Transformations of matricial α-Stieltjes non-negative definite sequences

Linear Algebra and its Applications 439 (2013) 3893–3933

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Linear Algebra and its Applications

www.elsevier.com/locate/laa

Transformations of matricial α-Stieltjesnon-negative definite sequences

Bernd Fritzsche, Bernd Kirstein ∗, Conrad Mädler

Mathematisches Institut, Universität Leipzig, Augustusplatz 10/11, 04109 Leipzig, Germany

a r t i c l e i n f o a b s t r a c t

Article history:Received 11 March 2013Accepted 1 October 2013Available online 5 November 2013Submitted by H. Bart

MSC:44A6047A57

Keywords:Hankel non-negative definite sequencesα-Stieltjes right-sided non-negative definitesequencesRight-sided α-shiftingCanonical Hankel parametrizationRight α-Stieltjes parametrizationChristoffel transformGeronimus transform

We investigate the inner structure of power moment sequences ofmatrix measures on the right semiaxis [α,+∞), where α is a givenreal number. To a given matrix sequence, we associate in a bijectiveway a new sequence of matrices, which we call the right α-Stieltjesparametrization. Thereby, one-to-one correspondences betweenpower moment sequences on the right semiaxis [α,+∞) withadditional properties and particular sequences of non-negativeHermitian matrices are established. We consider distinguishedtransformations of matrix sequences the study of which wassuggested by considering some natural transformations of matrixmeasures on an interval. A main theme is to describe the rightα-Stieltjes parametrization of the transformed sequence in termsof the right α-Stieltjes parametrization of the original sequence.

© 2013 Elsevier Inc. All rights reserved.

1. Introduction

This paper is a continuation of work done in the papers [27,33], where two truncated matricialpower moment problems on semiinfinite intervals made up one of the main topics. The startingpoint of studying such type of problems was the famous two parts memoir of Stieltjes [50,51]where the author’s investigations on questions for special continued fractions led him to the powermoment problem on the interval [0,+∞). A complete theory of the treatment of power momentproblems on semiinfinite intervals in the scalar case was developed by M.G. Kreın in collaboration

* Corresponding author.E-mail addresses: [email protected] (B. Fritzsche), [email protected] (B. Kirstein),

[email protected] (C. Mädler).

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3894 B. Fritzsche et al. / Linear Algebra and its Applications 439 (2013) 3893–3933

with A.A. Nudel’man (see [42, Section 10], [43], [44, Chapter V]). What concerns a modern operator-theoretical treatment of the power moment problems named after Hamburger and Stieltjes and itsinterrelations we refer the reader to Simon [48].

The matrix version of the classical Stieltjes moment problem was studied in Adamyan andTkachenko [1,2], Andô [5], Bolotnikov [7,8,10], Bolotnikov and Sakhnovich [11], Chen and Hu [15],Chen and Li [16], Dyukarev [24,25], Dyukarev and Katsnel’son [29,30], Hu and Chen [38]. The consid-erations of this paper deal with the case of a semiinfinite interval [α,+∞) and continue former workdone in [27,33].

In order to formulate the moment problems we are going to study, we first review some notation.Throughout this paper, let p and q be positive integers. Let C, R, N0, and N be the set of all complexnumbers, the set of all real numbers, the set of all non-negative integers, and the set of all positiveintegers, respectively. For every choice of α ∈ R∪ {−∞} and β ∈ R∪ {+∞}, let Zα,β be the set of allintegers k for which α � k � β holds. For all β ∈ R, let �β� := max{k ∈ Z: k � β}. If X is a non-emptyset, then X p×q stands for the set of all p × q matrices each entry of which belongs to X , and X p isshort for X p×1. If (X ,A) is a measurable space, then each countably additive mapping whose domainis A and whose values belong to the set C

q×q� of all non-negative Hermitian complex q × q matrices

is called a non-negative Hermitian q × q measure on (X ,A).Let BR be the σ -algebra of all Borel subsets of R. For all Ω ∈ BR \ {∅}, let BΩ be the σ -algebra

of all Borel subsets of Ω , let Mq�(Ω) be the set of all non-negative Hermitian q × q measures on

(Ω,BΩ) and, for all κ ∈ N0 ∪ {+∞}, let Mq�,κ (Ω) be the set of all σ ∈ Mq

�(Ω) such that theintegral

s(σ )j :=

∫Ω

t j σ(dt) (1.1)

exists for all j ∈ Z0,κ . If Ω ∈ BR \ {∅}, if κ ∈ N0 ∪ {+∞} and if (s j)κj=0 is a sequence of complex

q × q matrices, then let Mq�[Ω; (s j)

κj=0,=] be the set of all σ ∈ Mq

�,κ (Ω) with s(σ )j = s j for all

j ∈ Z0,κ . Two matricial power moment problems lie in the background of our considerations. Thefirst one is the following:

M[Ω; (s j)κj=0,=] Let Ω ∈ BR \ {∅}, let κ ∈ N0 ∪ {+∞}, and let (s j)

κj=0 be a sequence of complex

q × q matrices. Describe the set Mq�[Ω; (s j)

κj=0,=] of all σ ∈ Mq

�,κ (Ω) for which s(σ )j = s j is

fulfilled for all j ∈ Z0,κ .

The second matricial moment problem under consideration is a truncated one with an additionalinequality condition for the last prescribed moment:

M[Ω; (s j)mj=0,�] Let Ω ∈ BR \ {∅}, let m ∈N0, and let (s j)

mj=0 be a sequence of complex q × q matri-

ces. Describe the set Mq�[Ω; (s j)

mj=0,�] of all σ ∈Mq

�,m(Ω) for which sm − s(σ )m is non-negative

Hermitian and, in the case m > 0, moreover s(σ )j = s j is fulfilled for all j ∈ Z0,m−1.

The considerations of this paper are mostly concentrated on the case that the set Ω is a one-sidedclosed infinite interval of the real axis. To get deeper insight into this situation, we start with someobservations on the case Ω = R, which is connected to matricial versions of the classical Ham-burger moment problem. In this case, the above formulated matricial moment problems have beenintensively investigated since the 1980s (see, e.g. Bolotnikov [9], Chen and Hu [13,14], Dym [23], Ko-valishina [40,41], and [28]). In order to explain some criterions of solvability in the case Ω = R, weintroduce certain sets of sequences of complex q × q matrices, which are determined by the proper-ties of particular Hankel matrices built of them. For all n ∈ N0, let H�

q,2n (respectively, H>q,2n) be the

set of all sequences (s j)2nj=0 of complex q × q matrices such that the block Hankel matrix

Hn := [s j+k]nj,k=0 (1.2)

B. Fritzsche et al. / Linear Algebra and its Applications 439 (2013) 3893–3933 3895

is non-negative Hermitian (resp. positive Hermitian). Furthermore, let H�q,∞ (resp. H>

q,∞) be the set

of all sequences (s j)∞j=0 of complex q × q matrices such that, for all n ∈ N0, the sequence (s j)

2nj=0 be-

longs to H�q,2n (resp. H>

q,2n). The elements of the set H�q,2n (resp. H>

q,2n) are called Hankel non-negative

definite (resp. Hankel positive definite) sequences. For all n ∈ N0, let H�,eq,2n be the set of all sequences

(s j)2nj=0 of complex q × q matrices for which there are matrices s2n+1 ∈ C

q×q and s2n+2 ∈ Cq×q such

that (s j)2(n+1)j=0 belongs to H�

q,2(n+1) . Furthermore, for all n ∈ N0, we will use H�,eq,2n+1 to denote the

set of sequences (s j)2n+1j=0 of complex q × q matrices for which there is some s2n+2 ∈ C

q×q such that

(s j)2(n+1)j=0 belongs to H�

q,2(n+1) . For all m ∈ N0, the elements of the set H�,eq,m are called Hankel non-

negative definite extendable sequences. For technical reasons, we set H�,eq,∞ := H�

q,∞ . The solvability ofthe Hamburger moment problems under consideration can be characterized as follows.

Theorem 1.1. (See [28, Theorem 4.17], [32, Theorem 6.6].) Let κ ∈ N0 ∪ {+∞} and let (s j)κj=0 be a sequence

of complex q × q matrices. Then Mq�[R; (s j)

κj=0,=] = ∅ if and only if (s j)

κj=0 ∈H�,e

q,κ .

Theorem 1.2. (See, e.g., [13, Theorem 3.2] or [28, Theorem 4.16].) Let n ∈ N0 and let (s j)2nj=0 be a sequence of

complex q × q matrices. Then Mq�[R; (s j)

2nj=0,�] = ∅ if and only if (s j)

2nj=0 ∈H�

q,2n.

Now let Ω ∈ BR \ {∅}, κ ∈ N0 ∪ {+∞} and σ ∈ Mq�,κ (Ω). If the mapping μ : BR → C

q×q

is defined by μ(B) := σ(B ∩ Ω), then clearly μ ∈ Mq�,κ (R) and, taking (1.1) into account, we

see that, for all k ∈ Z0,κ , the equation s(μ)

k = s(σ )

k holds. Thus, if σ ∈ Mq�[Ω; (s j)

κj=0,=] (resp.

σ ∈ Mq�[Ω; (s j)

2nj=0,�]), then μ ∈ Mq

�[R; (s j)κj=0,=] (resp. μ ∈ Mq

�[R; (s j)2nj=0,�]). Hence, if the

set Mq�[Ω; (s j)

κj=0,=] (resp. Mq

�[Ω; (s j)2nj=0,�]) is non-empty, then (s j)

κj=0 ∈ H�,e

q,κ (resp. (s j)2nj=0 ∈

H�q,2n). The choice of Ω ∈ BR \ {∅} in the problems M[Ω; (s j)

κj=0,=] and M[Ω; (s j)

mj=0,�] leads us

consequently to special subclasses of the classes H�q,2n and H�,e

q,κ . In this paper, we continue the in-vestigations in [27,33], where the case of an interval Ω = [α,+∞) with arbitrarily given α ∈ R wasconsidered. In order to give a better motivation and formulation of the main aims of this paper, weare going to recall the characterizations of solvability of our moment problems, which were obtainedin [27]. This requires some preparations.

Let α ∈ C, κ ∈ N ∪ {+∞} and (s j)κj=0 be a sequence of complex p × q matrices. Let the sequence

(sα� j)κ−1j=0 be defined by

sα� j := −αs j + s j+1 (1.3)

for all j ∈ Z0,κ−1. Then (sα� j)κ−1j=0 is called the sequence generated from (s j)

κj=0 by right-sided α-shifting.

(An analogous left-sided version is discussed in [33, Definition 2.1].) It seems to be useful to statesome simple, but useful interrelation between the sequences (s j)

κj=0 and (sα� j)

κ−1j=0 .

Remark 1.3. Let α ∈ C, let κ ∈ N ∪ {+∞}, and let (s j)κj=0 be a sequence of complex p × q matrices.

For all j ∈ N0 and all k ∈ N with j + k � κ , it is readily checked that

s j+k = αks j +k−1∑l=0

αk−1−lsα� j+l.

Let K�q,0,α :=H�

q,0. For all n ∈ N, let

K�q,2n,α := {(s j)

2nj=0 ∈ H�

q,2n

∣∣ (sα� j)2(n−1)j=0 ∈ H�

q,2(n−1)

}.


For all m ∈N0, denote by Fq,m the set of all sequences (s j)mj=0 of matrices from C

q×q . Then we set

K�q,2n+1,α := {(s j)

2n+1j=0 ∈ Fq,2n+1

∣∣ {(s j)2nj=0, (sα� j)

2nj=0

}⊆ H�q,2n

}.

For all m ∈ N0, let K�,eq,m,α be the set of all sequences (s j)

mj=0 of complex q × q matrices for which

there exists a complex q × q matrix sm+1 such that (s j)m+1j=0 belongs to K�

q,m+1,α . Obviously, we have

K�,eq,2n,α = {(s j)

2nj=0 ∈ H�

q,2n

∣∣ (sα� j)2n−1j=0 ∈ H�,e

q,2n−1

}for all n ∈N and

K�,eq,2n+1,α = {(s j)

2n+1j=0 ∈ H�,e

q,2n+1

∣∣ (sα� j)2nj=0 ∈ H�

q,2n

}for all n ∈ N0. For all m ∈ N0, we call a sequence (s j)

mj=0 α-Stieltjes right-sided non-negative definite

(resp. α-Stieltjes right-sided non-negative definite extendable) if it belongs to K�q,m,α (resp. to K�,e

q,m,α).In the case α = 0, the sequence (s j)

mj=0 is also called Stieltjes non-negative definite (resp. Stieltjes

non-negative definite extendable). Note that left versions of these notions are also studied in [33, Defi-nition 1.3].

Remark 1.4. Let α ∈ R, m ∈ N0 and (s j)mj=0 ∈ K�

q,m,α (resp. K�,eq,m,α). Then we easily see that (s j)

lj=0 ∈

K�q,l,α (resp. K�,e

q,l,α) for all l ∈ Z0,m .

In view of Remark 1.4, for all α ∈ R, let K�q,∞,α be the set of all sequences (s j)

∞j=0 of com-

plex q × q matrices such that (s j)mj=0 belongs to K�

q,m,α for all m ∈ N0, and let K�,eq,∞,α := K�

q,∞,α .

The notation Iq stands for the identity matrix in Cq×q . Using the just introduced sets of se-

quences of complex q × q matrices, we are able to formulate the solvability criterions of the prob-lems M[[α,+∞); (s j)

mj=0,=] and M[[α,+∞); (s j)

mj=0,�], which were obtained in [27]:

Theorem 1.5. Let α ∈ R, let κ ∈ N0 ∪ {+∞}, and let (s j)κj=0 be a sequence of complex q × q matrices. Then

Mq�[[α,+∞); (s j)

κj=0,=] = ∅ if and only if (s j)

κj=0 ∈K�,e

q,κ,α .

In the case κ ∈ N0, a proof of Theorem 1.5 is given in [27, Theorem 1.3]. If κ = +∞, the assertedequivalence can be proved using the equation Mq

�[[α,+∞); (s j)∞j=0,=] = ⋂∞

m=0 Mq�[[α,+∞);

(s j)mj=0,=] and the matricial version of the Helly–Prohorov theorem (see [31, Satz 9]). We omit the

details of the proof, the essential idea of which is originated in [4, proof of Theorem 2.1.1].

Theorem 1.6. (See [27, Theorem 1.4].) Let α ∈ R, let m ∈ N0 , and let (s j)mj=0 be a sequence of complex

q × q matrices. Then Mq�[[α,+∞); (s j)

mj=0,�] = ∅ if and only if (s j)

mj=0 ∈K�

q,m,α .

The importance of Theorems 1.5 and 1.6 led us in [27,33] to a closer look at the propertiesof sequences of complex q × q matrices, which are α-Stieltjes right-sided non-negative definite orα-Stieltjes right-sided non-negative definite extendable. Guided by our former investigations on Han-kel non-negative definite sequences and Hankel non-negative definite extendable sequences, whichwere done in [28], we started in [27, Section 4] and [33] a thorough study of the structure ofα-Stieltjes right-sided non-negative definite sequences and α-Stieltjes right-sided non-negative defi-nite extendable sequences.

In this paper, we study several relevant transformations for sequences which preserve the classesof α-Stieltjes right-sided non-negative definite sequences and some of their interesting subclasses.Since α-Stieltjes right-sided non-negative definite sequences can be considered as power moment se-quences of non-negative Hermitian measures on the right semiaxis [α,+∞), these transformations


for sequences can be motivated and better understood if they are considered in connection with cor-responding transformations for non-negative Hermitian measures on intervals of the real line R. Forthe convenience of the reader, we summarize in Appendix B essential facts on non-negative Hermi-tian measures on R and their special transformations which are of particular interest for this paper.It is important that these transformations have an interesting additional property. Namely, roughlyspeaking, the images of measures with equal power moments have equal power moments again. Forthis reason, corresponding analogues of these transformations can also be introduced in structureswhich are in a one-to-one correspondence to power moment sequences such as block Jacobi matricesor orthogonal matrix polynomials. In the scalar case, such investigations have been done at severalplaces (see, e.g., Bueno and Marcellán [12], Derevyagin and Derkach [21], Yoon [53], Zhedanov [54]).We will specify matricial aspects of this topic in future work.

This paper is organized as follows: In Section 2, we discuss two parametrizations of sequencesof complex matrices, namely the canonical Hankel parametrization on the one hand and the rightα-Stieltjes parametrization on the other hand. We recall some essential facts on this objects andsome interrelations between them. This material is mostly taken from [32,33]. In Section 3, we sum-marize some facts on the class of completely degenerate α-Stieltjes right-sided non-negative definitesequences. In particular, we characterize these classes in terms of the right α-Stieltjes parametriza-tion. In Section 4, we investigate the binomial transform of sequences against to the background ofα-Stieltjes right-sided non-negative definite sequences. It should be mentioned that in [32] the au-thors have studied related questions for Hankel non-negative definite sequences. In Section 5, we con-sider an α-Stieltjes right-sided non-negative definite sequence together with the sequence generatedfrom it by right-sided α-shifting. In particular, we verify that the latter sequence is α-Stieltjes right-sided non-negative definite, too (see Proposition 5.1). Furthermore, we derive some interrelationsbetween the corresponding right α-Stieltjes parametrizations of the two sequences (see Theorem 5.3).In Section 6, we introduce a transformation for sequences of matrices which can be considered as con-verse to right-sided α-shifting. The main result is Proposition 6.4, where some connections betweenthe corresponding right α-Stieltjes parametrizations are indicated. In Section 7, we consider particularadditive perturbations of α-Stieltjes right-sided non-negative definite sequences which are motivatedby addition of matrix measures, which are concentrated in a single point ζ which is located at theleft-hand side of α. The main result is Theorem 7.5, which contains a complete description of theright α-Stieltjes parametrization of a sequence which is constructed from an α-Stieltjes right-sidednon-negative definite sequence by a special kind of additive perturbation at the point ζ located at theleft-hand side from α. In Section 8, we consider the case α = 0 and infinite 0-Stieltjes right-sided pos-itive definite sequences of q × q matrices. For such sequences (s j)

∞j=0 Yu.M. Dyukarev [26] considered

the moment problem M[[0,+∞); (s j)∞j=0,=]. In particular he studied the problem of indeterminacy.

In order to generalize the famous indeterminacy criterion for the scalar case which was found byStieltjes [50,51] he introduced a particular inner parametrization for infinite 0-Stieltjes right-sidedpositive definite sequences of q × q matrices which turns out to be a direct matricial generalizationof the classical parameters used by Stieltjes (see Proposition 8.30). The central theme of Section 8is the discussion of interrelations between Yu.M. Dyukarev’s parametrization and our right 0-Stieltjesparametrization introduced in Definition 2.2. In particular, we will show that there is a bijective corre-spondence between the two types of parametrizations (see Theorems 8.22 and 8.24). Furthermore, westate a principle of constructing infinite 0-Stieltjes right-sided positive definite sequences of q × q ma-trices outgoing from an ordered pair of infinite sequences of positive Hermitian q × q matrices (seeProposition 8.27).

2. Canonical Hankel and α-Stieltjes parametrizations

With later applications to the matrix version of the Hamburger moment problem in mind, aparticular inner parametrization, called canonical Hankel parametrization, for sequences of complexmatrices was developed in [28,32,35]. First we are going to recall the definition of the canonical Han-kel parametrization of a sequence of complex p × q matrices. To prepare this notion, we need somefurther matrices built from the given data. If A ∈ C

p×q , then the notation A† stands for the Moore–Penrose inverse of A (see also Appendix A).


Let κ ∈N0 ∪ {+∞} and let (s j)κj=0 be a sequence of complex p × q matrices. Then, for all j,k ∈N0

with j � k � κ , let

y〈s〉j,k := [s∗

j , s∗j+1, . . . , s∗

k

]∗and z〈s〉

j,k := [s j, s j+1, . . . , sk]. (2.1)

For all n ∈ N0 with 2n � κ , let

H 〈s〉n := [s j+k]n

j,k=0. (2.2)

If n ∈ N0 is such that 2n + 1 � κ , then let

K 〈s〉n := [s j+k+1]n

j,k=0, (2.3)

and for all n ∈N0 with 2n + 2 � κ , let

G〈s〉n := [s j+k+2]n

j,k=0. (2.4)

We use the notation L〈s〉0 := s0, and, for all n ∈ N with 2n � κ , let

L〈s〉n := s2n − z〈s〉

n,2n−1

(H 〈s〉

n−1

)†y〈s〉

n,2n−1. (2.5)

In cases in which it is clear which sequence (s j)κj=0 of complex p × q matrices is meant, we will write

y j,k , z j,k , Hn , Kn , Gn , and Ln instead of y〈s〉j,k , z〈s〉

j,k , H 〈s〉n , K 〈s〉

n , G〈s〉n , and L〈s〉

n , respectively. For all k ∈ N

with 2k − 1 � κ , let

Θ〈s〉k := z〈s〉

k,2k−1

(H 〈s〉

k−1

)†y〈s〉

k,2k−1, Σ〈s〉k := z〈s〉

k,2k−1

(H 〈s〉

k−1

)†K 〈s〉

k−1

(H 〈s〉

k−1

)†y〈s〉

k,2k−1

and, for all k ∈N with 2k � κ , let

M〈s〉k := z〈s〉

k,2k−1

(H 〈s〉

k−1

)†y〈s〉

k+1,2k, N〈s〉k := z〈s〉

k+1,2k

(H 〈s〉

k−1

)†y〈s〉

k,2k−1,

and let

Λ〈s〉k := M〈s〉

k + N〈s〉k − Σ

〈s〉k . (2.6)

Moreover, let 0p×q be the null matrix in Cp×q and let

Θ〈s〉0 := 0p×q, Σ

〈s〉0 := 0p×q, M〈s〉

0 := 0p×q,

N〈s〉0 := 0p×q, Λ

〈s〉0 := 0p×q. (2.7)

If it is clear which sequence (s j)κj=0 of complex p × q matrices is meant, then we will write Θk ,

Σk , Mk , Nk , and Λk for Θ〈s〉k , Σ

〈s〉k , M〈s〉

k , N〈s〉k , and Λ

〈s〉k , respectively. Now we are able to recall the

notion of canonical Hankel parametrization which was introduced in [28, Definition 2.28] and [35,Definition 2.3]. There one can find further details.

Definition 2.1. Let κ ∈ N ∪ {+∞} and let (s j)2κj=0 be a sequence of complex p × q matrices. For all

k ∈ Z1,κ , let Ck := s2k−1 − Λk−1, where Λk−1 was introduced in (2.7) and (2.6), and, for all k ∈ Z0,κ ,let Dk := Lk , where Lk was introduced in (2.5). Then the pair [(Ck)

κk=1, (Dk)

κk=0] is called the canonical

Hankel parametrization of (s j)2κj=0.

There is a one-to-one correspondence between a sequence (s j)2κj=0 from C

p×q and its canonical

Hankel parametrization [(Ck)κk=1, (Dk)

κk=0]. The original sequence (s j)

2κj=0 can be explicitly recovered

by its canonical Hankel parametrization [(Ck)κk=1, (Dk)

κk=0].

The considerations in [32, Proposition 2.10(b)] and [33, Proposition 3.4] show that the membershipof a sequence to the class H�

q,2n or some of its interesting subclasses can be nicely characterized in


terms of its canonical Hankel parametrization. This leads to the explicit construction of sequencesbelonging to prescribed subclasses of H�

q,2n .Assume now that α ∈ R. Guided by our former observations on Hankel non-negative definite

sequences, we were interested in finding an inner parametrization of sequences of complex matri-ces which particularly reflects the membership of a sequence to the class of α-Stieltjes right-sidednon-negative definite sequences or some of its interesting subclasses. Against to this background, weintroduced in [33, Definition 4.2] the notion of right α-Stieltjes parametrization of a sequence ofcomplex p × q matrices.

If α ∈ C, κ ∈ N ∪ {+∞}, and a sequence (s j)κj=0 of complex p × q matrices are given, then let

(t j)κ−1j=0 be defined by t j := sα� j and (1.3) for all j ∈ Z0,κ−1, and let

Hα�n := H 〈t〉n and Lα�n := L〈t〉

n

for all n ∈ N0 with 2n + 1 � κ , let Kα�n := K 〈t〉n for all n ∈ N0 with 2n + 2 � κ , and, for all j,k ∈ N0

with j � k � κ , let

yα� j,k := y〈t〉j,k and zα� j,k := z〈t〉

j,k.

Furthermore, let Θα�k := Θ〈t〉k for all k ∈ N0 with 2k � κ and let

Mα�k := M〈t〉k and Λα�k := Λ

〈t〉k for all k ∈ N0 with 2k � κ − 1,

where (t j)κ−1j=0 is defined by t j := sα� j and (1.3) for all j ∈ Z0,κ−1.

Definition 2.2. (See [33, Definition 4.2].) Let α ∈ C, let κ ∈ N0 ∪ {+∞}, and let (s j)κj=0 be a sequence

from Cp×q . Then the sequence (Q j)

κj=0 given by Q 2k := Lk for all k ∈N0 with 2k � κ and by Q 2k+1 :=

Lα�k for all k ∈N0 with 2k + 1 � κ is called the right α-Stieltjes parametrization of (s j)κj=0.

There is a one-to-one correspondence between a sequence (s j)κj=0 from C

p×q and its rightα-Stieltjes parametrization (Q j)

κj=0 (see [33, Remark 4.3]). In particular, the original sequence (s j)

κj=0

can be explicitly recovered from its right α-Stieltjes parametrization (Q j)κj=0.

Now we are going to state characterizations of particular classes of sequences of complexq × q matrices in terms of their canonical right α-Stieltjes parametrization. Before doing this, weintroduce some important subclasses of the class of α-Stieltjes right-sided non-negative definite se-quences. More precisely, we turn our attention to some subclass of K�

q,κ,α , which is characterized bystronger positivity properties. Let α ∈R. Then we set K>

q,0,α :=H>q,0. For all n ∈ N, let

K>q,2n,α := {(s j)

2nj=0 ∈ H>

q,2n

∣∣ (sα� j)2(n−1)j=0 ∈ H>

q,2(n−1)

}.

Furthermore, for all n ∈N0, let

K>q,2n+1,α := {(s j)

2n+1j=0 ∈ Fq,2n+1

∣∣ {(s j)2nj=0, (sα� j)

2nj=0

}⊆ H>q,2n

}.

If m ∈ N0 and (s j)mj=0 ∈ K>

q,m,α , then, for all l ∈ Z0,m , the sequence (s j)lj=0 belongs to K>

q,l,α . Thus, itis natural to denote by K>

q,∞,α the set of all sequences (s j)∞j=0 of complex q × q matrices such that

(s j)mj=0 belongs to K>

q,m,α for all m ∈ N0.

Here and in the following, for all A ∈ Cp×q , let N (A) be the null space of A and let R(A) be the

column space of A. Furthermore, let Cq×q> be the set of all positive Hermitian complex q × q matrices.

Theorem 2.3. (See [33, Theorem 4.12].) Let α ∈ R, κ ∈ N0 ∪ {+∞}, (s j)κj=0 be a sequence of complex

q × q matrices and (Q j)κj=0 be the right α-Stieltjes parametrization of (s j)

κj=0 . Then:


(a) s∗j = s j for all j ∈ Z0,κ if and only if Q ∗

j = Q j for all j ∈ Z0,κ .

(b) (s j)κj=0 ∈K�

q,κ,α if and only if the following two conditions hold true:

(I) Q j ∈ Cq×q� for all j ∈ Z0,κ .

(II) If κ � 2, then N (Q j) ⊆N (Q j+1) holds for all j ∈ Z0,κ−2 .

(c) (s j)κj=0 ∈K�,e

q,κ,α if and only if the following two conditions are fulfilled:

(III) Q j ∈ Cq×q� for all j ∈ Z0,κ .

(IV) If κ � 1, then N (Q j) ⊆N (Q j+1) for all j ∈ Z0,κ−1 .

(d) (s j)κj=0 belongs to K>

q,κ,α if and only if Q j ∈ Cq×q> for all j ∈ Z0,κ .

Since a sequence from Cq×q can be uniquely recovered from its right α-Stieltjes parametrization,

the results of Theorem 2.3 allow us to construct sequences belonging to prescribed subclasses ofα-Stieltjes right-sided non-negative definite sequences.

At the end of this section, we state some results on the right α-Stieltjes parametrization ofα-Stieltjes right-sided non-negative definite sequences.

Lemma 2.4. (See [33, Lemmas 6.7 and 6.9].) Let α ∈ R, κ ∈ N ∪ {+∞} and (s j)κj=0 ∈ K�

q,κ,α . Further, let(Q j)

κj=0 be the right α-Stieltjes parametrization of (s j)

κj=0 and let Q −1 := 0q×q. Then s2k+1 −Λk = Q 2k+1 +

(α Iq + Q 2k Q †2k−1)Q 2k and Q 2k+1 = s2k+1 −Λk − (α Iq + Lk Q †

2k−1)Lk for all k ∈N0 with 2k + 1 � κ , whereΛk is given via (2.6).

Lemma 2.5. (See [33, Lemma 6.19].) Let α ∈R, κ ∈ Z2,+∞∪{+∞} and (s j)κj=0 ∈K�

q,κ,α with right α-Stieltjesparametrization (Q j)

κj=0 . For all k ∈ N0 with 2k + 1 � κ − 1, then sα�2k+1 − Λα�k = Q 2k+2 + (α Iq +

Q 2k+1 Q †2k)Q 2k+1 .

3. Completely degenerate α-Stieltjes right-sided non-negative definite sequences

In this section, we recall the corresponding analogue of completely degenerate Hankel non-negative definite sequences for the class of α-Stieltjes right-sided non-negative definite sequences. Forthis reason, first we consider completely degenerate Hankel non-negative definite sequences, whichwere already discussed in [28,27,32,35]. First we recall the corresponding notions, which are basedon the matrices defined via (2.5). If n ∈ N0 and if (s j)

2nj=0 ∈ H�

q,2n , then (s j)2nj=0 is called completely

degenerate if Ln = 0. For all n ∈ N0, the set H�,cdq,2n of all completely degenerate sequences belonging

to H�q,2n is a subset of H�,e

q,2n (see [28, Corollary 2.14]). If m ∈ N0 and (s j)2mj=0 ∈ H�,e

q,2m are given, then

from [28, Proposition 2.13] one can easily see that (s j)2mj=0 belongs to H�,cd

q,2m if and only if there issome n ∈ Z0,m such that Ln = 0q×q . A Hankel non-negative definite sequence (s j)

∞j=0 of complex q × q

matrices is said to be completely degenerate if there is some non-negative integer n such that (s j)2nj=0

is a completely degenerate Hankel non-negative definite sequence. By H�,cdq,∞ we denote the set of

all completely degenerate Hankel non-negative definite sequences (s j)∞j=0 of complex q × q matrices.

A Hankel non-negative definite sequence (s j)∞j=0 of complex q × q matrices is called completely de-

generate of order n if the sequence (s j)2nj=0 is completely degenerate. By H�,cd,n

q,∞ we denote the set ofall Hankel non-negative definite sequences (s j)

∞j=0 of complex q × q matrices which are completely

degenerate of order n. If n ∈ N0 and (s j)∞j=0 ∈ H�,cd,n

q,∞ , then (s j)2mj=0 ∈ H�,cd

q,2m for all m ∈ Zn,∞ . For anarbitrarily given α ∈ R, now we introduce further subclasses of the class of completely degenerateHankel non-negative definite sequences. Let n ∈ N0. Then we define

K�,cdq,2n,α := K�

q,2n,α ∩H�,cdq,2n


and

K�,cdq,2n+1,α := {(s j)

2n+1j=0 ∈ K�

q,2n+1,α: (sα� j)2nj=0 ∈ H�,cd

q,2n

}.

Furthermore, let K�,cdq,∞,α be the set of all sequences (s j)

∞j=0 ∈ K�

q,∞,α for which there exists some

m ∈N0 such that (s j)mj=0 ∈K�,cd

q,m,α .Now we characterize the membership of a sequence to one of the just introduced classes in terms

of its right α-Stieltjes parametrization.

Proposition 3.1. (See [33, Proposition 5.10].) Let α ∈ R, let m ∈ N0 , let (s j)mj=0 be a sequence of complex

q × q matrices, and let (Q j)mj=0 be the right α-Stieltjes parametrization of (s j)

mj=0 . Then (s j)

mj=0 belongs to

K�,cdq,m,α if and only if the following three conditions are fulfilled:

(i) Q m = 0q×q.

(ii) If m � 1, then Q j ∈Cq×q� for all j ∈ Z0,m−1 .

(iii) If m � 2, then N (Q j) ⊆N (Q j+1) for all j ∈ Z0,m−2 .

Proposition 3.2. (See [33, Proposition 5.11].) Let α ∈ R, let (s j)∞j=0 be a sequence of complex q × q matrices,

and let (Q j)∞j=0 be the right α-Stieltjes parametrization of (s j)

∞j=0 . Then (s j)

∞j=0 belongs to K�,cd

q,∞,α if and onlyif there is an m ∈ N0 such that Q k = 0q×q for all k ∈ Zm,∞ and the conditions (ii) and (iii) of Proposition 3.1hold true.

4. Binomial transformation

For all γ , δ ∈ C, each κ ∈ N0 ∪ {+∞}, and each sequence (s j)κj=0 of complex p × q matrices, we

consider now the sequence (s( ,γ ,δ)

j )κj=0 given by

s( ,γ ,δ)

j :=j∑

l=0

(j

l

)δlγ j−lsl for all j ∈ Z0,κ . (4.1)

The sequence (s( ,γ ,δ)

j )κj=0 is called the (δ, γ )-binomial transform of the sequence (s j)κj=0. Against to the

background of Hankel non-negative definite sequences this construction was studied in [32, Section 4].In particular, it was verified in [32, Section 4] that the membership of a sequence to several subclassesof the class of Hankel non-negative definite sequences is invariant with respect to binomial transform.The main theme of this section is to derive analogous results for several subclasses of α-Stieltjesright-sided non-negative definite sequences.

Example 4.1. Let γ , δ, ζ ∈ C, A ∈ Cp×q and κ ∈ N0 ∪ {+∞}. For all j ∈ Z0,κ , let s j := ζ j A. Then

s( ,γ ,δ)

j = (γ + δζ ) j A for all j ∈ Z0,κ .

Example 4.2. Let γ , δ ∈ C, κ ∈ N0 ∪{+∞} and (s j)κj=0 be a sequence of complex p × q matrices. Then

s( ,γ ,0)

j = γ j s0 and s( ,0,δ)

j = δ j s j for all j ∈ Z0,κ .

Now we first summarize some more or less obvious, but useful properties of the sequence intro-duced by (4.1).

Remark 4.3. Let γ , δ ∈ C, κ ∈ N0 ∪ {+∞}, (s j)κj=0 be a sequence from C

p×q and m ∈ Z0,κ . Then⋂mj=0 N (s j) ⊆N (s( ,γ ,δ)

m ) and R(s( ,γ ,δ)m ) ⊆∑m

j=0 R(s j).


Remark 4.4. Let γ , δ ∈ C, κ ∈ N0 ∪ {+∞}, and n ∈ N. For all m ∈ Z1,n , let pm,qm ∈ N, let (s(m)j )κj=0 be

a sequence of complex pm × qm matrices, let Lm ∈Cp×pm , and let Rm ∈C

qm×q . Then

((n∑

m=1

Lms(m)j Rm

)( ,γ ,δ))κ

j=0

=n∑

m=1

Lm[((

s(m)j

)( ,γ ,δ))κj=0

]Rm.

Remark 4.5. Let γ , δ ∈ C, κ ∈ N0 ∪ {+∞} and n ∈ N. For all m ∈ Z1,n , let pm,qm ∈ N and let

(s(m)j )κj=0 be a sequence of complex pm × qm matrices. Then (diag[(s(m)

j )( ,γ ,δ)]nm=1)

κj=0 =

((diag[s(m)j ]n

m=1)( ,γ ,δ))κj=0.

Remark 4.6. Let γ , δ,λ,μ ∈ C, κ ∈ N0 ∪ {+∞}, and (s j)κj=0 be a sequence from C

p×q . From [32,

Lemma 4.6] we see then that ((s( ,γ ,δ))( ,λ,μ)

j )κj=0 = (s( ,μγ +λ,μδ)

j )κj=0.

Remark 4.7. Let γ , δ ∈ C, κ ∈ N∪ {+∞} and (s j)κj=0 be a sequence of complex p × q matrices. Let the

sequence (t j)κj=0 be defined by t j := s( ,γ ,δ)

j for all j ∈ Z0,κ and let the sequence (u j)κ−1j=0 be defined

by u j := δs j+1 for all j ∈ Z0,κ−1. Then (tγ � j)κ−1j=0 = (u( ,γ ,δ)

j )κ−1j=0 .

The next result indicates the relation between (1,α)-binomial transformation and right-sidedα-shifting.

Remark 4.8. Let α ∈ C, κ ∈ N ∪ {+∞} and (s j)κj=0 be a sequence of complex p × q matrices. Let the

sequence (t j)κj=0 be defined by t j := s( ,α,1)

j for all j ∈ Z0,κ and let the sequence (u j)κ−1j=0 be defined by

u j := s0� j for all j ∈ Z0,κ−1. Then from Remark 4.7 one can easily see that (tα� j)κ−1j=0 = (u( ,α,1)

j )κ−1j=0 .

Remark 4.9. Let γ , δ ∈C, let κ ∈ N0 ∪{+∞}, and let (s j)κj=0 be a sequence of complex p × q matrices.

Let (u j)κj=0 be given by u j := (−1) j s j for all j ∈ Z0,κ . For every choice of j ∈ Z0,κ , then s( ,γ ,−δ)

j =u( ,γ ,δ)

j .

The theme of our next consideration can be described as follows. We consider a sequence(s j)

κj=0 ∈ K�

q,κ,α . Let γ ∈ R and δ ∈ [0,+∞). Then we are going to prove in Theorem 4.12 that the

(δ, γ )-binomial transform (s( ,γ ,δ)

j )κj=0 of (s j)κj=0 belongs to the class K�

q,κ,γ +αδ . First we start withsome preliminary observations.

Remark 4.10. Let γ , δ ∈C and n ∈N0. Then the matrix

Bq,n(γ , δ) := [b(q)

j,k(γ , δ)]n

j,k=0

given by

b(q)

j,k(γ , δ) :={(k

j

)δ jγ k− j Iq, if 0 � j � k � n,

0q×q, if 0 � k < j � n

fulfills obviously det Bq,n(γ , δ) = δn(n+1)q

2 and [Bq,n(γ , δ)]∗ = [Bq,n(γ , δ)]T.


Lemma 4.11. Let γ , δ ∈C, κ ∈ N0 ∪ {+∞} and (s j)κj=0 be a sequence from C

p×q.

(a) For all n ∈ N0 with 2n � κ , the matrix H 〈s( ,γ ,δ)〉n admits the representation

H 〈s( ,γ ,δ)〉n = [B p,n(γ , δ)

]∗Hn[

Bq,n(γ , δ)]. (4.2)

In particular, det(H 〈s( ,γ ,δ)〉n ) = δn(n+1)q det Hn, rank(H 〈s( ,γ ,δ)〉

n ) � rank Hn and, in the case δ = 0, more-

over rank(H 〈s( ,γ ,δ)〉n ) = rank Hn.

(b) For all n ∈ N0 with 2n+1 � κ , the matrix K 〈s( ,γ ,δ)〉n admits the representation K 〈s( ,γ ,δ)〉

n = [B p,n(γ , δ)]∗ ×(γ Hn + δKn)[Bq,n(γ , δ)], which in particular implies det(K 〈s( ,γ ,δ)〉

n ) = δn(n+1)q det(γ Hn + δKn), the

inequality rank(K 〈s( ,γ ,δ)〉n ) � rank(γ Hn + δKn) and, in the case δ = 0, moreover rank(K 〈s( ,γ ,δ)〉

n ) =rank(γ Hn + δKn).

Proof. (a) In view of Remark 4.10, Eq. (4.2) follows from [32, Lemma 4.9]. The other assertions in (a)are consequences of (4.2) and Remark 4.10.

(b) Let n ∈ N0 with 2n+1 � κ . For all j ∈ Z0,2n , let t j := δs j+1. Then δKn = H 〈t〉n . Because of part (a)

and Remark 4.7, we have[B p,n(γ , δ)

]∗H 〈t〉

n[

Bq,n(γ , δ)]= [t( ,γ ,δ)

j+k

]nj,k=0 = [−γ s( ,γ ,δ)

j+k + s( ,γ ,δ)

j+k+1

]nj,k=0

= −γ H 〈s( ,γ ,δ)〉n + K 〈s( ,γ ,δ)〉

n

and, consequently, using part (a) again, then[B p,n(γ , δ)

]∗(γ Hn + δKn)

[Bq,n(γ , δ)

]= γ

[B p,n(γ , δ)

]∗Hn[

Bq,n(γ , δ)]+ [B p,n(γ , δ)

]∗H 〈t〉

n[

Bq,n(γ , δ)]

= γ H 〈s( ,γ ,δ)〉n + (−γ H 〈s( ,γ ,δ)〉

n + K 〈s( ,γ ,δ)〉n

)= K 〈s( ,γ ,δ)〉n .

The rest then follows from Remark 4.10. �Theorem 4.12. Let α,γ ∈ R, let δ ∈ [0,+∞), let κ ∈ N0 ∪ {+∞}, and let (s j)

κj=0 ∈ K�

q,κ,α with

right α-Stieltjes parametrization (Q j)κj=0 . Then (s( ,γ ,δ)

j )κj=0 ∈ K�q,κ,γ +αδ and (δ j Q j)

κj=0 is the right

(γ + αδ)-Stieltjes parametrization of (s( ,γ ,δ)

j )κj=0 .

Proof. Let n ∈N with 2n � κ . According to Lemma 4.11, we have

H 〈s( ,γ ,δ)〉n = [Bq,n(γ , δ)

]∗Hn[

Bq,n(γ , δ)]. (4.3)

Because of (s j)κj=0 ∈ K�

q,κ,α and 2n � κ , we get (s j)2nj=0 ∈ H�

q,2n and hence Hn ∈ C(n+1)q×(n+1)q� .

Thus, from (4.3) we see that H 〈s( ,γ ,δ)〉n ∈ C

(n+1)q×(n+1)q� , i.e., we have (s( ,γ ,δ)

j )2nj=0 ∈ H�

q,2n . Let

[(Ck)nk=1, (Dk)

nk=0] be the canonical Hankel parametrization of (s j)

2nj=0, let [(C

k)nk=1, (D

k)nk=0] be the

canonical Hankel parametrization of (s( ,γ ,δ)

j )2nj=0, and let (Q

j )κj=0 be the right (γ + αδ)-Stieltjes

parametrization of (s( ,γ ,δ)

j )κj=0. For all k ∈ Z0,n , from Definitions 2.2 and 2.1 and, according to

(s j)2nj=0 ∈H�

q,2n and [32, Theorem 4.18], we conclude

δ2k Q 2k = δ2k Lk = δ2k Dk = D

k = L〈s( ,γ ,δ)〉k = Q

2k. (4.4)

Now let n ∈N with 2n + 1 � κ . Since γ and δ belong to R, we get


K 〈s( ,γ ,δ)〉n = [Bq,n(γ , δ)

]∗(γ Hn + δKn)

[Bq,n(γ , δ)

](4.5)

from Lemma 4.11. Consequently, we obtain

−(γ + αδ)H 〈s( ,γ ,δ)〉n + K 〈s( ,γ ,δ)〉

n

= −(γ + αδ)[

Bq,n(γ , δ)]∗

Hn[

Bq,n(γ , δ)]+ [Bq,n(γ , δ)

]∗(γ Hn + δKn)

[Bq,n(γ , δ)

]= [Bq,n(γ , δ)

]∗[δ(−αHn + Kn)

][Bq,n(γ , δ)

]= δ[

Bq,n(γ , δ)]∗

Hα�n[

Bq,n(γ , δ)], (4.6)

where the 1st equation is due to (4.3) and (4.5), and the 3rd equation is due to (1.2) and (1.3). Be-cause of (s j)

κj=0 ∈K�

q,κ,α and 2n + 1 � κ , we have (sα� j)2nj=0 ∈H�

q,2n and hence Hα�n ∈C(n+1)q×(n+1)q� .

According to δ ∈ [0,+∞) and (4.6), we get then −(γ + αδ)H 〈s( ,γ ,δ)〉n + K 〈s( ,γ ,δ)〉

n ∈C(n+1)q×(n+1)q� , i.e.,

((s( ,γ ,δ)

)γ +αδ� j

)2nj=0 ∈ H�

q,2n. (4.7)

For all j ∈ Z0,2n , let r j := sα� j . Obviously, (r j)2nj=0 ∈H�

q,2n . Let [(Ek)nk=1, (Fk)

nk=0] and [(E

k)nk=1, (F

k)nk=0]

be the canonical Hankel parametrization of (r j)2nj=0 and (r( ,γ ,δ)

j )2nj=0, respectively. We consider an

arbitrary k ∈ Z0,n . For all j ∈ Z0,2n , let v j := δr( ,γ ,δ)

j and u j := δr j . In view of (2.5), (2.1), and (1.2),we get then

δL〈r( ,γ ,δ)〉k = L〈v〉

k . (4.8)

Because of Remark 4.4, we obtain u( ,γ ,δ)

j = v j for all j ∈ Z0,2n , which implies

L〈v〉k = L〈u( ,γ ,δ)〉

k . (4.9)

For all j ∈ Z0,2n , let t j := δs j+1. For all j ∈ Z0,2n , we have, in view of (1.3), then

u j = δr j = δsα� j = δ(−αs j + s j+1) = −αδs j + δs j+1 = −αδs j + t j (4.10)

and, in view of Remark 4.7, moreover (s( ,γ ,δ))γ � j = t( ,γ ,δ)

j . For every choice of j in Z0,2n , let w j :=(s( ,γ ,δ))γ +αδ� j . Taking into account (4.10), Remark 4.4, and (1.3), for all j ∈ Z0,2n , we then conclude

u( ,γ ,δ)

j = −αδs( ,γ ,δ)

j + t( ,γ ,δ)

j = −αδs( ,γ ,δ)

j + (s( ,γ ,δ))γ � j

= −(γ + αδ)s( ,γ ,δ)

j + s( ,γ ,δ)

j+1 = (s( ,γ ,δ))γ +αδ� j = w j,

which implies L〈u( ,γ ,δ)〉k = L〈w〉

k . For all k ∈ Z0,n , then from Definitions 2.2 and 2.1, from (r j)2nj=0 ∈H�

q,2nand [32, Theorem 4.18], from (4.8) and (4.9) we infer

δ2k+1 Q 2k+1 = δ2k+1Lα�k = δ2k+1L〈r〉k = δ2k+1 Fk = δF

k = δL〈r( ,γ ,δ)〉k

= L〈v〉k = L〈u( ,γ ,δ)〉

k = L〈w〉k = Q

2k+1. (4.11)

Because (s( ,γ ,δ)

j )2nj=0 ∈ H�

q,2n is proved for all n ∈ N with 2n � κ and since (4.7) is checked for all

n ∈ N with 2n + 1 � κ , we have (s( ,γ ,δ)

j )κj=0 ∈ K�q,κ,γ +αδ . Furthermore, (4.4) and (4.11) show that

(δ j Q j)κj=0 is the right (γ + αδ)-Stieltjes parametrization of (s( ,γ ,δ)

j )κj=0. �In order to give a corollary of Theorem 4.12, we recall a result which shows how one can compute

the rank of a block Hankel matrix by its right α-Stieltjes parametrization.


Lemma 4.13. (See [33, Lemma 4.11].) Let α ∈ R, let κ ∈ N0 ∪ {+∞}, and let (s j)κj=0 ∈ K�

q,κ,α with

right α-Stieltjes parametrization (Q j)κj=0 . Then rank Hn = ∑n

k=0 rank Q 2k and det Hn = ∏nk=0 det Q 2k

for all n ∈ N0 with 2n � κ . Moreover, if κ � 1, then rank Hα�n = ∑nk=0 rank Q 2k+1 and det Hα�n =∏n

k=0 det Q 2k+1 for all n ∈ N0 with 2n + 1 � κ .

Corollary 4.14. Let α,γ ∈ R, let δ ∈ (0,+∞), let κ ∈ N0 ∪ {+∞}, and let (s j)κj=0 ∈ K�

q,κ,α with

right α-Stieltjes parametrization (Q j)κj=0 . Let b j := s( ,γ ,δ)

j for all j ∈ Z0,κ and let (B j)κj=0 be the right

(γ + αδ)-Stieltjes parametrization of (b j)κj=0 . Then:

(a) N (B j) =N (Q j) and R(B j) =R(Q j) for all j ∈ Z0,κ .(b) det B j = δqj det Q j and rank B j = rank Q j for all j ∈ Z0,κ .

(c) det H 〈b〉n = δn(n+1)q det Hn and rank H 〈b〉

n = rank Hn for all n ∈ Z0, κ2

.

(d) If κ � 1 and t j := bγ +αδ� j for all j ∈ Z0,κ−1 , then det H 〈t〉n = δ[(n+1)2]q det Hα�n and rank H 〈t〉

n =rank Hα�n for all n ∈ Z0, κ−1

2.

Proof. (a), (b) According to Theorem 4.12, we have B j = δ j Q j for all j ∈ Z0,κ , which, in view ofδ ∈ (0,+∞), implies (a) and (b).

(c), (d) From Theorem 4.12 we also see that (b j)κj=0 ∈ K�

q,κ,γ +αδ , which, in view of Lemma 4.13and (b), shows that (c) and (d) hold true. �Example 4.15. Let α,γ ∈ R, let κ ∈ N0 ∪ {+∞} and let (s j)

κj=0 ∈ K�

q,κ,α with right α-Stieltjes

parametrization (Q j)κj=0. Let b j := s( ,γ ,0)

j , for all j ∈ Z0,κ , and let (B j)κj=0 be the right γ -Stieltjes

parametrization of (b j)κj=0. Theorem 4.12 shows that (b j)

κj=0 ∈ K�

q,κ,γ and, in view of Theorem 4.12,we have B j = δ j,0 Q 0 for all j ∈ Z0,κ , where δ j,k is the Kronecker delta. Further, Lemma 4.13 shows

that det H 〈b〉n = δn,0 det H0 and rank H 〈b〉

n = rank H0 for all n ∈N0 with 2n � κ . Suppose now κ � 1. Lett j := bγ � j for all j ∈ Z0,κ−1. Then we see from Lemma 4.13 and Corollary 4.14 that rank H 〈t〉

n = 0 forall n ∈ Z0, κ−1

2.

Corollary 4.16. Let α,β ∈ R, δ ∈ (0,+∞), κ ∈ N0 ∪ {+∞} and (s j)κj=0 be a sequence from C

q×q. Then

(s j)κj=0 ∈K�

q,κ,α if and only if (s( ,β−αδ,δ)

j )κj=0 ∈K�q,κ,β .

Proof. If (s j)κj=0 ∈K�

q,κ,α , then Theorem 4.12 yields (s( ,β−αδ,δ)

j )κj=0 ∈K�q,κ,β .

Conversely, now suppose (s( ,β−αδ,δ)

j )κj=0 ∈K�q,κ,β . Remark 4.6 shows that

(s( ,β−αδ,δ)

)( ,α− βδ, 1δ)

j = s( ,0,1)

j = s j for each j ∈ Z0,κ .

Since 1δ

∈ (0,+∞) holds, Theorem 4.12 implies (s j)κj=0 ∈K�

q,κ,α . �Now we derive for the classes K�,e

q,κ,α , K>q,κ,α and K�,cd

q,κ,α analogous results to Theorem 4.12.

Proposition 4.17. Let α,γ ∈ R, let δ ∈ [0,+∞), let m ∈ N0 , and let (s j)mj=0 ∈ K�,e

q,m,α . Then (s( ,γ ,δ)

j )mj=0 ∈

K�,eq,m,γ +αδ .

Proof. Use Theorem 4.12 and parts (b) and (c) of Theorem 2.3. �


Corollary 4.18. Let α,β ∈ R, δ ∈ (0,+∞), and m ∈ N0 . Further, let (s j)mj=0 be a sequence of complex

q × q matrices. Then (s j)mj=0 ∈K�,e

q,m,α if and only if (s( ,β−αδ,δ)

j )mj=0 ∈K�,e

q,m,β .

Proof. Proceed like in the proof of Corollary 4.16, using Proposition 4.17 instead of Theorem 4.12. �Proposition 4.19. Let α,γ ∈ R, let δ ∈ (0,+∞), let κ ∈ N0 ∪ {+∞}, and let (s j)

κj=0 ∈ K>

q,κ,α . Then

(s( ,γ ,δ)

j )κj=0 ∈K>q,κ,γ +αδ .

Proof. Use Theorem 4.12 and parts (b) and (d) of Theorem 2.3. �Corollary 4.20. Let α,β ∈ R, let δ ∈ (0,+∞), let κ ∈ N0 ∪ {+∞}, and let (s j)

κj=0 be a sequence of complex

q × q matrices. Then the sequence (s j)κj=0 belongs to K>


j )κj=0 ∈K>q,κ,β .

Proof. Proceed like in the proof of Corollary 4.16, using Proposition 4.19 instead of Theorem 4.12 and1δ

∈ (0,+∞). �Proposition 4.21. Let α,γ ∈ R, let δ ∈ [0,+∞), let κ ∈ N0 ∪ {+∞} and let (s j)

κj=0 ∈ K�,cd

q,κ,α . Then

(s( ,γ ,δ)

j )κj=0 ∈K�,cdq,κ,γ +αδ .

Proof. Use Theorem 4.12. �Corollary 4.22. Let α,β ∈ R, δ ∈ (0,+∞), κ ∈ N0 ∪ {+∞} and (s j)

κj=0 be a sequence of complex q × q ma-

trices. Then (s j)κj=0 belongs to K�,cd


j )κj=0 belongs to K�,cdq,κ,β .

Proof. Proceed like in the proof of Corollary 4.16, using Proposition 4.21 instead of Theorem 4.12 and1δ

∈ (0,+∞). �It should be mentioned that Bennett [6] studied in the scalar case several transformations for mo-

ment sequences of the types of Hamburger, Stieltjes or Hausdorff, which produce moment sequencesof the same type. One of them is the so-called binomial convolution. Let (s j)

∞j=0 and (t j)

∞j=0 be real

sequences. Then the sequence (u j)∞j=0 defined by

u j :=n∑

k=0

(n

k

)sktn−k

is called the binomial convolution of (s j)∞j=0 and (t j)

∞j=0. If (s j)

∞j=0 and (t j)

∞j=0 belong to K�

1,∞,0, then [6,

Proposition 2] yields that their binomial convolution (u j)∞j=0 also belongs to K�

1,∞,0.

5. Some interrelations between right-sided α-shifting and right α-Stieltjes parametrization

First we consider an α-Stieltjes right-sided non-negative definite sequence (s j)κj=0 and show by

application of some results on the binomial transform from Section 4 that the sequence (sα� j)κ−1j=0

generated from it by right-sided α-shifting is α-Stieltjes right-sided non-negative definite, too.

Proposition 5.1. Let α ∈R, κ ∈N∪ {+∞}, and (s j)κj=0 ∈K�

q,κ,α . Then (sα� j)κ−1j=0 belongs to K�

q,κ−1,α .


Proof. For all j ∈ Z0,κ−1, let u j := sα� j . Because of (s j)κj=0 ∈K�

q,κ,α , then

(u j)2� κ−1

2 �j=0 ∈ H�

q,2� κ−12 �, (5.1)

which, in particular, proves the assertion in the case κ = 1. Now suppose κ � 2. For all j ∈ Z0,κ−1, let

t j := s j+1 and, for all j ∈ Z0,κ−2, let v j := u j+1, w j := uα� j , and r j := s( ,−α,1)

j+2 . For all j ∈ Z0,κ−1, weconclude

w( ,−α,1)

j = −αu( ,−α,1)

j + v( ,−α,1)

j = −αu( ,−α,1)

j + (u( ,−α,1))−α� j

= −αu( ,−α,1)

j + [−(−α)u( ,−α,1)

j + u( ,−α,1)

j+1

]= u( ,−α,1)

j+1

= −αs( ,−α,1)

j+1 + t( ,−α,1)

j+1 = −αs( ,−α,1)

j+1 + (s( ,−α,1))−α� j+1

= −αs( ,−α,1)

j+1 + [−(−α)s( ,−α,1)

j+1 + s( ,−α,1)

j+2

]= s( ,−α,1)

j+2 = r j,

where the 1st equation is due to Remark 4.4, the 2nd equation is due to Remark 4.7, the 3rd equationis due to (1.3), the 5th equation is due to Remark 4.4, the 6th equation is due to Remark 4.7, and the7th equation is due to (1.3). In view of (4.1) and Remark 4.6, for all j ∈ Z0,κ−2, we have then

uα� j = w j = w( ,0,1)

j = (w( ,−α,1))( ,α,1)

j = r( ,α,1)

j . (5.2)

From Theorem 4.12 we get (s( ,−α,1)

j )κj=0 ∈ K�q,κ,0, which yields (s( ,−α,1)

j )2� κ

2 �j=0 ∈ H�

q,2� κ2 � , i.e.,

H 〈s( ,−α,1)〉� κ

2 � ∈ C(� κ

2 �+1)q×(� κ2 �+1)q

� . In view of (2.2) and (2.4), this implies that G〈s( ,−α,1)〉� κ

2 �−1 ∈ C� κ

2 �q×� κ2 �q

� ,

i.e., (s( ,−α,1)

j+2 )2� κ

2 �−2j=0 ∈ H�

q,2� κ2 �−2. Hence, (r j)

2� κ2 �−2

j=0 ∈ H�q,2� κ

2 �−2. From [32, Proposition 4.10(a)] we

obtain then (r( ,α,1)

j )2� κ

2 �−2j=0 ∈ H�

q,2� κ2 �−2 and, because of (5.2), thus (uα� j)

2� κ2 �−2

j=0 ∈ H�q,2� κ

2 �−2. Taking

additionally (5.1) into account, we obtain (sα� j)κ−1j=0 ∈K�

q,κ−1,α . �Now we are going to complement the result of Proposition 5.1 by computing the right α-Stieltjes

parametrization of (sα� j)κ−1j=0 . In order to realize this aim, we need the following observation.

Remark 5.2. Let α ∈R, κ ∈N0 ∪ {+∞} and (s j)κj=0 ∈K�

q,κ,α . From [35, Lemma 3.2] one can easily seethen that the following statements hold true:

(a) s j ∈ Cq×qH for all j ∈ Z0,κ and sα� j ∈C

q×qH for all j ∈ Z0,κ−1.

(b) s2k ∈Cq×q� for all k ∈ N0 with 2k � κ and sα�2k ∈C

q×q� for all k ∈ N0 with 2k + 1 � κ .

(c) N (s2k) ⊆N (s j) and R(s j) ⊆R(s2k) for all k ∈ Z0, κ2

and each j ∈ Z2k,2� κ2 �−1.

(d) N (sα�2k) ⊆N (sα� j) and R(sα� j) ⊆R(sα�2k) for all k ∈ Z0, κ−12

and each j ∈ Z2k,2� κ−12 �−1.

If the right α-Stieltjes parametrization of (s j)κj=0 ∈K�

q,κ,α is given, then the following result shows

that the right α-Stieltjes parametrization of (sα� j)κ−1j=0 can be computed recursively.

Theorem 5.3. Let α ∈ R, κ ∈ N ∪ {+∞}, and (s j)κj=0 ∈ K�

q,κ,α . Then the right α-Stieltjes parametrization

(S j)κ−1j=0 of (sα� j)

κ−1j=0 is given in terms of the right α-Stieltjes parametrization (Q j)

κj=0 of (s j)

κj=0 by S2k =

Q 2k+1 for all k ∈N0 with 2k � κ − 1 and, in the case κ � 2, by the recurrence formulas

S2k+1 = Q 2k+2 + Q 2k+1(

Q †2k − S†

2k−1

)Q 2k+1

for all k ∈ N0 with 2k + 1 � κ − 1 where S−1 := 0q×q.


Proof. From Definition 2.2 we get S2k = Lα�k = Q 2k+1 for all k ∈ N0 with 2k � κ − 1. Now supposeκ � 2. For all j ∈ Z0,κ−1, let t j := sα� j and, for all j ∈ Z0,κ−2, let u j := tα� j . From Definition 2.2, (2.5),

(2.1), (1.2), (1.3), the assumption (s j)κj=0 ∈ K�

q,κ,α , κ � 2, part (c) of Remark 5.2, and Remark A.1 one

can easily conclude that Q 2 + Q 1 Q †0 Q 1 = S1. Now suppose κ � 4. We consider an arbitrary k ∈ N with

2k + 2 � κ . Then κ − 1 ∈ N and 2k + 1 � κ − 1. From Proposition 5.1 we obtain (sα� j)κ−1j=0 ∈K�

q,κ−1,α .Hence, Lemmas 2.4 and 2.5 and Definition 2.2 yield

S2k+1 = sα�2k+1 − Λα�k − (α Iq + Lα�k S†2k−1

)Lα�k

= Q 2k+2 + (α Iq + Q 2k+1 Q †2k

)Q 2k+1 − (α Iq + Q 2k+1 S†

2k−1

)Q 2k+1

= Q 2k+2 + Q 2k+1(

Q †2k − S†

2k−1

)Q 2k+1. �

The notation Cq×qH stands for the set of all Hermitian complex q × q matrices. We will use the

Löwner semi-ordering in Cq×qH , i.e., we write A � B or B � A in order to indicate that A and B are

Hermitian complex matrices such that A − B is non-negative Hermitian.

Corollary 5.4. Let α ∈ R, let κ ∈ Z2,+∞ ∪{+∞}, and let (s j)κj=0 ∈K�

q,κ,α with right α-Stieltjes parametriza-

tion (Q j)κj=0 . Let (S j)

κ−1j=0 be the right α-Stieltjes parametrization of (sα� j)

κ−1j=0 . Then:

(a) S2k−1 � Q 2k � 0q×q for all k ∈ Z1, κ2

.(b) N (S2k−1) ⊆N (Q 2k) and R(Q 2k) ⊆R(S2k−1) for all k ∈ Z1, κ

2.

(c) 0 � det Q 2k � det S2k−1 and rank Q 2k � rank S2k−1 for all k ∈ Z1, κ2

.

(d) 0 � det Hn � (det s0)∏n

k=1 det S2k−1 = det(s0)det(α2 Hn−1 − 2αKn−1 + Gn−1) and rank Hn �rank s0 +∑n

k=1 rank S2k−1 = rank s0 + rank(α2 Hn−1 − 2αKn−1 + Gn−1) for all n ∈ Z1, κ2

.

Proof. (a) According to part (b) of Theorem 2.3, we get Q j ∈ Cq×q� for all j ∈ Z0,κ , which, in view of

Theorem 5.3, implies S1 = Q 2 + Q 1 Q †0 Q 1 = Q 2 + Q ∗

1 Q †0 Q 1 � Q 2 � 0q×q . In particular, (a) is proved

in the case κ � 3. Now suppose κ � 4. We have already shown that there exists an integer k ∈ N with2k � κ − 2 such that S2l−1 � Q 2l � 0q×q for all l ∈ Z1,k , which, in view of Remark A.2, yields that

Q †2k � Q †

2k Q 2k S†2k−1 Q 2k Q †

2k . According to part (b) of Theorem 2.3, we have Q j ∈Cq×q� for all j ∈ Z0,κ

and N (Q j) ⊆ N (Q j+1) for all j ∈ Z0,κ−2, which, in view of Remark A.1, implies Q 2k+1 Q †2k Q 2k =

Q 2k+1 and Q 2k Q †2k Q 2k+1 = Q 2k+1. Using Theorem 5.3, we obtain then

S2k+1 = Q 2k+2 + Q 2k+1 Q †2k Q 2k

(Q †

2k − S†2k−1

)Q 2k Q †

2k Q 2k+1

= Q 2k+2 + Q ∗2k+1

(Q †

2k − Q †2k Q 2k S†

2k−1 Q 2k Q †2k

)Q 2k+1 � Q 2k+2 � 0q×q.

Hence, (a) is proved by mathematical induction.(b), (c) Use (a).(d) Let n ∈ N be such that 2n � κ . The asserted inequalities can be proved using Lemma 4.13,

Definition 2.2, and (c). Let t j := sα� j for all j ∈ Z0,κ−1 and let u j := tα� j for all j ∈ Z0,κ−2. Ac-

cording to Proposition 5.1, we have then (t j)κ−1j=0 ∈ K�

q,κ−1,α , which, in view of Lemma 4.13, implies

rank H 〈u〉m =∑m

k=0 rank S2k+1 and det H 〈u〉m =∏m

k=0 det S2k+1 for all m ∈N0 with 2m + 1 � κ − 1. Using(1.2), (1.3), (2.3), and (2.4), we obtain furthermore

H 〈u〉n−1 = −αH 〈t〉

n−1 + K 〈t〉n−1 = −α(−αHn−1 + Kn−1) + (−αKn−1 + Gn−1)

= α2 Hn−1 − 2αKn−1 + Gn−1.

Hence, the asserted equations follow. �


The next result shows that the right α-Stieltjes parametrization of an arbitrarily given sequence(s j)

κj=0 ∈ K�

q,κ,α can be computed recursively, using s0 and the right α-Stieltjes parametrization of

(sα� j)κ−1j=0 .

Corollary 5.5. Let α ∈ R, κ ∈ N ∪ {+∞} and (s j)κj=0 ∈ K�

q,κ,α . Then the right α-Stieltjes parametrization

(Q j)κj=0 of (s j)

κj=0 is given in terms of s0 and the right α-Stieltjes parametrization (S j)

κ−1j=0 of (sα� j)

κ−1j=0 by

S−1 := 0q×q and the recurrence formulas

Q 2k ={

s0, if k = 0,

S2k−1 − S2k−2(Q †2k−2 − S†

2k−3)S2k−2, if k � 1

for all k ∈ N0 with 2k � κ and by Q 2k+1 = S2k for all k ∈N0 with 2k + 1 � κ .

Proof. Use Definition 2.2 and Theorem 5.3. �Now we consider an α-Stieltjes right-sided non-negative definite sequence (s j)

κj=0 and dis-

cuss some interrelations between the right α-Stieltjes parametrization and the canonical Hankelparametrization of some sequences generated from (s j)

κj=0.

Proposition 5.6. Let α ∈ R, let κ ∈ Z3,+∞ ∪ {+∞}, and let (s j)κj=0 ∈ K�


parametrization (Q j)κj=0 . Let [(Ck)

� κ2 �

k=1, (Dk)� κ

2 �k=0] be the canonical Hankel parametrization of (s j)

2� κ2 �

j=0 , let

(S j)κ−1j=0 be the right α-Stieltjes parametrization of (sα� j)

κ−1j=0 , and let [(Ek)

� κ−12 �

k=1 , (Fk)� κ−1

2 �k=0 ] be the canon-

ical Hankel parametrization of (sα� j)2� κ−1

2 �j=0 . Then:

(a) Q 2k = Dk � 0q×q for all k ∈ N0 with 2k � κ .(b) N (Dk) ⊆N (Ck+1) for all k ∈N0 with 2k + 2 � κ .(c) If α � 0, then Ck � Q 2k−1 for all k ∈ N with 2k � κ .(d) Q 2k+1 = S2k = Fk � 0q×q for all k ∈ N0 with 2k + 1 � κ .(e) N (Fk) ⊆N (Ek+1) for all k ∈ N0 with 2k + 2 � κ − 1.(f) If α � 0, then Ek � S2k−1 for all k ∈N with 2k � κ − 1.(g) S2k−1 � Q 2k for all k ∈ N with 2k � κ .

Proof. (a), (d) Use Definitions 2.2 and 2.1 and part (b) of Theorem 2.3.(b), (e) This follows from the definition of the set K�

q,κ,α [32, Proposition 2.10(b), Proposi-tion 2.15(b)], and part (a) of Remark A.1.

(c) This is a consequence of [33, part (a) of Corollary 6.12].(f) Use Proposition 5.1 and [33, part (a) of Corollary 6.12].(g) This follows from part (a) of Corollary 5.4. �

6. On inverse right-sided α-shifting of sequences of complex p × q matrices

In the preceding sections, the right-sided α-shifting of sequences of complex p × q matrices(see (1.3)) played an important role. Now we study a converse operation for sequences of com-plex p × q matrices. More precisely, given α ∈ C, A ∈ C

p×q , κ ∈ N0 ∪ {+∞}, and a sequence (s j)κj=0

from Cp×q , we look for a sequence (r j)

κ+1j=0 from C

p×q with r0 = A such that the sequence (rα� j)κj=0

which is generated from (r j)κ+1j=0 by right-sided α-shifting coincides with the original sequence

(s j)κj=0. Against to this background, we introduce the following operation for sequences of complex

p × q matrices.


Definition 6.1. Let α ∈ C, A ∈Cp×q , κ ∈N0 ∪ {+∞}, and (s j)

κj=0 be a sequence of complex p × q ma-

trices. Then the sequence (s〈α,A]j )κ+1

j=0 defined by

s〈α,A]j :=

{A, if j = 0,

α j A +∑ j−1l=0 α j−1−lsl, if j � 1,

for all j ∈ Z0,κ+1 (6.1)

is called the sequence generated from A and (s j)κj=0 by inverse right-sided α-shifting.

Remark 6.2. Let A ∈ Cp×q , κ ∈ N0 ∪ {+∞}, and (s j)

κj=0 be a sequence of complex p × q matrices.

Then s〈0,A]0 = A and, for all j ∈ Z1,κ+1, moreover s〈0,A]

j = s j−1.

The following result shows that the construction, which was introduced in Definition 6.1, is theright tool for our purposes.

Lemma 6.3. Let α ∈ C, A ∈ Cp×q, κ ∈ N0 ∪ {+∞} and (r j)

κ+1j=0 and (s j)

κj=0 be sequences of complex

p × q matrices. Then the following statements are equivalent:

(i) r0 = A and rα� j = s j for all j ∈ Z0,κ .

(ii) r j = s〈α,A]j for all j ∈ Z0,κ+1 .

Proof. “(i) ⇒ (ii)”: Use Definition 6.1 and Remark 1.3.“(ii) ⇒ (i)”: From (ii) and Definition 6.1 we get r0 = s〈α,A]

0 = A. Because of (1.3), (ii), and Defini-tion 6.1, we have

rα�0 = −αr0 + r1 = −αs〈α,A]0 + s〈α,A]

1 = −αA + α1 A +1−1∑l=0

α1−1−lsl = s0

and, in the case κ � 1, for all j ∈ Z1,κ , moreover

rα� j = −αr j + r j+1 = −αs〈α,A]j + s〈α,A]

j+1

= −α

(α j A +

j−1∑l=0

α j−1−lsl

)+ α j+1 A +

j∑l=0

α j−lsl = α j− j s j = s j .

Hence, (i) is fulfilled. �Let α ∈R, κ ∈N0 ∪ {+∞} and (s j)

κj=0 be a sequence of complex q × q matrices. Then let

B�α

[(s j)

κj=0

] := {E ∈Cq×q

∣∣ (s〈α,E]j

)κ+1j=0 ∈ K�

q,κ+1,α

}.

Now we consider a sequence (s j)κj=0 such that B�

α [(s j)κj=0] = ∅ and an A ∈ B�

α [(s j)κj=0]. Then our

aim is to compute the right α-Stieltjes parametrization of (s〈α,A]j )κ+1

j=0 .

Proposition 6.4. Let α ∈ R, let κ ∈N0 ∪{+∞}, let (s j)κj=0 be a sequence of complex q × q matrices such that

B�α [(s j)

κj=0] = ∅, and let A ∈ B�

α [(s j)κj=0]. Then the right α-Stieltjes parametrization (B j)

κ+1j=0 of (s〈α,A]

j )κ+1j=0

is given in terms of A and the right α-Stieltjes parametrization (Q j)κj=0 of (s j)

κj=0 by the recurrence formulas

B2k ={

A, if k = 0,

Q 2k−1 − Q 2k−2(B†2k−2 − Q †

2k−3)Q 2k−2, if k � 1

for all k ∈N0 with 2k � κ + 1 and by B2k+1 = Q 2k for all k ∈N0 with 2k + 1 � κ + 1, where Q −1 := 0q×q.


Proof. For all j ∈ Z0,κ+1, let r j := s〈α,A]j . Then (B j)

κ+1j=0 is the right α-Stieltjes parametrization of

(r j)κ+1j=0 . Because of A ∈ B�

α [(s j)κj=0], we have (r j)

κ+1j=0 ∈ K�

q,κ+1,α . From Lemma 6.3 we know r0 = Aand rα� j = s j for all j ∈ Z0,κ . Hence, (Q j)

κj=0 is the right α-Stieltjes parametrization of (rα� j)

κj=0. The

application of Corollary 5.5 completes the proof. �Corollary 6.5. Let α ∈ R, let κ ∈ N0 ∪ {+∞}, let (s j)

κj=0 be a sequence of complex q × q matrices such that

B�α [(s j)

κj=0] = ∅, and let A ∈ B�

α [(s j)κj=0]. Then the right α-Stieltjes parametrization (Q j)

κj=0 of (s j)

κj=0 is

given in terms of the right α-Stieltjes parametrization (B j)κ+1j=0 of (s〈α,A]

j )κ+1j=0 by Q 2k = B2k+1 for all k ∈ N0

with 2k � κ and by the recurrence formula

Q 2k+1 = B2k+2 + Q 2k(

B†2k − Q †

2k−1

)Q 2k

for all k ∈ N0 with 2k + 1 � κ where Q −1 := 0q×q.

Proof. Use Proposition 6.4. �7. Particular additive perturbations of α-Stieltjes right-sided non-negative definite sequences

In this section, we consider sequences of matrices which are constructed by particular additiveperturbations of α-Stieltjes right-sided non-negative definite sequences. As it will be discussed inAppendix B, these additive perturbations can be interpreted as power moment sequences of a matrixmeasure which is located in a single point.

Example 7.1. Let ζ ∈ R and A ∈ Cq×q� . Then (ζ j A)∞j=0 ∈K�

q,∞,ζ . Indeed, if s j := ζ j A for all n ∈ N0, then

Hn = [ζ j+k A]n

j,k=0 = (row[ζ k Iq

]nk=0

)∗A(row

[ζ k Iq

]nk=0

) ∈C(n+1)q×(n+1)q� ,

and, according to (1.3), we have sζ� j = −ζ s j + s j+1 = −ζ(ζ j A) + ζ j+1 A = 0q×q for all j ∈ N0 and

hence Hζ�n = 0(n+1)q×(n+1)q ∈C(n+1)q×(n+1)q� for all n ∈ N0.

Remark 7.2. Let α ∈ R, β ∈ [α,+∞), and κ ∈ N0 ∪ {+∞}. Then K�q,κ,β ⊆K�

q,κ,α .

Lemma 7.3. Let α, ζ ∈ R, let A ∈ Cq×q� , let κ ∈ N0 ∪ {+∞}, and let (s j)

κj=0 ∈ K�

q,κ,α . Then (s j + ζ j A)κj=0 ∈K�

q,κ,min{α,ζ } .

Proof. Because of (s j)κj=0 ∈ K�

q,κ,α and Remark 7.2, we have (s j)κj=0 ∈K�

q,κ,min{α,ζ } . From Example 7.1

we get (ζ j A)∞j=0 ∈ K�q,∞,ζ , which implies (ζ j A)κj=0 ∈ K�

q,κ,ζ . In view of Remark 7.2, we obtain then

(ζ j A)κj=0 ∈ K�q,κ,min{α,ζ } . Now, the application of Remark 7.2 and [33, Remark 2.6] yields the asser-

tion. �The main goal of this section is to compute, in the case that α ∈ R, (s j)

κj=0 ∈ K�

q,κ,α , A ∈ Cq×q�

and ζ ∈ (−∞,α] are given, the right ζ -Stieltjes parametrization of (s j + ζ j A)κj=0 in terms of the rightα-Stieltjes parametrization of (s j)

κj=0. In order to realize this aim, we need the following result.

Proposition 7.4. (See [33, Proposition 7.5].) Let α ∈ R, β ∈ [α,+∞), κ ∈ N0 ∪ {+∞} and (s j)κj=0 ∈ K�

q,κ,β .Then the right α-Stieltjes parametrization (Q j)

κj=0 of (s j)

κj=0 is given in terms of the right β-Stieltjes

parametrization (R j)κj=0 of (s j)

κj=0 by Q 2k = R2k for all k ∈ N0 with 2k � κ and, in the case κ � 1, by the


recurrence formulas Q 2k+1 = R2k+1 +[(β −α)Iq + R2k(R†2k−1 − Q †

2k−1)]R2k for all k ∈N0 with 2k + 1 � κ ,where Q −1 := 0q×q and R−1 := 0q×q.

Now we state the main result of this section.

Theorem 7.5. Let α ∈ R, ζ ∈ (−∞,α], A ∈ Cq×q� , κ ∈ N0 ∪ {+∞}, and (s j)

κj=0 ∈ K�

q,κ,α . Then the right

ζ -Stieltjes parametrization (P j)κj=0 of the sequence (s j + ζ j A)κj=0 is given in terms of A and the right

α-Stieltjes parametrization (Q j)κj=0 of (s j)

κj=0 by the recurrence formulas

P2k ={

Q 0 + A, if k = 0,

Q 2k + P2k−1(Q †2k−2 − P †

2k−2)P2k−1, if k � 1

for all k ∈N0 with 2k � κ and

P2k+1 = Q 2k+1 + [(α − ζ )Iq + Q 2k(

Q †2k−1 − P †

2k−1

)]Q 2k

for all k ∈N0 with 2k + 1 � κ where P−1 := 0q×q and Q −1 := 0q×q.

Proof. For all j ∈ Z0,κ , let p j := s j +ζ j A. From Definition 2.2 we get then P0 = L〈p〉0 = p0 = s0 +ζ 0 A =

L0 + A = Q 0 + A, which, in particular, proves the assertion in the case κ = 0.Now suppose κ � 1. Observe that (P j)

κj=0 is the right ζ -Stieltjes parametrization of (p j)

κj=0. For

all j ∈ Z0,κ , let v j := p( ,−ζ,1)

j . From Lemma 7.3 we obtain (p j)κj=0 ∈K�

q,κ,ζ . Theorem 4.12 yields then

(v j)κj=0 ∈ K�

q,κ,0 and that (P j)κj=0 is the right 0-Stieltjes parametrization of (v j)

κj=0. For all j ∈ Z0,κ ,

let b j := s( ,−ζ,1)

j . Let (S j)κ−1j=0 be the right 0-Stieltjes parametrization of (b0� j)

κ−1j=0 . According to Re-

mark 4.4 and Example 4.1, we have v0 = b0 + A and v j = b j for all j ∈ Z1,κ . Hence, in view of (1.3),we easily get v0� j = b0� j for all j ∈ Z0,κ−1, which implies that (S j)

κ−1j=0 is also the right 0-Stieltjes

parametrization of (v0� j)κ−1j=0 . Corollary 5.5 yields then the recurrence formulas

P2k ={

v0, if k = 0,

S2k−1 − S2k−2(P †2k−2 − S†

2k−3)S2k−2, if k � 1

for all k ∈ N0 with 2k � κ and P2k+1 = S2k for all k ∈ N0 with 2k + 1 � κ , where S−1 := 0q×q . Let

(B j)κj=0 be the right 0-Stieltjes parametrization of (b j)

κj=0. In view of (s j)

κj=0 ∈ K�

q,κ,α , Theorem 4.12

yields (b j)κj=0 ∈K�

q,κ,α−ζ . Because of (α−ζ ) ∈ [0,+∞), from Remark 7.2 we see that (b j)κj=0 ∈K�

q,κ,0.Hence, Theorem 5.3 provides us S2k = B2k+1 for all k ∈N0 with 2k � κ − 1 and, in the case κ � 2, therecurrence formulas

S2k+1 =⎧⎨⎩

B2 + B1 B†0 B1, if k = 0,

B2k+2 + B2k+1(B†2k − S†

2k−1)B2k+1, if k � 1

for all k ∈ N0 with 2k + 1 � κ − 1. Because of Remark 4.6 and Example 4.2, we have b( ,ζ,1)

j =(s( ,−ζ,1))

( ,ζ,1)

j = s( ,1·(−ζ )+ζ,1·1)

j = s( ,0,1)

j = s j for all j ∈ Z0,κ . Taking into account (b j)κj=0 ∈ K�

q,κ,0,Theorem 4.12 then shows that (B j)

κj=0 is the right ζ -Stieltjes parametrization of (s j)

κj=0. In view of

α ∈ [ζ,+∞) and (s j)κj=0 ∈ K�

q,κ,α , Proposition 7.4 provides us then B2k = Q 2k for all k ∈ N0 with2k � κ and the recurrence formulas

B2k+1 ={

Q 1 + (α − ζ )Q 0, if k = 0,

Q 2k+1 + [(α − ζ )Iq + Q 2k(Q † − B†)]Q 2k, if k � 1
2k−1 2k−1


for all k ∈ N0 with 2k + 1 � κ . Hence, we have P1 = S0 = B1 = Q 1 + (α − ζ )Q 0. Then, in view ofP0 = Q 0 + A, in the case κ = 1, the assertion is also proved.

Now suppose κ � 2. Then we easily get P2 = Q 2 + P1(Q †0 − P †

0)P1, which proves the assertion inthe case κ = 2.

Now suppose κ � 3. For all k ∈ N with 2k + 1 � κ , we have then

P2k+1 = S2k = B2k+1 = Q 2k+1 + [(α − ζ )Iq + Q 2k(

Q †2k−1 − B†

2k−1

)]Q 2k

= Q 2k+1 + [(α − ζ )Iq + Q 2k(

Q †2k−1 − S†

2k−2

)]Q 2k

= Q 2k+1 + [(α − ζ )Iq + Q 2k(

Q †2k−1 − P †

2k−1

)]Q 2k

and, for all k ∈ Z2,+∞ with 2k � κ , moreover

P2k = S2k−1 − S2k−2(

P †2k−2 − S†

2k−3

)S2k−2

= B2k + B2k−1(

B†2k−2 − S†

2k−3

)B2k−1 − P2k−1

(P †

2k−2 − S†2k−3

)P2k−1

= Q 2k + S2k−2(

Q †2k−2 − S†

2k−3

)S2k−2 − P2k−1

(P †

2k−2 − S†2k−3

)P2k−1

= Q 2k + P2k−1(

Q †2k−2 − S†

2k−3

)P2k−1 − P2k−1

(P †

2k−2 − S†2k−3

)P2k−1

= Q 2k + P2k−1(

Q †2k−2 − P †

2k−2

)P2k−1.

In the case κ � 3, the proof is also complete. �Corollary 7.6. Let α ∈R, ζ ∈ (−∞,α], A ∈C

q×q� , κ ∈N0 ∪{+∞}, and (s j)

κj=0 ∈K�


parametrization (Q j)κj=0 . Let (P j)

κj=0 be the right ζ -Stieltjes parametrization of (s j + ζ j A)κj=0 . Then:

(a) P2k � Q 2k � 0q×q for all k ∈ N0 with 2k � κ .(b) N (P2k) ⊆N (Q 2k) and R(Q 2k) ⊆R(P2k) for all k ∈N0 with 2k � κ .(c) 0 � det Q 2k � det P2k and rank Q 2k � rank P2k for all k ∈N0 with 2k � κ .(d) Let t j := s j + ζ j A for all j ∈ Z0,κ . Then 0 � det Hn � det H 〈t〉

n and rank Hn � rank H 〈t〉n for all n ∈ N0

with 2n � κ .(e) If κ � 2, then N (Q 2k) ⊆N (P2k+1) and R(P2k+1) ⊆R(Q 2k) for all k ∈N0 with 2k + 1 � κ − 1.(f) If κ � 2, then rank P2k+1 � rank Q 2k for all k ∈ N0 with 2k + 1 � κ − 1.(g) If κ � 1, then P2k+1 � Q 2k+1 + (α − ζ )Q 2k � Q 2k+1 � 0q×q and P2k+1 � (α − ζ )Q 2k � 0q×q for all

k ∈N0 with 2k + 1 � κ .(h) If κ � 1, then N (P2k+1) ⊆N (Q 2k+1) and R(Q 2k+1) ⊆R(P2k+1) for all k ∈ N0 with 2k + 1 � κ .(i) If κ � 1, then 0 � det Q 2k+1 � det P2k+1 , 0 � (α − ζ )q det Q 2k � det P2k+1 , and rank Q 2k+1 �

rank P2k+1 for all k ∈ N0 with 2k + 1 � κ .

(j) Let u j := t j�ζ for all j ∈ Z0,κ−1 . Then 0 � det Hα�n � det H 〈u〉n , 0 � (α − ζ )

(n+1)(n+2)q2 det Hn � det H 〈u〉

n ,

and rank Hα�n � rank H 〈u〉n for all n ∈N0 with 2n + 1 � κ .

Proof. Because of ζ ∈ (−∞,α] and Lemma 7.3, we have (s j + ζ j A)κj=0 ∈K�q,κ,ζ . According to part (b)

of Theorem 2.3, we get then

P j ∈Cq×q� and Q j ∈C

q×q� for each j ∈ Z0,κ (7.1)

and, in the case κ � 2, furthermore

N (Q j) ⊆ N (Q j+1) for each j ∈ Z0,κ−2. (7.2)

(e) Because of (7.1) and (7.2), we see from Theorem 7.5 that (e) holds true.(a) According to (7.1), Theorem 7.5, and A ∈ C

q×q� , we see that P0 = Q 0 + A � Q 0 � 0q×q . In par-

ticular, (a) is proved in the case κ ∈ {0,1}. Now suppose κ � 2. We have already shown that there is


an integer k ∈ N0 with 2k � κ − 2 such that P2l � Q 2l � 0q×q for all l ∈ Z0,k , which, in view of Re-

mark A.2, implies Q †2k � Q †

2k Q 2k P †2k Q 2k Q †

2k . From (e) we have N (Q 2k) ⊆ N (P2k+1) and R(P2k+1) ⊆R(Q 2k), which, in view of Remark A.1, implies P2k+1 Q †

2k Q 2k = P2k+1 and Q 2k Q †2k P2k+1 = P2k+1. Us-

ing Theorem 7.5, we obtain then

P2k+2 = Q 2k+2 + P2k+1(

Q †2k − P †

2k

)P2k+1

= Q 2k+2 + P2k+1 Q †2k Q 2k

(Q †

2k − P †2k

)Q 2k Q †

2k P2k+1

= Q 2k+2 + P∗2k+1

(Q †

2k − Q †2k Q 2k P †

2k Q 2k Q †2k

)P2k+1 � Q 2k+2 � 0q×q.

Hence, (a) is proved by induction.(b), (c) Use (a).(d) Because of ζ ∈ (−∞,α] and Lemma 7.3, we have (t j)

κj=0 ∈ K�

q,κ,ζ . Thus, from Lemma 4.13and (c) we obtain (d).

(f) Use (e).(g) According to (7.1), Theorem 7.5, and ζ ∈ (−∞,α], we conclude P1 = Q 1 + (α − ζ )Q 0 � Q 1 �

0q×q . In particular, (g) is proved if κ ∈ {1,2}. Now suppose κ � 3. We have already shown that thereis an integer k ∈ N0 with 2k + 1 � κ − 2 such that P2l+1 � Q 2l+1 + (α − ζ )Q 2l � Q 2l+1 � 0q×q for

all l ∈ Z0,k , which, in view of Remark A.2, implies Q †2k+1 � Q †

2k+1 Q 2k+1 P †2k+1 Q 2k+1 Q †

2k+1. Because of

(7.1) and (7.2), Remark A.1 imply Q 2k+2 Q †2k+1 Q 2k+1 = Q 2k+2 and Q 2k+1 Q †

2k+1 Q 2k+2 = Q 2k+2. UsingTheorem 7.5 and ζ ∈ (−∞,α], we obtain then

P2k+3 = Q 2k+3 + [(α − ζ )Iq + Q 2k+2(

Q †2k+1 − P †

2k+1

)]Q 2k+2

= Q 2k+3 + (α − ζ )Q 2k+2 + Q 2k+2 Q †2k+1 Q 2k+1

(Q †

2k+1 − P †2k+1

)Q 2k+1 Q †

2k+1 Q 2k+2

= Q 2k+3 + (α − ζ )Q 2k+2 + Q ∗2k+2

(Q †

2k+1 − Q †2k+1 Q 2k+1 P †

2k+1 Q 2k+1 Q †2k+1

)Q 2k+2

� Q 2k+3 + (α − ζ )Q 2k+2 � Q 2k+3 � 0q×q.

Hence, P2k+1 � Q 2k+1 + (α − ζ )Q 2k � Q 2k+1 � 0q×q is proved for all k ∈ Z0, κ−12

by mathematical

induction. Thus, from (7.1) and ζ ∈ (−∞,α] we also obtain P2k+1 � (α − ζ )Q 2k � 0q×q for all k ∈Z0, κ−1

2.

(h), (i) Use (g).(j) Since (t j)

κj=0 belongs to K�

q,κ,ζ , Lemma 4.13 and (i) yield (j). �

Corollary 7.7. Let α ∈ R, ζ ∈ (−∞,α), A ∈ Cq×q� , and κ ∈ Z2,+∞ ∪ {+∞}. Further, let (s j)

κj=0 ∈ K�

q,κ,α

with right α-Stieltjes parametrization (Q j)κj=0 . Let (P j)

κj=0 be the right ζ -Stieltjes parametrization of

(s j + ζ j A)κj=0 . Then:

(a) N (P2k+1) = N (Q 2k), R(Q 2k) = R(P2k+1) for all k ∈ N0 with 2k + 1 � κ − 1, and rank Q 2k =rank P2k+1 for all k ∈ N0 with 2k + 1 � κ − 1.

(b) Let t j := s j + ζ j A for all j ∈ Z0,κ and let u j := t j�ζ for all j ∈ Z0,κ−1 . Then rank Hn = rank H 〈u〉n for all

n ∈ N0 with 2n + 1 � κ − 1.

Proof. (a) Since ζ ∈ (−∞,α) is assumed, parts (a) and (g) of Corollary 7.6 yield N (P2k+1) ⊆N ((α − ζ )Q 2k) = N (Q 2k) and R(Q 2k) = R((α − ζ )Q 2k) ⊆ R(P2k+1) for all k ∈ N0 with 2k + 1 � κ ,which, in view of part (e) of Corollary 7.6, implies (a).

(b) Because of ζ ∈ (−∞,α) and Lemma 7.3, we have (t j)κj=0 ∈ K�

q,κ,ζ . Taking into accountLemma 4.13 and (a), we obtain (b). �


8. On some interrelations between the right 0-Stieltjes parametrization and the classical Stieltjesparameters for 0-Stieltjes right-sided positive definite sequences

In this section, we consider sequences belonging to the set K>q,∞,0 of infinite 0-Stieltjes right-sided

positive definite sequences of complex q × q matrices. Let (s j)∞j=0 ∈ K>

q,∞,0. Then Yu.M. Dyukarevstudied in [26] the moment problem M[[0,+∞); (s j)

∞j=0,=]. One of his main results (see [26, The-

orem 8]) is a generalization of a classical criterion due to Stieltjes [50,51] for the indeterminacy ofthis moment problem. In order to find an appropriate matricial version of Stieltjes’ indeterminacycriterion Yu.M. Dyukarev had to look for a convenient matricial generalization of the parameter se-quences which Stieltjes obtained from the consideration of particular continued fractions associatedwith the sequence (s j)

∞j=0. In this way, Yu.M. Dyukarev found an interesting inner parametrization

of sequences belonging to K>q,∞,0. The main theme of this section is to present some interrelations

between Yu.M. Dyukarev’s parametrization and our right 0-Stieltjes parametrization introduced inDefinition 2.2.

Remark 8.1. Let (s j)∞j=0 ∈ K>

q,∞,0. Then from the definition of the set K>q,∞,0 it is clear that the ma-

trices Hn and Kn are positive Hermitian and, in particular, invertible for all n ∈N0.

Let

vq,0 := Iq and vq,k :=[

Iq

0kq×q

](8.1)

for all k ∈ N. The following construction of a pair of sequences of q × q matrices associated with asequence (s j)

∞j=0 ∈K>

q,∞,0 goes back to Yu.M. Dyukarev [26, p. 77].

Definition 8.2. Let (s j)∞j=0 ∈K>

q,∞,0 and let

Mk :={

s−10 , if k = 0,

v∗q,k H−1

k vq,k − v∗q,k−1 H−1

k−1 vq,k−1, if k � 1

and

Lk :={

s0s−11 s0, if k = 0,

y∗0,k K −1

k y0,k − y∗0,k−1 K −1

k−1 y0,k−1, if k � 1

for all k ∈ N0. Then the ordered pair [(Lk)∞k=0, (Mk)

∞k=0] is called the Dyukarev–Stieltjes parametrization

(shortly DS-parametrization) of (s j)∞j=0.

Remark 8.3. Let (s j)∞j=0 ∈K>

q,∞,0 with DS-parametrization [(Lk)∞k=0, (Mk)

∞k=0]. In view of Definition 8.2,

(8.1), (1.2), (2.1), and (2.3), we can easily see then, that, for all k ∈N0, the matrix Mk only depends onthe matrices s0, . . . , s2k and that the matrix Lk only depends on the matrices s0, s1, . . . , s2k+1.

It should be mentioned that, for a given sequence (s j)∞j=0 ∈ K>

q,∞,0, Yu.M. Dyukarev [26] treatedthe moment problem M[[0,+∞); (s j)

∞j=0,=] by approximation through the sequence (M[[0,+∞);

(s j)kj=0,�])k∈N0 of truncated moment problems. One of his central results [26, Theorem 7] shows

that the resolvent matrices for the truncated moment problems can be multiplicatively decom-posed into elementary factors which are determined by the corresponding first sections of the DS-parametrization of (s j)

∞j=0.

To establish connections between the DS-parametrization and the right 0-Stieltjes parametrizationof 0-Stieltjes right-sided positive definite sequences of complex q × q matrices we now investigatea particular transformation in the set of ordered pairs of infinite sequences of positive Hermitiancomplex q × q matrices which will turn out to be a bijection.


Remark 8.4. Let (Ak)∞k=0 and (Bk)

∞k=0 be sequences of positive Hermitian complex q × q matrices. Let

the sequences (Xk)∞k=0 and (Yk)

∞k=0 be given by

Xk :=⎧⎨⎩

B−10 , if k = 0,

(→∏ k−1

j=0 B j A j)−∗B−1

k (→∏ k−1

j=0 B j A j)−1, if k � 1

(8.2)

and

Yk :=( k→∏

j=0

B j A j

)−∗Ak

( k→∏j=0

B j A j

)−1

(8.3)

for all k ∈ N0. Then from (8.2) and (8.3) it is clear that the matrices Xk and Yk are invertible for allk ∈N0. By straightforward computations we get X−1

0 Y0 = (B0 A0)−1, X0Y −1

0 = (B0 A0)∗ , and, further-

more,

X−1k Yk =

( k−1→∏j=0

B j A j

)( k→∏j=0

B j A j

)−1

, XkY −1k =

( k−1→∏j=0

B j A j

)−∗( k→∏j=0

B j A j

)∗

for all k ∈N. Thus, by induction, we obtain

m→∏j=0

X−1j Y j =

( m→∏j=0

B j A j

)−1

and

m→∏j=0

X j Y−1j =

( m→∏j=0

B j A j

)∗

for all m ∈N0.

Lemma 8.5. Let (Ak)∞k=0 and (Bk)

∞k=0 be sequences of positive Hermitian complex q × q matrices. Let the se-

quences (Xk)∞k=0 and (Yk)

∞k=0 be given by (8.2) and (8.3). For all k ∈ N0 , then Xk and Yk are positive Hermitian

complex q × q matrices. Furthermore,

Ak =( k→∏

j=0

X j Y−1j

)Yk

( k→∏j=0

X j Y−1j

)∗

and

Bk =⎧⎨⎩

X−10 , if k = 0,

(→∏ k−1

j=0 X−1j Y j)X−1

k (→∏ k−1

j=0 X−1j Y j)

∗, if k � 1

for all k ∈N0 .

Proof. Since Ak and Bk are positive Hermitian complex q × q matrices for all k ∈ N0, we can easilysee from (8.2) and (8.3), that Xk and Yk are positive Hermitian complex q × q matrices for all k ∈N0,

as well. According to (8.2) we have B0 = X−10 . Remark 8.4 yields

→∏ kj=0 B j A j = (

→∏ kj=0 X j Y

−1j )∗ for all

k ∈ N0 and→∏ k−1

j=0 B j A j = (→∏ k−1

j=0 X−1j Y j)

−1 for all k ∈ N which, in view of (8.3) and (8.2) completesthe proof. �Remark 8.6. Let (Xk)

∞k=0 and (Yk)


the sequences (Ak)∞k=0 and (Bk)

∞k=0 be given by


Ak :=( k→∏

j=0

X j Y−1j

)Yk

( k→∏j=0

X j Y−1j

)∗(8.4)

and

Bk :=⎧⎨⎩

X−10 , if k = 0,

(→∏ k−1

j=0 X−1j Y j)X−1

k (→∏ k−1

j=0 X−1j Y j)

∗, if k � 1(8.5)

for all k ∈ N0. Then from (8.4) and (8.5) it is clear that the matrices Ak and Bk are invertible for allk ∈N0. By straightforward computations we get B0 A0 = (X−1

0 Y0)−1, B0 A0 = (X0Y −1

0 )∗ , and, further-more,

Bk Ak =( k−1→∏

j=0

X−1j Y j

)( k→∏j=0

X−1j Y j

)−1

, Bk Ak =( k−1→∏

j=0

X j Y−1j

)−∗( k→∏j=0

X j Y−1j

)∗

for all k ∈N. Thus, by induction, we obtain

m→∏j=0

B j A j =( m→∏

j=0

X−1j Y j

)−1

and

m→∏j=0

B j A j =( m→∏

j=0

X j Y−1j

)∗

for all m ∈N0.

Lemma 8.7. Let (Xk)∞k=0 and (Yk)

∞k=0 be sequences of positive Hermitian complex q × q matrices. Let the se-

quences (Ak)∞k=0 and (Bk)

∞k=0 be given by (8.4) and (8.5). For all k ∈ N0 , then Ak and Bk are positive Hermitian

complex q × q matrices. Furthermore,

Xk =⎧⎨⎩

B−10 , if k = 0,

(→∏ k−1

j=0 B j A j)−∗B−1

k (→∏ k−1

j=0 B j A j)−1, if k � 1

and

Yk =( k→∏

j=0

B j A j

)−∗Ak

( k→∏j=0

B j A j

)−1

for all k ∈ N0 .

Proof. Since Xk and Yk are positive Hermitian complex q × q matrices for all k ∈ N0, we can easilysee from (8.4) and (8.5), that Ak and Bk are positive Hermitian complex q × q matrices for all k ∈ N0,

as well. According to (8.5) we have X0 = B−10 . Remark 8.6 yields

→∏ k−1j=0 X−1

j Y j = (→∏ k−1

j=0 B j A j)−1 for all

k ∈ N and→∏ k

j=0 X j Y−1j = (

→∏ kj=0 B j A j)

∗ for all k ∈ N0 which, in view of (8.5) and (8.4) completes theproof. �Proposition 8.8. By (8.2) and (8.3) a well-defined bijection is given on the set of ordered pairs of infinitesequences of positive Hermitian complex q × q matrices, the inverse of which is given by (8.4) and (8.5).

Proof. Combine Lemmas 8.5 and 8.7. �To express the DS-parametrization [(Lk)

∞k=0, (Mk)

∞k=0] of a sequence (s j)

∞j=0 ∈ K>

q,∞,0 in terms ofits right 0-Stieltjes parametrization (Q j)

∞j=0, we first derive representations of the matrices (Mk)

∞k=0


and (Lk)∞k=0 involving the system (dk)

∞k=0 of monic left orthogonal matrix polynomials and the left

system (ck)∞k=0 of matrix polynomials of the second kind with respect to (s j)

∞j=0, respectively. Using

factorization results for the matrix polynomials (dk)∞k=0 and (ck)

∞k=0 obtained in [19], we then get the

desired formulas.

Remark 8.9. Let (s j)∞j=0 ∈ H>

q,∞ . Then from the definition of the set H>q,∞ it is clear that the ma-

trix Hn is positive Hermitian and, in particular, invertible for all n ∈ N0.

Remark 8.10. Let n ∈ N and let (s j)2nj=0 be a sequence of complex q × q matrices. Then, in view of

(1.2) and (2.1), the matrix Hn admits the block representation

Hn =[

Hn−1 yn,2n−1zn,2n−1 s2n

]

and we see from (2.5) that Ln is the Schur complement of Hn−1 in Hn .

Remark 8.11. Let (s j)∞j=0 ∈ H>

q,∞ . Then, for all n ∈ N, we see, in view of Remarks 8.9 and 8.10 anda well-known formula for the inverse of a block matrix, that the matrices Hn−1, Hn , and Ln areinvertible and that the inverse of Hn admits the block representation

H−1n =

[H−1

n−1 + H−1n−1 yn,2n−1L−1

n zn,2n−1 H−1n−1 −H−1

n−1 yn,2n−1L−1n

−L−1n zn,2n−1 H−1

n−1 L−1n

].

For all n ∈N0, let Eq,n :C→ C(n+1)q×q be defined by

Eq,n(z) :=

⎡⎢⎢⎢⎣

z0 Iq

z1 Iq...

zn Iq

⎤⎥⎥⎥⎦ . (8.6)

Notation 8.12. Let (s j)∞j=0 ∈H>

q,∞ . Then let

dk := Yk Eq,k

for all k ∈N0, where

Yk :={

Iq, if k = 0,

[−zk,2k−1 H−1k−1, Iq], if k � 1

for all k ∈N0.

Remark 8.13. Using Remark 8.11, (8.6), and Choque Rivero [17, Proposition 4.4] we see immediatelythat (dk)

∞k=0 is the unique monic left orthogonal system of matrix polynomials with respect to (s j)

∞j=0.

Lemma 8.14. Let (s j)∞j=0 ∈H>

q,∞ . For all k ∈ N and all z ∈C, then

[Eq,k(z)

]∗H−1

k

[Eq,k(z)

]− [Eq,k−1(z)]∗

H−1k−1

[Eq,k−1(z)

]= [dk(z)]∗

L−1k

[dk(z)

].

Proof. Let k ∈ N. In view of Remark 8.11, we have

H−1k − diag

[H−1

k−1,0q×q]=

[−H−1k−1 yk,2k−1

]L−1

k

[−zk,2k−1 H−1k−1, Iq

].

Iq


From Remarks 8.10 and 8.9 we can easily see that H∗k−1 = Hk−1 and z∗

k,2k−1 = yk,2k−1. Taking addi-tionally into account (8.6) and Notation 8.12 we obtain then[

Eq,k(z)]∗

H−1k

[Eq,k(z)

]− [Eq,k−1(z)]∗

H−1k−1

[Eq,k−1(z)

]= [Eq,k(z)

]∗(H−1

k − diag[

H−1k−1,0q×q

])[Eq,k(z)

]= [Eq,k(z)]∗

Y ∗k L−1

k Yk[

Eq,k(z)]

= [dk(z)]∗

L−1k

[dk(z)

]for all z ∈ C. �Lemma 8.15. Let α ∈R and let (s j)

∞j=0 ∈K>

q,∞,α . For all k ∈N, then

dk(α) = (−1)k

k−1←∏j=0

Lα� j L−1j .

Proof. In view of Remark 8.18 and the definition of the set H>q,∞ we have (s j)

∞j=0 ∈ H>

q,∞ . Takinginto account (2.2), (2.3), and Remark 8.18, we obtain furthermore

det H 〈t〉n = det(αHn − Kn) = (−1)(n+1)q det(−αHn + Kn) = (−1)(n+1)q det Hα�n = 0

for all n ∈ N0, where the sequence (t j)∞j=0 is given by t j := αs j − s j+1 for all j ∈ N0. In view of

Notation 8.12, the application of [19, Theorem 4.10] yields then det L j = 0 for all j ∈N0 and

dk(α) =k−1←∏j=0

L〈t〉j L−1

j

for all k ∈N. Because of (1.3) we have furthermore t j = −sα� j for all j ∈N0. Taking into account (2.5),

(2.1), and (2.2), this implies L〈t〉j = −Lα� j for all j ∈ N0 which completes the proof. �

Notation 8.16. Let (s j)∞j=0 ∈H>

q,∞ . Then let ck : C→Cq×q be defined by

ck(z) :={0q×q, if k = 0,

Zk Eq,k−1(z), if k � 1

for all k ∈N0, where

Zk := Yk

[0q×kq

Sk−1

]

for all k ∈N and

Sm :=

⎡⎢⎢⎣

s0 0q×q . . . 0q×q

s1 s0 . . . 0q×q...

.... . .

...

sm sm−1 . . . s0

⎤⎥⎥⎦

for all m ∈N0.

Remark 8.17. From the construction of ck it is immediately clear that ck is the q × q matrix polynomialwhich is (s j)

2kj=0-associated with respect to dk in the sense of Choque Rivero [17, Definition 4.6].

Taking Remark 8.13 into account we see that (ck)∞k=0 is the left system of matrix polynomials of the

second kind with respect to (s j)∞j=0.


Remark 8.18. Let α ∈R and let (s j)∞j=0 ∈K>

q,∞,α . Then from the definition of the set K>q,∞,α it is clear

that the matrices Hn and Hα�n are positive Hermitian and, in particular, invertible for all n ∈ N0.

Lemma 8.19. Let α ∈R and let (s j)∞j=0 ∈K>

q,∞,α . For all k ∈ N, then

y∗0,k H−1

α�k y0,k − y∗0,k−1 H−1

α�k−1 y0,k−1

= [ck+1(α) + Lα�k L−1k ck(α)

]∗L−1α�k

[ck+1(α) + Lα�k L−1

k ck(α)].

Proof. Let k ∈ N. In view of Remark 8.18 we have (sα� j)∞j=0 ∈ H>

q,∞ . As in the proof of Lemma 8.14we obtain then

H−1α�k − diag

[H−1

α�k−1,0q×q]=

[−H−1α�k−1 yα�k,2k−1

Iq

]L−1α�k

[−zα�k,2k−1 H−1α�k−1, Iq

]and furthermore H∗

α�k−1 = Hα�k−1 and z∗α�k,2k−1 = yα�k,2k−1. Taking additionally into account (2.1)

and Notation 8.16 we get

y∗0,k H−1

α�k y0,k − y∗0,k−1 H−1

α�k−1 y0,k−1 = y∗0,k

(H−1

α�k − diag[

H−1α�k−1,0q×q

])y0,k

= y∗0,kY ∗

α�k L−1α�kYα�k y0,k,

where Yα�k := [−zα�k,2k−1 H−1α�k−1, Iq]. For all complex q × q matrix polynomials P with degree at

most k + 1 let

S(P ) :=k+1∑j=0

A js j,

where (A j)k+1j=0 is the unique sequence of complex q × q matrices such that P admits the representa-

tion P (z) =∑k+1j=0 z j A j for all z ∈ C. In view of (8.6), then

Yα�k y0,k = S(dα�k),

where dα�k := Yα�k Eq,k . According to [19, Proposition 6.8] we have det Lk = 0 and

(z − α)dα�k(z) = dk+1(z) + Lα�k L−1k dk(z)

for all z ∈C. Because of Remark 8.18 the sequence (s j)∞j=0 belongs to H>

q,∞ . Taking into account (2.2),(2.3), and Remark 8.18, we obtain furthermore

det H 〈t〉k−1 = det(αHk−1 − Kk−1) = (−1)kq det(−αHk−1 + Kk−1)

= (−1)kq det Hα�k−1 = 0,

where the sequence (t j)∞j=0 is given by t j := αs j − s j+1 for all j ∈ N0. In view of Notation 8.12, the

application of [19, Theorem 4.9] yields then

dk+1(α) = L〈t〉k L−1

k dk(α).

Because of (1.3) we have t j = −sα� j for all j ∈ N0. Taking into account (2.5), (2.1), and (2.2), this

implies L〈t〉k = −Lα�k . Thus,

(α − z)dα�k(z) = [L〈t〉k L−1

k dk(α) − L〈t〉k L−1

k dk(α)]− [dk+1(z) + Lα�k L−1

k dk(z)]

= fk+1(α) − fk+1(z)

holds true for all z ∈C, where fk+1 :C →Cq×q is defined by

fk+1(z) := dk+1(z) + Lα�k L−1dk(z).
k


Using [19, Lemma 6.5 and Remark 8.2] we can, in view of Notation 8.16, then conclude that

S(dα�k) = ck+1(α) + Lα�k L−1k ck(α)

holds true which completes the proof. �Lemma 8.20. Let α ∈ R and let (s j)

∞j=0 ∈K>

q,∞,α . For all k ∈ Z2,∞ then

ck(α) + Lα�k−1L−1k−1ck−1(α) = (−1)k−1Lk−1

k−2←∏j=0

L−1α� j L j.

Proof. In view of Remark 8.18 and the definition of the set H>q,∞ we have (s j)

∞j=0 ∈ H>

q,∞ . Taking(2.2), (2.3), and Remark 8.18 into account, we obtain furthermore

det H 〈t〉n = det(αHn − Kn) = (−1)(n+1)q det(−αHn + Kn) = (−1)(n+1)q det Hα�n = 0

for all n ∈ N0, where the sequence (t j)∞j=0 is given by t j := αs j − s j+1 for all j ∈ N0. In view of

Notation 8.16, the application of [19, Proposition 8.11] yields then det L〈t〉j = 0 for all j ∈ N0, det L j = 0

for all j ∈N, and

ck(α) − L〈t〉k−1L−1

k−1ck−1(α) = Lk−1

k−2←∏j=0

(L〈t〉

j

)−1L j

for all k ∈ Z2,∞ . Because of (1.3) we have furthermore t j = −sα� j for all j ∈ N0. Taking into account

(2.5), (2.1), and (2.2), this implies L〈t〉j = −Lα� j for all j ∈ N0 which completes the proof. �

Now we are going to express the DS-parametrization of a sequence (s j)∞j=0 ∈ K>

q,∞,0 in terms ofits right 0-Stieltjes parametrization.


q,∞,0 with right 0-Stieltjes parametrization (Q j)∞j=0. Then, in view of

part (d) of Theorem 2.3, the matrices Q j are positive Hermitian and, in particular, invertible for allj ∈ N0.

Theorem 8.22. Let (s j)∞j=0 ∈ K>

q,∞,0 with right 0-Stieltjes parametrization (Q j)∞j=0 and DS-parametrization

[(Lk)∞k=0, (Mk)

∞k=0]. Then

Lk =( k→∏

j=0

Q 2 j Q −12 j+1

)Q 2k+1

( k→∏j=0

Q 2 j Q −12 j+1

)∗(8.7)

and

Mk =⎧⎨⎩

Q −10 , if k = 0,

(→∏k−1

j=0 Q −12 j Q 2 j+1)Q −1

2k (→∏k−1

j=0 Q −12 j Q 2 j+1)

∗, if k � 1(8.8)

for all k ∈ N0 .

Proof. Using Remark 8.21, Definitions 2.2 and 8.2, and (1.3), we obtain

Q −10 = L−1

0 = s−10 = M0


and (Q 0 Q −1

1

)Q 1(

Q 0 Q −11

)∗ = Q 0 Q −11 Q 1 Q −∗

1 Q ∗0 = Q 0 Q −1

1 Q 0 = L0L−10�0L0 = s0s−1

1 s0 = L0.

Now let k ∈ N. From (8.6) and (8.1) we see then that Eq,k(0) = vq,k and Eq,k−1(0) = vq,k−1 hold truewhich, in view of Definition 8.2 and Lemma 8.14, implies

Mk = [Eq,k(0)]∗

H−1k

[Eq,k(0)

]− [Eq,k−1(0)]∗

H−1k−1

[Eq,k−1(0)

]= [dk(0)]∗

L−1k

[dk(0)

].

Using Lemma 8.15 and Definition 2.2 we get

dk(0) = (−1)k

k−1←∏j=0

L0� j L−1j = (−1)k

k−1←∏j=0

Q 2 j+1 Q −12 j .

Taking additionally into account Remark 8.21, we can thus conclude that

[dk(0)

]∗ = (−1)k

k−1→∏j=0

Q −∗2 j Q ∗

2 j+1 = (−1)k

k−1→∏j=0

Q −12 j Q 2 j+1

holds true. According to Definition 2.2 we have furthermore Q 2k = Lk . Thus, (8.8) is proved. From(2.2), (1.3), and (2.3) we see that H0�k = Kk and H0�k−1 = Kk−1 hold true which, in view of Defini-tion 8.2 and Lemma 8.19, implies

Lk = y∗0,k H−1

0�k y0,k − y∗0,k−1 H−1

0�k−1 y0,k−1

= [ck+1(0) + L0�k L−1k ck(0)

]∗L−1

0�k

[ck+1(0) + L0�k L−1

k ck(0)].

Using Lemma 8.20 and Definition 2.2 we get

ck+1(0) + L0�k L−1k ck(0) = (−1)k Lk

k−1←∏j=0

L−10� j L j = (−1)k Q 2k

k−1←∏j=0

Q −12 j+1 Q 2 j .

Taking additionally into account Remark 8.21, we can thus conclude that

[ck+1(0) + L0�k L−1

k ck(0)]∗ = (−1)k

( k−1→∏j=0

Q ∗2 j Q −∗

2 j+1

)Q ∗

2k

= (−1)k

( k−1→∏j=0

Q 2 j Q −12 j+1

)Q 2k Q −1

2k+1 Q 2k+1

= (−1)k

( k→∏j=0

Q 2 j Q −12 j+1

)Q 2k+1

holds true. According to Definition 2.2 we have furthermore Q 2k+1 = L0�k . Using Remark 8.21, thus(8.7) follows. �

Conversely, now we express the right 0-Stieltjes parametrization of a sequence (s j)∞j=0 ∈K>

q,∞,0 interms of its DS-parametrization.



∞k=0]. Then, in view of [26,

Theorem 7], the matrices Lk and Mk are positive Hermitian and, in particular, invertible for all k ∈ N0.


Theorem 8.24. Let (s j)∞j=0 ∈ K>

q,∞,0 with right 0-Stieltjes parametrization (Q j)∞j=0 and DS-parametrization

[(Lk)∞k=0, (Mk)

∞k=0]. Then

Q 2k =⎧⎨⎩

M−10 , if k = 0,

(→∏ k−1

j=0M jL j)−∗M−1

k (→∏ k−1

j=0M jL j)−1, if k � 1

(8.9)

and

Q 2k+1 =( k→∏

j=0

M jL j

)−∗Lk

( k→∏j=0

M jL j

)−1

(8.10)

for all k ∈ N0 .

Proof. Use Remark 8.21, Theorem 8.22, and Lemma 8.7. �Corollary 8.25. Let (s j)

∞j=0 ∈K>


∞k=0]. Then

Lk =( k−1→∏

j=0

M−1j L−1

j

)M−1

k

( k−1←∏j=0

L−1j M−1

j

)

and

L0�k =( k−1→∏

j=0

M−1j L−1

j

)M−1

k L−1k M−1

k

( k−1←∏j=0

L−1j M−1

j

)

for all k ∈ N.

Proof. Use Definition 2.2, Theorem 8.24, and Remark 8.23. �Corollary 8.25 was obtained in [18, Remark 4.4, formulas (4.38) and (4.39)] in a different way.Now we are able to express recursively a sequence (s j)

∞j=0 ∈ K>

q,∞,0 in terms of its DS-parametrization.

Proposition 8.26. Let (s j)∞j=0 ∈ K>


∞k=0]. Then s0 = M−1

0 and

s1 = (M0L0)−∗L0(M0L0)

−1 . Furthermore,

s2k = zk,2k−1 H−1k−1 yk,2k−1 +

( k−1→∏j=0

M jL j

)−∗M−1

k

( k−1→∏j=0

M jL j

)−1

and

s2k+1 = zk+1,2k K −1k−1 yk+1,2k +

( k→∏j=0

M jL j

)−∗Lk

( k→∏j=0

M jL j

)−1

for all k ∈ N.


Proof. From Definition 8.2 we obtain the assertions for s0 and s1. Let k ∈ N. Denote by (Q j)∞j=0 the

right 0-Stieltjes parametrization of (s j)∞j=0. Then from (2.5) and Definition 8.2 we get

s2k = zk,2k−1 H†k−1 yk,2k−1 + Q 2k. (8.11)

Taking into account that (s0� j)∞j=0 = (s j+1)

∞j=0 and applying (2.5) to the sequence (t j)

∞j=0 := (s0� j)

∞j=0

we obtain from (2.5) and Definition 8.2 analogously to (8.11) the relation

s2k+1 = zk+1,2k K †k−1 yk+1,2k + Q 2k+1. (8.12)

Now the combination of (8.11) with (8.2) and (8.12) with (8.3) yields the remaining assertions. �Proposition 8.26 shows that, in view of (2.1), (1.2), and (2.3), the sequence (s j)

∞j=0 can be re-

cursively reconstructed from the sequences (Lk)∞k=0 and (Mk)

∞k=0. In particular, the sequence (s j)

∞j=0

is uniquely determined by its DS-parametrization. This can also be seen from Remark 8.23, Theo-rem 8.24, Proposition 8.8, and the fact, that in view of part (d) of Theorem 2.3 the right 0-Stieltjesparametrization gives rise to a one-to-one correspondence between the set K>

q,∞,0 and the set ofinfinite sequences of positive Hermitian complex q × q matrices.

Conversely, inspired by Proposition 8.26 we are led to a possibility of constructing recursivelya sequence from K>

q,∞,0 on the basis of a pair [(Lk)∞k=0, (Mk)

∞k=0] of positive Hermitian complex

q × q matrices.

Proposition 8.27. Let (Lk)∞k=0 and (Mk)


the sequence (s j)∞j=0 be recursively defined by

s2k :=⎧⎨⎩

M−10 , if k = 0,

zk,2k−1 H−1k−1 yk,2k−1 + (

→∏ k−1j=0M jL j)

−∗M−1k (

→∏ k−1j=0M jL j)

−1, if k � 1

and

s2k+1 :=⎧⎨⎩

(M0L0)−∗L0(M0L0)

−1, if k = 0,

zk+1,2k K −1k−1 yk+1,2k + (

→∏ kj=0M jL j)

−∗Lk(→∏ k

j=0M jL j)−1, if k � 1

for all k ∈N0 . Then (s j)∞j=0 ∈K>

q,∞,0 and [(Lk)∞k=0, (Mk)

∞k=0] is the DS-parametrization of (s j)

∞j=0 .

Proof. Denote by (Q j)∞j=0 the right 0-Stieltjes parametrization of (s j)

∞j=0. Using Definition 2.2, (2.5),

(1.3), (2.1), (2.2), and (2.3), we get (8.9) and (8.10) for all k ∈ N0. Lemma 8.5 then shows that, forall j ∈ N0, the matrix Q j is positive Hermitian, which, in view of part (d) of Theorem 2.3, im-plies (s j)

∞j=0 ∈ K>

q,∞,0. From Lemma 8.5 we see furthermore that (8.7) and (8.8) hold true for all

k ∈ N0. Taking additionally Theorem 8.22 into account, this shows that [(Lk)∞k=0, (Mk)

∞k=0] is the DS-

parametrization of (s j)∞j=0. �

Proposition 8.28. Let [(Lk)∞k=0, (Mk)

∞k=0] be an ordered pair of sequences of matrices taken from C

q×q> . Then:

(a) Let

Q 2k :=⎧⎨⎩

M−10 , if k = 0,

(→∏ k−1

j=0M jL j)−∗M−1

k (→∏ k−1

j=0M jL j)−1, if k � 1

and


Q 2k+1 :=( k→∏

j=0

M jL j

)−∗Lk

( k→∏j=0

M jL j

)−1

for all k ∈ N0 . Then (Q j)∞j=0 is a sequence of matrices from C

q×q> .

(b) There is a unique sequence (s j)∞j=0 ∈ K>


∞k=0], namely the

unique sequence (s j)∞j=0 with right 0-Stieltjes parametrization (Q j)

∞j=0 .

Proof. (a) This follows from Lemma 8.5.(b) Let the sequence (s j)

∞j=0 be defined recursively as in Proposition 8.27 using the sequences

(Lk)∞k=0 and (Mk)

∞k=0. In view of Propositions 8.27 and 8.26, then (s j)

∞j=0 is the unique sequence

from K>q,∞,0 with DS-parametrization [(Lk)

∞k=0, (Mk)

∞k=0]. Taking into account the definition of the

sequences (Q j)∞j=0 and (s j)

∞j=0, (1.3), (2.1), (2.2), (2.3), (2.5), and Definition 2.2, we see furthermore

that (Q j)∞j=0 is the right 0-Stieltjes parametrization of (s j)

∞j=0 which means that (s j)

∞j=0 is the unique

sequence with right 0-Stieltjes parametrization (Q j)∞j=0. �

Now we are going to demonstrate that in the scalar case the DS-parametrization of a 0-Stieltjesright-sided positive definite sequence coincides with the classical parameters used by Stieltjes [50,51]to formulate his indeterminacy criterion. Before doing this we mention that M.G. Kreın was able tofind a mechanical interpretation for Stieltjes’ investigations on continued fractions (see Gantmacherand Kreın [36, Anhang 2] or Akhiezer [4, appendix]). Against to the background of his mechanicalinterpretation M.G. Kreın divided Stieltjes’ original parameters into two groups which play the rolesof lengths and masses, respectively. Now we want to recall the concrete definition of these parameters(see Kreın and Nudel’man [44, Chapter V, formula (6.1)]).

Definition 8.29. Let (s j)∞j=0 ∈K>

1,∞,0 and let �n := det Hn and �(1)n := det Kn for all n ∈N0. Let

lk := �2k

�(1)

k �(1)

k−1

and mk := (�(1)

k−1)2

�k�k−1

for all k ∈ N0, where �−1 := 1 and �(1)−1 := 1. Then the ordered pair [(lk)∞k=0, (mk)

∞k=0] is called the

Krein–Stieltjes parametrization (shortly KS-parametrization) of (s j)∞j=0.

Now we verify that for a sequence (s j)∞j=0 ∈ K>

1,∞,0 its DS-parametrization and its KS-parametri-zation coincide.

Proposition 8.30. Let (s j)∞j=0 ∈ K>

1,∞,0 with KS-parametrization [(lk)∞k=0, (mk)∞k=0] and DS-parametrization

[(Lk)∞k=0, (Mk)

∞k=0]. Then [(lk)∞k=0, (mk)

∞k=0] = [(Lk)

∞k=0, (Mk)

∞k=0].

Proof. Denote by (Q j)∞j=0 the right 0-Stieltjes parametrization of (s j)

∞j=0. In view of Definition 8.29,

(2.3), (1.3), Lemma 4.13, and Remark 8.21, we have

�n = det Hn =n∏

k=0

det Q 2k =n∏

k=0

Q 2k > 0

and

�(1)n = det Kn = det H0�n =

n∏k=0

det Q 2k+1 =n∏

k=0

Q 2k+1 > 0

for all n ∈ N0. Using Definition 8.29, Remark 8.21, and Theorem 8.22, we obtain then


m0 = (�(1)−1)

2

�0�−1= 12

Q 0 · 1= Q −1

0 = M0

and

l0 = �20

�(1)0 �

(1)−1

= Q 20

Q 1 · 1= (Q 0 Q −1

1

)Q 1(

Q 0 Q −11

)∗ = L0

and furthermore

mk = (�(1)

k−1)2

�k�k−1=(

�(1)

k−1

�k−1

)2�k−1

�k=( k−1→∏

j=0

Q −12 j Q 2 j+1

)Q −1

2k

( k−1→∏j=0

Q −12 j Q 2 j+1

)∗= Mk

and

lk = �2k

�(1)

k �(1)

k−1

=(

�k

�(1)

k

)2 �(1)

k

�(1)

k−1

=( k→∏

j=0

Q 2 j Q −12 j+1

)Q 2k+1

( k→∏j=0

Q 2 j Q −12 j+1

)∗= Lk

for all k ∈N. �Appendix A. On the Moore–Penrose inverse of a complex matrix

Let A ∈ Cp×q . Then (see, e.g., [22, Proposition 1.1.1]) there exists a unique matrix G ∈ C

q×p whichsatisfies the four equations AG A = A, G AG = G , (AG)∗ = AG and (G A)∗ = G A. This matrix G is calledthe Moore–Penrose inverse of A and is denoted by A†.

Remark A.1. Let A ∈ Cp×q and r, s ∈ N. Then:

(a) Let B ∈Cr×q . Then N (A) ⊆N (B) if and only if B A† A = B .

(b) Let C ∈Cp×s . Then R(C) ⊆R(A) if and only if A A†C = C .

Remark A.2. Let A, B ∈Cq×q with A � B � 0q×q . Then

Iq � A A† =√

A† A√

A† �√

A† B√

A† = (√B√

A†)∗(√

B√

A†),

which implies Iq � (√

B√

A† )(√

B√

A† )∗ , and hence

B† �√

B†(√

B√

A†)(√

B√

A†)∗√

B† = B† B A† B B†. �Appendix B. Some transformations of non-negative Hermitian measures

In this appendix, we summarize some facts on non-negative Hermitian q × q measures on mea-surable spaces. What concerns the integration theory of measurable matrix functions with respectto a non-negative Hermitian q × q measure, we refer to Rosenberg [47] (see also [22, Section 2.2]). If(Ω,A) is a measurable space, then let Mq

�(Ω,A) be the set of all non-negative Hermitian q × q mea-sures on (Ω,A). The following result describes the behavior of non-negative Hermitian q × q mea-sures under measurable transformations. Let BC be the set of all Borel subsets of C.

Proposition B.1. Let (Ω,A) and (Ω, A) be measurable spaces and μ ∈ Mq�(Ω,A). Further, let

T : Ω → Ω be an A-A-measurable mapping. Then T (μ) : A → Cq×q defined by [T (μ)]( A) := μ(T −1( A))

is a non-negative Hermitian measure which belongs to Mq�(Ω, A). Furthermore, if f : Ω → C is an


A-BC-measurable mapping, then f ∈ L1(Ω, A, T (μ);C) if and only if f ◦ T ∈ L1(Ω,A,μ;C). If f be-longs to L1(Ω, A, T (μ);C), then∫

A

f d[T (μ)

]=∫

T −1( A)

( f ◦ T )dμ for all A ∈ A. (B.1)

Proof. Obviously, μ := T (μ) belongs to Mq�(Ω, A). We consider an A-BC-measurable mapping

f : Ω → C. According to [34, Lemma B.1], the condition f ∈ L1(Ω, A, T (μ);C) holds true if andonly if:

(I) For all v ∈Cq , the function f belongs to L1(Ω, A, v∗[T (μ)]v;C).

Since T (v∗μv) = v∗[T (μ)]v holds true for all v ∈ Cq , from [20, Proposition 2.6.5] we see that (I) is

equivalent to:

(II) For all v ∈Cq , the function f ◦ T belongs to L1(Ω,A, v∗μv;C).

Using [34, Lemma B.1], we get that (II) is fulfilled if and only if f ◦ T ∈L1(Ω,A,μ;C) is true.For all j ∈ Z1,q , let e j := [δ j1, δ j2, . . . , δ jq]∗ , where δ jk := 1 for j = k and δ jk := 0 for j = k. Further-

more, for all j ∈ Z1,q and each k ∈ Z1,q , let v(0)

jk := 12 (e j + ek), v(1)

jk := 12 (e j − iek), v(2)

jk := 12 (e j − ek),

and v(3)

jk := 12 (e j + iek). We consider an arbitrary f ∈ L1(Ω, A, T (μ);C) and an arbitrary A ∈ A. For

every choice of j and k in Z1,q , from [22, Remark 1.1.1], [34, Lemma B.3], T (v∗μv) = v∗[T (μ)]v forall v ∈C

q , and [20, Proposition 2.6.5] we then get

e∗j

( ∫A

f d[T (μ)

])ek =

3∑l=0

il(

v(l)jk

)∗( ∫A

f d[T (μ)

])v(l)

jk

=3∑

l=0

il∫A

f d((

v(l)jk

)∗[T (μ)

]v(l)

jk

)=3∑

l=0

il∫A

f d[T((

v(l)jk

)∗μv(l)

jk

)]

=3∑

l=0

il∫

T −1( A)

( f ◦ T )d[(

v(l)jk

)∗μv(l)

jk

]

=3∑

l=0

il(

v(l)jk

)∗[ ∫T −1( A)

( f ◦ T )dμ

]v(l)

jk

= e∗j

[ ∫T −1( A)

( f ◦ T )dμ

]ek.

Consequently, (B.1) is also proved. �The mapping T (μ) constructed in Proposition B.1 is called the image measure of μ under T .In this appendix, we also study special Borel-measurable transformations of the real axis, which

turn out to be closely connected via power moments with the special transformations of sequences ofmatrices discussed above. We start with affine mappings. For all subsets Ω of C and all γ , δ ∈ C, letδΩ + γ := {δω + γ | ω ∈ Ω}. If Ω ∈ BR \ {∅}, μ ∈ Mq

�(Ω), γ , δ ∈ R, Ω ∈ BR with δΩ + γ ⊆ Ω and

Tδ,γ : Ω → Ω is defined by Tδ,γ (ω) := δω + γ , then we call the image measure μ( ,γ ,δ) of μ underTδ,γ the (δ, γ )-binomial transform of μ.


The next result shows that the binomial transform of measures is via power moments intimatelyrelated with the binomial transformation of sequences of matrices.

Lemma B.2. Let Ω ∈ BR \ {∅}, κ ∈ N0 ∪ {+∞}, (s j)κj=0 be a sequence of complex q × q matrices, μ ∈

Mq�[Ω; (s j)

κj=0,=], γ , δ ∈ R and Ω ∈ BR with δΩ + γ ⊆ Ω . Then μ( ,γ ,δ) ∈ Mq

�[Ω; (s( ,γ ,δ)

j )κj=0,=],where (s( ,γ ,δ)

j )κj=0 is defined in (4.1).

Proof. For all j ∈ Z0,κ , we have

s(μ( ,γ ,δ))

j =∫Ω

t j μ( ,γ ,δ)(dt) =∫Ω

t j [Tδ,γ (μ)](dt) =

∫T −1

δ,γ (Ω)

(δt + γ ) j [μ(dt)]

=j∑

l=0

(j

l

)δlγ j−l

∫Ω

t jμ(dt) =j∑

l=0

(j

l

)δlγ j−ls(μ)

l =j∑

l=0

(j

l

)δlγ j−lsl

= s( ,γ ,δ)

j ,

where the 1st and 5th equation is due to (1.1), the 3rd equation is due to Proposition B.1, the6th equation is due to μ ∈Mq

�[Ω; (s j)κj=0,=] and the 7th equation is due to (4.1). �

Now we consider measures on a fixed interval [α,+∞). We construct several operations of pro-ducing measures via distinguished densities.

Remark B.3. Let α ∈ R and σ ∈ Mq�,1([α,+∞)). Then, in view of the choice of σ , the integral∫

B(t −α)σ (dt) exists for all B ∈B[α,+∞) and μ :B[α,+∞) →Cq×q� defined by μ(B) := ∫B(t −α)σ (dt)

belongs to Mq�([α,+∞)).

If α ∈ R and σ ∈ Mq�,1([α,+∞)), then, in view of Remark B.3, we call the measure

σC :B[α,+∞) →Cq×q� defined by

σC(B) :=∫B

(t − α)σ (dt)

the Christoffel transform of σ . From the definition of σC we immediately see that σC({α}) = 0q×q .

Remark B.4. Let α ∈ R and σ ∈ Mq�,1([α,+∞)). Let σ := σ − [σ({α})]δα , where δα is the Dirac

measure on B[α,+∞) with unit mass in α. Then it is immediately checked that σ ∈ Mq�,1([α,+∞))

and σC = σC.

We verify now that the power moments of the Christoffel transform of a measure σ are generatedfrom the sequence of power moments of σ by right-sided α-shifting.

Lemma B.5. Let α ∈ R, κ ∈ N ∪ {+∞}, (s j)κj=0 be a sequence of complex q × q matrices and σ ∈

Mq�[[α,+∞); (s j)

κj=0,=]. Then σC belongs to the set Mq

�[[α,+∞); (sα� j)κ−1j=0 ,=], where (sα� j)

κ−1j=0 is

defined in (1.3).


Proof. By assumption we have∫[α,+∞)

t j σ(dt) = sk for all j ∈ Z0,κ . Thus, taking (1.3) into account,for all j ∈ Z0,κ , we get then∫

[α,+∞)

t j σC(dt) =∫

[α,+∞)

t j(t − α)σ (dt) = s j+1 − αs j = sα� j . �

Now we are going to study a special transformation for measures, which will turn out in somesense as converse to the Christoffel transform. In order to realize this aim, we introduce a particularsubclass of Mq

�([α,+∞)).

Notation B.6. Let α ∈ R. Then Mq�,−1([α,+∞)) denotes the set of all σ ∈ Mq

�([α,+∞)) for which

σ({α}) = 0q×q and the integral∫(α,+∞)

1t−α σ (dt) exists.

Remark B.7. Let α ∈ R, σ ∈Mq�,−1([α,+∞)) and A ∈C

q×q� . Furthermore, let δα be the Dirac measure

on B[α,+∞) with unit mass in α. Then, taking Notation B.6 into account, it is immediately checkedthat the mapping σG;A : B[α,+∞) →C

q×q defined by

σG;A(B) :=∫

B\{α}

1

t − ασ(dt) + (δα A)(B)

belongs to Mq�([α,+∞)) and satisfies the relations σG;A({α}) = A and σG;A([α,+∞)) =∫

(α,+∞)1

t−α σ (dt) + A. The mapping σG;A is called the Geronimus transform of σ with parameter A.In particular, σG;0q×q is called the Geronimus transform of σ .

Now we state some interrelations between Christoffel transform and Geronimus transform.

Proposition B.8. Let α ∈ R and σ ∈ Mq�,1([α,+∞)). Then the Christoffel transform σC of σ belongs to

Mq�,−1([α,+∞)). Furthermore, the Geronimus transform (σC)G;σ({α}) of σC with parameter σ({α}) fulfills

(σC)G;σ({α}) = σ .

Proof. Remark B.3 provides σC({α}) = 0q×q . Thus, from the choice of σ and the definition of σC wesee that σC ∈Mq

�,−1([α,+∞)). Let B ∈ B[α,+∞) . Then from the definition of σC we get∫B\{α}

1

t − ασC(dt) =

∫B\{α}

1

t − α(t − α)σ (dt) = σ

(B \ {α})

={

σ(B \ {α}), if α ∈ B,

σ (B), if α /∈ B.

Furthermore, [δα(B)][σ({α})] = σ({α}) if α ∈ B and [δα(B)][σ({α})] = 0q×q if α /∈ B . Thus,

(σC)G;σ ({α})(B) =∫

B\{α}

1

t − ασC(dt) + [δα(B)

][σ({α})]= σ(B). �

Proposition B.9. Let α ∈ R, σ ∈ Mq�,−1([α,+∞)) and A ∈ C

q×q� . Then σG;A ∈ Mq

�,1([α,+∞)) and(σG;A)C = σ − δασ ({α}).

Proof. From the choice of σ and the definition of σG;A we see that σG;A ∈Mq�,1([α,+∞)). For each

B ∈ B[α,+∞) , we get from the definitions of the corresponding transforms then


(σG;A)C(B) =∫B

(t − α)σG;A(dt) =∫

B\{α}(t − α)σG;A(dt)

=∫

B\{α}(t − α)

1

t − ασ(dt) = σ

(B \ {α})= σ(B) − [δα(B)

][σ({α})]. �

We obtain now that the power moments of a measure and the power moments of its Geronimustransform are related via the operation of inverse right-sided α-shifting of sequences, which wasintroduced in (6.1).

Proposition B.10. Let α ∈ R, κ ∈ N0 ∪ {+∞}, (s j)κj=0 be a sequence from C

q×q, and A ∈ Cq×q� . Fur-

ther let σ ∈ Mq�,−1([α,+∞)) ∩ Mq

�[[α,+∞); (s j)κj=0,=] and let M := ∫

(α,+∞)1

t−α σ (dt). Then σG;A ∈Mq

�[[α,+∞); (s〈α,M]j + α j A)κ+1

j=0 ,=], where (s〈α,M]j )κ+1

j=0 is defined in (6.1).

Proof. In view of (1.1), Remark B.7 and (6.1), we obtain

s(σG;A)

0 = σG;A([α,+∞)

)= M + A = s〈α,M]j + α0 A. (B.2)

The choice of σ implies σ ∈ Mq�[[α,+∞); (s j)

κj=0,=] and σ({α}) = 0q×q . Thus, for all l ∈ Z0,κ , we

get

sl = s(σ )

l =∫

[α,+∞)

tl σ(dt) =∫

(α,+∞)

tl σ(dt). (B.3)

Let j ∈ Z1,κ+1. Further, let t ∈ (α,+∞). In the case α = 0, we get

t j

t − α= α j

t − α+

j−1∑l=0

α j−1−ltl. (B.4)

Formula (B.4) holds true in the case α = 0 as well. Because of σ ∈Mq�,−1([α,+∞))∩Mq

�[[α,+∞);(s j)

κj=0,=] and (B.4), the integral

∫(α,+∞)

t j

t−α σ (dt) exists. Since (B.4), (B.3) and (6.1) hold true, therelation ∫

(α,+∞)

t j

t − ασ(dt) =

∫(α,+∞)

α j

t − ασ(dt) +

j−1∑l=0

α j−1−l∫

(α,+∞)

tl σ(dt)

= α j M +j−1∑l=0

α j−1−lsl = s〈α,M]j

follows. Thus, in view of (1.1) and Remark B.7, we see that

s(σG;A)

j =∫

(α,+∞)

t j

t − ασ(dt) +

∫{α}

t j (δα A)(dt) = s〈α,M]j + α j A. (B.5)

Using (B.2) and (B.5), we get σG;A ∈Mq�[[α,+∞); (s〈α,M]

j + α j A)κ+1j=0 ,=]. �

We indicate now that the investigations of Section 7 can be interpreted in the language of mea-sures by additive perturbations generated by point masses. The next consideration contains a powermoments interpretation of the sequence of matrices, which arises in the result of the perturbation ofthe original sequence of matrices in Section 7.


Remark B.11. Let α, ζ ∈ R, σ ∈ Mq�([α,+∞)) and A ∈ C

q×q� . Let β := min{α, ζ } and δζ be the Dirac

measure defined on B[β,+∞) with unit mass at ζ . We then easily see that μ : B[β,+∞) → Cq×q� de-

fined by μ(B) := σ(B ∩ [α,+∞)) + [δζ (B)]A belongs to Mq�([β,+∞)).

If α, ζ ∈ R, σ ∈ Mq�([α,+∞)), A ∈ C

q×q� , β := min{α, ζ } and δζ denotes the Dirac measure de-

fined on B[β,+∞) with unit mass at ζ , then, in view of Remark B.11, we call σU;ζ,A :B[β,+∞) → Cq×q�

defined by

σU;ζ,A(B) := σ(

B ∩ [α,+∞))+ [δζ (B)

]A

the Uvarov transform of σ with parameters ζ and A.

Remark B.12. Let α, ζ ∈ R, κ ∈ N0 ∪ {+∞}, σ ∈ Mq�[[α,+∞); (s j)

κj=0,=] and A ∈ C

q×q� . Let β :=

min{α, ζ }. In view of [34, Lemma B.3], then σU;ζ,A ∈Mq�[[β,+∞); (s j + ζ j A)κj=0,=].

At the end of this appendix, we add some historical remarks concerning the scalar case. Thetransformations of measures studied here have analogues as transformation of quasi-definite linearfunctionals L defined on the linear space of polynomials PR with real coefficients. If L is a linearfunctional on PR , then for each k ∈ N0 we set sk := L(xk) and sk is said to be the k-th moment ofL. The functional L is then called quasi-definite (resp. positive definite) if for each n ∈ N0 the Hankelmatrix Hn := [s j+k]n

j,k=0 is non-singular (resp. positive definite). If L is such a quasi-definite linearfunctional on PR , then there exists a unique sequence of monic orthogonal polynomials with respectto L. Several authors (see, e.g., Bueno and Marcellán [12], Derevyagin and Derkach [21], Yoon [53],Zhedanov [54]) studied special transformations which preserve the class of quasi-definite linear func-tionals on PR . In this way, the names Christoffel, Geronimus, Darboux and others were used forthese transformations. We considered special transformations which preserve positive definite linearfunctionals. However, we preferred to formulate all in the language of measures with finite power mo-ments. Such measures generate obviously positive definite linear functionals on PR . A central problemis to describe how the unique system of monic orthogonal polynomials of the image functional canbe computed in terms of the unique system of monic orthogonal polynomials of the preimage func-tional and the parameters of the concrete transformation. This leads to a whole variety of challengingquestions including lower-upper triangular factorization of Jacobi matrices (see, e.g., [3,12,21,37,39,45,46,49,52–54]).

References

[1] V.M. Adamyan, I.M. Tkachenko, Solution of the Stieltjes truncated matrix moment problem, Opuscula Math. 25 (2005)5–24.

[2] V.M. Adamyan, I.M. Tkachenko, General solution of the Stieltjes truncated matrix moment problem, in: Operator Theoryand Indefinite Inner Product Spaces, in: Oper. Theory Adv. Appl., vol. 163, Birkhäuser, Basel, 2006, pp. 1–22.

[3] M. Adler, P. van Moerbeke, Darboux transforms on band matrices, weights, and associated polynomials, Int. Math. Res. Not.2001 (2001) 935–984.

[4] N.I. Akhiezer, The Classical Moment Problem and Some Related Questions in Analysis, Hafner Publishing Co., New York,1965, translated by N. Kemmer.

[5] T. Andô, Truncated moment problems for operators, Acta Sci. Math. (Szeged) 31 (1970) 319–334.[6] G. Bennett, Hausdorff means and moment sequences, Positivity 15 (2011) 17–48.[7] V.A. Bolotnikov, Descriptions of solutions of a degenerate moment problem on the axis and the halfaxis, Teor. Funktsiı

Funktsional. Anal. i Prilozhen. 50 (1988) 25–31.[8] V.A. Bolotnikov, Degenerate Stieltjes moment problem and associated J -inner polynomials, Z. Anal. Anwend. 14 (1995)

441–468.[9] V.A. Bolotnikov, On degenerate Hamburger moment problem and extensions of nonnegative Hankel block matrices, Integral

Equations Operator Theory 25 (1996) 253–276.[10] V.A. Bolotnikov, On a general moment problem on the half axis, Linear Algebra Appl. 255 (1997) 57–112.[11] V.A. Bolotnikov, L.A. Sakhnovich, On an operator approach to interpolation problems for Stieltjes functions, Integral Equa-

tions Operator Theory 35 (1999) 423–470.

http://refhub.elsevier.com/S0024-3795(13)00621-6/bib4D5232313535363435s1




















[12] M.I. Bueno, F. Marcellán, Darboux transformation and perturbation of linear functionals, Linear Algebra Appl. 384 (2004)215–242.

[13] G.N. Chen, Y.J. Hu, The truncated Hamburger matrix moment problems in the nondegenerate and degenerate cases, andmatrix continued fractions, Linear Algebra Appl. 277 (1998) 199–236.

[14] G.N. Chen, Y.J. Hu, The Nevanlinna–Pick interpolation problems and power moment problems for matrix-valued functions.III. The infinitely many data case, Linear Algebra Appl. 306 (2000) 59–86.

[15] G.N. Chen, Y.J. Hu, A unified treatment for the matrix Stieltjes moment problem in both nondegenerate and degeneratecases, J. Math. Anal. Appl. 254 (2001) 23–34.

[16] G.N. Chen, X.Q. Li, The Nevanlinna–Pick interpolation problems and power moment problems for matrix-valued functions,Linear Algebra Appl. 288 (1999) 123–148.

[17] A.E. Choque Rivero, The resolvent matrix for the Hausdorff matrix moment problem expressed in terms of orthogonalmatrix polynomials, Complex Anal. Oper. Theory 7 (2013) 927–944.

[18] A.E. Choque Rivero, Stieltjes matrix n-convergent, Linear Algebra Appl. (2013), submitted for publication.[19] A.E. Choque Rivero, C. Mädler, On Hankel-positive definite perturbations of Hankel-positive definite sequences and interre-

lations to orthogonal polynomials, Complex Anal. Oper. Theory (2013), to be reviewed.[20] D.L. Cohn, Measure Theory, Birkhäuser, Boston, MA, 1980.[21] M.S. Derevyagin, V.A. Derkach, Darboux transformations of Jacobi matrices and Padé approximation, Linear Algebra Appl.

435 (2011) 3056–3084.[22] V.K. Dubovoj, B. Fritzsche, B. Kirstein, Matricial Version of the Classical Schur Problem, Teubner-Texte Math. (Teubner Texts

in Mathematics), vol. 129, B.G. Teubner Verlagsgesellschaft mbH, Stuttgart, 1992, with German, French and Russian sum-maries.

[23] H. Dym, On Hermitian block Hankel matrices, matrix polynomials, the Hamburger moment problem, interpolation andmaximum entropy, Integral Equations Operator Theory 12 (1989) 757–812.

[24] Yu.M. Dyukarev, The Stieltjes matrix moment problem, Deposited in VINITI (Moscow) at 22.03.81, No. 2628-81, 1981,Manuscript, 37 pp.

[25] Yu.M. Dyukarev, Multiplicative and additive Stieltjes classes of analytic matrix-valued functions and interpolation problemsconnected with them. II, Teor. Funktsiı Funktsional. Anal. i Prilozhen. (1982) 40–48, 127.

[26] Yu.M. Dyukarev, Indeterminacy criteria for the Stieltjes matrix moment problem, Mat. Zametki 75 (2004) 71–88.[27] Yu.M. Dyukarev, B. Fritzsche, B. Kirstein, C. Mädler, On truncated matricial Stieltjes type moment problems, Complex Anal.

Oper. Theory 4 (2010) 905–951.[28] Yu.M. Dyukarev, B. Fritzsche, B. Kirstein, C. Mädler, H.C. Thiele, On distinguished solutions of truncated matricial Hamburger

moment problems, Complex Anal. Oper. Theory 3 (2009) 759–834.[29] Yu.M. Dyukarev, V.È. Katsnel’son, Multiplicative and additive Stieltjes classes of analytic matrix-valued functions and inter-

polation problems connected with them. I, Teor. Funktsiı Funktsional. Anal. i Prilozhen. (1981) 13–27, 126.[30] Yu.M. Dyukarev, V.È. Katsnel’son, Multiplicative and additive Stieltjes classes of analytic matrix-valued functions, and inter-

polation problems connected with them. III, Teor. Funktsiı Funktsional. Anal. i Prilozhen. (1984) 64–70.[31] B. Fritzsche, B. Kirstein, Schwache Konvergenz nichtnegativ hermitescher Borelmaße, Wiss. Z. Karl-Marx-Univ. Leipzig

Math.-Natur. Reihe 37 (1988) 375–398.[32] B. Fritzsche, B. Kirstein, C. Mädler, On Hankel nonnegative definite sequences, the canonical Hankel parametrization, and

orthogonal matrix polynomials, Complex Anal. Oper. Theory 5 (2011) 447–511.[33] B. Fritzsche, B. Kirstein, C. Mädler, On a special parametrization of matricial α-Stieltjes one-sided non-negative def-

inite sequences, in: Interpolation, Schur Functions and Moment Problems. II, in: Oper. Theory Adv. Appl., vol. 226,Birkhäuser/Springer, Basel AG, Basel, 2012, pp. 211–250.

[34] B. Fritzsche, B. Kirstein, C. Mädler, On matrix-valued Herglotz–Nevanlinna functions with an emphasis on particular sub-classes, Math. Nachr. 285 (2012) 1770–1790.

[35] B. Fritzsche, B. Kirstein, C. Mädler, T. Schwarz, On a Schur-type algorithm for sequences of complex p × q-matrices and itsinterrelations with the canonical Hankel parametrization, in: Interpolation, Schur Functions and Moment Problems. II, in:Oper. Theory Adv. Appl., vol. 226, Birkhäuser/Springer, Basel AG, Basel, 2012, pp. 117–192.

[36] F.R. Gantmacher, M.G. Kreın, Oszillationsmatrizen Oszillationskerne und kleine Schwingungen mechanischer Systeme, Wis-senschaftliche Bearbeitung der deutschen Ausgabe: Alfred Stöhr, Math. Lehrbüch. Monogr. Abt. 1, Band V, Akademie-Verlag,Berlin, 1960, English version: Oscillation Matrices and Kernels and Small Vibrations of Mechanical Systems, revised ed.,AMS Chelsea Publishing, Providence, RI, 2002, translation based on the 1941 Russian original, edited and with a preface byAlex Eremenko.

[37] F.A. Grünbaum, L. Haine, Bispectral Darboux transformations: an extension of the Krall polynomials, Int. Math. Res. Not.1997 (1997) 359–392.

[38] Y.J. Hu, G.N. Chen, A unified treatment for the matrix Stieltjes moment problem, Linear Algebra Appl. 380 (2004) 227–239.[39] D.H. Kim, K.H. Kwon, D.W. Lee, F. Marcellán, Compatible pairs of orthogonal polynomials, Bull. Korean Math. Soc. 36 (1999)

779–797.[40] I.V. Kovalishina, J -expansive matrix-valued functions, and the classical problem of moments, Akad. Nauk Armjan. SSR Dokl.

60 (1975) 3–10.[41] I.V. Kovalishina, Analytic theory of a class of interpolation problems, Izv. Akad. Nauk SSSR Ser. Mat. 47 (1983) 455–497.[42] M.G. Kreın, The ideas of P.L. Cebyšev and A.A. Markov in the theory of limiting values of integrals and their further devel-

opment, Uspehi Matem. Nauk (N.S.) 6 (1951) 3–120.[43] M.G. Kreın, The description of all solutions of the truncated power moment problem and some problems of operator

theory, Mat. Issled. 2 (1967) 114–132.













http://refhub.elsevier.com/S0024-3795(13)00621-6/bib4352s1

http://refhub.elsevier.com/S0024-3795(13)00621-6/bib43524Ds1

http://refhub.elsevier.com/S0024-3795(13)00621-6/bib43524Ds1


















































[44] M.G. Kreın, A.A. Nudel’man, The Markov moment problem and extremal problems, in: Ideas and Problems of and TheirFurther Development, in: Transl. Math. Monogr., vol. 50, Amer. Math. Soc., Providence, RI, 1977, translated from Russian byD. Louvish.

[45] V.B. Matveev, Darboux transformation and the explicit solutions of differential-difference and difference–difference evolu-tion equations. I, Lett. Math. Phys. 3 (1979) 217–222.

[46] V.B. Matveev, M.A. Salle, Differential-difference evolution equations. II. Darboux transformation for the Toda lattice, Lett.Math. Phys. 3 (1979) 425–429.

[47] M. Rosenberg, The square-integrability of matrix-valued functions with respect to a non-negative Hermitian measure, DukeMath. J. 31 (1964) 291–298.

[48] B. Simon, The classical moment problem as a self-adjoint finite difference operator, Adv. Math. 137 (1998) 82–203.[49] V.P. Spiridonov, A.S. Zhedanov, Discrete Darboux transformations, the discrete-time Toda lattice, and the Askey–Wilson

polynomials, Methods Appl. Anal. 2 (1995) 369–398.[50] T.J. Stieltjes, Recherches sur les fractions continues, Ann. Fac. Sci. Toulouse Sci. Math. Sci. Phys. 8 (1894) J1–J122.[51] T.J. Stieltjes, Recherches sur les fractions continues [Suite et fin], Ann. Fac. Sci. Toulouse Sci. Math. Sci. Phys. 9 (1895)

A5–A47.[52] V.B. Uvarov, The connection between systems of polynomials that are orthogonal with respect to different distribution

functions, Ž. Vycisl. Mat. Mat. Fiz. 9 (1969) 1253–1262.[53] G.J. Yoon, Darboux transforms and orthogonal polynomials, Bull. Korean Math. Soc. 39 (2002) 359–376.[54] A.S. Zhedanov, Rational spectral transformations and orthogonal polynomials, J. Comput. Appl. Math. 85 (1997) 67–86.




















Transformations of matricial α-Stieltjes non-negative definite sequences

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