516-2522.[2] R. Bhatia, Matrix Analysis, Springer-Verlag, New York, 1997.[3] R. Bhatia and F. Kittaneh, On the singular values of a product of operators, SIAM J. MatrixAnal. Appl. 11 (1990) 272-277.[4] R. Bhatia, Positive Definite Matrices, Princeton Press, 2007.[5] R. Bhatia and F. Kittaneh, The matrix arithmetic-geometric mean inequality revisited, LinearAlgebra Appl. 428 (2008) 2177-2191.[6] O. Hirzallah and F. Kittaneh, Inequalities for sums and direct sums of Hilbert space operators,Linear Algebra Appl. 424 (2007) 71-82.[7] F. Kittaneh, Notes on some inequalities for Hilbert space operators, Publ. RIMS Kyoto Univ.24 (1988) 283-293.[8] Y. Tao, More results on singular value inequalities of matrices, Linear Algebra Appl. 416(2006) 724–729.[9] X. Zhan, Matrix Inequalities, Springer-Verlag, Berlin, 2002.Department of Mathematics, Faculty of Mathematical Sciences, University of Mazan-daran, P. O. Box 47416-1468, Babolsar, Iran.Mathematics, School of Engineering and Science Victoria University, PO Box 14428,Melbourne City, MC 8001, Australia.School of Computational and Applied Mathematics, University of the Witwater-srand, Private Bag 3, Johannesburg 2050, South Africa.E-mail address: taghavi@umz.ac.ir, vahid.darvish@vu.edu.au, m.nazari@stu.umz.ac.ir,sever.dragomir@vu.edu.au

Some results on singular value inequalities of compact operators in Hilbert space

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