Kyungpook Mathematical Journal 2018; 58(4): 637-650
On [m,C]-symmetric Operators
Department of Mathematics, Kanagawa University, Hiratsuka 259-1293, Japan, e-mail : chiyom01@kanagawa-u.ac.jp, Department of Mathematics and Statistics, Sejong University, Seoul 05006, Korea, e-mail : jieunlee7@sejong.ac.kr and jieun7@ewhain.net, Department of Mathematics, Tohoku Medical and Pharmaceutical University, Sendai 981-8558, Japan, e-mail : tanahasi@tohoku-mpu.ac.jp, Meguro-ku Nakane 11-10-201, Tokyo 152-0031, Japan, e-mail : juntomi@med.email.ne.jp
*Corresponding Author.
Received: August 23, 2017; Revised: January 12, 2018; Accepted: October 23, 2018; Published online: December 23, 2018.

This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License (http://creativecommons.org/licenses/by-nc/3.0/) which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract

In this paper first we show properties of isosymmetric operators given by M. Stankus [13]. Next we introduce an [m,C]-symmetric operator T on a complex Hilbert space ℋ. We investigate properties of the spectrum of an [m,C]-symmetric operator and prove that if T is an [m,C]-symmetric operator and Q is an n-nilpotent operator, respectively, then T +Q is an [m+2n − 2, C]-symmetric operator. Finally, we show that if T is [m,C]-symmetric and S is [n,D]-symmetric, then TS is [m+n−1, CD]-symmetric.

Keywords: Hilbert space, linear operator, conjugation, 𝑚-isometry, 𝑚-symmetric operator.
1. Introduction

Let ℋ be a complex Hilbert space with the inner product 〈,〉 and B(ℋ) be the set of bounded linear operators on ℋ. Let ℕ be the set of all natural numbers. For the study of Jordan operators, J.W. Helton ([9] and [10]) introduced an operator TB(ℋ) which satisfies

$αm(T):=∑j=0m(-1)j(mj)T*m-jTj=0 (m∈ℕ).$

In particular, if T is normal, then αm(T) = (T* − T)m. An operator TB(ℋ) is said to be an m-symmetric operator if αm(T) = 0. Hence T is 1-symmetric if and only if T is Hermitian. It is well known that if T is m-symmetric, then T is n-symmetric for all nm. The concept of m-symmetric operators is little strong. For example, if T is m-symmetric, then σ(T) ⊂ ℝ (cf.[10]). And T is Hermitian even if T is 2-symmetric. Also if T is normal and m-symmetric, then T is Hermitian due to the fact that T* − T is normal and nilpotent, that is, T* − T = 0.

Recently, C. Gu and M. Stankus ([8]) showed interesting properties of m-symmetric operators. On the other hand, for m ∈ ℕ, an operator TB(ℋ) is said to be an m-isometric operator if

$βm(T):=∑j=0m(-1)j(mj)T*m-jTm-j=0.$

It is well known that if T is m-isometric, then T is n-isometric for all nm. In 1995, J. Agler and M. Stankus [1] introduced an m-isometric operator and showed many important results of such an operator. If T is an invertible m-isometric operator and m is even, then T is (m–1)-isometric. But if T is m-symmetric and m is even, then T is always (m–1)-symmetric by Theorem 3.4 of [12]. For every odd number m, there exists an invertible m-isometric operator T which is not (m–1)-isometric (see Theorem 1 in [5]).

Throughout this paper, let I be the identity operator on ℋ and m, n be natural numbers. An operator QB(ℋ) is said to be a nilpotent operator of order n if Qn = 0 and Qn–1 ≠ 0. For a subset A ⊂ ℂ, let A* = { : zA }. Let σ(T) and σp(T) be the spectrum and the point spectrum of TB(ℋ), respectively. The approximate point spectrum of T is defined by σa(T) := { z ∈ ℂ : T − zI is not bounded below}, and the surjective spectrum of T is defined by σs(T) := { z ∈ ℂ : T − zI is not surjective}. It is known that σ(T) = σa(T)∪ σs(T), σa(T)* = σs(T*), and σs(T)* = σa(T*).

2. Isosymmetric Operators

First we show the following result of m-symmetric operators.

### Proposition 2.1

Let TB(ℋ). Then the following statements hold;

T is a 2-symmetric operator if and only if T is Hermitian.

Let T be an m-symmetric operator. For ab and non-zero vectors x, y ∈ ℋ, if Tx = ax and Ty = by, thenx, y〉 = 0.

Let T be an m-symmetric operator. For ab and sequences {xk}, {yk} of unit vectors of ℋ, if (T − a)xk → 0 and (T − b)yk → 0, then$limk→∞〈xk,yk〉=0$.

Proof

(a) If T is Hermitian, then it is obvious that T is 2-symmetric. If T is 2-symmetric, then T is 1-symmetric from [12, Theorem 3.4] and so it is Hermitian. (b) Since a, bσ(T), it follows from [10] that a, b are real numbers. Hence it holds

$0=〈αm(T)x,y〉=(b-a)m·〈x,y〉.$

Since ab, we have 〈x, y〉 = 0.

(c) By similar arguments of the proof of (b), a, b are real numbers and it holds

$0=limk→∞〈αm(T)xk,yk〉=(b-a)m·limk→∞〈xk,yk〉.$

Since ab, we have $limk→∞〈xk,yk〉=0$.

### Definition 1

For an operator TB(ℋ), we define γm,n(T) by

$γm,n(T)=∑j=0m(-1)j(mj)T*m-jαn(T)Tm-j=∑k=0n(-1)k(nk)T*n-kβm(T)Tk.$

Then T is said to be (m, n)-isosymmetric if γm,n(T) = 0.

It is easy to see that

$γm+1,n(T)=T*γm,n(T)T-γm,n(T) and γm,n+1(T)=T*γm,n(T)-γm,n(T)T.$

Hence if T is (m, n)-isosymmetric, then T is (m′, n′)-isosymmetric for all n′n and n′n. M. Stankus proved the following properties.

### Proposition 2.2.([13, Corollary 30])

Let T be (m, n)-isosymmetric.

If σ(T) ⊂ {x ∈ ℝ : |x| > 1} or σ(T) ⊂ {x ∈ ℝ : |x| < 1}, then T is n-symmetric.

If σ(T) ⊂ {e : 0 < θ < π} or σ(T) ⊂ {e : π < θ < 2π}, then T is m-isometric.

For a, b ∈ ℂ and non-zero vectors x, y ∈ ℋ, if Tx = ax, Ty = by, then it holds that

$〈γm,n(T)x,y〉=〈(∑j=0m(-1)j(mj)T*m-jαn(T)Tm-j)x,y〉 =(ab¯-1)m(a-b¯)n〈x,y〉.$

Hence we have the following theorem.

### Theorem 2.3

Let T be (m, n)-isosymmetric and x, y be unit vectors and xk, yk be sequences of unit vectors of ℋ.

If Tx = ax, Ty = by, ab and ab̄, thenx, y〈 = 0.

If (T − a)xk → 0, (T − b)yk → 0 (k → ∞), ab and ab̄, then$limk→∞〈xk,yk〉=0$.

### Theorem 2.4

Let T be (m, n)-isosymmetric.

Then Tk is (m, n)-isosymmetric for any k ∈ ℕ.

If T is invertible, then T−1is (m, n)-isosymmetric.

Proof

(1) Note that for k ∈ ℕ, the following equation holds;

$(ykxk-1)m(yk-xk)n=((yx-1) (yk-1xk-1+yk-2xk-2+⋯+1))m·((y-x) (yk-1+yk-2x+⋯+xk-1))n=∑ℓ=0m(k-1)∑j=0n(k-1)λℓμjym(k-1)-ℓyn(k-1)-j(yx-1)m(y-x)nxjxm(k-1)-ℓ$

where λ and μj are some constants. From this, we have

$γm,n(Tk)=∑ℓ=0m(k-1)∑j=0n(k-1)λℓμjT*m(k-1)-ℓ+n(k-1)-jγm,n(T)Tj+m(k-1)-ℓ.$

Hence Tk is (m, n)-isosymmetric.

(2) Assume that T is invertible. Since

$0=T*-m-nγm,n(T)T-m-n =∑j=0m(-1)j(mj)T*-m-nT*m-jαn(T)Tm-jT-m-n =∑j=0m(-1)j(mj)T*-n-jαn(T)T-n-j =∑j=0m(-1)j(mj)T*-j(T*-nαn(T)T-n)T-j ={∑j=0m(-1)j(mj)T*-j·(αn(T-1))·T-j=γm,n(T-1) (m is even)∑j=0m(-1)j(mj)T*-j·(-αn(T-1))·T-j=-γm,n(T-1) (m is odd),$

it follows that T−1 is (m, n)-isosymmetric.

Operators T and S are said to be doubly commuting if TS = ST and TS* = S*T. From the equation

$((y1+y2) (x1+x2)-1)m((y1+y2)-(x1+x2))n=∑j=0n∑i+l+h=m(nj) (mi,l,h) (y1+y2)iy2l(y1x1-1)h(y1-x1)n-j(y2-x2)jx1lx2i,$

if T and S are doubly commuting, then it holds

$γm,n(T+S)=∑j=0n∑i+l+h=m(nj) (mi,l,h)·(T*+S*)iS*lγh,n-j(T)αj(S)TlSi.$

### Theorem 2.5

Let T be (m, n)-isosymmetric and let Q be a nilpotent operator of order k. If T and Q are doubly commuting, then T +Q is (m+2k −2, n+2k −1)- isosymmetric.

Proof

From equation (2.1), it holds

$γm+2k-2,n+2k-1(T+Q)=∑j=0n+2k-1∑i+l+h=m+2k-2(n+2k-1j) (m+2k-2i,l,h)·(T*+Q*)iQ*lγh,n+2k-1-j(T)αj(Q)TlQi.$

(1) If j ≥ 2k or ik or lk, then αj(Q) = 0 or Qi = 0 or Q*l = 0, respectively.

(2) If j ≤ 2k −1 and ik −1 and lk − 1, then h = m+2k − 2 − i − lm and n + 2k − 1 − jn + 2k − 1 (2k − 1) = n, i.e., γh,n+2k−1−j(T) = 0.

By (1) and (2) we have γm+2k−2,n+2k−1(T + Q) = 0. Therefore T + Q is (m + 2k − 2, n + 2k − 1)-isosymmetric.

Note that the equation

$(y1y2x1x2-1)m·(y1y2-x1x2)n=∑k=0m∑j=0n(mk) (nj)y1j+k(y1x1-1)m-k(y1-x1)n-j(y2x2-1)k(y2-x2)jx1kx2n-j.$

From this, if T and S are doubly commuting, then it holds

$γm,n(TS)=∑k=0m∑j=0n(mk) (nj)T*j+kγm-k,n-j(T)·γk,j(S)TkSn-j.$

### Theorem 2.6

Let T be (m, n)-isosymmetric and let S be m′-isometric and n′-symmetric. If T and S are doubly commuting, then TS is (m+m′ −1, n+n′ −1)- isosymmetric.

Proof

From equation (2.2), it holds

$γm+m′-1,n+n′-1(TS)=∑k=0m+m′-1∑j=0n+n′-1(n+n′-1j) (m+m′-1k)T*j+k·γm+m′-1-k,n+n′-1-j(T)·γk,j(S)·TkSn-j.$

(1) If km′ or jn′, then γk,j(S) = 0.

(2) If km′ −1 and jn′ − 1, then m+m′ − 1 − km and n + n′ − 1 − jn, i.e., γm+m′−1−k,n+n′−1−j(T) = 0.

By (1) and (2) we have γm+m′−1,n+n′−1(TS) = 0. Hence it completes the proof.

For a complex Hilbert space ℋ, let ℋ⊗ℋ denote the completion of the algebraic tensor product of ℋ and ℋ endowed a reasonable uniform cross-norm. For operators TB(ℋ) and SB(ℋ), TSB(ℋ ⊗ ℋ) denote the tensor product operator defined by T and S. Note that TS = (TI)(IS) = (IS)(TI).

### Theorem 2.7

Let T be (m, n)-isosymmetric and let S be m′-isometric and n′-symmetric. Then TS is (m + m′ − 1, n + n′ − 1)-isosymmetric.

Proof

It is clear that if T is (m, n)-isosymmetric, then TI is (m, n)-isosymmetric and if S is m′-isometric and n′-symmetric, then IS is m′-isometric and n′-symmetric. Since TI and IS are doubly commuting, it follows from Theorem 2.6 that TS is (m + m′ − 1, n + n′ − 1)-isosymmetric. Hence it completes the proof.

3. Conjugation and Example

In this section, we introduce [m,C]-symmetric operators and provide several examples. An antilinear operator C on ℋ is said to be a conjugation if C satisfies C2 = I and 〈Cx,Cy〉= 〈y, x〉 for all x, y ∈ ℋ. An operator TB(ℋ) is said to be complex symmetric if CTC = T* for some conjugation C.

### Definition 2

For an operator TB(ℋ) and a conjugation C, we define the operator αm(T;C) by

$αm(T;C)=∑j=0m(-1)j(mj)CTm-jC·Tj.$

An operator TB(ℋ) is said to be an [m,C]-symmetric operator if αm(T;C) = 0.

Hence if T is complex symmetric and [m,C]-symmetric, then T is m-symmetric. It holds that

$CTC·αm(T;C)-αm(T;C)·T=αm+1(T;C).$

Moreover, if T is [m,C]-symmetric, then T is [n,C]-symmetric for every natural number n (≥ m) and ker(αm−1(T;C)) (m ≥ 2) is an invariant subspace for T.

### Example 3.1

Let ℋ = ℂ2 and let C be a conjugation on ℋ given by $C(xy)=(y¯x¯)$ for x, y ∈ ℂ.

If $T=(i11-i)$ on C2, then T is not Hermitian and $CTC=(i11-i)=T$.

Hence T is [1, C]-symmetric.

Hence, in this case, σ(T) = {0} due to the fact that T is nilpotent.

Let $S=(i22-i)$ on C2. Then S is not Hermitian and $CSC=(i22-i)$ and CSC = SS*. Therefore, S is [1, C]-symmetric. Furthermore, σ(S) = {1,−1}.

If $R=(112i12i2)$ on C2, then

$CR2C-2CRC·R+R2=(154-32i-32i34)-2(940094)+(3432i32i154)=0.$

Hence R is [2, C]-symmetric. It is easy to see that R is not [1, C]-symmetric. Moreover, $σ(R)={32}$.

Let $W=(2i11-2i)$ on C2. Then it is easy to see that CWC = WW*.

Hence W is [1, C]-symmetric and $σ(W)={3i,-3i}$.

### Example 3.2

Let ℋ = ℓ2, let ${en}n=1∞$ be the natural basis of ℋ and let C : ℋ → ℋ be a conjugation given by

$C(∑n=1∞xnen)=∑n=1∞xn¯en$

where {xn} is a sequence in C with $∑n=1∞|xn|2<∞$ and Cen = en.

If UB(ℋ) is the unilateral shift on ℓ2, then it is easy to compute U = CUC and so U is a [1, C]-symmetric operator with (unit disk).

Let W be the weighted shift given by Wen = αnen+1, where $αn={2i (n=1)n+1ni (n≥2)$.

Then

$(CW2C-2CWCW+W2)en=[(αn¯-αn)αn+1¯-αn(αn+1¯-αn+1)]en$

for all n ≥ 1. Hence W is [2, C]-symmetric operator.

An operator A ∈ ℒ(ℋ) is n-Jordan if A = T + N where T is self-adjoint, N is nilpotent of order [$n+12$], and TN = NT where [k] denotes the integer part of k.

### Example 3.3

Let C be a conjugation C on ℋ. Suppose that A = T + N is an n-Jordan operator where T = T* = CTC, N is nilpotent of order [$n+12$], TN = NT, and CN = NC. Then A is [n,C]-symmetric for the conjugation C. Indeed, since T = T* = CTC, TN = NT, and CN = NC, it follows that

$∑j=0n(-1)j(nj)CAn-jC·Aj=∑j=0n(-1)j(nj) (T+N)n-j·(T+N)j=0$

which means that A is an n-symmetric operator from [12, Theorem 3.2]. Hence A is [n,C]-symmetric.

4. [m,C]-symmetric Operators

Let C be a conjugation on ℋ. Then C satisfies ||Cx|| = ||x|| and C(αx) = ᾱ·Cx for all x ∈ ℋ and all α ∈ ℂ. Moreover, since C2 = I, it follows that (CTC)* = CT*C and (CTC)n = CTnC for every positive integer n (see [7] for more details).

We now provide properties of [m,C]-symmetric operators.

### Theorem 4.1

Let TB(ℋ) and let C be a conjugation on. Then the following assertions hold;

T is an [m,C]-symmetric operator if and only if so is T*.

If T is an [m,C]-symmetric operator, then Tk is [m,C]-symmetric for any k ∈ ℕ.

If T is an [m,C]-symmetric operator and invertible, then T−1is [m,C]- symmetric.

If T is a [2, C]-symmetric operator, then ker(T) ⊂ ker(T2)∩C(ker(T2)).

Proof

(a) Since T is [m,C]-symmetric, it follows that αm(T;C) = 0. Therefore,

$0=C(αm(T;C))*C={αm(T*;C) (m is even)-αm(T*;C) (m is odd).$

Hence T* is [m,C]-symmetric. The converse implication holds in a similar way. (b) Note that

$(ak-bk)m=((a-b) (ak-1+ak-2b+⋯+bk-1))m=∑j=0m(k-1)λjam(k-1)-j(a-b)mbj,$

where λj are some coefficients (j = 0, ...,m(k − 1)). This implies that

$αm(Tk;C)=∑j=0m(k-1)λjCTm(k-1)-jC·αm(T;C)·Tj=0.$

Hence Tk is [m,C]-symmetric.

(c) Since T is [m,C]-symmetric, it follows that αm(T;C) = 0 and therefore

$0=CT-mC·αm(T;C)·T-m={αm(T-1;C) (m is even)-αm(T-1;C) (m is odd).$

Hence T−1 is [m,C]-symmetric.

(d) It is clear ker(T) ⊂ ker(T2). If T is [2, C]-symmetric and x ∈ ker(T), then

$CT2Cx=2CTCTx-T2x=0$

and hence T2Cx = 0. Thus Cx ∈ ker(T2) and so xC(ker(T2)). Hence we get ker(T) ⊂ ker(T2)∩C(ker(T2)).

### Lemma 4.2

For TB(ℋ), a conjugation C, and two complex numbers λ, μ, it holds

$αm(T;C)=∑n1+n2+n3=m(-1)n2(mn1,n2,n3) (CTC-λI)n1(T-μI)n2(λ-μ)n3.$

In particular, for λ ∈ ℂ we have

$αm(T;C)=∑j=0m(-1)j(mj) (CTC-λI)m-j(T-λI)j.$
Proof

Using the multinomial formula, it holds

$αm(T;C)=(y-x)m|y=CTC,x=T=([y-λ]-[x-μ]+[λ-μ])m|y=CTC,x=T=∑n1+n2+n3=m(-1)n2(mn1,n2,n3) (y-λ)n1(x-μ)n2(λ-μ)n3|y=CTC,x=T=∑n1+n2+n3=m(-1)n2(mn1,n2,n3) (CTC-λI)n1(T-μI)n2(λ-μ)n3.$

Equation (4.1) follows from λ = μ in the first formula.

By Lemma 2.7 of [3], for TB(ℋ) and two complex numbers λ, μ, it holds

$βm(T)=∑n1+n2+n3=m(-1)n2(mn1,n2,n3) (T*-μ¯I)n1Tn1μ¯n2(T-λI)n2(λμ¯-1)n3.$

We investigate properties of spectra of [m,C]-symmetric operators. In [11], S. Jung, E. Ko and J. E. Lee proved the following result.

### Proposition 4.3.([11, Lemma 3.21])

If C is a conjugation onand TB(ℋ), then σ(CTC) = σ(T)*, σp(CTC) = σp(T)*, σa(CTC) = σa(T)* and σs(CTC) = σs(T)*.

### Theorem 4.4

Let TB(ℋ) be an [m,C]-symmetric operator where C is a conjugation on. Then σp(T) = σp(T)*, σa(T) = σa(T)*, σs(T) = σs(T)* and σ(T) = σ(T)*.

Proof

Let zσa(T). Then there exists a sequence {xn} of unit vectors such that (Tz)xn → 0 (n→∞). By equation (4.1) it holds

$αm(T;C)=∑j=0m(-1)j(mj) (CTC-zI)m-j(T-zI)j.$

Hence we have 0 = limn→∞αm(T;C)xn = limn→∞ (CTCz)m xn. Therefore, it is easy to see zσa(CTC). Hence σa(T) ⊂ σa(CTC). Since σa(CTC) = σa(T)* by Proposition 4.3, this means σa(T) ⊂ σa(T)*. Hence σa(T)*σa(T)** = σa(T) and so σa(T) = σa(T)*.

Since T* is also an [m,C]-symmetric operator by Theorem 4.1, we have σa(T*) = σa(T*)*. Hence σs(T) = σs(T)* and σ(T) = σa(T) ∪ σs(T) = σa(T)*σs(T)* = σ(T)*. From the above proof it is clear that σp(T) = σp(T)*.

For T,SB(ℋ), a pair (T,S) of operators is said to be a C-doubly commuting pair if TS = ST and CSC · T = T · CSC for a conjugation C.

### Lemma 4.5

Let (T,S) be a C-doubly commuting pair where C is a conjugation on. Then it holds

$αm(T+S;C)=∑j=0m(mj)αj(T;C)·αm-j(S;C).$
Proof

From the assumption, it holds T ·CSjC = CSjC ·T and S·CTjC = CTjC ·S for every j ∈ ℕ. It is clear that equation (4.2) holds for m = 1. Assume that equation (4.2) holds for m. Then by (3.1) we have

$αm+1(T+S;C)=C(T+S)C·αm(T+S;C)-αm(T+S;C)·(T+S)=∑j=0m(mj) (CTC+CSC)·αj(T;C)·αm-j(S;C)-∑j=0m(mj)αj(T;C)·αm-j(S;C)·(T+S)=∑j=0m(mj) (CTC·αj(T;C)-αj(T;C)·T)αm-j(S;C)+∑j=0m(mj)αj(T;C) (CSC·αm-j(S;C)-αm-j(S;C)·S)=∑j=0m(mj)αj+1(T;C)·αm-j(S;C)+∑j=0m(mj)αj(T;C)·αm+1-j(S;C)=∑j=0m+1(m+1j)αj(T;C)·αm+1-j(S;C).$

Hence equation (4.2) holds for any m ∈ ℕ.

Therefore we have the following theorem.

### Theorem 4.6

Let TB(ℋ) be an [m,C]-symmetric operator and let SB(ℋ) be an [n,C]-symmetric operator where C is a conjugation on. If (T,S) is a C-doubly commuting pair, then T + S is an [m + n − 1, C]-symmetric operator.

Proof

By Lemma 4.5, it holds

$αm+n-1(T+S;C)=∑j=0m+n-1(m+n-1j) αj(T;C)·αm+n-1-j(S;C).$

(i) If 0 ≤ jm − 1, then m+ n − 1 − jm+ n − 1 − (m − 1) = n. Therefore we have

αm+n−1−j(S;C) = 0.

(ii) If jm, then αj(T;C) = 0.

Hence we get αm+n−1(T + S;C) = 0 and so T + S is [m+ n − 1, C]-symmetric.

### Theorem 4.7

Let C be a conjugation on. If Q is a nilpotent operator of order n, then Q is a [2n − 1, C]-symmetric operator.

Proof

It holds

$α2n-1(Q;C)=∑j=02n-1(-1)j(2n-1j)CQ2n-1-jC·Qj.$

(i) If 0 ≤ jn − 1, then 2n − 1 − j ≥ 2n − 1 − (n − 1) = n. Hence Q2n−1−j = 0.

(ii) If jn, then Qj = 0.

Therefore α2n−1(Q;C) = 0 and hence Q is [2n − 1, C]-symmetric.

### Corollary 4.8

Let TB(ℋ) be an [m,C]-symmetric operator and let QB(ℋ) be a nilpotent operator of order n where C is a conjugation on. If (T,Q) is a C-doubly commuting pair, then T + Q is an [m+2n − 2, C]-symmetric operator.

Proof

The proof follows from Theorems 4.6 and 4.7.

### Example 4.9

Let Cn be the conjugation on Cn defined by

$Cn(z1,z2,⋯,zn):=(z1¯,z2¯,⋯,zn¯).$

Assume that Rn is an n × n matrix as follows;

$Rn=aIn+Qn=(a00⋯00a0⋯0⋮⋱⋱⋱⋮000⋱0000⋯a)+(0b0⋯000b⋯0⋮⋱⋱⋱⋮000⋱b000⋯0)$

for a, bC. Since Qn is nilpotent of order n, it follows that Corollary 4.8 that Rn is a [2n − 1, Cn]-symmetric operator.

### Lemma 4.10

If (T,S) is a C-doubly commuting pair where C is a conjugation on ℋ, then it holds

$αm(TS;C)=∑j=0m(mj)αj(T;C)·Tm-j·CSjC·αm-j(S;C).$
Proof

It is easy to see that equation (4.3) holds for m = 1. Assume that equation (4.3) holds for m. Then by (3.1) we obtain

$αm+1(TS;C)=(CTSC)·αm(TS;C)-αm(TS;C)·TS=CTC·CSC∑j=0m(mj)αj(T;C)·Tm-j·CSjC·αm-j(S;C)-∑j=0m(mj)αj(T;C)·Tm+1-j·CSjC·αm-j(S;C)·S=∑j=0m(mj) (CTC·αj(T;C)-αj(T;C)·T)Tm-j·CSj+1C·αm-j(S;C)+∑j=0m(mj)αj(T;C)·Tm+1-j·CSjC·(CSC·αm-j(S;C)-αm-j(S;C)·S)=∑j=0m(mj)αj+1(T;C)·Tm-j·CSj+1C·αm-j(S;C)+∑j=0m(mj)αj(T;C)·Tm+1-j·CSjC·αm+1-j(S;C)=∑j=0m+1(m+1j)αj(T;C)·Tm+1-j·CSjC·αm+1-j(S;C).$

Hence equation (4.3) holds for any m ∈ ℕ.

### Theorem 4.11

Let T be an [m,C]-symmetric operator and let S be an [n,C]- symmetric operator where C is a conjugation on. If (T,S) is a C-doubly commuting pair, then TS is an [m + n − 1, C]-symmetric operator.

Proof

Since (T,S) is a C-doubly commuting pair, it follows from equation (4.3) that

$αm+n-1(TS;C)=∑j=0m+n-1(m+n-1j)αj(T;C)·Tm+n-1-j·CSjC·αm+n-1-j(S;C).$

(i) If 0 ≤ jm − 1, then m+ n − 1 − jm+ n − 1 − (m − 1) = n. Therefore we get αm+n−1−j(S;C) = 0.

(ii) If mj, then αj(T;C) = 0.

Therefore αm+n−1(TS;C) = 0. Hence TS is [m + n − 1, C]-symmetric.

### Corollary 4.12

Let C be a conjugtion on. If T = T1I and S = IS1where T1and S1are [m,C]-symmetric, then TS is [2m − 1, C]-symmetric.

Proof

Since T1 and S1 are [m,C]-symmetric, it follows that T = T1I and S = IS1 are [m,C]-symmetric. In addition, we know that (T,S) is a C-doubly commuting pair. Therefore, TS is [2m − 1, C]-symmetric by Theorem 4.11.

In [6], B. Duggal proved the following proposition.

### Proposition 4.13

Let T and S be an m-isometric operator and an n-isometric operator, respectively. Then TS is an (m + n − 1)-isometric operator.

Similarly, we show the following result.

### Theorem 4.14

Let T be an [m,C]-symmetric operator and let S be an [n,D]- symmetric operator where C and D are conjugations on. Then TS is an [m + n − 1, CD]-symmetric operator on ℋ ⊗ ℋ.

Proof

Since C and D are conjugations on ℋ, it follows from [4] that CD is a conjugation on ℋ ⊗ ℋ. If T is [m,C]-symmetric and S is [n,D]-symmetric, it is easy to see that TI is [m,CD]-symmetric and IS is [n,CD]-symmetric on ℋ ⊗ ℋ, respectively. Also it is clear that (TI, IS) is a CD-doubly commuting pair. Since TS = (TI)(IS), it follows from Theorem 4.11 that (TI)(IS) = TS is [m + n − 1, CD]-symmetric.

Acknowledgements

This paper is dedicated to the memory of Professor Takayuki Furuta with cordial appreciation. This is partially supported by Grant-in-Aid Scientific Research No.15K04910. The second author was supported by Basic Science Research Program through the National Research Foundation of Korea(NRF) funded by the Ministry of Education, Science and Technology(2016R1A2B4007035).

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