Siberian Mathematical Journal, Vol. 52, No. 1, pp. 41–53, 2011Original Russian Text Copyright c© 2011 Voronin A. F.

A METHOD FOR DETERMINING THE PARTIALINDICES OF SYMMETRIC MATRIX FUNCTIONS

A. F. Voronin UDC 517.544

Abstract: We propose a method for determining the partial indices of matrix functions with somesymmetries. It rests on the canonical factorization criteria of the author’s previous articles. We showthat the method is efficient for the symmetric classes of matrix functions: unitary, hermitian, orthogonal,circular, symmetric, and others. We apply one of our results on the partial indices of Hermitian matrixfunctions and find effective well-posedness conditions for a generalized scalar Riemann problem (theMarkushevich problem).

Keywords: factorization, Riemann problem, symmetric matrix function, partial index

Introduction. In order to state the problem we introduce some notation and constructions. LetLn×m stand for the space of n×m matrix functions with entries in L1(R), where R is the extended realline R = (−∞,∞) ∪ {±∞}. Denote by Ff the Fourier image

Ff(p) =

∞∫

−∞f(t)eipt dt, p ∈ R,

of f ∈ Ln×m; by Wn×n, the Wiener algebra of continuous matrix functions of the form c+ Ff , where cis a constant matrix of size n, and f ∈ Ln×n; by Wn×n

+ (Wn×n− ), the subalgebra of Wn×n consisting of

the matrix functions of the form c+ Ff with f(t) = 0 for t < 0 (t > 0).Given some algebra A, denote by GA the group of invertible elements of A. Say that G ∈ GWn×n

admits standard factorization whenever there exists an expression as the matrix product

G(x) = G+(x)D(x)G−(x), x ∈ R, (0.1)

where G± ∈ GWn×n± , which are called factors, and D(x) is the diagonal matrix function

D(x) = diag

((x− ix+ i

)κ1, . . . ,

(x− ix+ i

)κn)

with the integers κ1 ≥ κ2 ≥ · · · ≥ κn called the partial indices of G. Refer to

κ := Ind detG(t) ≡ 1

2πΔR arg detG(x) =

n∑j=1

κj

as the total index of G.For the sake of simplicity, in this article we consider only the case of the total index of G equal to

zero (κ = 0). We can similarly consider a more general case in which the total index is a multiple of thesize of G.

If κ1 = κ2 = · · · = κn = 0, and so D = I is the identity matrix, then G = G+G− is a canonicalfactorization of G.

The author was partially supported by the Russian Foundation for Basic Research (Grant 10–01–00384) andthe Integration Project of the Siberian Division of the Russian Academy of Sciences (Grant 2009–93).

Novosibirsk. Translated from Sibirskiı Matematicheskiı Zhurnal, Vol. 52, No. 1, pp. 54–69, January–February, 2011.

Original article submitted March 25, 2010.

0037-4466/11/5201–0041

c©

41

By the Riemann problem (factorization problem) we understand the problem of decomposing G ∈GWn×n into the matrix product (0.1).

Owing to the wide range of its applications, the Riemann problem is one of the most topical problemsof complex analysis. The Wiener–Hopf system of equations [1, 2], characteristic systems of singularintegral equations with the Cauchy kernel [3], and convolution equations and systems on a finite interval[4, 5] are directly related to the Riemann problem. The Riemann problem is used in the inverse scatteringproblems of quantum theory [6, 7] and is one of the main tools in the inverse scattering method [8–10].Moreover, it is applied to filtration problems [11, pp. 361–362], probability theory [12], and some branchesof the theory of partial differential equations [13].

An important unsolved question in the theory of the Riemann problem is to calculate the invariantsof the problem: the partial indices of G(x). This article addresses to the problem. Here we proposean efficient method for determining the partial indices of G ∈ GWn×n provided that the latter enjoyscertain symmetry.

It is known how to calculate the partial indices of triangular and rational matrix functions (thefactorization theory of these matrix functions is well understood), as well as in the case that one of thefactors is a rational matrix function; see [14]. Moreover, it is shown in [15] that every Hermitian positivedefinite matrix function admits the canonical factorization (all its partial indices are equal to zero).

In this article we find sufficient conditions for the existence of canonical factorization for symmetricclasses of matrix functions: unitary, hermitian, orthogonal, circular, symmetric, and others. The matrixfunctions under study are mainly of size 2. By way of example, we apply one of our results (on the partialindices of Hermitian matrix functions) and find new effective well-posedness conditions for a generalizedscalar Riemann problem (the Markushevich problem).

The proofs of theorems in Sections 2–6 follow the same scheme (proof by contradiction). In Section 7we generalize this scheme and recast it as a method for determining the partial indices of symmetricmatrix functions of arbitrary size.

1. Preliminaries. The following is well known (see [2, Theorems 7.2 and 7.3; 16, Chapter 1, § 5]for instance):

Theorem 1. Take G ∈ GWn×n. Then G(x) admits standard factorization (0.1).Each standard factorization of G(x) is of the form

G(x) = G+(x)D(x) G−(x), x ∈ R, (1.1)

whereG± ∈ GWn×n

± , G+(x) = G+(x) Ω+(x), G−(x) = Ω−(x)G−(x). (1.2)

If all partial indices are equal,κ1 = κj , j = 2, . . . , n, (1.3)

then Ω− = Ω−1+ is an arbitrary constant nondegenerate matrix.

If (1.3) is violated then Ω± are rational matrix functions in GWn×n± with

(1) ω+jk(x) ≡ 0 for κk > κj ,

(2) ω+jk(x) = constj for κk = κj ,

(3) ω+jk(x) is an arbitrary polynomial in (x− i)/(x+ i) of degree ≤ κj − κk for κk < κj ,

where ω+jk(x) (j, k = 1, . . . , n) are the entries of Ω+(x), and

Ω−(x) = D−1(x)Ω−1+ (x)D(x), x ∈ R.

Theorem 1 implies that if (1.3) is violated then the general form of Ω+(x) is

Ω+(x) =

⎛⎜⎝Q1 ∗ . . . ∗0 Q2 . . . ∗0 . . . Qj ∗0 0 . . . Qm

⎞⎟⎠ , (1.4)

42

where Qj are arbitrary constant nondegenerate square matrices whose size is equal to the multiplicitymj of κj (j = 1, . . . ,m ≤ n). The asterisks mark the blocks whose entries are arbitrary polynomials in(x− i)/(x+ i) of appropriate degrees.

Suppose that

G ∈ GWn×n, κ ≡n∑j=1

κj = 0. (1.5)

Assume that at least one partial index of G(x) is nonzero. Given this matrix, denote by Kn the set(tuple) of its partial indices. Therefore, Kn = {κ1, κ2, . . . , κn} with κ1 ≥ κ2 ≥ · · · ≥ κn and

n∑j=1

|κj | = 0. (∗)

Denote by P+(R; Kn) the class of rational matrix functions (1.4) corresponding to Kn. For a fixedx0 ∈ R put

P+(Kn) := {Ω+(x0) : Ω+ ∈ P+(R; Kn)}.

It follows from (1.4) that the class of constant matrices P+(Kn) is independent of x0. Indeed, byconstruction Qj , j = 1, . . . ,m, are independent of x0, while the blocks ∗ are arbitrary constant matrices(of appropriate size) for every x0 ∈ R. Consequently, to every tuple Kn there corresponds a class ofconstant matrices P+(Kn).

It follows from (1.4) that

I, V ∈ P+(Kn), I1 /∈ P+(Kn)

for every tuple Kn, where V ∈ Tn is the class of all upper triangular constant nondegenerate matricesof size n, and I1 is the matrix of size n consisting of zeros with the exception of the secondary diagonal,consisting of ones. For n = 2,

I1 =

(0 11 0

).

The second part of Theorem 1 and (1.4) imply

Proposition 1. If (1.5) and (∗) are fulfilled then

(i) G+(x) := G+(x)Ω+(x) is a factor for arbitrary Ω+ ≡(ω+kj

)in P+(R; Kn) with ω+

n1 = 0;

(ii) P+(Kn) ⊇ Tn for every Kn.If in addition κk = κj for j = k (j, k = 1, . . . , n) then Ω±(x) ∈ Tn for every fixed x ∈ R. In particular,

the general form of Ω+(x) for n = 2 is

Ω+(x) =

(c1 P (x)0 c2

),

where c1 and c2 are arbitrary constants (with c1c2 = 0), and P (x) is an arbitrary polynomial in (x −i)/(x+ i) of degree at most κ1 − κ2.

Observe an important property of P+(Kn). For every n > 1 the number of the distinct tuplesof the partial indices Kn and the number of the distinct classes of the rational matrices P+(R; Kn)are unbounded in general. But the number of the distinct classes of the constant matrices P+(Kn)corresponding to the distinct tuples Kn is bounded for every n > 1, which follows from the second partof Theorem 1 (and (1.4)). For instance, for n = 2 every tuple K2 corresponds to only one class of constantmatrices P+(K2), which coincides with the class T2 of upper triangular matrices; for n = 3 every tupleK3 corresponds to only one of the three distinct classes P+(K3).

Let us present two canonical factorization criteria.

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Criterion 1. Suppose that (1.5) is fulfilled. Put

(ω+kj(x)

):= G−1

+ (x)G+(x), (1.6)

where G+ and G+ are the factors in (1.1) and (0.1) respectively. Then ω+n1 ≡ const. Moreover, if ω+

n1 = 0then (1.1) and (0.1) are canonical factorizations.

The converse also holds: if G(x) admits the canonical factorization (0.1) then there exists anothercanonical factorization (1.1) such that ω+

n1 = const = 0 in (1.6).

Proof of Criterion 1. The structure of Ω+ in (1.4) implies that ω+n1(x) is independent of x

for every factorization (canonical or not). The rest of the proof repeats that of Criterion 1 in [17] bycontradiction.

The next criterion is straightforward from the second part of Theorem 1.

Criterion 2. Suppose that (1.5) and (1.6) are fulfilled. Moreover, suppose that (∗) is fulfilled. Thenif there exists x0 ∈ R such that

(ω+kj(x0)

)/∈ P+(Kn) for every tuple Kn then the assumption on the

validity of (∗) is false, and (1.1) and (0.1) are canonical factorizations.

We can see that in order to apply these criteria we should have a priori information on the product

of the factors G−1+ (x) and G+(x) at some point x0 ∈ R, which in general is a problem since G+ and G+

are unknown. We can however obtain some properties of these factors.

Proposition 2. Suppose that (1.5) is fulfilled. Then if for some x0 ∈ R the constant matrix G(x0)is unitary then we may assume that the factors G±(x0) in (0.1) are also unitary matrices. Otherwise, ifG(x0) is not unitary then we may assume that in (0.1) only G+(x0) and D(x0) are unitary matrices.

Proof of Proposition 2. Assume that G admits no canonical factorization (otherwise the claim

follows directly from the second part of Theorem 1). Then Theorem 1 implies that G+(x) := G+(x)Ω+(x),where Ω+ is an arbitrary matrix in P+(R; Kn), is a factor in (1.1). By the Schur triangulation theorem[18, Chapter 2, § 6] we can express G+(x0) as the product G+(x0) = UV , where U is a unitary matrix,

and V ∈ Tn. According to Proposition 1(ii), we can put Ω+(x0) = V −1; then G+(x0) = U .Directly from (0.1) we have

G−(x0) = D−1(x0)G−1+ (x0)G(x0).

Then the matrices G(x0) and G−(x0) are either both unitary or are not both unitary since D(x0) andG+(x0) are unitary by construction. The proof of Proposition 2 is complete.

The following obvious statement generalizes the results below for symmetric matrix functions toarbitrary matrix functions with the zero total index.

Proposition 3. Take G ∈ GWn×n with n > 1. If G(x) admits canonical factorization then so doesthe matrix M(x) := A+(x)G(x)A−(x) with A± ∈ GWn×n

± .

2. The class of matrix functions

KL1 := {(mjk) ≡M ∈ GW 2×2 : M(x) = −A+(x)M−1(x)A+(x), x ∈ R, where A+ ∈ GW 2×2+ }.

Theorem 2. Take M ∈ KL1 with Ind detM(x) = 0. If there exists x0 ∈ R such that

A+(x0)M−1(x0) = D2I1, where D2 = diag(c1,−c2) with c1, c2 > 0, (2.1)

then M(x) admits canonical factorization.

Proof. Proceed by contradiction. Assume that M admits no canonical factorization. Then byTheorem 1 for M(x) we have the regular factorization

M(x) = M+(x)D(x)M−(x), x ∈ R, (2.2)

44

with

M± ∈ GW 2×2± , D(x) = diag(pκ1 , p−κ1), p =

x− ix+ i

, κ1 > 0.

Inverting both sides of (2.2) and taking conjugates, we obtain

M−1(x) = M−1− (x)D(x)M−1

+ (x) (2.3)

since D(x) = D−1(x). Inserting (2.2) and (2.3) into the defining equality of KL1, we obtain

M+(x)D(x)M−(x) = −A+(x)M−1− (x)D(x)M−1

+ (x)A+(x), x ∈ R. (2.4)

By Theorem 1, this equality of two regular factorizations of M(x) implies the existence of Ω+ ∈P+(R; K2) satisfying

M+(x)Ω+(x) = A+(x)M−1− (x), x ∈ R. (2.5)

From this, using the equality

M−1− (x) = M−1(x)M+(x)D(x),

which follows from (2.2), we obtain

M+(x)Ω+(x) = A+(x)M−1(x)M+(x)D(x). (2.6)

Multiply (2.6) for x = x0 by D(x0) on the right and put (tkj) := Ω+(x0)D(x0). By (2.1),

(bkj) ≡M+(x0)(tkj) = D2I1M+(x0). (2.7)

This directly implies that

b11 ≡ m+11t11 = c1m

+21, b21 ≡ m+

21t11 = −c2m+11, (2.8)

since (tkj) is an upper triangular nondegenerate matrix by construction, and M+(x0) ≡(m+kj

). Then

(2.8) and the nondegeneracy condition for M+(x0) yield m+21 = 0 and m+

11 = 0. Taking the ratio of thetwo equalities of (2.8), we find

m+11

m+21

= −c1m+21

c2m+11

.

An elementary rearrangement yields the impossible chain of relations

c2|m+11|2 = −c1|m+

21|2 = 0

since cj > 0 for j = 1, 2. Hence, the assumption that M(x) admits no canonical factorization is false.The proof of Theorem 2 is complete.

As an interesting application of Theorem 2 consider one of its obvious corollaries.

Corollary 1. Take G ∈ GW 2×2 with Ind detG(x) = 0 such that G(x) is a circular matrix satisfying

G(x) = −G−1(x), x ∈ R.

If there exists x0 ∈ R such that G(x0) = I1 diag(1,−1) then G(x) admits canonical factorization.

Proof. Put A+ := I and M(x) := G(x). Then we can verify directly that M(x) satisfies thehypotheses of Theorem 2.

3. The class of matrix functions KL2 := {M ∈ GW 2×2 : M(x) = M−1(−x), x ∈ R}.

45

Theorem 3. Take M ∈ KL2 with

Ind detM(x) = 0, M(∞) = c1I, where c1 = ±1. (3.1)

Then G(x) admits canonical factorization.

The proof of Theorem 3 is similar to that of Theorem 2. Assume that M admits no canonicalfactorization. Theorem 1 implies that M(x) admits a regular factorization (2.2). In both sides of (2.2)take the inverse with reflection; it leaves M(x) invariant since M ∈ KL2, and we obtain directly

M(x) = M−1− (−x)D(x)M−1

+ (−x) (3.2)

since D(x) = D−1(−x). By (2.2) and (3.2),

M+(x)D(x)M−(x) = M−1− (−x)D(x)M−1

+ (−x), x ∈ R. (3.3)

By Theorem 1, this equality of two regular factorizations of M(x) implies the existence of Ω+ ∈P+(R; K2) satisfying

M+(x)Ω+(x) = M−1− (−x), x ∈ R. (3.4)

By the obvious equalityM−1

− (−x) = M(x)M+(−x)D(−x)

we deduce from (3.4) thatM+(x)Ω+(x) = M(x)M+(−x)D(−x). (3.5)

For x = ∞ this yieldsΩ+(∞) = M−1

+ (∞)M(∞)M+(∞)

sinceM±(∞) = M±(−∞), D(±∞) = I. (3.6)

Thus, Ω+(∞) = Ω−1+ (∞). Hence, the diagonal entries of the triangular matrix Ω+ are equal to ±1.

Verify now that there exists a solution M+ ∈ GW 2×2+ to the matrix equation

c1M+(x) = M(x)M+(−x)D(−x) (3.7)

for M+ ∈ GW 2×2+ ; formally, (3.7) results from (3.5) with Ω+(x) = c1I. Decompose (3.7) into two special

Riemann boundary value problems for the vector functions V +1 (±x) and V +

2 (±x):

c1V+1 (x) = p−κ1M(x)V +

1 (−x), x ∈ R, (3.8)

c1V+2 (x) = pκ1M(x)V +

2 (−x), x ∈ R, (3.9)

where V +j (x) =

(m+

1j(x),m+2j(x)

)Tfor j = 1, 2 are column vectors, T stands for transposition, and

M+ ≡(m+ik

).

By the assumption that a regular factorization (2.2) exists, for D = I there exists a nontrivial solutionto (3.5) with some matrix Ω+ ∈ P+(R; K2) whose diagonal elements are equal to ±1. Consequently,there exists a nontrivial solution to the Riemann problem (3.8) since Ω+(x) is an upper triangular matrix.

In order to prove the solvability of the special Riemann problem (3.9) consider the Riemann problem

U+(x) = c1pκ1M(x)U−(x), U± =

(u±1 , u

±2

)T, u±1 , u

±2 ∈W 1×1

± . (3.10)

This problem has a nontrivial solution [16, Theorem 5.6] since the partial indices of c1pκ1M(x) are

nonnegative by construction. Then the first equality in (3.10) yields

U−(−x) = c1pκ1M(x)U+(−x) (3.11)

46

since M(x) = M−1(−x) and c1 = ±1. Put V +2 (x) := U+(x)+U−(−x). Adding the first equality in (3.10)

to (3.11), we obtain (3.9).Therefore, (3.8) and (3.9) yield

c1Q+(x)D(x) = M(x)Q+(−x),

where Q+(x) consists of the column vectors V +1 (x) and V +

2 (x). Hence, Q+ ∈ W 2×2+ . It remains to show

that Q+ ∈ GW 2×2+ ; i.e., the last equality is a regular factorization of M(x).

Inserting the expression for M(x) in (2.2) into the last matrix equality, we obtain

c1Q+(x)D(x) = M+(x)D(x)M−(x)Q+(−x).

Elementary rearrangements yield

D(x)M−(x)Q+(−x) = c1M−1+ (x)Q+(x)D(x).

Thus, by the analytic continuation theorem and the generalized Liouville theorem [1, Chapter 1, § 3.1],we have M−1

+ (x)Q+(x) = ω+(x) ∈ P+(R; K2) in exactly the same fashion as while proving Theorem 7.3

in [2]. Consequently, Q+ = M+ω+ ∈ GW 2×2

+ .For x = ∞, taking (3.6) and (3.1) into account, we obtain from (3.7) the contradictory chain of

relations c1I = M(∞) = c1I. Hence, the assumption that M(x) admits no canonical factorization isfalse. The proof of Theorem 3 is complete.

Let us present an example of matrix functions, also with certain symmetry, for which Theorem 3helps us establish the existence of canonical factorization.

Proposition 1. Take

M ∈ GW 2×2, M(x) =

(m11(x) −m12(x)m12(−x) m22(x)

), M(∞) = ±

√m(∞)I1, (3.12)

where mjj(x) = mjj(−x), j = 1, 2, put m(x) = detM(x), and denote by√m(x) the principal value of

the square root of m. Then M(x) admits the canonical factorization

M(x) = M+(x)AMT+ (−x), x ∈ R, (3.13)

whereM+ ∈ GW 2×2

+ , MT+ (x) ≡ I2M

T+(x)I2, I2 = diag(1,−1),

and A = AT is a constant nondegenerate matrix of size 2.

Proof. Put

G(x) :=1√m(x)

I1M(x). (3.14)

By assumption, m(x) = 0 for x ∈ R; thus, Indm(x) = 0 since m is an even function. The Wiener–Levitheorem (see [19, Theorems W and L, Chapter 1, § 1] for instance) yields the existence of l ∈ L1(R)

such that√m(x) = 1 + F l(x), and so

√m(x) ∈ GW 1×1. Consequently,

√m(x) admits the canonical

factorization [2]: √m(x) = m+(x)m+(−x), x ∈ R, m+ ∈ GW 1×1

+ . (3.15)

The hypotheses of the theorem together with (3.14) and (3.15) imply that G(x) satisfies the hypothe-ses of Theorem 3. Indeed, direct calculations yield

G−1(−x) =√m(x)M−1(−x)I1 =

1√m(x)

(m22(x) m12(−x)−m12(x) m11(x)

)I1 = G(x),

√m(∞)G(∞) = I1M(∞) = ±

√m(∞)I (G(∞) = ±I).

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Hence, Theorem 3 implies that G(x) admits canonical factorization. Then so does M(x) in view of (3.14)and (3.15). We have

M(x) = M+(x)M−(x), x ∈ R,where M± ∈ GW 2×2

± .We can see that the matrices of the form (3.12) constitute the class of matrix functions

KL3 := {M ∈ GW 2×2 : M(x) = MT (−x), x ∈ R}closed under the composition of transposition T and reflection. Hence, for every matrix M(x) ofclass KL3 the equality

M+(x)M−(x) = MT− (−x)MT

+ (−x), x ∈ R,of two canonical factorizations is valid. Then the second part of Theorem 1 implies that there existsa constant nondegenerate matrix A such that

M+(x)A = MT− (−x), A−1M−(x) = MT

+ (−x), x ∈ R.The last two equalities imply (3.13). The proof of Proposition 1 is complete.

Put

KL4 := {G ∈ GW 2×2 : G(x) = M(−x)I2, x ∈ R, where M ∈ LK3, I2 = diag(1,−1)}.It is easy to see that the class KL4 is closed under transposition with reflection.

Proposition 1 and the definitions of KL3 and KL4 imply

Corollary 2. Take M ∈ KL3 or M ∈ KL4 with m(x) = detM(x). If M(∞) = ±√m(∞) I1 or

M(∞) = ±i√m(∞) I1I2 then M(x) admits canonical factorization.

Consider a convolution integral equation of the second kind on a finite segment. It is known [4] tobe equivalent to the Riemann problem with the matrix coefficient

G(x) = − 1

Λ−(x)

(1 −eixbFk−(x)

e−ixbFk+(x) 1−Fk(x)

), (3.16)

where k(t) = 0, t /∈ (−b, b), k ∈ L1(−b, b), and k±(t) = θ(±t)k(t), t ∈ R, Λ−(x) = 1−Fk−(x).

Lemma 1. Suppose that k(t) = k(−t) for t ∈ R and Λ−(x) = 0 for x ∈ R. Then all partial indicesof G(x) in (3.16) are equal to κ1 = Ind Λ−(−x) ≥ 0.

Proof. Put d(x) := detG(x). Then

d(x) =Λ−(−x)

Λ−(x), Ind d(x) = 2κ1 ≥ 0

since Fk∓(−x) = Fk±(x) ∈W 1×1± .

Put G1(x) := −Λ−(−x)G(x). The total index of G1(x) is equal to zero since

detG1(x) = Λ−(−x)Λ−(x).

It is easy to see that G1 ∈ KL3. Corollary 2 implies that all partial indices of G1(x) are equal to zero.Hence, the partial indices of G(x) are equal to κ1 by construction. The proof of the lemma is complete.

Note that Lemma 1 also appears in [20].

4. The class of Hermitian matrix functions. Consider to begin with the matrix functions ofsize 2. It is easy to show that the Hermitian matrix functions of class GW 2×2 are of the general form

G(x) =

(g11(x) g12(x)

g12(x) g22(x)

), (4.1)

where gkj ∈W 1×1, Imgjj(x) = 0, k, j = 1, 2, and d(x) := detG(x) = 0, x ∈ R.Verify that Proposition 1 implies

48

Theorem 4. Suppose that the entries of (4.1) are additionally subject to the restrictions

gjj(x) = gjj(−x), g12(x) = −g12(−x), x ∈ R. (4.2)

IfG(∞) = ±

√d(∞)I1 (4.3)

then all partial indices of G in (4.1) are equal to zero.

Indeed, putmjj(x) := gjj(x), j = 1, 2, m12(x) := −g12(x).

By (4.2) and (4.3) it is easy to see that the matrix M ≡ (mkj) resulting from G satisfies the hypothesesof Proposition 1. Hence, Proposition 1 implies that M(x) and, consequently, G(x) admit canonicalfactorizations.

Theorem 4 complements Theorem 5 of [15] in the case n = 2 and detM(∞) < 0.

Theorem 5. Suppose that

M ∈ GWn×n, n > 1, M(x) = M∗(x) ≡MT (x), x ∈ R.

If there exists x0 ∈ R such that all eigenvalues ofM(x0) are of the same sign thenM(x) admits canonicalfactorization.

We present a proof of Theorem 5, basing on a canonical factorization criterion, at the end of thissection. Note that a proof of Theorem 5 in [15] is different and does not carry over to Hermitian matrixfunctions.

Take the Wiener algebra W0 of continuous functions of the form Ff , where f ∈ L1(R), and thesubalgebra W0+ (W0−) of W0 consisting of the functions Ff with f(t) = 0 for t < 0 (for t > 0).

By way of application of Theorem 4 we find new effective conditions for the solvability of a generalizedscalar Riemann problem (the Markushevich problem). For the simplicity of presentation we consider theproblem on the contour coinciding with R.

We study the boundary value problem of finding a piecewise analytic function ϕ ∈ W0 from theboundary condition on the real line R:

ϕ+(x) = a(x)ϕ−(x) + b(x)ϕ−(x) + c(x), x ∈ R, (4.4)

witha, b ∈W 1×1, c ∈W0, a(x) = 0, x ∈ R. (4.5)

The bibliographies on this problem appeared in [21] and [22, Chapter 5, § 39.3]. Presently the mainresults in the study of this problem, its stability conditions, and the numbers of linearly independentsolutions and conditions for solvability are obtained in the case

|a(x)| ≥ |b(x)|, x ∈ R, (4.6)

or reduce to it.In [21] the problem (4.4), (4.5) is reduced to the (equivalent) Riemann problem for a piecewise

analytic vector Φ(x) = (Φ1(x),Φ2(x))T with Φ1,Φ2 ∈W0 and

Φ+(x) = G(x)Φ−(x) + g(x), x ∈ R, Φ±j ∈W0±, j = 1, 2. (4.7)

HereG ∈ GW 2×2, g(x) = (g1(x), g2(x))T , gj ∈W0, j = 1, 2, (4.8)

Φ±1 (x) = ϕ±(x), Φ±

2 (x) = ϕ∓(x).

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The matrix G(x) and the vector g(x) are defined as

G(x) =1

a(x)I2

(|a(x)|2 − |b(x)|2 b(x)

b(x) −1

)≡ 1

a(x)I2G1(x) (4.9)

with

I2 = diag(1,−1), g1(x) =a(x)c(x)− b(x)c(x)

a(x), g2(x) = − c(x)

a(x),

G1(x) =

(|a(x)|2 − |b(x)|2 b(x)

b(x) −1

). (4.10)

From (4.10) and (4.5) we can see that

G1 ∈ GW 2×2, G1(x) = G∗1, det G1(x) = −|a(x)|2 = 0, x ∈ R.

Ifa(−x) = a(x), b(−x) = −b(x) (4.11)

then Theorem 4 holds for G1. Hence, the partial indices of G1(x) are equal to zero. Consequently, thepartial indices of G are equal to zero as well since the Cauchy index of an even function a(x), x ∈ R, isequal to zero. Therefore, we have

Theorem 6. Suppose that (4.11) is fulfilled. Then the generalized Riemann problem (4.4), (4.5)(the Markushevich problem) is well-posed: a solution exists, is unique, and stable in the norm of W0

with respect to the coefficients a, b, and c.

Remark. It is not difficult to state a theorem similar Theorem 6 for the boundary value problem

a(x)ϕ+(x) + b(x)ϕ+(x) = c(x)ϕ−(x) + d(x)ϕ−(x) + f(x), x ∈ R, (4.12)

naturally generalizing (4.4). This problem is equivalent [21] to the Riemann boundary value problemwith a Hermitian matrix of size 2; hence, Theorem 4 applies to (4.12).

Proof of Theorem 5. We may assume that M(x0) = diag(λ1, . . . , λn), where λ1, . . . , λn are theeigenvalues of the constant matrix M(x0). Indeed, every Hermitian matrix is unitarily similar to thediagonal matrix consisting of its eigenvalues. Consequently,

M(x0) = U diag(λ1, . . . , λn)U∗.

Hence, it suffices to prove the theorem for the Hermitian matrix U−1M(x)(U−1)∗.The rest of the proof goes by contradiction. Assume that M admits no canonical factorization.

Theorem 1 implies that for M(x) we have the regular factorization

M(x) = M+(x)D(x)M−(x), x ∈ R, (4.13)

with

M± ∈ GWn×n± , D(x) = diag(pκ1 , . . . , pκn), p =

x− ix+ i

, κ1 ≥ κ2 ≥ · · · ≥ κn. (4.14)

The hypotheses of the theorem imply that the determinant of M(x) is a real function. Hence, the totalindex of M(x) is equal to zero, κ = 0. Since M is a Hermitian matrix, (4.13) yields

M(x) = M∗(x) = M∗−(x)I1D1(x)I1M

∗+(x) (4.15)

with D1(x) = diag(p−κn , . . . , p−κ1) since D∗(x) = I1D1(x)I1. The right-hand sides of (4.13) and (4.15)are regular factorizations of M(x) by construction. Then the uniqueness theorem for partial indices (see[2, Theorem 7.1] for instance) yields D(x) = D1(x), and so κ1 = −κn, . . . , κn = −κ1. Therefore,

M(x) = M+(x)D(x)M−(x) = M∗−(x)I1D(x)I1M

∗+(x). (4.16)

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By Theorem 1, this equality of two regular factorizations of M(x) implies the existence of a matrixfunction Ω+ ∈ P+(R; Kn) satisfying

M+(x)Ω+(x) = M∗−(x)I1. (4.17)

From this for x = x0 by the obvious equalities

I1 = I−11 , M∗

−(x) = M(x)(M∗+)−1(x)(D∗)−1(x), (D∗)−1(x) = D(x)

we obtain(akj) := Ω+(x0)I1D

−1(x0) = M−1+ (x0)M(x0)

(M−1

+

)∗(x0) =: (bkj). (4.18)

Let us calculate ann and bnn. The left-hand side of (4.18) yields ann = 0 since ω+n1 = 0 (see Proposi-

tion 1(i)). On the other hand, the right-hand side of (4.18) yields

bnn =n∑j=1

∣∣m+nj

∣∣2λj = 0,

where(m+kj

)= M−1

+ (x0). Therefore, we have the contradictory chain 0 = ann = bnn = 0, which completes

the proof.

5. The class of unitary matrix functions.

Theorem 7. Take G ∈ GWn×n with n > 1. If

G(x) = G⊥(−x), x ∈ R (G⊥ ≡ (GT )−1) (5.1)

then G(x) admits canonical factorization.

Note that (5.1) holds for even unitary matrix functions. Theorem 7 is announced in [23] and provedin [17] by the same method as Theorems 2, 3, and 5 of this article.

For orthogonal matrix functions Theorem 7 directly implies

Theorem 8. Take G ∈ GWn×n with n > 1 and ImG = 0. If G(x) = G⊥(−x), x ∈ R, then G(x)admits canonical factorization.

6. The class of matrix functions

KL5 := {M ∈ GW 2×2 : M(x) = −A+(x)M(−x)A−(x), x ∈ R,where A± ∈ GW 2×2

± , A−1± (−x) = −A±(x)}.

Theorem 9. Take M ∈ KL5 with Ind detM(x) = 0. If

A+(∞) = I2I1 ≡(

0 1−1 0

)(6.1)

then G(x) admits canonical factorization.

The proof follows the same scheme, and we omit the details. Under the assumption that M(x) admitsno canonical factorization, the hypotheses of the theorem imply that

M+(x)Ω+(x) = A+(x)M+(−x), x ∈ R, (6.2)

with Ω+ ∈ P+(R; K2) since

M(x) = M+(x)D(x)M−(x) = −A+(x)M+(−x)D(x)M−(−x)A−(x).

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By the obvious equality M+(∞) = M+(−∞) and (6.1), we deduce from (6.2) for x = ∞ that

(bkj) := M+(∞)Ω+(∞) = I2I1M+(∞). (6.3)

The matrix Ω+ is upper triangular (see Proposition 1), and the entrywise expression of (6.3) impliesdirectly that

b11 ≡ c1m+11 = m+

21, b21 ≡ c1m+21 = −m+

11, (6.4)

where(m+kj

)= M+(∞). It is easy to see that the equalities in (6.4) are contradictory since c1 = 0 and

detM+(∞) = 0. Consequently, the assumption that M(x) admits no canonical factorization is false. Theproof of Theorem 9 is complete.

7. Conclusion. The proofs of all our theorems on the existence of canonical factorization (Theo-rems 2, 3, 5, 7, and 9) follow the same scheme. Deviating from this scheme and using natural general-ization and formalization, we extract a method for determining the partial indices of symmetric matrixfunctions. We proceed to describe the resulting method.

The capabilities of the method. This method either establishes that the matrix function underconsideration, which enjoys certain symmetry, admits canonical factorization, and then all its partialindices are equal to zero, or it keeps the question of the existence of canonical factorization open.

In this article we showed that the method is effective on the symmetric classes of matrix functions:unitary, hermitian, symmetric, circular, orthogonal, and others. In all these classes we select the sub-classes of matrix functions admitting canonical factorization. For the simplicity of arguments in mostclasses under study the size of the matrices was taken to be 2.

The domain of applicability of the method. Suppose that a matrix function G(x) satisfies(1.5) and enjoys the following symmetry: there exists an operator F : GWn×n → GWn×n such that

FG(x) = G(x), x ∈ R. (7.1)

The operator F acts on a factorization of G(x) (on the right-hand side of (0.1)) according to one of thefollowing two formulas: either

FG(x) = F+1 G−(x)D(x)F−

1 G+(x), (7.2)

where F±1 G∓ ∈ GWn×n

± , or

FG(x) = F+1 G+(x)D(x)F−

1 G−(x), (7.21)

where F±1 G± ∈ GWn×n

± . These expressions involve some operators F±1 : Wn×n → Wn×n. It is natural

to assume that G, F , and F±1 are known, whereas G± and D are not. We are required to establish the

validity of the identity D ≡ I.

The scheme of the algorithm of the method. To begin with, assume that (7.21) holds. Put

G±(x) := F±1 G±(x). Then (7.21) becomes

FG(x) = G+(x)D(x)G−(x). (7.3)

From (7.1) and (7.3) we obtain the equality for two regular factorizations of G(x):

G+(x)D(x)G−(x) = G+(x)D(x)G−(x). (7.4)

Form the matrix (ω+kj(x)

):= G−1

+ (x)G+(x)

as in the canonical factorization criteria. The final formula for(ω+kj(x)

)is

(ω+kj(x)

)= G−1

+ (x)F+1 G+(x) =:

(b+kj(x)

), x ∈ R. (7.5)

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Proceed to analyze the right-hand side of (7.5), aiming to establish the existence of x0 ∈ R such thatb+n1(x0) = 0 or (b+kj(x0)) /∈ P+(Kn) for every tuple of indices Kn satisfying (∗). If we find x0 ∈ R with

this property then according to Criteria 1 and 2 all partial indices of G(x) are equal to zero. Otherwise,when we fail to find x0 ∈ R, the question of the existence of canonical factorization remains open.

Suppose now (7.2). Put G±(x) := F±1 G∓(x). Then (7.2) becomes (7.3). From (7.1) and (7.3) we

obtain (7.4). Form the matrix (ω+kj(x)) := G−1

+ (x)G+(x). Finally, we have a formula similar to (7.5):

(ω+kj(x)) = G−1

+ (x)F+1 G−(x) := (b+kj(x)), x ∈ R, (7.6)

where G−(x) = D−1(x)G−1+ (x)G(x). We inspect (7.6) in the same fashion as (7.5).

We should note that the inspection of the right-hand sides of (7.5) and (7.6) might simplify if weestablish some property of the constant matrix G+(x0), where x0 is a fixed point of R. For instance, ifG+(x0) ∈ Tn then we can calculate b+n1(x0) since in this case we may assume that G+(x0) = I; the lastequality follows from Proposition 1(i), (ii).

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A. F. Voronin

Sobolev Institute of Mathematics, Novosibirsk, Russia

E-mail address: voronin@math.nsc.ru

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