Semicommutators and Zero Product of Block Toeplitz Operators with Harmonic Symbols

Abstract

We completely characterize the finite rank semicommutators, commutators, and zero product of block Toeplitz operators 𝑇_𝐹 and 𝑇_𝐺 with 𝐹,𝐺∈ℎ^∞⊗𝑀_𝑛×𝑛 on the vector valued Bergman space 𝐿²_𝑎(𝔻,ℂ^𝑛).

1. Introduction

Let 𝔻 be the open unit disk in the complex plane ℂ, and let 𝑑𝐴 be the normalized Lebesgue area measure on 𝔻. Denote by 𝐿^∞(𝔻,𝑑𝐴) and 𝐿²(𝔻,𝑑𝐴) the space of essential bounded measurable functions and the space of square integral functions with respect to 𝑑𝐴, respectively. The Bergman space 𝐿²_𝑎 consists of all analytic functions in 𝐿²(𝔻,𝑑𝐴). Let 𝐿²(𝔻,ℂ^𝑛)=𝐿²(𝔻,𝑑𝐴)⊗ℂ^𝑛 and 𝐿²_𝑎(𝔻,ℂ^𝑛)=𝐿²_𝑎⊗ℂ^𝑛, where ⊗ means the Hilbert space tensor product. Let 𝑀_𝑛×𝑛 be the set of all 𝑛×𝑛 complex matrixes in the complex plane ℂ, and let 𝐿^∞(𝔻,𝑑𝐴)⊗𝑀_𝑛×𝑛 be the space of matrix valued essential bounded Lebesgue measurable functions on 𝔻.

Given a matrix valued function Φ(𝑧)=[𝜑_𝑖𝑗(𝑧)]_𝑛×𝑛∈𝐿^∞(𝔻,𝑑𝐴)⊗𝑀_𝑛×𝑛, the Toeplitz operator 𝑇_Φ and the Hankel operator 𝐻_Φ on 𝐿²_𝑎(𝔻,ℂ^𝑛) with symbol Φ are defined by 𝑇_Φℎ=𝑃(Φℎ) and 𝐻_Φℎ=(𝐼−𝑃)(Φℎ), respectively, where 𝑃 is the orthogonal projection from 𝐿²(𝔻,ℂ^𝑛) onto 𝐿²_𝑎(𝔻,ℂ^𝑛). Then the operators 𝑇_Φ and 𝐻_Φ have the following matrix representations: 𝑇_Φ=𝑇_𝜑11⋯𝑇_𝜑1𝑛𝑇⋮⋮_𝜑𝑛1⋯𝑇_𝜑𝑛𝑛,𝐻_Φ=𝐻_𝜑11⋯𝐻_𝜑1𝑛𝐻⋮⋮_𝜑𝑛1⋯𝐻_𝜑𝑛𝑛. (1)

The problem of the finite rank semicommutator of two Toeplitz operators on the Hardy space has been completely solved in [1, 2]. The analogous problems on the Bergman space have been characterized in [3–5]. In the case of vector valued Hardy space, Gu and Zheng [6] have studied the problems of compact semicommutators and commutators of two block Toeplitz operators.

In this paper, we study the related problems of the two block Toeplitz operators 𝑇_𝐹 and 𝑇_𝐺 with 𝐹,𝐺∈ℎ^∞⊗𝑀_𝑛×𝑛 on the vector valued Bergman space 𝐿²_𝑎(𝔻,ℂ^𝑛), where ℎ^∞ denotes the space of all bounded harmonic functions on 𝔻. Our main idea here is to reduce the problems on 𝐿²_𝑎(𝔻,ℂ^𝑛) to the problems of the finite sum of some Toeplitz operator products or some Hankel operator products on the Bergman space 𝐿²_𝑎. In the following, we recall two useful theorems. For every 𝑓∈ℎ^∞, denote by 𝑓₊ (or 𝑓₋=𝑓−𝑓₊) the analytic part (or the coanalytic part) of 𝑓.

Theorem 1 (see [7]). Suppose that 𝑓_𝑗 and 𝑔_𝑗 are bounded harmonic functions on 𝔻 for 𝑗=1,…,𝑘. Then the following are equivalent:(1)∑^𝑘_𝑗=1𝐻^∗_𝑔𝑗𝐻_𝑓𝑗 has finite rank,(2)∑^𝑘_𝑗=1𝐻^∗_𝑔𝑗𝐻_𝑓𝑗=0,(3)𝜎(𝑓₁,…,𝑓_𝑘;𝑔₁,…,𝑔_𝑘)≡0, where 𝜎(𝑓₁,…,𝑓_𝑘;𝑔₁,…,𝑔_𝑘)=Δ[∑^𝑘_𝑖=1(𝑓_𝑖)₋(𝑔_𝑖)₊]=(1−|𝑧|²)²∑^𝑘_𝑖=1(𝑓_𝑖)^′₋(𝑔_𝑖)^′₊.

Theorem 2 (see [8]). Let 𝑢₁,…𝑢_𝑛,𝑣₁,…,𝑣_𝑛 be bounded harmonic functions on 𝔻, and 𝑥₁,…,𝑥_𝑛, 𝑦₁,…,𝑦_𝑛∈𝐿²_𝑎. Then ∑^𝑛_𝑗=1𝑇_𝑢𝑗𝑇_𝑣𝑗=∑^𝑛_𝑗=1𝑥_𝑗⊗𝑦_𝑗 if and only if the following two conditions hold: (a)∑^𝑛_𝑗=1𝑢_𝑗𝑣_𝑗=(1−|𝑧|²)²∑^𝑛_𝑗=1𝑥_𝑗𝑦_𝑗; (b)∑^𝑛_𝑗=1[𝑃𝑢_𝑗−𝑢_𝑗(0)][𝑃𝑣_𝑗−𝑣_𝑗(0)]=0.

2. Semicommutators of Block Toeplitz Operators

In this section, we discuss the finite rank semicommutators and commutators of the block Toeplitz operators 𝑇_𝐹, 𝑇_𝐺 on 𝐿²_𝑎(𝔻,ℂ^𝑛) with 𝐹,𝐺∈ℎ^∞⊗𝑀_𝑛×𝑛. In the following part of this paper, 𝐸_𝑗 denote the matrix unit with (𝑗,𝑗) th entry equal to one and all others equal to zero.

Theorem 3. Let 𝐹(𝑧)=𝐹₊(𝑧)+𝐹₋(𝑧), and let 𝐺(𝑧)=𝐺₊(𝑧)+𝐺₋(𝑧)∈ℎ^∞⊗𝑀_𝑛×𝑛 such that 𝐹₊(0)=𝐺₋(0)=0_𝑛×𝑛. Then the following statements are equivalent: (1)𝑇_𝐹𝐺−𝑇_𝐹𝑇_𝐺 has finite rank; (2)𝑇_𝐹𝐺=𝑇_𝐹𝑇_𝐺; (3)for each 1≤𝑗≤𝑛, there exist a matrix 𝑀_𝑗∈𝑀_𝑛×𝑛 and a permutation matrix 𝑄_𝑗, such that 𝑀_𝑗𝑄_𝑗𝐺₋𝐸_𝑗=0_𝑛×𝑛 and 𝐹₊𝑄_𝑗(𝐼+𝑀_𝑗)=0_𝑛×𝑛.

Proof. (1)⇒(2) Note that 𝑇_𝐹𝐺−𝑇_𝐹𝑇_𝐺=𝐻^∗_𝐹^∗𝐻_𝐺. If 𝐻^∗_𝐹^∗𝐻_𝐺 has a finite rank, then ∑^𝑛_𝑝=1𝐻^∗_𝑓𝑖𝑝𝐻_𝑔𝑝𝑗 has a finite rank for 1≤𝑖,𝑗≤𝑛. By Theorem 1, we have that ∑^𝑛_𝑝=1𝐻^∗_𝑓𝑖𝑝𝐻_𝑔𝑝𝑗=0 for 1≤𝑖,𝑗≤𝑛. Hence, 𝑇_𝐹𝐺=𝑇_𝐹𝑇_𝐺.
(2)⇒(3) If 𝑇_𝐹𝐺−𝑇_𝐹𝑇_𝐺=𝐻^∗_𝐹^∗𝐻_𝐺=0, then [𝐻^∗_𝐹^∗𝐻_𝐺]_𝑖,𝑗=∑^𝑛_𝑝=1𝐻^∗_𝑓𝑖𝑝𝐻_𝑔𝑝𝑗=0, for 1≤𝑖,𝑗≤𝑛. By Theorem 1, we get (1−|𝑧|²)²∑^𝑛_𝑝=1[(𝑓_𝑖𝑝)^′₊(𝑔_𝑝𝑗)^′₋]=0 on 𝔻. Therefore, ∑^𝑛_𝑝=1[(𝑓_𝑖𝑝)^′₊(𝑔_𝑝𝑗)^′₋]=0 on 𝔻 for 1≤𝑖,𝑗≤𝑛.
It is known that if 𝐻(𝑧,𝑤) is holomorphic in 𝔻×𝔻 and 𝐻(𝑧,𝑧)=0, then 𝐻(𝑧,𝑤)≡0 for (𝑧,𝑤)∈𝔻×𝔻 in several complex variables. So we can “complexificate” the above identity and have ∑^𝑛_𝑝=1[(𝑓_𝑖𝑝)^′₊(𝑧)(𝑔_𝑝𝑗)^′₋(𝑤)]=0 for 1≤𝑖,𝑗≤𝑛, (𝑧,𝑤)∈𝔻×𝔻. Therefore, 0=^𝑛_𝑝=1𝑓_𝑖𝑝^′₊(𝑢)𝑔_𝑝𝑗^′₋(𝑣=_{𝑧0𝑛𝑝=1})𝑓_𝑖𝑝^′₊(𝑢)𝑔_𝑝𝑗^′₋(𝑣=)𝑑𝑢_𝑛𝑝=1𝑓_𝑖𝑝+(𝑧)𝑔_𝑝𝑗^′₋(𝑣=_{𝑤0𝑛𝑝=1})𝑓_𝑖𝑝+(𝑧)𝑔_𝑝𝑗^′₋(𝑣=)𝑑𝑣_𝑛𝑝=1𝑓_𝑖𝑝+(𝑧)𝑔_𝑝𝑗−(𝑤), (2) which means that 𝐹₊(𝑧)𝐺₋(𝑤)=0_𝑛×𝑛 on 𝔻×𝔻.
Note that det 𝐹₊(𝑧) is the analytic function of 𝑧 and det𝐺₋(𝑤) is the coanalytic function of 𝑤. If det𝐹₊(𝑧) is not identically zero, then 𝐺₋=0_𝑛×𝑛; that is, 𝐺 is analytic. If det𝐺₋(𝑤) is not identically zero, then 𝐹₊=0_𝑛×𝑛; that is, 𝐹 is coanalytic.
In the following, we deal with the case that det𝐹₊≡0 and det𝐺₋≡0. Suppose 𝐺₋𝐸_𝑗=(𝑔^𝑗₁,…,𝑔^𝑗_𝑛)^𝑇 for 1≤𝑗≤𝑛. Since det𝐺₋≡0, 𝐺₋𝐸_𝑗 is linearly dependent. Let 𝑄_𝑗 be a permutation matrix such that 𝑄_𝑗𝐺₋𝐸_𝑗=𝑔^𝑗_𝜎(1),…,𝑔^𝑗_𝜎𝑘0,…,𝑔^𝑗_𝜎(𝑛)^𝑇, (3) where {𝑔^𝑗_𝜎(1),…,𝑔^𝑗_𝜎(𝑘0)} is maximally linearly independent of {𝑔^𝑗_𝜎(1),…,𝑔^𝑗_𝜎(𝑛)}. And 𝑔^𝑗_𝜎𝑘0+1=𝑎¹₁𝑔^𝑗_𝜎(1)+⋯+𝑎¹_𝑘0𝑔^𝑗_𝜎𝑘0,⋮𝑔^𝑗_𝜎(𝑛)=𝑎^𝑛−𝑘0₁𝑔^𝑗_𝜎(1)+⋯+𝑎^𝑛−𝑘0_𝑘0𝑔^𝑗_𝜎𝑘0, (4) with the matrix representation 0⋯00⋯00⋮⋮0𝑎⋮⋮0⋯0⋯₁₁⋯𝑎¹_𝑘0𝑎⋮⋮_{𝑛−𝑘01}⋯𝑎^𝑛−𝑘0_𝑘0𝑔−1⋯0⋮⋮0⋯−1^𝑗𝜎(1)⋮𝑔^𝑗_𝜎𝑘0⋮𝑔^𝑗_{𝜎()𝑛−1}𝑔^𝑗_𝜎(𝑛)=0_𝑛×𝑛, (5)
that is, 𝑀_𝑗𝑄_𝑗𝐺₋𝐸_𝑗=0_𝑛×𝑛.
Since 𝐹₊(𝑧)[𝐺₋(𝑤)𝐸_𝑗]=0, we have [𝐹₊(𝑧)𝑄_𝑗][𝑄_𝑗𝐺₋(𝑤)𝐸_𝑗]=0_𝑛×𝑛. Denote 𝐹₊=[𝑓_𝑖𝑗]_𝑛×𝑛. It follows that 𝑓_𝑘𝜎(1)𝑔^𝑗_𝜎(1)+⋯+𝑓_{𝑘𝜎(𝑘0)}𝑔^𝑗_𝜎𝑘0+⋯+𝑓_{𝑘𝜎(𝑛)}𝑔^𝑗_𝜎(𝑛)=0 (6) for 1≤𝑘≤𝑛. Note that 𝑔^𝑗_{𝜎(𝑘0+𝑙)}=𝑎^𝑙₁𝑔^𝑗_𝜎(1)+⋯+𝑎^𝑙_𝑘0𝑔^𝑗_𝜎(𝑘0), 1≤𝑙≤𝑛−𝑘₀. Therefore 𝑓_𝑘𝜎(1)𝑔^𝑗_𝜎(1)+⋯+𝑓_{𝑘𝜎(𝑘0)}𝑔^𝑗_𝜎𝑘0+^𝑛−𝑘0_𝑙=1𝑎^𝑙₁𝑔^𝑗_𝜎(1)+⋯+𝑎^𝑙_𝑘0𝑔^𝑗_𝜎𝑘0𝑓_{𝑘𝜎(𝑘0+𝑙)}=0, (7) and hence, (𝑓_𝑘𝜎(1)+∑^𝑛−𝑘0_𝑙=1𝑎^𝑙₁𝑓_{𝑘𝜎(𝑘0+𝑙)})𝑔^𝑗_𝜎(1)+(𝑓_𝑘𝜎(2)+∑^𝑛−𝑘0_𝑙=1𝑎^𝑙₂𝑓_{𝑘𝜎(𝑘0+𝑙)})𝑔^𝑗_𝜎(2)+⋯+(𝑓_{𝑘𝜎(𝑘0)}+∑^𝑛−𝑘0_𝑙=1𝑎^𝑙_𝑘0𝑓_{𝑘𝜎(𝑘0+𝑙)})𝑔^𝑗_𝜎(𝑘0)=0. Since {𝑔^𝑗_𝜎(1),…,𝑔^𝑗_𝜎(𝑘0)} is linearly independent, we have 𝑓_𝑘𝜎(1)+^𝑛−𝑘0_𝑙=1𝑎^𝑙₁𝑓_{𝑘𝜎(𝑘0+𝑙)}⋮𝑓=0,_{𝑘𝜎(𝑘0)}+^𝑛−𝑘0_𝑙=1𝑎^𝑙_𝑘0𝑓_{𝑘𝜎(𝑘0+𝑙)}=0, (8) with the matrix representation 𝑓_1𝜎(1)⋯𝑓_1𝜎(𝑘0)⋯𝑓_1𝜎(𝑛)𝑓_2𝜎(1)⋯𝑓_2𝜎(𝑘0)⋯𝑓_2𝜎(𝑛)𝑓⋮⋮⋮⋮⋮⋮_{𝑛−1𝜎(1)}⋯𝑓_{𝑛−1𝜎(𝑘0)}⋯𝑓_{𝑛−1𝜎(𝑛)}𝑓_𝑛𝜎(1)⋯𝑓_{𝑛𝜎(𝑘0)}⋯𝑓_{𝑛𝜎(𝑛)}×𝑎1…00⋯0⋮⋮⋮⋮0⋯10⋯0₁₁⋯𝑎¹_𝑘0𝑎0⋯0⋮⋮⋮⋮_{𝑛−𝑘01}⋯𝑎^𝑛−𝑘0_𝑘00⋯0=0_𝑛×𝑛, (9)
that is 𝐹₊𝑄_𝑗(𝐼+𝑀_𝑗)=0_𝑛×𝑛.
In the following, we prove (3)⇒(1).
Note that 𝑇_𝐹𝑇_𝐺−𝑇_𝐹𝐺=𝑇_{(𝐹++𝐹−)}𝑇_{(𝐺++𝐺−)}−𝑇_{(𝐹++𝐹−)(𝐺++𝐺−)}=𝑇_{(𝐹+𝐺++𝐹−𝐺−+𝐹−𝐺+)}+𝑇_𝐹+𝐺−−𝑇_{(𝐹+𝐺++𝐹−𝐺−+𝐹−𝐺+)}−𝑇_𝐹+𝐺−=𝑇_𝐹+𝑇_𝐺−−𝑇_𝐹+𝐺−. (10) By the hypothesis, we have 𝑇_𝐹+𝑇_{𝐺−𝐸𝑗}=𝑇_{𝐹+𝑄𝑗(𝐼+𝑀𝑗−𝑀𝑗)𝑄𝑗}𝑇_{𝐺−𝐸𝑗}=𝑇_{𝐹+𝑄𝑗(𝐼+𝑀𝑗)𝑄𝑗}𝑇_{𝐺−𝐸𝑗}−𝑇_{𝐹+𝑄𝑗𝑀𝑗𝑄𝑗}𝑇_{𝐺−𝐸𝑗}=𝑇_{𝐹+𝑄𝑗(𝐼+𝑀𝑗)𝑄𝑗}𝑇_{𝐺−𝐸𝑗}−𝑇_{𝐹+𝑄𝑗}𝑇_{𝑀𝑗𝑄𝑗𝐺−𝐸𝑗}=0_𝑛×𝑛,𝐹₊𝐺₋𝐸_𝑗=𝐹₊𝑄_𝑗𝐼+𝑀_𝑗−𝑀_𝑗𝑄_𝑗𝐺₋𝐸_𝑗=𝐹₊𝑄_𝑗𝐼+𝑀_𝑗𝑄_𝑗𝐺₋𝐸_𝑗−𝐹₊𝑄_𝑗𝑀_𝑗𝑄_𝑗𝐺₋𝐸_𝑗=0_𝑛×𝑛. (11) Hence, we obtain the desired result.

Remark 4. If the assumption 𝐹₊(0)=𝐺₋(0)=0_𝑛×𝑛 is removed and 𝐹₊,𝐺₋ are replaced by 𝐹^′₊,𝐺^′₋ in statement (3), respectively, then the above theorem still holds.

Corollary 5. Let 𝐹(𝑧)=𝐹₊(𝑧)+𝐹₋(𝑧), and let 𝐺(𝑧)=𝐺₊(𝑧)+𝐺₋(𝑧)∈ℎ^∞⊗𝑀_𝑛×𝑛 such that 𝐹₊(0)=𝐺₋(0)=0_𝑛×𝑛. Suppose that |det𝐹₊|+|det𝐺₋| is not identically zero. Then the following are equivalent:(1)𝑇_𝐹𝐺−𝑇_𝐹𝑇_𝐺 has finite rank; (2)𝑇_𝐹𝐺=𝑇_𝐹𝑇_𝐺; (3)if det𝐹₊ is not identically zero, then 𝐺 is analytic; if det𝐺₋ is not identically zero, then 𝐹 is coanalytic.

Proof. The proof is obvious.

Let Φ=𝐹−𝐺00,Ψ=𝐺0𝐹0. (12) As in [6], the commutator 𝑇_𝐹𝑇_𝐺−𝑇_𝐺𝑇_𝐹 can be reduced to the semicommutator 𝑇_ΦΨ−𝑇_Φ𝑇_Ψ.

It is easy to check that 𝑇_ΦΨ−𝑇_Φ𝑇_Ψ=𝑇_{𝐹𝐺−𝐺𝐹}0−𝑇00^𝐹−𝑇_𝐺𝑇00^𝐺0𝑇_𝐹0=𝑇_{𝐹𝐺−𝐺𝐹}0−𝑇00^𝐹𝑇_𝐺−𝑇_𝐺𝑇_𝐹0.00 (13) If 𝑇_𝐹𝑇_𝐺=𝑇_𝐺𝑇_𝐹, Theorem 2 implies that 𝐹𝐺=𝐺𝐹. Therefore, 𝑇_Φ𝑇_Ψ=𝑇_ΨΦ. Combining Theorem 3, we get the following explicit theorem.

Theorem 6. Suppose that 𝐹=𝐹₊+𝐹₋∈ℎ^∞⊗𝑀_𝑛×𝑛 and 𝐺=𝐺₊+𝐺₋∈ℎ^∞⊗𝑀_𝑛×𝑛. Then 𝑇_𝐹𝑇_𝐺=𝑇_𝐺𝑇_𝐹 if and only if 𝐹 and 𝐺 satisfy the following two conditions: (1)𝐹𝐺=𝐺𝐹; (2)for each 1≤𝑗≤𝑛, there exist permutation matrixes 𝑄_𝑗 and 𝕄_𝑗∈𝑀_2𝑛×2𝑛 such that 𝕄_𝑗𝑄_𝑗^𝐺′−_𝐹′−𝐸_𝑗=0_𝑛×𝑛 and [𝐹^′₊−𝐺^′₊]𝑄_𝑗(𝐼+𝕄_𝑗)=0_𝑛×𝑛.

3. Zero Product of Two Block Toeplitz Operators

In this section, we discuss the zero product of two block Toeplitz operators 𝑇_𝐹 and 𝑇_𝐺 with 𝐹,𝐺∈ℎ^∞⊗𝑀_𝑛×𝑛 on 𝐿²_𝑎(𝔻,ℂ^𝑛).

Theorem 7. Let 𝐹(𝑧)=𝐹₊(𝑧)+𝐹₋(𝑧) and 𝐺(𝑧)=𝐺₊(𝑧)+𝐺₋(𝑧)∈ℎ^∞⊗𝑀_𝑛×𝑛 such that 𝐹₊(0)=𝐺₋(0)=𝐹₋(0)=𝐺₊(0)=0_𝑛×𝑛. Then 𝑇_𝐹𝑇_𝐺=0_𝑛×𝑛 if and only if (1)𝐹𝐺=0_𝑛×𝑛; (2)for each 1≤𝑗≤𝑛, there exist a matrix 𝑁_𝑗∈𝑀_𝑛×𝑛 and a permutation matrix 𝑄¹_𝑗, such that 𝑁_𝑗𝑄¹_𝑗𝐺₋𝐸_𝑗=0_𝑛×𝑛 and 𝐹₊𝑄¹_𝑗(𝐼+𝑁_𝑗)=0_𝑛×𝑛.

Proof. By Theorem 2, it is easy to know that ∑^𝑛_𝑝=1𝑓_𝑖𝑝(𝑧)𝑔_𝑝𝑗(𝑧)=0 for 1≤𝑖,𝑗≤𝑛; that is, 𝐹𝐺=0_𝑛×𝑛. Since 𝑇_𝐹𝑇_𝐺=𝑇_𝐹𝑇_𝐺−𝑇_𝐹𝐺=0_𝑛×𝑛, we obtain the desired result by Theorem 3.

Corollary 8. Let 𝐹(𝑧)=𝐹₊(𝑧)+𝐹₋(𝑧) and let 𝐺(𝑧)=𝐺₊(𝑧)+𝐺₋(𝑧)∈ℎ^∞⊗𝑀_𝑛×𝑛 such that 𝐹₊(0)=𝐺₋(0)=𝐹₋(0)=𝐺₊(0)=0_𝑛×𝑛. (1)If neither det𝐺₊ nor det𝐺₋ is identically zero, then 𝑇_𝐹𝑇_𝐺=0 if and only if 𝐹=0_𝑛×𝑛. (2)If neither det𝐹₊ nor det𝐹₋ is identically zero, then 𝑇_𝐹𝑇_𝐺=0 if and only if 𝐺=0_𝑛×𝑛.

Proof. (1) We only need to prove the necessity. By the proof of Theorem 3, we have 𝐹₊(𝑧)𝐺₋(𝑤)=0_𝑛×𝑛 on 𝔻×𝔻. By the theorem given above, we know that 𝐹𝐺=[𝐹₊+𝐹₋][𝐺₊+𝐺₋]=0_𝑛×𝑛. Taking Laplace transform, it follows that 𝐹^′₋𝐺^′₊=0_𝑛×𝑛. As in Theorem 3, we can prove that 𝐹₋(𝑤)𝐺₊(𝑧)=0_𝑛×𝑛 on 𝔻×𝔻. “Complexificate” the identity 𝐹𝐺=0_𝑛×𝑛, and then [𝐹₊(𝑧)+𝐹₋(𝑤)][𝐺₊(𝑧)+𝐺₋(𝑤)]=0_𝑛×𝑛. Hence, 𝐹₊(𝑧)𝐺₊(𝑧)=−𝐹₋(𝑤)𝐺₋(𝑤) on 𝔻×𝔻. Since 𝐹₊(0)=𝐺₋(0)=0_𝑛×𝑛, we have 𝐹₊𝐺₊=𝐹₋𝐺₋=0_𝑛×𝑛.
It is clear that det𝐺₊(𝑧) is the analytic function of 𝑧 and det𝐺₋(𝑤) is the coanalytic function of 𝑤. If det𝐺₊(𝑧) is not identically zero, then 𝐹₋≡0_𝑛×𝑛 and 𝐹₊≡0_𝑛×𝑛. Therefore, 𝐹=0_𝑛×𝑛. If det𝐺₋(𝑤) is not identically zero, then 𝐹₋≡0_𝑛×𝑛 and 𝐹₊≡0_𝑛×𝑛. Therefore, 𝐹=0_𝑛×𝑛.
(2) The proof is similar to the proof of (1).

Acknowledgments

This research is supported by NSFC (11271059, 11201438), Research Fund for the Doctoral Program of Higher Education of China, Shandong Province Young Scientist Research Award Fund (BS2012SF031), and the Fundamental Research Funds for the Central Universities (201213011).