# J-Self-Adjoint Projections in Krein Spaces

Authors: Xiao-Ming Xu; Yile Zhao
Journal: Journal of Function Spaces
 (2020)

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## Abstract

Letℋ be a Krein space with fundamental symmetry J. Starting with a canonical block-operator matrix representation of J, we study the regular subspaces of ℋ. We also present block-operator matrix representations of the J-self-adjoint projections for the regular subspaces of ℋ, as well as for the regular complements of the isotropic part in a pseudo-regular subspace of ℋ.


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## Body

## 1. Introduction

Throughout this paper, letℋ be a separable complex Hilbert space with the inner product 〈⋅, ⋅〉 and let ℬℋ be the algebra of all bounded linear operators on ℋ. A contraction in ℬℋ is an operator Q in ℬℋ such that ∥Q∥ ≤ 1. For an operator T∈ℬℋ, T∗, σ(T), σP(T), ℛT, and NT denote the adjoint, the spectrum, the point spectrum, the range, and the null space of T, respectively. An operator T in ℬℋ is said to be self-adjoint if T = T∗.An operatorJ in ℬℋ is said to be a symmetry (or self-adjoint involution) if J = J∗ = J−1. If the symmetry J is nonscalar, then(1)x,y=Jx,y,which defines an indefinite inner product on ℋ, and ℋ,J is called a Krein space ([1–3]).ForT∈ℬℋ, the J-adjoint operator of T is the unique operator T♯ in ℬℋ satisfying(2)Tx,y=x,T♯y,for all x,y∈ℋ. It is easy to see that T♯ = JT∗J.Definition 1.
IfT∈ℬℋ, then (a) T is J-normal if TT♯ = T♯T; (b) T is J-self-adjoint if T♯ = T; (c) T is J-positive if JT ≥ 0; (d) T is J-negative if JT ≤ 0.
An idempotent inℬℋ is called a projection. A projection is normal if and only if it is self-adjoint. However, there exist J-normal projections which are not J-self-adjoint (see [4]).
For a subspaceS of the Krein space ℋ, S⊥ denotes the J-orthocomplement of S in ℋ, that is,(3)S⊥=h∈ℋ:h,s=0for allsinS.
It is obvious thatS⊥ equals the usual orthocomplement JS⊥. Let S0:=S∩S⊥ be the isotropic part of S. If S0=0, then S is said to be J-nondegenerate. Otherwise, S is said to be J-degenerate.Definition 2.
IfS is a subspace of the Krein space ℋ, then (a) S is positive if PSJPS≥0, where PS is the orthogonal projection of ℋ onto S; (b) S is uniformly positive if PSJPS≥εPS for some ϵ > 0; (c) S is regular if ℋ=S+˙S⊥; (d) S is pseudoregular if S and the algebraic sum S+S⊥ are closed.
It is well known that a subspaceS of ℋ is regular if and only if it is the range of a (unique) J-self-adjoint projection in ℬℋ (see [2]). Therefore, there is a one-to-one correspondence between regular subspaces of ℋ and J-self-adjoint projections in ℬℋ. It is also proved in [5] that a closed subspace S is regular if and only if PSJPS2≥εPS for some ϵ > 0. In consequence, a closed subspace S is uniformly positive if and only if it is regular and positive, and due to Proposition 4 in [6], this is the case if and only if it is a regular subspace with a J-positive projection.
Pseudoregular subspaces are important since they enable to generalize some Pontryagin space arguments to general Krein space (see [7]). Pseudoregular subspaces and its properties have been studied extensively by many authors (see [4, 7, 8]). In [4], the authors proved that a closed subspace S of ℋ is pseudoregular if and only if it is the range of a J-normal projection in ℬℋ. They also showed in the same paper that a pseudoregular subspace S admits infinitely many J-normal projections onto it, unless S is regular. In [8], Giribet et al. gave a block-operator matrix representation of the fundamental symmetry J depending on a pseudoregular subspace S of ℋ, and from here on, they characterized the J-self-adjoint projections for the regular complements of S0 in S.
In this paper, we give a new block-operator matrix representation of the fundamental symmetryJ related to a closed subspace S of ℋ. This offers an improvement over the result in [8], since we do not need to impose the assumption of the pseudoregularity of S. We also study the J-self-adjoint projections for the regular subspaces of ℋ, as well as for the regular complements of the isotropic part in a pseudoregular subspace of ℋ.
The paper is organized as follows. In Section2, we give a block-operator matrix representation of the fundamental symmetry J. Therein, we also characterize the regular subspaces of ℋ and present a block-operator matrix representation of the J-self-adjoint projections for the regular subspaces of ℋ. In Section 3, we study the pseudoregular subspaces of ℋ. If S is a pseudoregular subspace of ℋ and ℒ is a regular complement of S0 in S, we give a block-operator matrix representation of the J-self-adjoint projection onto ℒ.


## 2.J-Self-Adjoint Projections for the Regular Subspaces

Letℋ be a Krein space with fundamental symmetry J. Then, J+:=I+J/2 and J−:=I−J/2 are mutually annihilating orthogonal projections. Denote ℋ+=ℛJ+ and ℋ−=ℛJ−. We have fundamental decomposition ℋ=ℋ+⊕ℋ−.LetS be a closed subspace of the Krein space ℋ, and let S0 be its isotropic part. Denote ℋ1=S∩ℋ+, ℋ2=S∩ℋ−, ℋ3=S0, ℋ4=S⊖ℋ1⊕ℋ2⊕ℋ3, ℋ5=S⊥⊖JS0⊕S⊥∩ℋ+⊕S⊥∩ℋ−, ℋ6=JS0, ℋ7=S⊥∩ℋ+, and ℋ8=S⊥∩ℋ−. It is easy to check that ℋi, 1 ≤ i ≤ 8, are pairwise orthogonal subspaces of ℋ. The operators in this paper are frequently treated as block-operator matrices with respect to the space decomposition:(4)ℋ=ℋ1⊕ℋ2⊕ℋ3⊕ℋ4⊕ℋ5⊕ℋ6⊕ℋ7⊕ℋ8.For a pseudoregular subspaceS of the Krein space ℋ, a block-operator matrix representation of the fundamental symmetry J was obtained with the space decomposition ℋ=S0⊕S⊖S0⊕S⊥⊖JS0⊕JS0 in [8]. We continue the study of the block-operator matrix representation of the fundamental symmetry J, but we do not impose the assumption of the pseudoregularity of the subspace S.Theorem 1.
Letℋ be a Krein space with fundamental symmetry J, and let S be a closed subspace of ℋ. Then, J has the operator matrix representation:(5)J=I1⊕−I2⊕000U0QI4−Q21/2V00V∗I4−Q21/2−V∗QV0U∗000⊕I7⊕−I8,with respect to space decomposition (4), where Ii is the identity operator on the corresponding space ℋi, i = 1, 2, 4, 7, 8, U is an isometric isomorphism from ℋ6 onto ℋ3, V is an isometric isomorphism from ℋ5 onto ℋ4, and Q is a self-adjoint contraction on ℋ4 with 0, ±1 ∉σP(Q).Proof.
It is clear thatℋ1 and ℋ7 are subspaces of ℋ+ and ℋ2 and ℋ8 are subspaces of ℋ−. So Jℋ+=Iℋ+ implies Jℋ1=Iℋ1 and Jℋ7=Iℋ7 and Jℋ−=−Iℋ− implies Jℋ2=−Iℋ2 and Jℋ8=−Iℋ8. Since Jℋ3=ℋ6, we have PℋiJℋ3=0 for i ≠ 6. Moreover, since Jℋ6=JJℋ3=J2ℋ3=Iℋ3=ℋ3, we get PℋiJℋ6=0 for i ≠ 3. Noting that J is self-adjoint, then J has the operator matrix representation:(6)J=I1⊕−I2⊕000J360J44J4500J45∗J550J36∗000⊕I7⊕−I8,with respect to space decomposition (4), where Ii is the identity operator on the corresponding space ℋi, i = 1, 2, 7, 8, J36∈ℬℋ6,ℋ3, J45∈ℬℋ5,ℋ4, and J44 and J55 are self-adjoint contractions on ℋ4 and ℋ5, respectively:LetJ36 = U. Since J2 = I, it is easy to see that UU∗ = I3 and U∗U = I6. Thus, U is an isometric isomorphism from ℋ6 onto ℋ3.LetJ44 = Q. If ξ∈NI4−Q and x = (0, 0, 0, ξ, 0, 0, 0, 0)T, then(7)Jx=0,0,0,Qξ,J45∗ξ,0,0,0T=0,0,0,ξ,J45∗ξ,0,0,0T.
Since ∥J∥ = 1, J45∗ξ=0 and Jx = x. It follows that x∈ℋ+, and hence, x∈S∩ℋ+=ℋ1. This implies ξ = 0. Thus, 1 ∉ σP(Q). Analogously, −1 ∉ σP(Q) and ±1 ∉ σP(J55).
Moreover, ifξ∈NQ and x = (0, 0, 0, ξ, 0, 0, 0, 0)T, then(8)x,y=Jx,y=Qξ,y+J45∗ξ,y=Qξ,y=0,for all y∈S. It follows that x∈S⊥, and hence, x∈S∩S⊥=S0=ℋ3. So ξ = 0, and hence, 0 ∉ σP(Q).
LetJ′=Pℋ4⊕ℋ5Jℋ4⊕ℋ5, that is,(9)J′=QJ45J45∗J55:ℋ4⊕ℋ5⟶ℋ4⊕ℋ5.
Then,J′ is a symmetry. So we have(10)J′+=12I4⊕I5+J′=12I4+QJ45J45∗I5+J55≥0,and by Proposition 5 in [9], there exists a contraction V from ℋ5 into ℋ4 such that(11)J45=I4+Q1/2VI5+J551/2.
Then, by a direct calculation, equationJ′+2=J′+ implies(12)I4+Q2+I4+Q1/2VI5+J55V∗I4+Q1/2=2I4+Q,I4+Q3/2VI5+J551/2+I4+Q1/2VI5+J553/2=2I4+Q1/2VI5+J551/2,I5+J551/2V∗I4+QVI5+J551/2+I5+J552=2I5+J55.
Noting that −1 ∉σP(Q) and −1 ∉ σP(J55), it follows that(13)iI4−Q=VI5+J55V∗,ii−QV=VJ55,iiiV∗I4+QV=I5−J55.
By (11) and (ii) of (13), we obtain(14)J45=I4+Q1/2VI5+J551/2=I4+Q1/2I4−Q1/2V=I4−Q21/2V.
By (i) and (ii) of (13), (I4 − Q) (VV∗ − I4) = 0, and since I4 − Q is injective, VV∗ = I4. By (ii) and (iii) of (13), (V∗V − I5) (I5 − J55) = 0, and since I5 − J55 is a self-adjoint operator with dense range in ℋ5, V∗V = I5. Thus, V is an isometric isomorphism from ℋ5 onto ℋ4, and by (ii) of (13) again,(15)J55=−V∗QV.
Now we see that(16)J′=QI4−Q21/2VV∗I4−Q21/2−V∗QV:ℋ4⊕ℋ5⟶ℋ4⊕ℋ5,and J has the asserted operator matrix.Lemma 1 (see [5]).
A closed subspaceS is regular if and only if PSJPS2≥ϵPS for some ϵ > 0. In this case, the J-self-adjoint projection onto S is determined as PSJPS†J, where PSJPS† stands for the Moore–Penrose inverse of PSJPS.Theorem 2.
Letℋ be a Krein space with fundamental symmetry J, and let S be a closed subspace of ℋ. Write J in (5) with respect to space decomposition (4). Then, S is regular if and only if ℋ3=0 and 0 ∉ σ(Q). In this case, ℋ6=0 and with respect to the space decomposition ℋ=ℋ1⊕ℋ2⊕ℋ4⊕ℋ5⊕ℋ7⊕ℋ8, J and the J-self-adjoint projection E onto S have operator matrix representations:(17)J=I1⊕−I2⊕QI4−Q21/2VV∗I4−Q21/2−V∗QV⊕I7⊕−I8and(18)E=I1⊕I2⊕I4Q−1I4−Q21/2V00⊕0⊕0,respectively.Proof.
It is clear thatPS has the operator matrix representation:(19)PS=I1⊕I2⊕I3⊕I4⊕0⊕0⊕0⊕0,with respect to space decomposition (4), and by a direct calculation,(20)PSJPS=I1⊕−I2⊕0⊕Q⊕0⊕0⊕0⊕0.
Then, by Lemma1, S is regular if and only if ℋ3=0 and Q2 ≥ ϵI4 for some ϵ > 0, and since Q is self-adjoint, this is the case if and only if ℋ3=0 and 0 ∉ σ(Q).
IfS is regular, then ℋ6=Jℋ3=0, and hence, ℋ=ℋ1⊕ℋ2⊕ℋ4⊕ℋ5⊕ℋ7⊕ℋ8. By Theorem 1, J has the operator matrix representation given in (17), and since PS=I1⊕I2⊕I4⊕0⊕0⊕0 with respect to ℋ=ℋ1⊕ℋ2⊕ℋ4⊕ℋ5⊕ℋ7⊕ℋ8, we have(21)PSJPS†=I1⊕−I2⊕Q⊕0⊕0⊕0†=I1⊕−I2⊕Q−1⊕0⊕0⊕0.
Then, by Lemma1, we get(22)E=PSJPS†J=I1⊕I2⊕I4Q−1I4−Q21/2V00⊕0⊕0,with respect to ℋ=ℋ1⊕ℋ2⊕ℋ4⊕ℋ5⊕ℋ7⊕ℋ8.
Recall that a closed subspaceS of ℋ is uniformly positive if and only if it is a regular subspace with a J-positive projection in ℬℋ. We give a characterization of the uniformly positive subspaces of the Krein space ℋ.Corollary 1.
Letℋ be a Krein space with fundamental symmetry J, and let S be a closed subspace of ℋ. Write J in (5) with respect to space decomposition (4). Then, S is uniformly positive if and only if ℋ2=0, ℋ3=0, and Q is a positive operator in ℬℋ4 with 0 ∉ σ(Q). In this case, the block-operator matrix representation of the J-positive projection E onto S is given by(23)E=I1⊕I4Q−1I4−Q21/2V00⊕0⊕0,with respect to the space decomposition ℋ=ℋ1⊕ℋ4⊕ℋ5⊕ℋ7⊕ℋ8.Proof (Necessity).
Suppose that the closed subspaceS is uniformly positive. Then, S is regular, and suppose that E is the J-self-adjoint projection onto S. By Theorem 2, ℋ3=0, 0 ∉ σ(Q), and with respect to ℋ=ℋ1⊕ℋ2⊕ℋ4⊕ℋ5⊕ℋ7⊕ℋ8, J and E have operator matrix representations (17) and (18), respectively. By a direct calculation, we have(24)JE=I1⊕−I2⊕QI4−Q21/2VV∗I4−Q21/2V∗I4−Q2Q−1V⊕0⊕0,and since JE ≥ 0, we see that ℋ2=0 and Q ≥ 0.Sufficiency.
Suppose thatℋ2=0, ℋ3=0, and Q is a positive operator in ℬℋ4 with 0 ∉ σ(Q). Then, by Theorem 1, J has the operator matrix representation:(25)J=I1⊕QI4−Q21/2VV∗I4−Q21/2−V∗QV⊕I7⊕−I8,with respect to ℋ=ℋ1⊕ℋ4⊕ℋ5⊕ℋ7⊕ℋ8, and by Theorem 2, S is regular and the J-self-adjoint projection E onto S has the operator matrix representation:(26)E=I1⊕I4Q−1I4−Q21/2V00⊕0⊕0,with respect to ℋ=ℋ1⊕ℋ4⊕ℋ5⊕ℋ7⊕ℋ8. It follows that(27)JE=I1⊕QI4−Q21/2VV∗I4−Q21/2V∗I4−Q2Q−1V⊕0⊕0=I1⊕Q1/20V∗I4−Q21/2Q−1/20Q1/2I4−Q21/2Q−1/2V00⊕0⊕0≥0,and hence, S is uniformly positive.Remark 1.
LetS be a subspace of the Krein space ℋ. Then, S is said to be negative (resp., uniformly negative) if PSJPS≤0 (resp., PSJPS≤−εPS for some ϵ > 0). If more S is closed, then S is uniformly negative if and only if it is regular and negative, or equivalently, if and only if it is a regular subspace with a J-negative projection. Arguing as in Corollary 1, we can also give a characterization of the uniformly negative subspaces of the Krein space ℋ. With the notation as in Corollary 1, a closed subspace S is uniformly negative if and only if ℋ1=0, ℋ3=0, and −Q is a positive operator in ℬℋ4 with 0 ∉ σ(Q). In this case, the block matrix representation of the J-negative projection F onto S is given by(28)F=I2⊕I4Q−1I4−Q21/2V00⊕0⊕0,with respect to the space decomposition ℋ=ℋ2⊕ℋ4⊕ℋ5⊕ℋ7⊕ℋ8.
In the end of this section, we give an alternative proof of Theorem 2.3 in [5].Corollary 2.
LetS be a regular subspace of ℋ with the J-self-adjoint projection E. Then, E is uniquely written as E = E1 + E2 with E1, a J-positive projection, and E2, a J-negative projection satisfying E1E2=E1E2∗=0.Proof.
WriteJ and E in (17) and (18), respectively. Since Q is a self-adjoint operator in ℬℋ4, there are unique positive operators Q+ and Q− in ℬℋ4 such that Q = Q+ − Q− and Q+Q− = Q−Q+ = 0 (see [10]). Note that 0 ∉ σ(Q), Q+, and Q− have closed ranges in ℋ4, and I4=PℛQ++PℛQ−. Let(29)E1=I1⊕0⊕PℛQ+Q+†PℛQ+−Q+21/2V00⊕0⊕0,E2=0⊕I2⊕PℛQ−Q−†PℛQ−−Q−21/2V00⊕0⊕0,with respect to ℋ=ℋ1⊕ℋ2⊕ℋ4⊕ℋ5⊕ℋ7⊕ℋ8. Then, E1 and E2 are projections, and Q+Q− = Q−Q+ = 0 implies E = E1 + E2 and E1E2=E1E2∗=0.
Moreover, since(30)JE1=I1⊕0⊕Q+PℛQ+−Q+21/2VV∗PℛQ+−Q+21/2V∗Q+†PℛQ+−Q+2V⊕0⊕0=I1⊕0⊕XX∗⊕0⊕0≥0,where X=Q+1/20V∗PℛQ+−Q+21/2Q+†1/20, we see that E1 is J-positive. Similarly, we have JE2 ≤ 0, and hence, E2 is J-negative.
Now, we prove the uniqueness ofE1 and E2. Since JE1 and JE2 are self-adjoint operators in ℬℋ, we have(31)JE1−JE2=−JE1JE2∗=−JE1E2∗J=0,−JE2JE1=−JE2JE1∗=−JE2E1∗J=0.
Noting thatJE = JE1 − (−JE2), where JE1 and −JE2 are the positive operators in ℬℋ, we see that JE1 and −JE2 are the positive part and the negative part of the self-adjoint operator JE, respectively. So JE1 and −JE2, and hence, E1 and E2 are unique.


## 3.J-Self-Adjoint Projections for the Pseudoregular Subspaces

In this section, we study the pseudoregular subspaces ofℋ. If S is a pseudoregular subspace of ℋ and ℒ is a complement of S0 in S, we present a block-operator matrix representation of the J-self-adjoint projection onto ℒ.Lemma 2.
Letℋ be a Krein space with fundamental symmetry J, and let S be a closed subspace of ℋ. Then, the following statements are equivalent:(a)
S is pseudoregular(b)
There exists a regular subspaceℳ such that S=S0+˙ℳ, where +˙ denotes the direct [⋅, ⋅]-orthogonal sum(c)
IfS=S0+˙T, then T is regular(d)
S⊖S0 is regular(e)
The operatorQ in (5) is invertibleProof.
Due to Proposition 4.1 in [4], (a) ⟺ (b) ⟺ (c).(c) ⟹ (d): since S=S0+˙S⊖S0, this is immediate.(d) ⟹ (b): since S⊖S0=ℋ1⊕ℋ2⊕ℋ4 and JS0=ℋ6,(32)x,y=Jx,y=0,for allx∈S0 and y∈S⊖S0. Thus, S=S0+˙S⊖S0, and hence, (d) ⟹ (b) is clear.(d) ⟺ (e): arguing as Theorem 2, S⊖S0 is regular if and only if the operator Q in (5) is invertible.
IfS be a pseudoregular subspace of ℋ and ℒ be a complement of S0 in S, then by Lemma 2, ℒ is a regular subspace. The following theorem gives a block-operator matrix representation of the J-self-adjoint projection onto ℒ.Theorem 3.
Letℋ be a Krein space with fundamental symmetry J, and let S be a pseudoregular subspace of ℋ. Write J in (5) with respect to space decomposition (4). If ℒ is a complement of S0 in S, then the J-self-adjoint projection E onto ℒ has the operator matrix representation:(33)E=I10000E16000I2000E2600UE16∗−UE26∗0UE46∗QUE46∗I4−Q21/2VUE16∗E16−UE26∗E26+UE46∗QE4600000I4Q−1I4−Q21/2VE460000000000000000000000000000000000,with respect to the space decomposition ℋ=⊕i=18ℋi, where Ei6∈ℬℋ6,ℋi, i = 1, 2, 4.Proof.
Ifℒ is a complement of S0 in S, then ℒ is regular. Suppose that E is the J-self-adjoint projection onto ℒ. Since ℒ⊆S, we have PS⊥=PS⊥Pℒ⊥, and hence,(34)PS⊥E=PS⊥Pℒ⊥E=PS⊥Pℒ⊥E=PS⊥0=0.
Moreover, sinceS0⊆S⊥⊆ℒ⊥=NE, we get ES0=0. Noting that ⊕i=58ℋi=S⊥ and ℋ3=S0, then E has the operator matrix representation:(35)E=E11E120E14E15E16E17E18E21E220E24E25E26E27E28E31E320E34E35E36E37E38E41E420E44E45E46E47E4800000000000000000000000000000000,with respect to the space decomposition ℋ=⊕i=18ℋi, where Eij∈ℬℋj,ℋi.
LetP=Pℋ1⊕ℋ2⊕ℋ4Eℋ1⊕ℋ2⊕ℋ4. By a direct calculation, P2=Pℋ1⊕ℋ2⊕ℋ4E2ℋ1⊕ℋ2⊕ℋ4, and since E2 = E, it follows that P2 = P. Thus, P is a projection on ℋ1⊕ℋ2⊕ℋ4. Furthermore, since ℋ1⊕ℋ2⊕ℋ4=S⊖S0 and S0⊆NPS⊖S0∩NE, we have(36)ℛP=PS⊖S0ES⊖S0=PS⊖S0ES=PS⊖S0ℒ=PS⊖S0S=S⊖S0=ℋ1⊕ℋ2⊕ℋ4.
SoP is the identity on ℋ1⊕ℋ2⊕ℋ4, and hence,(37)E=I1000E15E16E17E180I200E25E26E27E28E31E320E34E35E36E37E38000I4E45E46E47E4800000000000000000000000000000000.
Then,(38)JE=I1000E15E16E17E180−I200−E25−E26−E27−E2800000000000QQE45QE46QE47QE48000XXE45XE46XE47XE48U∗E31U∗E320U∗E34U∗E35U∗E36U∗E37U∗E380000000000000000,where X=V∗I4−Q21/2, and since JE is self-adjoint, we have(39)E15=0,U∗E31=E16∗,E17=0,E18=0,E25=0,U∗E32=−E26∗,E27=0,E28=0,QE45=I4−Q21/2V,U∗E34=E46∗Q,QE47=0,QE48=0,U∗E35=E46∗I4−Q21/2V,U∗E37=0,U∗E38=0.
SinceU is an isometric isomorphism from ℋ6 onto ℋ3 and 0 ∉ σ(Q), it also follows that(40)E31=UE16∗,E32=−UE26∗,E45=Q−1I4−Q21/2V,E34=UE46∗Q,E47=0,E48=0,E35=UE46∗I4−Q21/2V,E37=0,E38=0.
Now, we see that(41)E=I10000E16000I2000E2600UE16∗−UE26∗0UE46∗QUE46∗I4−Q21/2VE3600000I4Q−1I4−Q21/2VE460000000000000000000000000000000000.
Moreover, sinceE2 = E, we have(42)E36=UE16∗E16−UE26∗E26+UE46∗QE46.
Thus,E has the asserted operator matrix.Corollary 3.
Letℋ be a Krein space with fundamental symmetry J, and let S be a pseudoregular subspace of ℋ. Write J in (5) with respect to space decomposition (4). Then, the J-self-adjoint projection E onto S⊖S0 has the operator matrix representation:(43)E=I1⊕I2⊕0⊕I4Q−1I4−Q21/2V00⊕0⊕0⊕0,with respect to space decomposition (4).Proof.
By Theorem3, the J-self-adjoint projection E onto S⊖S0 has operator matrix representation (33) with respect to space decomposition (4). Moreover, since PS0=0⊕0⊕I3⊕0⊕0⊕0⊕0⊕0, the equation PS0E=0 implies(44)0000000000000000UE16∗−UE26∗0UE46∗QUE46∗I4−Q21/2VUE16∗E16−UE26∗E26+UE46∗QE46000000000000000000000000000000000000000000=0.
It follows thatUE16∗=0,−UE26∗=0, and UE46∗Q=0. Noting that U is an isometric isomorphism from ℋ6 onto ℋ3 and Q is a self-adjoint contraction with dense range in ℋ4, we get that E16 = 0, E26 = 0, and E46 = 0. Thus, E has the asserted operator matrix.



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