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On properties of k-quasi-class A(n) operators

Abstract

Let n and k be positive integers; an operator TB(H) is called a k-quasi-class A(n) operator if T k ( | T 1 + n | 2 1 + n | T | 2 ) T k 0, which is a common generalization of class A and class A(n) operators. In this paper, firstly we prove some basic structural properties of this class of operators, showing that if T is a k-quasi-class A(n) operator, then the nonzero points of its point spectrum and joint point spectrum are identical, the eigen-spaces corresponding to distinct eigenvalues of T are mutually orthogonal, the nonzero points of its approximate point spectrum and joint approximate point spectrum are identical; secondly we consider the tensor products for k-quasi-class A(n) operators, giving a necessary and sufficient condition for TS to be a k-quasi-class A(n) operator when T and S are both nonzero operators.

MSC:47B20, 47A63.

1 Introduction

Let be a separable complex Hilbert space and be the set of complex numbers. Let B(H) denote the C -algebra of all bounded linear operators acting on . Recall that TB(H) is called p-hyponormal for p>0 if ( T T ) p ( T T ) p 0 [1]; when p=1, T is called hyponormal. T is called paranormal if T x 2 T 2 xx for all xH [2, 3]. T is called normaloid if T n = T n for all nN (equivalently, T=r(T), the spectral radius of T). In order to discuss the relations between paranormal and p-hyponormal and log-hyponormal operators (T is invertible and log T TlogT T ), Furuta et al. [4] introduced a very interesting class of operators: class A defined by | T 2 | | T | 2 0, where |T|= ( T T ) 1 2 , which is called the absolute value of T and they showed that class A is a subclass of paranormal and contains p-hyponormal and log-hyponormal operators. Recently Yuan and Gao [5] introduced class A(n) (i.e., | T 1 + n | 2 1 + n | T | 2 ) operators and n-paranormal operators (i.e., T 1 + n x 1 1 + n Tx for every unit vector xH) for some positive integer n. For more interesting properties on class A(n) and n-paranormal operators, see [68].

Let , be complex Hilbert spaces and HK the tensor product of , ; i.e., the completion of the algebraic tensor product of , with the inner product x 1 y 1 , x 2 y 2 = x 1 , x 2 y 1 , y 2 for x 1 , x 2 H, y 1 , y 2 K. Let TB(H) and SB(K). TSB(HK) denotes the tensor product of T and S; i.e., (TS)(xy)=TxSy for xH, yK.

Definition 1.1 TB(H) is called a k-quasi-class A(n) operator for positive integers n and k if

T k ( | T 1 + n | 2 1 + n | T | 2 ) T k 0.

In general, the following implications hold:

p-hyponormalclass Aclass A(n)k-quasi-class A(n).

In this paper, firstly we prove some basic structural properties of this class of operators, showing that if T is a k-quasi-class A(n) operator, then the nonzero points of its point spectrum and joint point spectrum are identical, the eigen-spaces corresponding to distinct eigenvalues of T are mutually orthogonal, the nonzero points of its approximate point spectrum and joint approximate point spectrum are identical; secondly we consider the tensor products for k-quasi-class A(n) operators, giving a necessary and sufficient condition for TS to be a k-quasi-class A(n) operator when T and S are both nonzero operators.

2 The basic properties for k-quasi-class A(n) operators

In the following lemma, we study the matrix representation of a k-quasi-class A(n) operator with respect to the direct sum of ran ( T k ) ¯ and its orthogonal complement.

Lemma 2.1 Let TB(H) be a k-quasi-class A(n) operator for positive integers n and k, and let T= ( T 1 T 2 0 T 3 ) on H= ran ( T k ) ¯ ker T k be 2×2 matrix expression. Assume that ran T k is not dense, then T 1 is a class A(n) operator on ran ( T k ) ¯ and T 3 k =0. Furthermore, σ(T)=σ( T 1 ){0}.

Proof Consider the matrix representation of T with respect to the decomposition H= ran ( T k ) ¯ ker T k : T= ( T 1 T 2 0 T 3 ) . Let P be the orthogonal projection of onto ran ( T k ) ¯ . Then T 1 =TP=PTP. Since T is a k-quasi-class A(n) operator, we have

P ( | T 1 + n | 2 1 + n | T | 2 ) P0.

Then

| T 1 1 + n | 2 1 + n = ( ( T P ) ( 1 + n ) ( T P ) ( 1 + n ) ) 1 1 + n = ( P | T 1 + n | 2 P ) 1 1 + n P| T 1 + n | 2 1 + n P

by Hansen’s inequality [9]. On the other hand

| T 1 | 2 = T 1 T 1 =P T TP=P | T | 2 PP| T 1 + n | 2 1 + n P.

Hence

| T 1 1 + n | 2 1 + n | T 1 | 2 .

That is, T 1 is a class A(n) operator on ran ( T k ) ¯ .

For any x=( x 1 , x 2 )H,

T 3 k x 2 , x 2 = T k ( I P ) x , ( I P ) x = ( I P ) x , T k ( I P ) x =0,

which implies T 3 k =0.

Since σ(T)G=σ( T 1 )σ( T 3 ), where is the union of the holes in σ(T), which happen to be a subset of σ( T 1 )σ( T 3 ) by [[10], Corollary 7], σ( T 3 )=0, and σ( T 1 )σ( T 3 ) has no interior points, we have σ(T)=σ( T 1 ){0}.

In [6], Yuan and Ji introduced (n,k)-quasiparanormal operators. TB(H) is called a (n,k)-quasiparanormal operator for positive integers n and k if

T 1 + n ( T k x ) 1 1 + n T k x n 1 + n T ( T k x )

for xH. □

In the following we give the relations between (n,k)-quasiparanormal and k-quasi-class A(n) operators.

Theorem 2.2 Let T be a k-quasi-class A(n) operator for positive integers n and k. Then T is a (n,k)-quasiparanormal operator.

To give a proof of Theorem 2.2, the following famous inequality is needed.

Lemma 2.3 (Hölder-McCarthy’s inequality [11])

Let A0. Then the following assertions hold:

  1. (1)

    A r x,x A x , x r x 2 ( 1 r ) for r>1 and all xH.

  2. (2)

    A r x,x A x , x r x 2 ( 1 r ) for r[0,1] and all xH.

Proof of Theorem 2.2 Suppose that T is k-quasi-class A(n) operator. Then

T k ( | T 1 + n | 2 1 + n | T | 2 ) T k 0.

Let xH. Then by Hölder-McCarthy’s inequality, we have

T k + 1 x 2 = T k | T | 2 T k x , x T k | T 1 + n | 2 1 + n T k x , x = ( T ( 1 + n ) T 1 + n ) 1 1 + n T k x , T k x T ( 1 + n ) T 1 + n T k x , T k x 1 1 + n T k x 2 ( 1 1 1 + n ) = T 1 + n ( T k x ) 2 1 + n T k x 2 n n + 1 .

So we have

T k + 1 x T 1 + n ( T k x ) 1 1 + n T k x n n + 1 ,

hence T is a (n,k)-quasiparanormal operator. □

Remark We give an example which is (n,k)-quasiparanormal, but not k-quasi-class A(n).

Example 2.4 Let T= ( 1 0 1 0 ) B( l 2 l 2 ). Then T is (n,k)-quasiparanormal, but not k-quasi-class A(n).

By simple calculation we have

T k | T 1 + n | 2 1 + n T k = ( 2 1 1 + n 0 0 0 ) and T k | T | 2 T k = ( 2 0 0 0 ) .

Hence T is not k-quasi-class A(n). However, for all μ>0,

T k ( T ( 1 + n ) T 1 + n ( 1 + n ) μ n T T + n μ 1 + n ) T k = ( 2 [ 1 ( 1 + n ) μ n + n μ 1 + n ] 0 0 0 ) .

By arithmetic-geometric mean inequality, we have

1(1+n) μ n +n μ 1 + n 0
(2.1)

for all μ>0. Therefore T is (n,k)-quasiparanormal by [[6], Lemma 2.2].

Theorem 2.5 Let TB(H) be a k-quasi-class A(n) operator for positive integers k and n. If MH is an invariant subspace of T, then the restriction T | M is also a k-quasi-class A(n) operator.

Proof Let P be the orthogonal projection of onto , and let T 1 =T | M . Then T k P=P T k P and T 1 =PTP | M . Since T is a k-quasi-class A(n) operator, we have

P T k | T n + 1 | 2 n + 1 T k PP T k | T | 2 T k P.

Since

P T k | T n + 1 | 2 n + 1 T k P = P T k P | T n + 1 | 2 n + 1 P T k P = P T k P ( T ( n + 1 ) T n + 1 ) 1 n + 1 P T k P P T k ( P T ( n + 1 ) T n + 1 P ) 1 n + 1 T k P = P T k ( ( P T P ) n + 1 ( P T P ) n + 1 ) 1 n + 1 T k P = ( T 1 k | T 1 n + 1 | 2 n + 1 T 1 k 0 0 0 )

by Hansen’s inequality and

P T k | T | 2 T k P=P T k P T TP T k P= ( T 1 k | T 1 | 2 T 1 k 0 0 0 ) ,

we have

( T 1 k | T 1 n + 1 | 2 n + 1 T 1 k 0 0 0 ) P T k | T n + 1 | 2 n + 1 T k PP T k | T | 2 T k P= ( T 1 k | T 1 | 2 T 1 k 0 0 0 ) ,

that is, T 1 is also a k-quasi-class A(n) operator. □

In the following, we shall show that if T is a k-quasi-class A(n) operator, then the nonzero points of its point spectrum and joint point spectrum are identical, the eigen-spaces corresponding to distinct eigenvalues of T are mutually orthogonal, the nonzero points of its approximate point spectrum and joint approximate point spectrum are identical.

Theorem 2.6 Let TB(H) be a k-quasi-class A(n) operator for positive integers n and k. If λ0 and (Tλ)x=0 for some xH, then ( T λ ) x=0.

Proof We may assume that x0. Let M 0 be a span of {x}. Then M 0 is an invariant subspace of T and

T= ( λ T 2 0 T 3 ) on H= M 0 M 0 .
(2.2)

Let P be the orthogonal projection of onto M 0 . It suffices to show that T 2 =0 in (2.2). Since T is a k-quasi-class A(n) operator and x= T k ( x λ k ) ran ( T k ) ¯ , we have

P ( | T n + 1 | 2 n + 1 | T | 2 ) P0.
(2.3)

We remark

P| T 2 | 2 P=P T T TTP=P T P T TPTP= ( | λ | 4 0 0 0 ) .

Then by Hansen’s inequality and (2.3), we have

( | λ | 2 0 0 0 ) = ( P ( | T n + 1 | 2 n + 1 ) n + 1 P ) 1 n + 1 P| T n + 1 | 2 n + 1 PP | T | 2 P=P T TP= ( | λ | 2 0 0 0 ) .

Hence we may write

| T n + 1 | 2 n + 1 = ( | λ | 2 A A B ) .

We have

( | λ | 4 0 0 0 ) = ( P | T n + 1 | 2 P ) 2 n + 1 P | T n + 1 | 2 n + 1 | T n + 1 | 2 n + 1 P = ( 1 0 0 0 ) ( | λ | 2 A A B ) ( | λ | 2 A A B ) ( 1 0 0 0 ) = ( | λ | 4 + A A 0 0 0 ) .

This implies A=0 and | T n + 1 | 2 = ( | λ | 2 ( n + 1 ) 0 0 B n + 1 ) . On the other hand, by simple calculation we have

| T n + 1 | 2 = ( | λ | 2 ( n + 1 ) λ ¯ n + 1 i = 0 n λ i T 2 T 3 n i λ n + 1 ( i = 0 n λ i T 2 T 3 n i ) | i = 0 n λ i T 2 T 3 n i | 2 + | T 3 n + 1 | 2 ) .

Hence

i = 0 n λ i T 2 T 3 n i =0
(2.4)

and

B=| T 3 n + 1 | 2 n + 1 .

Since T is a k-quasi-class A(n) operator, by simple calculation we have

0 T k ( | T n + 1 | 2 n + 1 | T | 2 ) T k = ( 0 λ ¯ k + 1 T 2 T 3 k λ k + 1 T 3 k T 2 D ) ,

where D=λ T 3 k T 2 ( i = 0 k 1 λ i T 2 T 3 k 1 i )+[ λ ¯ ( i = 0 k 1 λ i T 2 T 3 k 1 i ) T 2 + T 3 k ( | T 3 n + 1 | 2 n + 1 | T 2 | 2 | T 3 | 2 )] T 3 k is a positive operator. Recall that ( X Y Y Z ) 0 if and only if X,Z0 and Y= X 1 2 W Z 1 2 for some contraction W. Thus we have

T 2 T 3 k =0
(2.5)

by λ0. By (2.4) and (2.5), we have T 2 =0. This completes the proof. □

Corollary 2.7 Let TB(H) be a k-quasi-class A(n) operator for positive integers n and k. Then the following assertions hold:

  1. (1)

    σ j p (T){0}= σ p (T){0}.

  2. (2)

    If (Tλ)x=0, (Tμ)y=0, and λμ, then x,y=0.

Proof (1) Clearly by Theorem 2.6.

  1. (2)

    Without loss of generality, we assume μ0. Then we have ( T μ ) y=0 by Theorem 2.6.

Thus we have μx,y=x, T y=Tx,y=λx,y. Since λμ, x,y=0. □

Theorem 2.8 Let TB(H) be a k-quasi-class A(n) operator for positive integers n and k. Then σ j a (T){0}= σ a (T){0}.

To prove Theorem 2.8, we need the following auxiliary results.

Lemma 2.9 (see [12])

Let be a complex Hilbert space. Then there exists a Hilbert space such that HK and a map φ:B(H)B(K) such that:

  1. (1)

    φ is a faithful -representation of the algebra B(H) on .

  2. (2)

    φ(A)0 for any A0 in B(H).

  3. (3)

    σ a (T)= σ a (φ(T))= σ p (φ(T)) for any TB(H).

Lemma 2.10 (see [13])

Let φ:B(H)B(K) be Berberian’s faithful -representation. Then σ j a (T)= σ j p (φ(T)).

Proof of Theorem 2.8 Let φ:B(H)B(K) be Berberian’s faithful -representation of Lemma 2.9. In the following, we shall show that φ(T) is also a k-quasi-class A(n) operator for positive integers n and k. In fact, since T is a k-quasi-class A(n) operator, we have

( φ ( T ) ) k ( | ( φ ( T ) ) n + 1 | 2 n + 1 | φ ( T ) | 2 ) ( φ ( T ) ) k = φ ( T k ( | T n + 1 | 2 n + 1 | T | 2 ) T k ) by Lemma 2.9(1) 0 by Lemma 2.9(2) .

Hence we have

σ a ( T ) { 0 } = σ a ( φ ( T ) ) { 0 } by Lemma 2.9(3) = σ p ( φ ( T ) ) { 0 } by Lemma 2.9(3) = σ j p ( φ ( T ) ) { 0 } by Corollary 2.7(1) = σ j a ( T ) { 0 } by Lemma 2.10 .

The proof is complete. □

Lemma 2.11 (see [5, 14])

If T satisfies ker(Tλ)ker ( T λ ) for some complex λ, then ker(Tλ)=ker ( T λ ) n for any positive integer n.

An operator is said to have finite ascent if ker T n =ker T n + 1 for some positive integer n.

Theorem 2.12 Let TB(H) be a k-quasi-class A(n) operator for positive integers n and k. Then Tλ has finite ascent for all complex number λ.

Proof By Theorem 2.2, we see that T is a (n,k)-quasiparanormal operator. So Tλ has finite ascent for all complex number λ by [[6], Theorem 4.1]. □

3 Tensor products for k-quasi-class A(n) operators

Let TS denote the tensor product on the product space HK for nonzero TB(H) and SB(K). The operation of taking tensor products TS preserves many properties of TB(H) and SB(K), but by no means all of them. For example the normaloid property is invariant under tensor products, the spectraloid property is not (see [[15], pp.623 and 631]); and TS is normal if and only if T and S are normal [16, 17]; however, there exist paranormal operators TB(H) and SB(K) such that TS is not paranormal [18]. Duggal [19] showed that for nonzero TB(H) and SB(K), TS is p-hyponormal if and only if T, S are p-hyponormal. This result was extended to p-quasihyponormal operators, class A operators, -class A operators, log-hyponormal operators and class A(s,t) operators ( ( | T | t | T | 2 s | T | t ) t s + t | T | 2 t , s,t>0) in [2023], respectively. The following theorem gives a necessary and sufficient condition for TS to be a k-quasi-class A(n) operator when T and S are both nonzero operators.

Theorem 3.1 Let TB(H) and SB(K) be nonzero operators. Then TSB(HK) is a k-quasi-class A(n) operator if and only if one of the following assertions holds:

  1. (1)

    T k + 1 =0 or S k + 1 =0.

  2. (2)

    T and S are k-quasi-class A(n) operators.

Proof It is clear that TS is a k-quasi-class A(n) operator if and only if

( T S ) k ( | ( T S ) 1 + n | 2 1 + n | T S | 2 ) ( T S ) k 0 T k ( | T 1 + n | 2 1 + n | T | 2 ) T k S k | S 1 + n | 2 1 + n S k + T k | T | 2 T k S k ( | S 1 + n | 2 1 + n | S | 2 ) S k 0 T k | T 1 + n | 2 1 + n T k S k ( | S 1 + n | 2 1 + n | S | 2 ) S k + T k ( | T 1 + n | 2 1 + n | T | 2 ) T k S k | S | 2 S k 0 .

Therefore the sufficiency is clear.

To prove the necessary. Suppose that TS is a k-quasi-class A(n) operator. Let xH and yK be arbitrary. Then we have

T k ( | T 1 + n | 2 1 + n | T | 2 ) T k x , x S k | S 1 + n | 2 1 + n S k y , y + T k | T | 2 T k x , x S k ( | S 1 + n | 2 1 + n | S | 2 ) S k y , y 0 .
(3.1)

It suffices to prove that if (1) does not hold, then (2) holds. Suppose that T k + 1 0 and S k + 1 0. To the contrary, assume that T is not a k-quasi-class A(n) operator, then there exists x 0 H such that

T k ( | T 1 + n | 2 1 + n | T | 2 ) T k x 0 , x 0 =α<0

and

T k | T | 2 T k x 0 , x 0 =β>0.

From (3.1) we have

α S k | S 1 + n | 2 1 + n S k y , y +β S k ( | S 1 + n | 2 1 + n | S | 2 ) S k y , y 0

for all yK, that is,

(α+β) S k | S 1 + n | 2 1 + n S k y , y β S k | S | 2 S k y , y
(3.2)

for all yK. Therefore S is a k-quasi-class A(n) operator. From Lemma 2.1 we can write S= ( S 1 S 2 0 S 3 ) on K= ran ( S k ) ¯ ker S k , where S 1 is a class A(n) operator. Let P be the orthogonal projection of onto ran ( S k ) ¯ . By the proof of Lemma 2.1, we have

( | S 1 1 + n | 2 1 + n 0 0 0 ) = ( ( S P ) ( 1 + n ) ( S P ) ( 1 + n ) ) 1 1 + n = ( P | S 1 + n | 2 P ) 1 1 + n P| S 1 + n | 2 1 + n P.

So we have

(α+β) S k | S 1 1 + n | 2 1 + n S k y , y (α+β) S k | S 1 + n | 2 1 + n S k y , y β S k | S | 2 S k y , y

for all yK by (3.2). Hence,

(α+β) | S 1 1 + n | 2 1 + n η , η β | S | 2 η , η =β | S 1 | 2 η , η
(3.3)

for all η ran ( S k ) ¯ .

Taking the supremum over all η ran ( S k ) ¯ , we have

(α+β) | S 1 1 + n | 1 1 + n 2 β S 1 2
(3.4)

by (3.3). Since self-adjoint operators are normaloid, we have

| S 1 1 + n | 1 1 + n 1 + n = ( | S 1 1 + n | 1 1 + n ) 1 + n = S 1 1 + n S 1 1 + n .
(3.5)

Hence we have

| S 1 1 + n | 1 1 + n S 1 .
(3.6)

By (3.4) and (3.6) we have

(α+β) S 1 2 β S 1 2 .

This implies that S 1 =0. Since S k + 1 y= S 1 S k y=0 for all yH, we have S k + 1 =0. This contradicts the assumption S k + 1 0. Hence T must be a k-quasi-class A(n) operator. A similar argument shows that S is also a k-quasi-class A(n) operator. The proof is complete. □

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Acknowledgements

This research is supported by the National Natural Science Foundation of China (11301155), (11271112), the Natural Science Foundation of the Department of Education, Henan Province (2011A110009), (13B110077), the Youth Science Foundation of Henan Normal University and the new teachers Science Foundation of Henan Normal University (No. qd12102).

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Li, X., Gao, F. On properties of k-quasi-class A(n) operators. J Inequal Appl 2014, 91 (2014). https://doi.org/10.1186/1029-242X-2014-91

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