The F α,β -transformation for sequences of complex matrices

In this section, we consider a particular transformation of sequences of matrices, which will be the elementary step of a particular algorithm. First we introduce a particular matrix polynomial and list some simple observations on it.

Notation 8.1. For each n∈N₀, let T_q,n := [δ_j,k+1I_q]ⁿ_j,k=0.

In particular, we haveT_q,0 =O_q×q and T_q,n =^hO_I^q×nq_{nq O}^O_nq×q^{q×q i}for all n∈N. For eachn∈N₀, the block matrix T_q,n has the shape of the first of the two matrices in Notation 3.1 with the matrix O_q×q in the q×q block main diagonal. Thus, det(I_(n+1)q−zT_q,n) = 1 holds true for allz∈C. Hence, the following matrix-valued function is well defined:

Notation 8.2. For alln∈N₀, let Rq,n:C→C(n+1)q×(n+1)q be given by R_q,n(z) := (I_(n+1)q−zT_q,n)⁻¹.

By virtue of T_q,nⁿ⁺¹ =O(n+1)q×(n+1)q, it is readily checked thatR_q,n(z) =^Pn_ℓ=0z^ℓT_q,n^ℓ holds true for allz∈C. In particular,R_q,n is a matrix polynomial of degreenwith invertible values, admitting the following block representation:

Example 8.3. Let n ∈ N₀, let z ∈ C, and let the sequence (s_j)ⁿ_j=0 be given by s_j := z^jI_q. Using Notation 3.1, thenS_n=R_q,n(z) and S_n= [R_q,n(z)]^∗.

Referring to the classes of particular block triangular matrices introduced in Notation A.19, we can conclude from Example 8.3:

Remark 8.4. Let n ∈ N₀ and let z ∈ C. In view of Notation A.19, then Rq,n(z) is a block Toeplitz matrix belonging toL_p,n and [R_q,n(z)]^∗ is a block Toeplitz matrix belonging to U_q,n. In particular, detR_q,n(z) = 1, by virtue of Remark A.21.

Given a sequence (s_j)^κ_j=0 of complex p×q matrices and a bounded interval [α, β] of R we have introduced in Notation 7.1 three particular sequences associated with (s_j)^κ_j=0. These three sequences correspond to the intervals [α,∞), (−∞, β], and [α, β], respectively. Now we slightly modify each of these sequences. We start with the case of the interval [α,∞).

Definition 8.5. Let (s_j)^κ_j=0 be a sequence of complex p×q matrices. Further let a₋₁ := s₀ and, in the caseκ≥1, let (a_j)^κ−1_j=0 be given by Notation 7.1. Then we call the sequence (a_j)^κ_j=0 given bya_j :=a_j−1 the [α,∞)-modification of (s_j)^κ_j=0. For each matrixX_k=X_k^hsi built from the sequence (s_j)^κ_j=0, we denote by X_α,k,•^′ := X_k^hai the corresponding matrix built from the sequence (a_j)^κ_j=0 instead of (s_j)^κ_j=0.

In the classical case α = 0, the sequences (sj)^κ_j=0 and (aj)^κ_j=0 coincide. For an arbitrary α∈R, the sequence (s_j)^κ_j=0is reconstructible from (a_j)^κ_j=0as well. The corresponding relations can be written in a convenient form, using the particular block Toeplitz matrices introduced in Notations 3.1 and 8.2:

Remark 8.6. If (sj)^κ_j=0 is a sequence of complexp×q matrices, then S^′_α,m,• = [Rp,m(α)]⁻¹S_m andS^′

α,m,• =S_m[R_q,m(α)]^−∗ for all m∈Z_0,κ.

Using [21, Lem. 4.6], we obtain the analogous relation between the corresponding reciprocal sequences associated via Definition 3.5:

Lemma 8.7 (cf. [21, Rem. 4.7]). Let (sj)^κ_j=0 be a sequence of complex p×q matrices.

Denote by (r_j)^κ_j=0 the reciprocal sequence associated to (a_j)^κ_j=0. For all m ∈ Z_0,κ, then S^hrim =R_q,m(α)S^♯_m and S^hri_m =S^♯

m[R_p,m(α)]^∗.

The analogue of Definition 8.5 for the interval (−∞, β] looks as follows:

Definition 8.8. Let (s_j)^κ_j=0 be a sequence of complexp×q matrices. Further let b₋₁:=−s0

and, in the caseκ≥1, let (b_j)^κ−1_j=0 be given by Notation 7.1. Then we call the sequence (b_j)^κ_j=0 given by b_j := b_j−1 the (−∞, β]-modification of (s_j)^κ_j=0. For each matrix X_k = X_k^hsi built from the sequence (s_j)^κ_j=0, we denote by X_•,k,β^′ :=X_k^hbi the corresponding matrix built from the sequence (b_j)^κ_j=0 instead of (s_j)^κ_j=0.

In particular, if β= 0, then (b_j)^κ_j=0 coincides with the sequence (−s_j)^κ_j=0. For an arbitrary β∈R, the sequence (s_j)^κ_j=0is reconstructible from (b_j)^κ_j=0as well. The corresponding relations can be written in a convenient form using the particular block Toeplitz matrices introduced in Notations 3.1 and 8.2:

Remark 8.9. If (s_j)^κ_j=0is a sequence of complexp×q matrices, thenS^′_•,m,β =−[Rp,m(β)]⁻¹S_m andS^′

•,m,β =−S_m[R_q,m(β)]^−∗ for allm∈Z_0,κ.

By virtue of Remark 3.14, we can conclude from Lemma 8.7 the analogous relation between the reciprocal sequences associated via Definition 3.5 to (sj)^κ_j=0 and to (bj)^κ_j=0:

Lemma 8.10. Let (s_j)^κ_j=0 be a sequence of complex p×q matrices. Denote by (r_j)^κ_j=0 the reciprocal sequence associated to(b_j)^κ_j=0. Then S^hri_m =−Rq,m(β)S^♯_m andS^hri_m =−S^♯

m[R_p,m(β)]^∗ for all m∈Z_0,κ.

In addition to the [α,∞)-modification (aj)^κ_j=0 and the (−∞, β]-modification (bj)^κ_j=0, we introduce, in view of Notation 7.1, the corresponding construction associated to the sequence (c_j)^κ−2_j=0:

Definition 8.11. Let (s_j)^κ_j=0 be a sequence of complex p×q matrices and let c₋₂:=−s₀. In the caseκ≥1, letc₋₁ := (α+β)s₀−s₁. Then we call the sequence (c_j)^κ_j=0 given byc_j :=c_j−2 the [α, β]-modification of (s_j)^κ_j=0. For each matrix X_k =X_k^hsibuilt from the sequence (s_j)^κ_j=0, we denote by X_α,k,β^′ :=X_k^hci the corresponding matrix built from the sequence (cj)^κ_j=0 instead of (s_j)^κ_j=0.

Remark 8.12. It is readily checked that the sequence (c_j)^κ_j=0 coincides with the [α,∞)-modification of (bj)^κ_j=0 as well as with the (−∞, β]-modification of (a_j)^κ_j=0.

Using Remarks 8.9 and 8.6 and Lemmata 8.10 and 8.7, resp., we obtain in particular:

Remark 8.13. If (s_j)^κ_j=0 is a sequence of complex p×q matrices, then S^′_α,m,β =

−[Rp,m(α)]⁻¹[R_p,m(β)]⁻¹S_m and S^′

α,m,β =−S_m[R_q,m(β)]^−∗[R_q,m(α)]^−∗ for allm∈Z_0,κ. Using the Cauchy product given in Notation 3.15 and the reciprocal sequence associated to the (−∞, β]-modification of (aj)^κ−1_j=0, we introduce now a transformation of sequences of complex matrices, which in the context of the moment problem on the interval [α, β] turns out to play the same role as theH-transform for the moment problem on R:

Definition 8.14. Supposeκ≥1. Let (s_j)^κ_j=0be a sequence of complexp×qmatrices. Denote by (g_j)^κ−1_j=0 the (−∞, β]-modification of (aj)^κ−1_j=0 and by (x_j)^κ−1_j=0 the Cauchy product of (b_j)^κ−1_j=0 and (g_j^♯)^κ−1_j=0. Then we call the sequence (t_j)^κ−1_j=0 given by t_j :=−a0s^†₀x_ja₀ theFα,β-transform of (s_j)^κ_j=0.

Since, in the classical caseα= 0 andβ = 1, the sequence (a_j)^κ−1_j=0 coincides with the shifted sequence (s_j+1)^κ−1_j=0, the F0,1-transform is given by t_j = −s1s^†₀x_js₁ with the Cauchy product (xj)^κ−1_j=0 of (bj)^κ−1_j=0 and (g^♯_j)^κ−1_j=0, where the sequence (bj)^κ−1_j=0 is given bybj =sj−sj+1 and the

sequence (g_j)^κ−1_j=0 is given by g₀ =−s₁ and by g_j =s_j−s_j+1 forj∈Z_1,κ−1.

Remark 8.15. Assume κ ≥ 1 and let (s_j)^κ_j=0 be a sequence of complex p×q matrices with Fα,β-transform (t_j)^κ−1_j=0. Then one can see from Remark 3.6 that for each k ∈ Z_0,κ−1, the matrix t_k is built from the matrices s₀, s₁, . . . , s_k+1. In particular, for all m ∈ Z_1,κ, the Fα,β-transform of (s_j)^m_j=0 coincides with (t_j)^m−1_j=0 .

The main goal of this section can be described as follows: Let κ ≥ 1 and let (s_j)^κ_j=0 ∈ F_q,κ,α,β^< . Denote by (t_j)^κ−1_j=0 the F_α,β-transform of (s_j)^κ_j=0. Then we are going to prove that (tj)^κ−1_j=0 ∈ F_{q,κ−1,α,β}^< . Our main strategy to realize this aim is based on using appropriate identities for block Hankel matrices associated with the sequences (s_j)^κ_j=0, (t_j)^κ−1_j=0 and also the three sequences (a_j)^κ−1_j=0, (b_j)^κ−1_j=0 and (c_j)^κ−2_j=0 built from (s_j)^κ_j=0 via Notation 7.1. Clearly, we will concentrate on the finite sections of the sequences (s_j)^κ_j=0. There will be two different cases, namely the sections (s_j)²ⁿ_j=0on the one hand and the sections (s_j)²ⁿ⁺¹_j=0 on the other hand.

We start with a collection of identities which will be later used in many proofs.

Lemma 8.16. Let n∈N₀ and let z, w∈C. Using Notation 8.2, then

[R_p,n(z)]⁻¹[R_p,n(w)]⁻¹ = [R_p,n(w)]⁻¹[R_p,n(z)]⁻¹, (8.1) [Rp,n(z)]⁻¹Rp,n(w) =Rp,n(w)[Rp,n(z)]⁻¹, (8.2) R_p,n(z)R_p,n(w) =R_p,n(w)R_p,n(z). (8.3) Proof. The identity (8.1) is an immediate consequence of Notation 8.2. Formula (8.2) follows from (8.1) and implies (8.3).

Remark 8.17. Letℓ, m∈N. Using Notations 4.5 and 8.1, direct computations give us:

(a) ∆_q,ℓ,m∆^∗_q,ℓ,m=I_mq⊕O_ℓ×ℓ and ∇q,ℓ,m∇^∗_q,ℓ,m=O_ℓ×ℓ⊕I_mq. (b) ∆^∗_q,ℓ,m∆_q,ℓ,m=I_mq and ∇^∗_q,ℓ,m∇_q,ℓ,m=I_mq.

(c) If n∈N, then ∆^∗_q,1,n∇q,1,n =T_q,n−1 and ∇q,1,n∆^∗_q,1,n =T_q,n.

Lemma 8.18. Let (sj)^κ_j=0 be a sequence of complex p×q matrices.

(a) Let m∈Z_1,κ. In view of Notations 3.1 and 4.5, then

∆^∗_p,m,1S_m =s₀∆^∗_q,m,1, S_m∆_q,m,1 = ∆_p,m,1s₀, (8.4)

∆^∗_p,1,mS_m =S_m−1∆^∗_q,1,m, S_m∆_q,1,m= ∆_p,1,mS_m−1, (8.5) S_m∇q,1,m=∇p,1,mS_m−1, and ∇^∗_p,1,mS_m =S_m−1∇^∗_q,1,m. (8.6) (b) Let n∈Z_0,κ. In view of Notation 8.1, then T_p,nS_n=S_nT_q,n and S_nT_q,n^∗ =T_p,n^∗ S_n. (c) Let n∈Z_0,κ and let z∈C. In view of Notation 8.2, then

[R_p,n(z)]⁻¹S_n=S_n[R_q,n(z)]⁻¹, S_n[R_q,n(z)]^−∗= [R_p,n(z)]^−∗S_n, (8.7) R_p,n(z)S_n=S_nR_q,n(z), and S_n[R_q,n(z)]^∗= [R_p,n(z)]^∗S_n. (8.8) Proof. (a) The identities (8.4)–(8.6) follow from Notation 4.5 and Remark 4.3.

(b) In view ofT_p,0 =O_p×p and T_q,0 =O_q×q, the case n= 0 is trivial. Supposen≥1. Using Remark 8.17(c), (8.5), and (8.6), we obtain

T_p,nS_n=∇p,1,n∆^∗_p,1,nS_n=∇p,1,nS_n−1∆^∗_q,1,n =S_n∇q,1,n∆^∗_q,1,n =S_nT_q,n and

S_nT_q,n^∗ =S_n∆_q,1,n∇^∗_q,1,n = ∆_p,1,nS_n−1∇^∗_q,1,n= ∆_p,1,n∇^∗_p,1,nS_n=T_p,n^∗ S_n.

(c) The identities (8.7) follow from Notation 8.2 and (b). The identities (8.8) follow from (8.7).

Remark 8.19. LetA be a complex p×q matrix. Then Lemma 8.18 yields:

(a) Let m ∈N. Denote by hhAiim the block diagonal matrix built via (3.4) fromA. In view of Notations 4.5 and 8.1, then

∆^∗_p,m,1hhAiim =A∆^∗_q,m,1, hhAiim∆q,m,1 = ∆p,m,1A, (8.9)

∆^∗_p,1,mhhAiim =hhAiim−1∆^∗_q,1,m, hhAiim∆_q,1,m= ∆_p,1,mhhAiim−1,

hhAii_m∇_q,1,m=∇_p,1,mhhAii_m−1, and ∇^∗_p,1,mhhAii_m =hhAii_m−1∇^∗_q,1,m. (8.10) (b) Letn∈N₀. Denote byhhAiin the block diagonal matrix built via (3.4) from A. In view

of Notation 8.1, then T_p,nhhAii_n=hhAii_nT_q,n and hhAii_nT_q,n^∗ =T_p,n^∗ hhAii_n.

(c) Letn∈N₀and letz∈C. Denote byhhAiin the block diagonal matrix built via (3.4) from A. In view of Notation 8.2, then

[Rp,n(z)]⁻¹hhAiin=hhAiin[Rq,n(z)]⁻¹, hhAiin[Rq,n(z)]^−∗= [Rp,n(z)]^−∗hhAiin, (8.11) R_p,n(z)hhAiin=hhAiinR_q,n(z), and hhAiin[R_q,n(z)]^∗= [R_p,n(z)]^∗hhAiin. (8.12) Remark 8.20. Let A be a complex p×q matrix and let n ∈ N₀. Denote by hhAiin the block diagonal matrix built via (3.4) fromA.

(a) It holds (hhAiin)^∗ =hhA^∗iin and, by virtue of Remark A.8, moreover (hhAiin)^†=hhA^†iin.

(b) Letr ∈N, letB ∈C^q×r, and letλ∈C. ThenhhλAiin=λhhAiinand hhABiin =hhAiinhhBiin. Now we turn our attention to the [α,∞)- and (−∞, β]-modification and to the [α, β]-modification of the Fα,β-transform and the interplay of these sequences and their re-ciprocal sequences with forming Cauchy products:

Lemma 8.21. Let (s_j)^κ_j=0 be a sequence of complex p×q matrices and assume κ≥1. Let the sequences (a_j)^κ−1_j=0 and (b_j)^κ−1_j=0 be given by Notation 7.1. Denote by (t_j)^κ−1_j=0 the Fα,β-transform of (s_j)^κ_j=0 and by (f_j)^κ−1_j=0 the [α,∞)-modification of (b_j)^κ−1_j=0.

(a) Denote by (g_j)^κ−1_j=0 the (−∞, β]-modification of (a_j)^κ−1_j=0, by (x_j)^κ−1_j=0 the Cauchy product of (f_j)^κ−1_j=0 and (g^♯_j)^κ−1_j=0, and by (u_j)^κ−1_j=0 the [α,∞)-modification of (t_j)^κ−1_j=0. For all j ∈ Z_0,κ−1, then u_j =−a0s^†₀x_ja₀.

(b) Denote by (y_j)^κ−1_j=0 the Cauchy product of (b_j)^κ−1_j=0 and (a^♯_j)^κ−1_j=0 and by (v_j)^κ−1_j=0 the (−∞, β]-modification of (t_j)^κ−1_j=0. For all j∈Z_0,κ−1, then v_j =−a0s^†₀y_ja₀.

(c) Denote by (z_j)^κ−1_j=0 the Cauchy product of (f_j)^κ−1_j=0 and (a^♯_j)^κ−1_j=0 and by (w_j)^κ−1_j=0 the [α, β]-modification of (t_j)^κ−1_j=0. For all j∈Z_0,κ−1, then w_j =−a0s^†₀z_ja₀.

Proof. We consider an arbitrary k∈Z_0,κ−1. Denote by (h_j)^κ−1_j=0 the reciprocal sequence asso-ciated to (g_j)^κ−1_j=0. In view of Definition 8.14 and Remark 3.17, then

S^hti_k =hh−a0s^†₀iikS^hbi_k S^hhi_k hha0iik. (8.13) According to Remark 8.6, we have

[R_p,k(α)]⁻¹S^hbi_k =S^hf_kⁱ. (8.14) Denote by (r_j)^κ−1_j=0 the reciprocal sequence associated to (a_j)^κ−1_j=0. Then Lemma 8.10 yields

−[R_q,k(β)]⁻¹S^hhi_k =S^hri_k . (8.15) (a) According to Remark 8.6, we have S^hui_k = [R_p,k(α)]⁻¹S^hti_k . By virtue of (8.13), (8.11), (8.14), and Remark 3.17, then

S^hui_k = [R_p,k(α)]⁻¹hh−a₀s^†₀ii_kS^hbi_k S^hhi_k hha₀ii_k=hh−a₀s^†₀ii_kS^hf_kⁱS^hhi_k hha₀ii_k=hh−a₀s^†₀ii_kS^hxi_k hha₀ii_k. follows. Consequently,u_k=−a0s^†₀x_ka₀ holds true.

(b) According to Remark 8.9, we have S^hvi_k =−[R_p,k(β)]⁻¹S^hti_k . By virtue of (8.13), (8.11), (8.7), (8.15), and Remark 3.17, we get then

S^hvi_k =−[Rp,k(β)]⁻¹hh−a0s^†₀iikS^hbi_k S^hhi_k hha0iik=hh−a0s^†₀iikS^hbi_k S^hri_k hha0iik =hh−a0s^†₀iikS^hyi_k hha0iik. Consequently,v_k=−a₀s^†₀y_ka₀ holds true.

(c) According to Remark 8.13, we have S^hwi_k = −[Rp,k(α)]⁻¹[R_p,k(β)]⁻¹S^hti_k . By virtue of (8.13), (8.11), (8.7), (8.14), (8.15), and Remark 3.17, then

S^hwi_k =−[Rp,k(α)]⁻¹[R_p,k(β)]⁻¹hh−a0s^†₀iikS^hbi_k S^hhi_k hha0iik

=hh−a₀s^†₀ii_kS^hf_kⁱS^hri_k hha₀ii_k=hh−a₀s^†₀ii_kS^hzi_k hha₀ii_k follows. Consequently,w_k=−a0s^†₀z_ka₀ holds true.

In the remaining part of this section, we will derive representations for the block Hankel ma-trices built from theF_α,β-transform of a sequence via Notation 4.1, (7.1), and (7.2). Therefore, we first supplement Notation 5.18:

Notation 8.22. Let κ, τ ∈ N₀∪ {∞} and let (sj)^κ_j=0 and (tj)^τ_j=0 be two sequences of complex p×q matrices. Denote by (r_j)^κ_j=0 the reciprocal sequence associated to (s_j)^κ_j=0. For each m∈N₀ with m≤min{κ, τ}, then let

D^hs,ti_m :=hhs₀ii_mS^hri_m S^hti_mhht^†₀ii_m+hhI_p−s₀s^†₀t₀t^†₀ii_m and

D^hs,ti

m :=hht^†₀iimS^hti

mS^hri

m hhs0iim+hhIq−t^†₀t0s^†₀s0iim.

In particular, ifp=q and if the sequence (t_j)^τ_j=0 is given by t_j :=δ_j0I_q, then D^hs,tim =D^hsim

and D^hs,ti_m = D^hsi_m. Referring to the classes of particular block triangular matrices introduced in Notation A.19, by virtue ofr₀ =s^♯₀=s^†₀ and Remark 3.17, we have:

Remark 8.23. Let κ, τ ∈ N₀ ∪ {∞} and let (s_j)^κ_j=0 and (t_j)^τ_j=0 be two sequences from C^p×q. For allm∈Z_0,κ∩Z_0,τ, thenD^hs,tim is a block Toeplitz matrix belonging to L_p,m andD^hs,ti_m is a block Toeplitz matrix belonging toU_q,m.

Remark 8.24. Letκ, τ ∈N₀∪ {∞}and let (s_j)^κ_j=0 and (t_j)^τ_j=0 be two sequences fromC^p×q. In view of Remarks 8.23 and A.21, for all m∈Z_0,κ∩Z_0,τ, then detD^hs,tim = 1 and detD^hs,ti_m = 1.

By virtue of Proposition 3.13, Remark 8.20(a), and Remark A.5, the following two remarks can be verified:

Remark 8.25. Let (s_j)^κ_j=0 be a sequence of complexp×q matrices and let the sequence (t_j)^κ_j=0 be given byt_j :=s^∗_j. For all m∈Z_0,κ, thenD^∗_m=D^hti_m.

Remark 8.26. Letκ, τ ∈N₀∪ {∞}and let (s_j)^κ_j=0and (t_j)^τ_j=0 be two sequences fromC^p×q. Let the sequences (u_j)^κ_j=0and (v_j)^τ_j=0be given byu_j :=s^∗_j andv_j :=t^∗_j, resp. For allm∈Z_0,κ∩Z_0,τ, then (D^hs,tim )^∗=D^hu,vi_m .

Using Remark 8.20(b) and Remark A.12, we obtain the following two remarks:

Remark 8.27. Let (s_j)^κ_j=0 be a sequence of complex p×q matrices and let m ∈ Z_0,κ. For all B ∈ C^p×u with R(B) ⊆ R(s0), then D_mhhBiim = hhs0iimS^♯_mhhBiim. For all C ∈ C^v×q with N(s₀)⊆ N(C), furthermorehhCiimD_m=hhCiimS^♯

mhhs0iim.

Remark 8.28. Let κ, τ ∈ N₀ ∪ {∞}, let (sj)^κ_j=0 and (t_j)^τ_j=0 be two sequences of complex p×qmatrices, and letm∈N₀ withm≤min{κ, τ}. Denote by (rj)^κ_j=0 the reciprocal sequence associated to (s_j)^κ_j=0. For all B ∈ C^p×u with R(B) ⊆ R(t0)∩ R(s0), then D^hs,tim hhBiim = hhs₀ii_mS^hrim S^htimhht^†₀Bii_m. For allC∈C^v×qwithN(t₀)∪N(s₀)⊆ N(C), furthermorehhCii_mD^hs,ti_m = hhCt^†₀iimS^hti_mS^hri_m hhs0iim.

Lemma 8.29. Suppose κ ≥1. Let (s_j)^κ_j=0 ∈ Dp×q,κ. Denote by (t_j)^κ−1_j=0 the Fα,β-transform of (s_j)^κ_j=0. Then t₀ =d₁, where d₁ is given by Definition 7.17.

Proof. Denote by (g_j)^κ−1_j=0 the (−∞, β]-modification of (a_j)^κ−1_j=0, by (h_j)^κ−1_j=0 the reciprocal se-quence associated to (g_j)^κ−1_j=0, and by (x_j)^κ−1_j=0 the Cauchy product of (b_j)^κ−1_j=0 and (h_j)^κ−1_j=0. Then h₀ =g^†₀ = (−a0)^†. Consequently, x₀ =b₀h₀ =−b0a^†₀. Taking into account Remark 7.26, we get thent₀ =−a0s^†₀x₀a₀ =a₀s^†₀b₀a^†₀a₀ =b₀s^†₀a₀a^†₀a₀ =d₁.

From Lemma 8.29 we obtain for theF_α,β-transform (tj)^κ−1_j=0 of a sequence (sj)^κ_j=0 belonging to Dp×q,κ in particular H₀^hti =d1. To obtain in Proposition 8.38 below, a convenient formula for the block Hankel matricesH_n^hti for an arbitraryn∈Nwith 2n+ 1≤κ, we need a series of auxiliary results:

Lemma 8.30. Suppose κ ≥ 1. Let (sj)^κ_j=0 ∈ Dp×q,κ and let the sequence (hj)^κ−1_j=0 be given by h_j := s_j+1. Then S_mhhs^†₀ii_mS^hhim = S^hhim hhs^†₀ii_mS_m and S_mhhs^†₀ii_mS^hhi_m = S^hhi_m hhs^†₀ii_mS_m for all m∈Z_0,κ−1.

Proof. Consider an arbitrary m ∈ Z_0,κ−1. According to Notation 3.1, we have S^hhi_m T_q,m = S_m−hhs0iim=T_p,mS^hhi_m . Because of (s_j)^κ_j=0∈ Dp×q,κ, furthermoreR(hj)⊆ R(s0) andN(s₀)⊆ N(h_j) hold true for all j ∈ Z_0,κ−1. By virtue of Remark A.12, we get then hhs0s^†₀iimS^hhim = S^hhim = S^hhim hhs^†₀s₀iim. Consequently, using additionally Remark 8.20(b) and Remark 8.19(b), we obtain thus

S_mhhs^†₀iimS^hhi_m =S^hhi_m T_q,m+hhs0iim

hhs^†₀iimS^hhi_m

=S^hhi_m T_q,mhhs^†₀iimS^hhi_m +hhs0s^†₀iimS^hhi_m =S^hhi_m hhs^†₀iimT_p,mS^hhi_m +S^hhi_m hhs^†₀s₀iim

=S^hhi_m hhs^†₀iim

Tp,mS^hhi_m +hhs0iim

=S^hhi_m hhs^†₀iimS_m. The second identity can be checked analogously.

Observe that, with the sequence (h_j)^κ−1_j=0 given in Lemma 8.30, we haveS_α,m,• =−αSm+S^hhim , S_α,m,• =−αS_m+S^hhi_m ,S_•,m,β =βS_m−S^hhim , and S_•,m,β =βS_m−S^hhi_m for all m ∈Z_0,κ−1. In view of Remark 7.2, we obtain:

Remark 8.31. Supposeκ≥1. Let (sj)^κ_j=0 be a sequence of complexp×q matrices. Denote by (f_j)^κ−1_j=0 the [α,∞)-modification of (bj)^κ−1_j=0 and by (g_j)^κ−1_j=0 the (−∞, β]-modification of (aj)^κ−1_j=0. Then f₀ =b0 and g₀ =−a0. For allj ∈Z_1,κ−1, furthermore f_j =cj−1=g_j.

Remark 8.32. Supposeκ≥1. Let (s_j)^κ_j=0 be a sequence of complexp×q matrices. Denote by (fj)^κ−1_j=0 the [α,∞)-modification of (bj)^κ−1_j=0 and by (gj)^κ−1_j=0 the (−∞, β]-modification of (aj)^κ−1_j=0.

From Remarks 8.31 and 7.3 we can conclude that f₀−g₀ = δs₀ and f_j −g_j = O_p×q for all j∈Z_1,κ−1.

Lemma 8.33. Suppose κ ≥ 2. Let (s_j)^κ_j=0 be a sequence of complex p×q matrices and let n ∈ Z_0,κ−2. Denote by (f_j)^κ−1_j=0 the [α,∞)-modification of (b_j)^κ−1_j=0 and by (g_j)^κ−1_j=0 the (−∞, β]-modification of (a_j)^κ−1_j=0. For all k∈Z_1,κ−n−1, then

S_α,n,•hhs^†₀iiny_k,k+n^hfi −S^hf_nⁱhhs^†₀iiny_α,k,k+n,• =βy_α,0,n,•s^†₀a_k−1−δhhs0s^†₀iiny_α,k,k+n,• (8.16) and

S_•,n,βhhs^†₀iiny^hgi_k,k+n−S^hgi_n hhs^†₀iiny_•,k,k+n,β =δhhs0s^†₀iiny_•,k,k+n,β−αy_•,0,n,βs^†₀b_k−1. (8.17) Proof. By way of example, we only show (8.17). First consider an arbitraryk∈Z_1,κ−1. In view of Remarks 8.31 and 7.2, thenf₀ =b₀ and g_k =c_k−1 =−αb_k−1+b_k. In particular, b₀s^†₀g_k =

−αb0s^†₀b_k−1+f₀s^†₀b_k. Using Remark 8.32, consequently,b₀s^†₀g_k−g0s^†₀b_k=−αb0s^†₀b_k−1+δs₀s^†₀b_k, implying (8.17) in the casen= 0.

Now assume κ ≥ 3 and n ≥ 1. Consider an arbitrary k ∈ Z_1,κ−n−1. From Remark 8.32 Remark 8.20(b) we inferS^hf_nⁱ−S^hgi_n =δhhs0iin. Hence, taking into account Remark 8.20(b), we get

S^hgi_n hhs^†₀iiny_•,k,k+n,β =S^hf_nⁱhhs^†₀iiny_•,k,k+n,β−δhhs0s^†₀iiny_•,k,k+n,β.

The application of Remark 8.6 to the sequence (b_j)^κ−1_j=0 yields furthermore S^hfnⁱ = [R_p,n(α)]⁻¹S_•,n,β, which by virtue of (8.7) and (8.11) implies Taking into account (8.19) and (8.18), then (8.17) follows.

Lemma 8.34. Suppose κ ≥ 2. Let (s_j)^κ_j=0 be a sequence of complex p×q matrices and of the matrix on the left-hand side of (8.20). In view of Remark 4.3, we get

S_•,n,βhhs^†₀iin∇p,1,ny_k+1,k+n^hgi =S_•,n,β

Consequently,

X=−g₀s^†₀b_k and Y =S_•,n−1,βhhs^†₀ii_n−1y^hgi_k+1,k+n−y^hgi_1,ns^†₀b_k−S^hgi_n−1hhs^†₀ii_n−1y•,k+1,k+n,β. According to Remark 8.31, thenX =a0s^†₀b_k. The application of Lemma 8.33 withn−1 instead ofnand k+ 1 instead ofk implies

S_•,n−1,βhhs^†₀iin−1y^hgi_k+1,k+n−S^hgi_n−1hhs^†₀iin−1y•,k+1,k+n,β

=δhhs0s^†₀iin−1y•,k+1,k+n,β−αy_{•,0,n−1,β}s^†₀b_k. Hence, Y = δhhs0s^†₀iin−1y•,k+1,k+n,β −(y^hgi_1,n +αy_{•,0,n−1,β})s^†₀b_k. Thus, (8.20) holds true, since Remarks 8.31 and 7.2 yield

y^hgi_1,n+αy_{•,0,n−1,β}=y^hci_0,n−1+αy_{•,0,n−1,β} =y_•,1,n,β.

Lemma 8.35. Suppose κ ≥ 2. Let (s_j)^κ_j=0 ∈ Dp×q,κ. Denote by (g_j)^κ−1_j=0 the (−∞, β]-modification of (a_j)^κ−1_j=0. For all n∈Z_1,κ−1, then

S_•,n,βhhs^†₀iin∇p,1,ny_1,n^hgi−S^hgi_n hhs^†₀iiny_•,0,n,β =y_•,0,n,βs^†₀a₀. (8.21) Proof. Letn∈Z_1,κ−1. Obviously, fork= 0 the matrix on the left-hand side of (8.20) coincides with the matrix on the left-hand side of (8.21). Thus, taking into account Lemma 8.34 and the block representation y_•,0,n,β =_y•,1,n,β^b0 , it is sufficient to prove that a0s^†₀b0 =b0s^†₀a0 and δhhs0s^†₀iiny_•,1,n,β−y_•,1,n,βs^†₀b₀ =y_•,1,n,βs^†₀a₀ are valid. Obviously,a₀s^†₀b₀ =b₀s^†₀a₀ is fulfilled, by virtue of Lemma 7.26. Because of (s_j)^κ_j=0 ∈ Dp×q,κ and Remark A.2, we haveR(bj) ⊆ R(s0) and N(s₀) ⊆ N(b_j) for allj ∈Z_0,κ−1. Using Remark A.12, we infer then δhhs0s^†₀iiny_•,1,n,β = δy_•,1,n,β = y_•,1,n,βs^†₀(δs0). By virtue of Remark 7.3, hence δhhs0s^†₀iiny_•,1,n,β −y_•,1,n,βs^†₀b0 = y_•,1,n,βs^†₀(δs₀−b₀) =y_•,1,n,βs^†₀a₀.

Lemma 8.36. Suppose κ ≥ 3. Let (s_j)^κ_j=0 ∈ L^<,e_q,κ,β. Denote by (g_j)^κ−1_j=0 the (−∞, β]-modification of (a_j)^κ−1_j=0. For all n∈Nwith 2n+ 1≤κ, then

S_•,n,βhhs^†₀iin∇q,1,nG^hgi_n−1−S^hgi_n hhs^†₀iinH_•,n,β∇q,1,n=

"

a₀s^†₀z_•,1,n,β

δL_•,n,β+y_•,1,n,βb^†₀d₁b^†₀z_•,1,n,β

#

. (8.22) Proof. Consider an arbitrary n ∈ N with 2n + 1 ≤ κ. According to Lemma 8.34, we have (8.20) for all k ∈ Z_1,n. Taking into account G^hgi_n−1 = [y_2,n+1^hgi , y_3,n+2^hgi , . . . , y_n+1,2n^hgi ] as well as H_•,n,β∇q,1,n = [y_•,1,n+1,β, y_•,2,n+2,β, . . . , y_•,n,2n,β] and G_•,n−1,β = [y_•,2,n+1,β, y_•,3,n+2,β, . . . , y_{•,n+1,2n,β}], thesenequations can be subsumed as columns in the single equation

S_•,n,βhhs^†₀ii_n∇_q,1,nG^hgi_n−1−S^hgi_n hhs^†₀ii_nH_•,n,β∇_q,1,n=

"

a0s^†₀z_•,1,n,β

δhhs0s^†₀iinG_•,n−1,β−y_•,1,n,βs^†₀z_•,1,n,β

# .

Using Remark 6.7, we conclude that the sequence (s_j)²ⁿ⁺¹_j=0 belongs toL^<,e_q,2n+1,β and, because of (6.2), hence belongs to H^<,e_q,2n+1. According to Proposition 5.8, thus (s_j)²ⁿ⁺¹_j=0 ∈ Dq×q,2n+1.

By virtue of Remark A.2, then R(bj) ⊆ R(s0) follows for all j ∈ Z_0,2n. Consequently, Re-mark A.12(a) yieldss₀s^†₀b₀ =b₀ and hhs0s^†₀iinG_•,n−1,β =G_•,n−1,β. In view of Remark 4.11, we get thereforeδhhs0s^†₀iinG_•,n−1,β=δ(L_•,n,β+y_•,1,n,βb^†₀z_•,1,n,β) and, hence,

δhhs0s^†₀iinG_•,n−1,β−y_•,1,n,βs^†₀z_•,1,n,β =δL_•,n,β+y_•,1,n,β(δb^†₀−s^†₀)z_•,1,n,β.

In view of (6.2), we have (b_j)²ⁿ_j=0 ∈ H_q,2n^< . Consequently, Proposition 5.9 yields (b_j)²ⁿ_j=0 ∈ D˜q×q,2n. Using Remark A.12, then we infer b₀b^†₀z_•,1,n,β = z_•,1,n,β and y_•,1,n,βb^†₀b₀ = y_•,1,n,β. Taking into account s₀s^†₀b₀=b₀, we have, by virtue of Remarks 7.3 and 7.26, furthermore

b₀(δb^†₀−s^†₀)b₀=δb₀−b₀s^†₀b₀ =δs₀s^†₀b₀−b₀s^†₀b₀= (δs₀−b₀)s^†₀b₀=a₀s^†₀b₀ =d₁. Consequently, we obtain (8.22) from

y_•,1,n,β(δb^†₀−s^†₀)z_•,1,n,β =y_•,1,n,βb^†₀b₀(δb^†₀−s^†₀)b₀b^†₀z_•,1,n,β =y_•,1,n,βb^†₀d₁b^†₀z_•,1,n,β. Lemma 8.37. Let n ∈ N and let (sj)²ⁿ⁺¹_j=0 ∈ F_q,2n+1,α,β^< . Denote by (gj)²ⁿ_j=0 the (−∞, β]-modification of (a_j)²ⁿ_j=0. Then

S_•,n,β∇q,1,nhhs^†₀iin−1∇^∗_q,1,nH_n^hgi−S^hgi_n hhs^†₀iinH_•,n,β

=y_•,0,n,βb^†₀d₁b^†₀z_•,0,n,β+δ∇q,1,nL_•,n,β∇^∗_q,1,n. (8.23) Proof. Let

S_•,n,β∇q,1,nhhs^†₀iin−1∇^∗_q,1,nH_n^hgi−S^hgi_n hhs^†₀iinH_•,n,β = [A, B]

be the block representation of the matrix on the left-hand side of (8.23) with (n+ 1)q×qblock A. In view of Remark 4.9, we have

∇^∗_q,1,nH_n^hgi= [y^hgi_1,n, G^hgi_n−1] and H_•,n,β = [y_•,0,n,β, H_•,n,β∇q,1,n].

Consequently, we obtain

A=S_•,n,β∇q,1,nhhs^†₀iin−1y_1,n^hgi−S^hgi_n hhs^†₀iiny_•,0,n,β and

B =S_•,n,β∇q,1,nhhs^†₀iin−1G^hgi_n−1−S^hgi_n hhs^†₀iinH_•,n,β∇q,1,n.

Since Proposition 7.11 shows that (s_j)²ⁿ⁺¹_j=0 ∈ Dq×q,2n+1, Lemma 8.35 in combination with (8.10) yields

A=S_•,n,βhhs^†₀iin∇q,1,ny^hgi_1,n−S^hgi_n hhs^†₀iiny_•,0,n,β =y_•,0,n,βs^†₀a₀.

Because of [23, Prop. 9.2], the sequence (s_j)²ⁿ⁺¹_j=0 belongs to L^<,e_q,2n+1,β. Thus, Lemma 8.36 in combination with (8.10) shows thatB coincides with the block matrix on the right-hand side of (8.22). Hence, we have

[A, B] =







b₀s^†₀a₀ a₀s^†₀z_•,1,n,β

y_•,1,n,βs^†₀a0 δL_•,n,β+y_•,1,n,βb^†₀d₁b^†₀z_•,1,n,β





=M+δ∇q,1,nL_•,n,β∇^∗_q,1,n

where

M :=

"

b₀s^†₀a₀ a₀s^†₀z_•,1,n,β y_•,1,n,βs^†₀a₀ y_•,1,n,βb^†₀d₁b^†₀z_•,1,n,β

# .

From Remark 7.26 we infer b₀b^†₀d₁b^†₀b₀ = b₀s^†₀a₀. As in the proof of Lemma 8.36, we can concludeb₀b^†₀z_•,1,n,β=z_•,1,n,β andy_•,1,n,βb^†₀b₀=y_•,1,n,β. Taking into account Remark 7.26, we obtain henceb₀b^†₀d₁b^†₀z_•,1,n,β =a₀s^†₀z_•,1,n,β andy_•,1,n,βb^†₀d₁b^†₀b₀ =y_•,1,n,βs^†₀a₀. Consequently,

M =

"

b0b^†₀d₁b^†₀b0 b0b^†₀d₁b^†₀z_•,1,n,β y_•,1,n,βb^†₀d₁b^†₀b₀ y_•,1,n,βb^†₀d₁b^†₀z_•,1,n,β

#

=y_•,0,n,βb^†₀d₁b^†₀z_•,0,n,β follows, implying (8.23).

We are now able to basically reduce the block Hankel matrix Hn^hti built from the Fα,β-transform (t_j)²ⁿ_j=0 of a sequence (s_j)²ⁿ⁺¹_j=0 ∈ F_q,2n+1,α,β^< to a block diagonal matrix con-sisting of the matricesd₁ and L_•,n,β given via (7.8) and Notation 4.10, resp. In the following proof, we make use of theH-transformation introduced in Definition 5.16:

Proposition 8.38. Let n ∈ N and let (s_j)²ⁿ⁺¹_j=0 ∈ F_q,2n+1,α,β^< with F_α,β-transform (t_j)²ⁿ_j=0. Then the block Hankel matrix Hn^hti admits the representations

H_n^hti=R_q,n(β)hha₀ii_nS^†_α,n,•(y_•,0,n,βb^†₀d₁b^†₀z_•,0,n,β+δ∇_q,1,nL_•,n,β∇^∗_q,1,n)S^†_α,n,•hha₀ii_n[R_q,n(β)]^∗ (8.24) and

H_n^hti=R_q,n(β)D^ha,bi_n ·diag(d₁, δD_•,n−1,βL_•,n,βD_•,n−1,β)·D^ha,bi

n [R_q,n(β)]^∗. (8.25) In particular, rankHn^hti= rankd₁+ rankL_•,n,β and detHn^hti =δ^nqdet(d₁) det(L_•,n,β).

Proof. Denote by (gj)²ⁿ_j=0 the (−∞, β]-modification of (aj)²ⁿ_j=0, by (hj)²ⁿ_j=0 the reciprocal se-quence associated to (g_j)²ⁿ_j=0, and by (x_j)²ⁿ_j=0 the Cauchy product of (b_j)²ⁿ_j=0 and (h_j)²ⁿ_j=0. In view of Definition 8.14, we have t_j = −a0s^†₀x_ja₀ for all j ∈ Z_0,2n. In particular, Hn^hti=hh−a0s^†₀iinHn^hxihha0iin. According to Proposition 4.2, we have

H_n^hxi=H_n^hbiS^hhi

n + (O_q×q⊕S^hbi_n−1)H_n^hhi=H_•,n,βS^hhi

n + (O_q×q⊕S_•,n−1,β)H_n^hhi. The application of Theorem 4.7 to the sequence (g_j)²ⁿ_j=0 yields furthermore

H_n^hhi=y_0,n^hhi∆^∗_q,n,1+ ∆_q,n,1z_0,n^hhi−S^hhi_n H_n^hgiS^hhi_n . Since obviously (O_q×q⊕S_•,n−1,β)∆_q,n,1=O_(n+1)q×q holds true, we get then

H_n^hti=hh−a₀s^†₀ii_n^hH_•,n,βS^hhi_n + (O_q×q⊕S_•,n−1,β)(y_0,n^hhi∆^∗_q,n,1−S^hhi_n H_n^hgiS^hhi_n )ⁱhha₀ii_n. (8.26) Applying Remark 8.9 to the sequence (a_j)²ⁿ_j=0, we obtain

S^hgi_n =−[Rq,n(β)]⁻¹S_α,n,• and S^hgi_n =−S_α,n,•[R_q,n(β)]^−∗. (8.27)

Denote by (rj)²ⁿ_j=0the reciprocal sequence associated to (aj)²ⁿ_j=0. The application of Lemma 8.10 to the sequence (a_j)²ⁿ_j=0 yields

S^hhi_n =−Rq,n(β)S^hri_n and S^hhi_n =−S^hri_n [R_q,n(β)]^∗. (8.28) According to Proposition 7.11(b), the sequence (a_j)²ⁿ_j=0 belongs toDq×q,2n. Applying Proposi-tion 3.20 and Lemma 3.21 to the sequence (a_j)²ⁿ_j=0, we get then

S^†_α,n,• =S^hri_n , S^†

α,n,• =S^hri

n (8.29)

and

S_α,n,•S^†_α,n,• =hha0a^†₀iin=S_α,n,•S^†

α,n,•, S^†_α,n,•S_α,n,• =hha^†₀a₀iin=S^†

α,n,•S_α,n,•. (8.30) In view of Corollary 3.10 it followsh_jg₀g^†₀=h_j for all j∈Z_0,2n. Thus,y^hhi_0,ng₀g^†₀ =y^hhi_0,n. Using (8.7) and (8.27)–(8.30), we can conclude furthermore

S^hgi_n S^hhi_n =hha₀a^†₀ii_n and S^hgi

n S^hhi

n =hha₀a^†₀ii_n. (8.31) By virtue of Remark 8.31 and (8.9), hence

y^hhi_0,n∆^∗_q,n,1=y^hhi_0,ng₀g^†₀∆^∗_q,n,1 =y_0,n^hhia₀a^†₀∆^∗_q,n,1=y^hhi_0,n∆^∗_q,n,1hha0a^†₀iin=y^hhi_0,n∆^∗_q,n,1S^hgi_n S^hhi_n (8.32) follows. In view of Remark 8.20(b), (8.11), (8.30), and (8.27), we obtain

hh−a0s^†₀iin=−Rq,n(β)[R_q,n(β)]⁻¹hha0a^†₀a₀s^†₀iin=−Rq,n(β)hha0iinhha^†₀a₀iin[R_q,n(β)]⁻¹hhs^†₀iin

=−Rq,n(β)hha0iinS^†_α,n,•S_α,n,•[Rq,n(β)]⁻¹hhs^†₀iin=Rq,n(β)hha0iinS^†_α,n,•S^hgi_n hhs^†₀iin. (8.33) Since the combination of the second equations in (8.28) and (8.29) yields moreover S^hhi_n =

−S^†_α,n,•[R_q,n(β)]^∗, we infer from (8.26), (8.32), and (8.33), using additionally (8.12), then H_n^hti =hh−a0s^†₀iin

hH_•,n,β+ (O_q×q⊕S_•,n−1,β)(y^hhi_0,n∆^∗_q,n,1S^hgi

n −S^hhi_n H_n^hgi)ⁱS^hhi

n hha0iin

=R_q,n(β)hha₀ii_nS^†_α,n,•

×^hS^hgi_n hhs^†₀iin(Oq×q⊕S_•,n−1,β)(S^hhi_n H_n^hgi−y^hhi_0,n∆^∗_q,n,1S^hgi

n )−S^hgi_n hhs^†₀iinH_•,n,βⁱ

×S^†

α,n,•hha0iin[R_q,n(β)]^∗.

(8.34)

According to Remarks 4.3 and 4.9, we have ∆^∗_q,n,1S^hgi_n = ∆^∗_q,n,1Hn^hgi. Thus, S^hhi_n H_n^hgi−y_0,n^hhi∆^∗_q,n,1S^hgi_n = (S^hhi_n −y^hhi_0,n∆^∗_q,n,1)H_n^hgi. Because of Remark 4.3 and Remark 8.17(a), furthermore

Oq×q⊕S_•,n−1,β=S_•,n,β∇q,1,n∇^∗_q,1,n and

S^hhi_n −y_0,n^hhi∆^∗_q,n,1 =Oq×q⊕S^hhi_n−1 =∇q,1,nS^hhi_n−1∇^∗_q,1,n.

hold true. By virtue of Remark 8.17(b) and (8.6), we obtain

(O_q×q⊕S_•,n−1,β)(S^hhi_n −y_0,n^hhi∆^∗_q,n,1) =S_•,n,β∇q,1,nS^hhi_n−1∇^∗_q,1,n=S_•,n,βS^hhi_n ∇q,1,n∇^∗_q,1,n, and, therefore,

(Oq×q⊕S_•,n−1,β)(S^hhi_n H_n^hgi−y^hhi_0,n∆^∗_q,n,1S^hgi

n ) =S_•,n,βS^hhi_n ∇q,1,n∇^∗_q,1,nH_n^hgi.

According to Proposition 7.11(a), the sequence (s_j)²ⁿ⁺¹_j=0 belongs to D_q×q,2n+1. From Lemma 8.30 and Remark A.1, we thus conclude S_α,n,•hhs^†₀iinS_•,n,β = S_•,n,βhhs^†₀iinS_α,n,•. In view of the first equations in (8.27), (8.7), and (8.11), then S^hgi_n hhs^†₀iinS_•,n,β =S_•,n,βhhs^†₀iinS^hgi_n follows. In combination with the first equation in (8.31) and (8.10), we get then

S^hgi_n hhs^†₀iinS_•,n,βS^hhi_n ∇q,1,n∇^∗_q,1,n =S_•,n,βhhs^†₀iinS^hgi_n S^hhi_n ∇q,1,n∇^∗_q,1,n

=S_•,n,βhhs^†₀iinhha0a^†₀iin∇q,1,n∇^∗_q,1,n=S_•,n,β∇q,1,nhhs^†₀iin−1∇^∗_q,1,nhha0a^†₀iin. Taking into account Remark 7.2 and (a_j)²ⁿ_j=0 ∈ D_q×q,2n, we get from Remark A.2 furthermore R(cj)⊆ R(a0) andN(a₀)⊆ N(c_j) for allj ∈Z_0,2n−1. By virtue of Remarks 8.31 and A.12(a), thena₀a^†₀g_j =g_j follows for allj∈Z_0,2n. In particular, hha0a^†₀iinHn^hgi =Hn^hgi. Consequently,

S^hgi_n hhs^†₀ii_n(O_q×q⊕S_•,n−1,β)(S^hhi_n H_n^hgi−y^hhi_0,n∆^∗_q,n,1S^hgi_n )

=S^hgi_n hhs^†₀ii_nS_•,n,βS^hhi_n ∇_q,1,n∇^∗_q,1,nH_n^hgi=S_•,n,β∇_q,1,nhhs^†₀ii_n−1∇^∗_q,1,nH_n^hgi. (8.35) Lemma 8.37 yields moreover (8.23). Substituting (8.35) into (8.34), we can then use (8.23) to conclude (8.24). From Remark 4.3 it is readily seen that S_•,n,β∆_q,n,1 = y_•,0,n,β and

∆^∗_q,n,1S_•,n,β =z_•,0,n,β hold true. Taking additionally into account (8.9), consequently y_•,0,n,βb^†₀d₁b^†₀z_•,0,n,β =S_•,n,β∆_q,n,1b^†₀d₁d^†₁d₁b^†₀∆^∗_q,n,1S_•,n,β

=S_•,n,βhhb^†₀d₁iin∆_q,n,1d^†₁∆^∗_q,n,1hhd1b^†₀iinS_•,n,β. (8.36) Because of Propositions 7.10 and 7.9, the sequence (b_j)²ⁿ_j=0 belongs to H^<,e_q,2n. According to Proposition 5.8, hence (b_j)²ⁿ_j=0 ∈ Dq×q,2n. Using Remark A.12, then

b₀b^†₀b_j =b_j and b_jb^†₀b₀=b_j for all j∈Z_0,2n (8.37) follow. Because of Remark 4.11, in particular

hhb0b^†₀iin−1L_•,n,βhhb^†₀b₀iin−1 =L_•,n,β. (8.38) In view of Notation 5.20, we have furthermore Ξ^hbi_n−1,2n = O, by virtue of (8.37). Denote by (h_j)²ⁿ⁻²_j=0 the H-transform of (b_j)²ⁿ_j=0. Taking into account (b_j)²ⁿ_j=0 ∈ D_q×q,2n, we can conclude

from Proposition 5.21 then the representations

H_n−1^hhi =hhb₀ii_n−1S^†_•,n−1,βL_•,n,βS^†

•,n−1,βhhb₀ii_n−1 (8.39) and

H_n−1^hhi =D_•,n−1,βL_•,n,βD_•,n−1,β. (8.40)

The application of Lemma 3.21 to the sequence (bj)²ⁿ_j=0 yields moreover S_•,n−1,βS^†_•,n−1,β =hhb₀b^†₀ii_n−1 =S_•,n−1,βS^†

•,n−1,β

and

S^†_•,n−1,βS_•,n−1,β =hhb^†₀b₀iin−1 =S^†

•,n−1,βS_•,n−1,β. In combination with (8.38), (8.39), and Remark 8.20(b), we infer

L_•,n,β =hhb0b^†₀iin−1L_•,n,βhhb^†₀b₀iin−1 =S_•,n−1,βS^†_•,n−1,βL_•,n,βS^†

•,n−1,βS_•,n−1,β

=S_•,n−1,βS^†_•,n−1,βS_•,n−1,βS^†_•,n−1,βL_•,n,βS^†

•,n−1,βS_•,n−1,βS^†

•,n−1,βS_•,n−1,β

=S_•,n−1,βhhb^†₀b₀iin−1S^†_•,n−1,βL_•,n,βS^†

•,n−1,βhhb0b^†₀iin−1S_•,n−1,β

=S_•,n−1,βhhb^†₀iin−1H_n−1^hhi hhb^†₀iin−1S_•,n−1,β.

(8.41)

Denote by (h_j)²ⁿ_j=0 the H-parameter sequence of (bj)²ⁿ_j=0. Because of (b_j)²ⁿ_j=0 ∈ H^<,e_q,2n, we get from Theorem 5.24 in particular h₂ = h0. Using Remarks 5.14 and 7.21, we infer then h₀ =L_•,1,β =B₃. According to Remark 7.21, we have furthermore

A₁=L_α,0,• =a₀ and B₁ =L_•,0,β =b₀. (8.42)

In view of (b_j)²ⁿ_j=0 ∈ H^<,e_q,2n, by virtue of Proposition 5.23 the sequence (h_j)²ⁿ⁻²_j=0 belongs to H_q,2n−2^<,e . Hence, Proposition 5.8 implies (h_j)²ⁿ⁻²_j=0 ∈ Dq×q,2n−2. According to Proposition 7.25, we have R(B3) ⊆ R(B2) and N(B₂) ⊆ N(B₃), whereas Proposition 7.24 yields R(B2) ⊆ R(d1) and N(d1) ⊆ N(B2). Consequently, R(hj) ⊆ R(h0) =R(B3) ⊆ R(B2) ⊆ R(d1) and, analogously, N(d₁)⊆ N(h_j) follow for all j∈Z_0,2n−2. In view of Remark A.12, hence

hhd₁d^†₁ii_n−1H_n−1^hhi hhd^†₁d₁ii_n−1=H_n−1^hhi . (8.43) Using Remark 8.20(b), (8.6), and (8.10), we infer from (8.41) then

∇q,1,nL_•,n,β∇^∗_q,1,n=∇q,1,nS_•,n−1,βhhb^†₀d₁d^†₁iin−1H_n−1^hhi hhd^†₁d₁b^†₀iin−1S_•,n−1,β∇^∗_q,1,n

=S_•,n,βhhb^†₀d1iin∇q,1,nhhd^†₁iin−1H_n−1^hhi hhd^†₁iin−1∇^∗_q,1,nhhd1b^†₀iinS_•,n,β.

(8.44) According to Proposition 7.24 and (8.42), we haveR(d1) =R(a0)∩R(b0) andN(a0)∪N(b0) = N(d₁). By virtue of (8.29) and Remark 8.28, thus

D^ha,bi_n hhd1iin=hha0iinS^†_α,n,•S_•,n,βhhb^†₀d₁iin

and

hhd1iinD^ha,bi_n =hhd1b^†₀iinS_•,n,βS^†_α,n,•hha0iin. In combination with (8.36) and (8.44), we get then

hha₀ii_nS^†_α,n,•(y_•,0,n,βb^†₀d₁b^†₀z_•,0,n,β+δ∇_q,1,nL_•,n,β∇^∗_q,1,n)S^†_α,n,•hha₀ii_n

=hha0iinS^†_α,n,•S_•,n,βhhb^†₀d₁iin

×∆_q,n,1d^†₁∆^∗_q,n,1+δ∇q,1,nhhd^†₁iin−1H_n−1^hhi hhd^†₁iin−1∇^∗_q,1,nhhd1b^†₀iinS_•,n,βS^†

α,n,•hha0iin

=D^ha,bi_n hhd1iin

∆_q,n,1d^†₁∆^∗_q,n,1+δ∇q,1,nhhd^†₁iin−1H_n−1^hhi hhd^†₁iin−1∇^∗_q,1,nhhd1iinD^ha,bi

n .

Furthermore, by virtue of (8.9), we have hhd1iin∆q,n,1d^†₁∆^∗_q,n,1hhd1iin = ∆q,n,1d₁d^†₁d₁∆^∗_q,n,1 =

Hence, we obtain (8.25) from (8.24). Taking into account δ > 0, the formulas for rankHn^hti

and detHn^hti follow, in view of Remarks 8.4, 8.24, and 5.19, from (8.25).

In the following, we use the equivalence relation “∼” introduced in Notation A.24:

Corollary 8.39. Let n∈Nand let (s_j)²ⁿ⁺¹_j=0 ∈ F_q,2n+1,α,β^< with Fα,β-transform (t_j)²ⁿ_j=0. Then L^hti_n =δR_q,n−1(β)D^ha,bi_n−1D_•,n−1,βL_•,n,βD_•,n−1,βD^ha,bi_n−1[R_q,n−1(β)]^∗ (8.45) and, in particular, rankL^hti_n = rankL_•,n,β, detL^hti_n =δ^nqdetL_•,n,β, and L^hti_n ∼δL_•,n,β.

Proof. From Notation 8.22 and Remark 4.3 we see that D^ha,bi_n =

In view of Example 8.3 and Remark 4.3, we have R_q,n(β) = L^hti_n =F/A. Consequently, (8.45) follows. Taking into account δ > 0, the formulas for rank and determinant ofL^hti_n can be obtained, in view of Remarks 8.4, 8.24, and 5.19, from (8.45).

Analogously, L^hti_n ∼δL_•,n,β follows from (8.45), using Remarks 8.4, 8.23, 5.19, and A.20 and Notation A.24.

Now we turn our attention to the block Hankel matrices built via (7.1) from the F_α,β-transform (t_j)²ⁿ⁺¹_j=0 of a sequence (s_j)²ⁿ⁺²_j=0 belonging to F_q,2n+2,α,β^< . We see that the matrix−αHn^hti+Kn^hti can be basically traced back toH_α,n,β:

Proposition 8.40. Let n ∈ N₀ and let (s_j)²ⁿ⁺²_j=0 ∈ F_q,2n+2,α,β^< with Fα,β-transform (t_j)²ⁿ⁺¹_j=0 . Let the sequence (u_j)²ⁿ_j=0 be given by u_j :=−αt_j+t_j+1. Then

H_n^hui=δR_q,n(β)hha0iinS^†_α,n,•H_α,n,βS^†_α,n,•hha0iin[R_q,n(β)]^∗ (8.46) and

H_n^hui=δRq,n(β)Dα,n,•H_α,n,βD_α,n,•[Rq,n(β)]^∗. (8.47) In particular, rankHn^hui = rankH_α,n,β, detHn^hui=δ^(n+1)qdetH_α,n,β, and Hn^hui∼δH_α,n,β. Proof. Denote by (u_j)²ⁿ⁺¹_j=0 the [α,∞)-modification of (tj)²ⁿ⁺¹_j=0 . In view of Definition 8.5, then Kn^hui = Hn^hui. Denote by (f_j)²ⁿ⁺¹_j=0 the [α,∞)-modification of (bj)²ⁿ⁺¹_j=0 , by (g_j)²ⁿ⁺¹_j=0 the (−∞, β]-modification of (aj)²ⁿ⁺¹_j=0 , by (h_j)²ⁿ⁺¹_j=0 the reciprocal sequence associated to (g_j)²ⁿ⁺¹_j=0 , and by (x_j)²ⁿ⁺¹_j=0 the Cauchy product of (f_j)²ⁿ⁺¹_j=0 and (h_j)²ⁿ⁺¹_j=0 . According to Lemma 8.21(a), thenu_j =−a0s^†₀x_ja₀ for all j∈Z_0,2n+1. In particular, K_n^hui =hh−a0s^†₀iinK_n^hxihha0iin. Because of Proposition 4.2, we have Kn^hxi = Kn^hfiS^hhi_n +S^hfnⁱKn^hhi. The application of Theorem 4.8 to the sequence (g_j)²ⁿ⁺¹_j=0 yields furthermore K_n^hhi =−S^hhin K_n^hgiS^hhi_n . From Remark 8.31 one can easily seeK_n^hfi=H_α,n,β =K_n^hgi. Consequently,

H_n^hui=K_n^hui =hh−a0s^†₀iinK_n^hxihha0iin=hh−a0s^†₀iin(H_α,n,βS^hhi

n −S^hf_nⁱS^hhi_n H_α,n,βS^hhi

n )hha0iin. (8.48) Applying Lemma 8.10 to the sequence (a_j)²ⁿ⁺¹_j=0 , we obtain (8.28), where (r_j)²ⁿ⁺¹_j=0 is the re-ciprocal sequence associated to (a_j)²ⁿ⁺¹_j=0 . The application of Remark 8.6 to (b_j)²ⁿ⁺¹_j=0 and of Remark 8.9 to (a_j)²ⁿ⁺¹_j=0 provides us

S^hf_nⁱ = [R_q,n(α)]⁻¹S_•,n,β (8.49) and (8.27). According to Proposition 7.11, we have (a_j)²ⁿ⁺¹_j=0 ∈ D_q×q,2n+1. The application of Proposition 3.20 and Lemma 3.21 to the sequence (a_j)²ⁿ⁺¹_j=0 yields then (8.29) and (8.30), resp.

By virtue of Remark 8.20(b), (8.11), (8.30), and (8.27), we get (8.33). Combining the second equations in (8.28) and (8.29), we obtain furthermoreS^hhi_n =−S^†_α,n,•[R_q,n(β)]^∗. Consequently, using (8.33) and (8.12), we infer from (8.48) that

H_n^hui=R_q,n(β)hha₀ii_nS^†_α,n,•(S^hgi_n hhs^†₀ii_n−S^hgi_n hhs^†₀ii_nS^hf_nⁱS^hhi_n )H_α,n,βS^hhi_n hha₀ii_n

=Rq,n(β)hha0iinS^†_α,n,•(S^hgi_n hhs^†₀iinS^hf_nⁱS^hhi_n −S^hgi_n hhs^†₀iin)H_α,n,βS^†

α,n,•hha0iin[Rq,n(β)]^∗. (8.50) According to Proposition 7.11, we have (s_j)²ⁿ⁺²_j=0 ∈ Dq×q,2n+2. Using Lemma 8.30 and

α,n,•hha0iin[Rq,n(β)]^∗. (8.50) According to Proposition 7.11, we have (s_j)²ⁿ⁺²_j=0 ∈ Dq×q,2n+2. Using Lemma 8.30 and