A Toeplitz-like operator with rational matrix symbol having poles on the unit circle: Invertibility and Riccati equations

This paper is a continuation of the work on unbounded Toeplitz-like operators $T_\Om$ with rational matrix symbol $\Om$ initiated in Groenewald et. al (Complex Anal. Oper. Theory 15, 1(2021)), where a Wiener-Hopf type factorization of $\Om$ is obtained and used to determine when $T_\Om$ is Fredholm and compute the Fredholm index in case $T_\Om$ is Fredholm. Due to the high level of non-uniqueness and complicated form of the Wiener-Hopf type factorization, it does not appear useful in determining when $T_\Om$ is invertible. In the present paper we use state space methods to characterize invertibility of $T_\Om$ in terms of the existence of a stabilizing solution of an associated nonsymmetric discrete algebraic Riccati equation, which in turn leads to a pseudo-canonical factorization of $\Om$ and concrete formulas of $T_\Om^{-1}$.


Introduction
In two recent papers [10,11], we explored the matrix analogue of an unbounded Toeplitz-like operator that was investigated in [7,8,9] for scalar rational symbols with poles on the unit circle T. While many of the classical operator theory topics, like Fredholmness, invertibility and spectrum, are well understood in the scalar case, the case of matrix symbols appears to be more intricate.In [10] a Wiener-Hopf type factorization was obtained, from which the Fredholm index can be determined, in case the symbol has no zeroes on T. However, this Wiener-Hopf type factorization has a high level of non-uniqueness and, unlike in the classical case, generally does not lead to a diagonalization of the symbol.As a result, although some further Fredholm characteristics can be determined from the Wiener-Hopf type factorization [11], it does not seem to be an adequate tool to compute the dimensions of the kernel and the cokernel, nor does it seem to give a clear characterization of invertibility.In the present paper we take a different approach to the question of invertibility of this Toeplitz-like operator, using Riccati equations and pseudo-canonical factorization.
Unbounded Toeplitz operators appeared first in a paper of Hartman and Wintner [12] in 1950, but only became an active topic with the seminal paper of Sarason [16] in connection to truncated Toeplitz operators; see [15] for how unbounded Toeplitz operators with matrix symbols come into play.More recently, kernels of unbounded Toeplitz operators appeared in the study of nearly backward shift invariant subspaces and Toeplitz inverses [3,4].
Next we introduce some notation, which is required to define our Toeplitzlike operator and state our main results.We write Rat for the space of rational functions, Rat(T) for the functions in Rat that only have poles on the unit circle T, Rat 0 (T) for the strictly proper functions in Rat(T), P for the space of polynomials and for positive integers k we indicate the polynomials of degree at most k by P k , extending it to all integers by setting P k = {0} in case k ≤ 0. For positive integers m and n, we indicate the spaces of m × n matrices with entries from these function spaces by Rat m×n , Rat 0 (T) m×n , etc.In the case of vector functions, when n = 1, we will just write m instead of m × 1.For 1 < p < ∞, L p and H p denote the Lebesgue space and Hardy space, respectively, and K p is the standard complement of H p in L p .With L p m , H p m and K p m we indicate the spaces of vectors of length m with entries from L p and H p , respectively.
Let Ω ∈ Rat m×m with possibly poles on T and det Ω ≡ 0, and let 1 < p < ∞.We then define the Toeplitz-like operator T Ω (H p m → H p m ) by Dom(T Ω ) = f ∈ H p m : Ωf = h + η where h ∈ L p m (T), and η ∈ Rat m 0 (T) , T Ω f = Ph with P the Riesz projection of L p m (T) onto H p m . (1.1) By the Riesz projection, P, we mean the projection of L p m onto H p m , as discussed in [14, pages 149-153].
As usual, Ω ∈ Rat m×m has a pole at z 0 ∈ C ∪ {∞}, if any of its entries has a pole at z 0 .In case det Ω ≡ 0, a zero of Ω is a pole of its inverse Ω −1 (z) := Ω(z) −1 .It is not necessarily the case that the zeroes of Ω correspond to the zeroes of det Ω, and Ω can have both a pole and a zero at the same point z 0 ∈ C ∪ {∞}.It was proved in [10] that T Ω is Fredholm if and only if Ω has no zeroes on T. In particular, for T Ω to be invertible, that is, bijective from Dom(T Ω ) onto H p m , it is necessary that Ω has no zeroes on T.
In the classical case, invertibility of Toeplitz operators with rational symbols can be studied via Riccati equations and canonical factorizations associated with state space realizations of the symbol, cf., [1,2].In the present paper, we follow the approach of [6].Let Ω ∈ Rat m×m and assume Ω is given by a minimal state space realization of the form with R 0 ∈ C m×m and with A, B, C and α, β, γ matrices of appropriate sizes such that A has all its eigenvalues in the open unit disc D and α has all its eigenvalues in the closed unit disc D, that is, A is stable and α is semi-stable (in [6] α is also stable).Minimality in this setting means that one cannot find a representation for Ω of this form with A and α matrices of smaller size; equivalently, the triples (C, A, B) and (γ, α, β) both provides observable and controllable discrete-time linear systems, cf., [13].Despite the fact that Ω has poles on T, so that α has eigenvalues on T, there is a fairly direct analogue of the canonical factorization result of [6].
Theorem 1.1.Let Ω ∈ Rat m×m with det Ω ≡ 0 and no zeroes on T and assume that Ω is given by the minimal realization (1.2) with A stable and α semi-stable.Then the following are equivalent: (i) There exists a matrix Q such that R 0 − γQB is invertible, Q satisfies the nonsymmetric discrete algebraic Riccati equation: and such that the matrices are both stable.
(ii) Ω has a right pseudo-canonical factorization Ω(z) = Ψ(z)Θ(z), i.e., Ψ and Θ are m × m rational matrix functions with det Ψ ≡ 0 and det Θ ≡ 0 and such that Θ has poles only outside or on the unit circle T, Θ −1 has poles only outside T, Ψ has poles only inside or on the unit circle T, and Ψ −1 has poles only inside the unit circle T.
3) such that A • and α • are stable, then a right pseudo-canonical factorization is obtained as follows: Let δ and D be invertible matrices such that δD = R 0 − γQB and set Then Ω(z) = Ψ(z)Θ(z) holds with Ψ and Θ defined as and the inverses of Ψ and Θ are given by Furthermore, the solution Q of the Riccati equation (1.3) so that A • and α • are stable is unique.Finally, the realizations in (1.6) and (1.7) for Θ, Ψ, Θ −1 and Ψ −1 are minimal.
The above result is proved in Section 2 and is essentially obtained by specifying Theorem 1.1 of [6] for the function Ω r defined by Ω r (z) = Ω(rz) (1.8) for 1 < r small enough, so that Ω r does not have poles and zeroes on T.More generally, for any function f , scalar-, vector-or matrix-valued, and scalar r > 0 we write f r for the function f r (z) = f (rz) defined for z ∈ C for which rz is in the domain of f .In order to characterize invertibility of T Ω more is required than what is in [6], we want to compare invertibility of the unbounded operator T Ω with invertibility of the bounded Toeplitz operator T Ωr .Note that invertibility in both cases means bijectivity on its domain of definition.Hence, the inverse of T Ω will be bounded.For r > 1, we define the annulus (1.9) It turns out that T Ω and T Ωr can be compared, not only with respect to invertibility, but even with respect to their Fredholm properties, in case they are Fredholm.
Theorem 1.2.Let Ω ∈ Rat m×m with det Ω ≡ 0 and assume that Ω has no zeroes on T. Define Ω r by (1.8).Let r 0 > 1 be such that Ω has no zeroes in the annulus A r0 and no poles in A r0 \ T. Then T Ω is Fredholm and for each 1 < r < r 0 , T Ωr is bounded and Fredholm and we have dim Ker T Ω = dim Ker T Ωr and codim Ran T Ω = codim Ran T Ωr .
In particular, T Ω is invertible if and only if T Ωr is invertible for some (and hence all) 1 < r < r 0 .
We shall prove Theorem 1.2 in Section 3. Our second main result together with [6], shows that invertibility of T Ω is equivalent to items (i) and (ii) in Theorem 1.1.With some further work we derive, in Section 4 below, formulas for the inverse of T Ω , as given in the following result.
Theorem 1.3.Let Ω ∈ Rat m×m with det Ω ≡ 0 and no zeroes on T and assume that Ω is given by the minimal realization (1.2) with A stable and α semi-stable.Then T Ω is invertible if and only if the Riccati equation (1.3) has a solution Q such that A • and α • in (1.4) are stable, or, equivalently, if Ω has a pseudocanonical factorization Ω(z) = Ψ(z)Θ(z) as in item (ii) of Theorem 1.1.In that case, the inverse of T Ω is the bounded operator given by (1.10)

Ω
has a block matrix representation T −1 Ω i,j with respect to the standard block basis of H p m that is given by where For the final result we present in this introduction we restrict to the case where p = 2, since it relies on a result from [6], which is proved only for p = 2.
Let Ω ∈ Rat m×m be given by the minimal realization with A stable and α semistable and assume A and α are of size s × s and t × t, respectively.We define the observability operator for the pair (C, A) as and the controllability operator for the pair (α, β) as Due to the semi-stability of α, C α,β need not be bounded, but it is the case that the subspace ).This will be proved in Lemma 4.2 in Section 4 below, where we will also prove the following proposition.
Proposition 1.4.Consider the case p = 2. Let Ω ∈ Rat m×m with det Ω ≡ 0 and such that T Ω is invertible.Assume that Ω is given by the minimal realization (1.2) with A stable and α semi-stable.Then the solution Q of the algebraic Riccati equation (1.3) that makes A • and α • in (1.4) stable is given by The formula for Q is analogous to that in [6], where poles on T are not allowed, but requires more attention since C α,β is not necessarily bounded.To see that the right hand side in (1.14) is well defined, we point out that O C,A maps C s into D r for r > 1 small enough, while T −1 Ω maps D r into D r , again for r > 1 small enough, which is contained in the domain of C α,β .That T −1 Ω maps D r into D r follows from Proposition 3.1 below.
We conclude this introduction with a brief overview of the remainder of the paper.In Section 2 we apply the main result of [6] to the function Ω r in (1.8), and translate back to the state space realization of Ω, leading to a proof of Theorem 1.1.In the next section we investigate the relation between T Ω and T Ωr , and give a proof of Theorem 1.2.The work of Sections 2 and 3, is then combined in Section 4 to prove Theorem 1.3 as well as Proposition 1.4.

Riccati equation, canonical factorization and inversion for Ω r
Suppose that Ω ∈ Rat m×m is given by the minimal realization formula (1.2), that is: with A being stable and α being semistable.It is then clear that Ω r defined by (1.8) admits the state space realization As in Theorem 1.1, let r 0 > 1 be such that Ω has no zeroes in the annulus A r0 and no poles in A r0 \ T, with A r0 as defined in (1.9).Then, for 1 < r < r 0 , Ω r has no poles and no zeroes on T. Therefore, the results of [6] apply to Ω r and its realization (2.2)-(2.3).Note that the paper [6] only considers the case of Toeplitz operators on H 2 m .However, since invertibility of Toeplitz operators on H p m with rational matrix symbols can be characterized in terms of their Wiener-Hopf factorizations, which are independent of p, invertibility on H p m is independent of the value of p.We now specify the main result of [6] to Ω r , together with some supplementary observations, in the next proposition.This result is subsequently used to prove Theorem 1.1.Proposition 2.1.Let Ω ∈ Rat m×m be given by the realization (2.1) with A stable and α semi-stable, so that Ω r is given by the realization (2.2)-(2.3).Let r 0 > 1 be such that Ω has no zeroes in the annulus A r0 and no poles in A r0 \ T. For 1 < r < r 0 the following are equivalent: (i) T Ωr is invertible.
(ii) There exists a matrix Q such that R 0 − γQB is invertible, Q satisfies the nonsymmetric discrete algebraic Riccati equation: and rA • and r −1 α • are stable, with A • and α • given by (1.4).
Moreover, the solution Q of the Riccati equation (2.4) such that rA • and r −1 α • are stable is unique and independent of r, i.e., for each 1 < r < r 0 one obtains the same solution Q in item (ii).Furthermore, if Q is as in (ii), then a canonical factorization of Ω r is obtained with Θ (r) = Θ r and Ψ (r) = Ψ r , where Θ and Ψ are defined as in Theorem 1.1, and Θ r and Ψ r are defined according to (1.8).
In case one of items (i)-(iii) holds, and hence all, we have Since Ω r has no poles on T, since 1 < r < r 0 , Theorem 1.1 of [6] applies to Ω r and its realization (2.2), leading to the equivalence of variations of (i)-(iii) in terms of the matrices in the realizations.Technically, Theorem 1.1 of [6] does not contain an item about the invertibility of T Ωr , but that invertibility of T Ωr is equivalent to the two items in the theorem follows by the discussion preceding the theorem, and this is also where the formula for T −1  Ωr in terms of the canonical factors appears.It thus remains to show that the statements of items (ii) and (iii) in terms of the realization matrices of Ω r correspond to the statements concerning the Riccati solutions and canonical factorization from Theorem 1.1 of [6], respectively.
We start with item (ii).From Theorem 1.1 of [6], and the preceding paragraphs, we obtain that invertibility of T Ωr is equivalent to the existence of a matrix Q r such that R 0 − γQ r B is invertible, that satisfies the Riccati equation and such that are both stable, corresponding to the claim of item (ii).Moreover, the matrix Q r with these properties is unique.It follows that the Riccati equation that Q r solved is independent of r, but it is less straightforward that the condition of having rA • and α• r stable does not introduce a dependency on r; in particular, A • and α • in the above formulas may depend on r.To see that this is not the case, we note that the matrix Q r can be obtained as the limit of a Riccati difference equation associated with the finite section method for T Ωr , as discussed in Section 4 of [6].Indeed, since Ω r is continuous on T and we assume T Ωr to be invertible, for N large enough the N -th section of T Ω,r , i.e., the Toeplitz block matrix with R j,r the j-th Fourier coefficient of Ω r , will be invertible, and the matrices Ωr,N O Cr,Ar,N with C βr,αr,N and O Cr,Ar,N given by solve the Riccati difference equation and Q N,r converges to Q r as N → ∞.Hence, in order to see that Q r is independent of r, it suffices to show that Q N,r is independent of r.Note that the Fourier coefficients of Ω r are given by where T Ω,N is as in (2.5) with R n,r = R n,1 , where R n,1 is defined according to (2.7) with r = 1.This shows that for large N , also T Ω,N invertible and Combining these identities with (2.8), it follows that is indeed independent of r, and consequently, Q r is also independent of r.
It remains to prove the equivalence of (ii) (or (i)) and (iii),and that (iii) can be achieved as described in the proposition, and the formulas for T −1  Ωr .The equivalence of (ii) and (iii), in fact, follows directly from Theorem 1.1 in [6].We now show that the formulas for the canonical factors from [6] lead to the factorization of Ω r using Θ and Ψ from Theorem 1.1.By the formulas in [6], the canonical factorization of Ω r is given by Ω r (z) = Ψ (r) (z)Θ (r) (z) where we factor R 0 − γQ r B = δD, as claimed, and set

The formula for T −1
Ωr now follows simply from the text preceding Theorem 1.1 in [6].
Using the equivalence of (ii) and (iii) in Proposition 2.1 it is easy to prove our second main result.
Proof of Theorem 1.1.Since Ω r , Ψ r and Θ r are rational matrix functions, the factorization Ω r (z) = Ψ r (z)Θ r (z) for some 1 < r < r 0 implies that also Ω(z) = Ψ(z)Θ(z) as well as the formulas for the inverses of Ψ and Θ. Hence (iii) in Proposition 2.1 is equivalent to (ii) in Theorem 1.1.
Next we show that the realizations of Θ and Ψ in (1.6) and of Θ −1 and Ψ −1 in (1.7) are minimal.Note that since the realization of Ω is minimal and A is stable and α is semi-stable, the McMillan degree, deg(Ω), of Ω is equal to the sum of the sizes of A and α, say s and t, respectively.From the formulas of Θ and Ψ it is clear that deg(Θ) ≤ s and deg(Ψ) ≤ t.On the other hand, since the McMillan degree is sublogarithmic, we have Hence we have equality in each step, which implies deg(Θ) = s and deg(Ψ) = t, in other words, the realizations of Θ and Ψ are minimal.By the observation right after Proposition 7.2 in [1], it follows that the realization for Θ (respectively Ψ) is minimal if and only if the realization of Θ −1 (respectively Ψ −1 ) is minimal.Hence, also the realizations of Θ −1 and Ψ −1 are minimal.
Since the solution Q = Q r of (2.4) used to construct rA • and r −1 α • is independent of r, it follows that the solution Q in item (ii) in Proposition 2.1 is such that rA • and r −1 α • are stable for all 1 < r < r 0 , so that A • is stable and α • is semi-stable.
Thus, from the equivalence of (ii) and (iii) in Proposition 2.1 it follows that we get the equivalence of (i) and (ii) in Theorem 1.1, except that at this stage we only get α • to be semi-stable.To see that α • is in fact stable, note that Ω(z) = Ψ(z)Θ(z) implies that Ψ(z) −1 = Θ(z)Ω(z) −1 .Since A is stable, Θ has no poles on T, and Ω −1 has no poles on T because Ω is assumed to have no zeroes on T. Therefore, Ψ −1 has no poles on T either, which implies, by minimality of the realization of Ψ −1 , that α • has no eigenvalues on T. Hence α • is stable.

Fredholmness of T Ω versus Fredholmness of T Ωr
In this section we prove Theorem 1.2.The proof relies heavily on the connection between T Ω and T Ωr , with Ω r as in (1.8), as explained in the next result.
Proposition 3.1.Let Ω ∈ Rat m×m with det Ω ≡ 0 and assume that Ω has no zeroes on T. Let r 0 > 1 be such that Ω has no zeroes in the annulus A r0 and no poles in A r0 \ T. Define Then for each 1 < r < r 0 we have Moreover, T Ω maps D r into D r , the inverse image T −1 Ω (D r ) of D r under T Ω lies in D r , and we have Proof.Let 1 < r < r 0 .We start by proving (3.2), except for the first inclusion.The second inclusion is trivial.If f ∈ D r , then f has an analytic extension to rD, so that, in particular, f is analytic on D.
The argument to show that H m (D) ⊂ Dom(T Ω ) is similar to that in the scalar case [9,Theorem 6.2].Note that each f ∈ H m (D) is also in D r ′ for some 1 < r ′ sufficiently close to 1. Hence it suffices to show that D r ⊂ Dom(T Ω ).Let f ∈ D r .Since Ω is rational, Ωf is meromorphic on rD with finitely many poles, each of finite multiplicity.Computing the residues of the poles of each of the entries in the vector function Ωf it is easy to write Ωf in the form g + ρ with g ∈ L p m and ρ ∈ Rat m 0 (T), showing that f ∈ Dom(T Ω ).Hence, we have proved the second inclusion.
Next we show that T Ω maps D r into D r and that (3.3) holds.Let f ∈ D r .Hence f has an analytic extension to rD and |f (z)| p is integrable on rT, that is, f ∈ H p m (rT).By (3.2), f ∈ Dom(T Ω ) and hence Ωf = g + ρ for some g ∈ L p m and ρ ∈ Rat m 0 (T).Write g = g + + g − with g + ∈ H p m and g − ∈ K p m , so that T Ω f = g + .Since f is analytic on rD and Ω and ρ are rational with no poles in A r \ T, it follows that g = Ωf − ρ must be analytic in A r \ T with the poles on T all having finite multiplicity.However, g ∈ L p m , and hence cannot have poles of finite multiplicity on T. Thus g is analytic in A r .Hence, g + is analytic on rD and g − on C \ r −1 D. Since r < r 0 , by an argument similar to that in the first part of the proof, it follows that g + ∈ H p m (rT) and g − ∈ K p m (r −1 T).This implies that g +,r (z) = g + (rz) and g −,r (z) = g − (rz) define functions in H p m and K p m , respectively.In particular, g + ∈ D r and it follows that T Ω maps D r into D r .Moreover, ρ is a rational matrix function with poles only in T, so that ρ r only has poles inside D and we obtain that ρ r ∈ K p m .Note further that on T Ω r (z)f r (z) = Ω(rz)f (rz) = g(rz) + ρ(rz) = g r (z) + ρ r (z) = g +,r (z) + g −,r (z) + ρ r (z).
Therefore, we have Finally, we show that T −1 Ω (D r ) ⊂ D r .Since 0 ∈ D r , this proves in particular that Ker T Ω ⊂ D r and hence the first inclusion of (3.2).To prove the inclusion, we require the Wiener-Hopf type factorization from [10, Theorems 1.1 and 1.2], namely Ω can be factored as for some integer k ≥ 0, Ω + , Ω • , Ω − ∈ Rat m×m and P 0 ∈ P m×m such that Ω − and Ω −1 − are both minus functions (i.e., no poles outside D), Ω + and Ω −1 + are both plus functions (i.e., no poles in D), Ω • = Diag m j=1 (φ j ) with φ j ∈ Rat(T) having no zeroes and having roots only on T (in [10, Theorems 1.1], φ j ∈ Rat can have zeroes on T, but this cannot occur since Ω has no zeroes on T), and P 0 a lower triangular polynomial with det(P 0 (z)) = z N for some integer N ≥ 0. It then follows from Theorem 1.3 in [10] that The latter two inclusions follow from the the fact that T Ω− and T Ω+ are invertible with inverses along with the argument from the previous paragraph applied to showing that D r is an invariant subspace for these two operators; for the latter, note that Ω − and Ω + do not have zeroes and poles on the annulus A r .Hence it remains to show that 0 (T) of the form η j = r j /q j ∈ Rat 0 (T) with q j the denominator of the j-th diagonal element of Ω 0 .Write has no poles on the annulus A r and no zeroes on A r \ T. Since Ξ −1 , f , h + and h − don't have poles on T, neither can Ξ −1 η.It follows that Ξ −1 η, Ξ −1 h + and Ξ −1 h − are all analytic on A r .Therefore, f is analytic on A r .However, f ∈ H p m , so that f in fact is analytic on rT.Using that Ξ −1 is rational, g + ∈ D r , g − and ρ are continuous on rT it follows that f is p-integrable on rT, and hence f ∈ D r .Lemma 3.2.Let Ω ∈ Rat m×m with det Ω ≡ 0. Then T Ω is Fredholm if and only if Ω has no zeroes on T. Assume T Ω is Fredholm and let r 0 > 1 be such that Ω has no zeroes in the annulus A r0 and no poles in A r0 \ T. Then T Ωr is bounded and Fredholm for each 1 < r < r 0 .
Proof.For 1 < r < r 0 , by definition of r 0 it is clear that T Ωr has no poles and no zeroes on T, so that T Ωr is bounded and Fredholm.For T Ω the result is not included in [10] but follows from the results proved there.Indeed, consider a Wiener-Hopf type decomposition of Ω as in the proof of Proposition 3.1, e.g., as in (3.4).Since Ω −1 + is a plus function and Ω −1 − is a minus function, they do not have poles on T. Also, since det P 0 (z) = z N for some integer N ≥ 0, P −1 0 as a function in Rat m×m can only have a pole at 0, so that P 0 also does not have zeroes on T. This shows that the zeroes of Ω on T correspond to the zeroes of Ω • .Since Ω • is a diagonal matrix function, its zeroes correspond to the zeroes of its diagonal elements, which are all on T, by construction.Since Ω has no zeroes on T, this implies that the numerators of the diagonal elements of Ω • are constant, assuming the numerators and denominators are co-prime.The statement for T Ω now follows from the fact that T Ω is Fredholm if and only if the numerators (assuming co-primeness) in Ω • are constant, according to Theorem 1.4 in [10].
As a consequence of Proposition 3.1, it is easy to show that the dimensions of the kernels of T Ω and T Ωr are the same.Corollary 3.3.Let Ω ∈ Rat m×m with det Ω ≡ 0 and assume that Ω has no zeroes on T. Let r 0 > 1 be such that Ω has no zeroes in the annulus A r0 and no Proof of Theorem 1.3.First assume T Ω is invertible.By Theorem 1.2, it follows that T Ωr is invertible for all 1 < r < r 0 , with r 0 as in Theorem 1.2.It then follows from Proposition 2.1 that a solution Q to the Riccati equation (1.3) exists, and reasoning as in the proof of Theorem 1.1 if follows that for this solution Q the matrices A • and α • in (1.4) are stable.
Conversely, if the Riccati equation ( 1.3) has a solution Q such that A • and α • are stable, then Theorem 1.1 provides a pseudo-canonical factorization of Ω, and, due to the stability of A • and α • , this factorization extends to a canonical factorization of Ω r for r > 1 small enough.It then follows from Proposition 2.1 that T Ωr is invertible, and, consequently, that T Ω is invertible, by Theorem 1.2.
From the factorization T Ω = T Ψ T Θ and the boundedness and boundedly invertibility of T Θ , obtained in Lemma 4.1, it follows that invertibility of T Ω corresponds to invertibility of T Ψ and, moreover, Since Θ, Θ −1 and Ψ −1 have no poles on T, the Toeplitz operators T Θ , T Θ −1 and T Ψ −1 are bounded and their block matrix representations are well understood.It remains to show that the block matrix representation of T −1 Ψ and T Ψ −1 are the same, as this would prove that these bounded operators coincide on the subspace of polynomials P m and equality would follow from their boundedness and the denseness of P m in H p m .Indeed, once it is proved that T and the block matrix representations of T Θ −1 and T Ψ −1 .Hence, we need to show that the matrix entries with respect to the standard (block) basis of H p m of T Ψ −1 T Ψ and T Ψ T Ψ −1 are I m on the diagonal and 0 elsewhere.The block matrix representation of T Ψ , in terms of the realization (1.2), is given by (4.1).The realization formula of Ψ −1 in item (ii) of Theorem 1.1, together with the stability of α • shows that the block matrix representation of T Ψ −1 is an upper block triangular Toeplitz matrix that is determined by its first block row, which is given by δ The case where j < i follows directly because the matrix representations of T Ψ −1 and T Ψ are both block upper triangular, and the case j = i follows because the block diagonal elements are each others inverses.Hence, it remains to consider the case where j > i.For this purpose, notice that For j > i we have If j = i + 1, then the summation in the middle term of the right-hand side is empty, and it is easy to see that the right-hand side collapses to 0 by a direct application of the first identity in (4.2).For j > i + 1, using the first identity in (4.2), we see that Inserting this formula back into the formula for [T Ψ −1 T Ψ ] ij , it follows that in the first summation in (4.3) the term k = 0 is added, while in the second summation the term k = j − i − 1 is added, so that [T Ψ −1 T Ψ ] ij = 0, as claimed.A similar computation, using the second identity of (4.2), shows that the block matrix representation of T Ψ T Ψ −1 also corresponds to the block matrix representation of I H p m .Finally, we turn to the proof of the last result in the introduction.Proposition 1.4 is stated for p = 2, but the lemma which we require for the proof also works for p = 2.For r > 1, define the invertible linear map Ω also maps D r into D r .In particular, the operator ΥT −1 Ω Υ −1 is a well-defined linear map on H p m , which is closed by [18,Problem III.5.7], and hence bounded by the closed graph theorem.To see that T −1 Ωr = ΥT −1 Ω Υ −1 , note that if f ∈ H p m , then Υ −1 f and T Ω Υ −1 f are in D r so that m and ℓ 2 m in the usual way.According to [6], the matrix Q in Proposition 2.1 is given by Q = C αr,βr T −1 Ωr O Cr,Ar .It was already noted in Proposition 2.1 that Q is independent of the value of 1 < r < r 0 .That also Q = C α,β T −1 Ω O C,A now follows directly from the identities derived in Lemma 4.2.

Υ 1 . 4 . 2 . 1 Ω Υ − 1 .
: D r → H p m , Υ : f → f r , with inverse Υ −1 : H p m → H p m , Υ −1 : f → f r −Lemma Let Ω ∈ Rat m×m with det Ω ≡ 0 be given by the minimal realization (1.2) with A stable and α semi-stable.Define r 0 as in Theorem 1.2, O C,A as in (1.12) and C α,β as in (1.13), and define O Cr,Ar and C αr,βr analogously, where C r , A r , α r , β r are defined as in (2.3) and 1 < r < r 0 .Then O C,A , O Cr,Ar and C αr,βr are bounded, the range of O C,A is contained in D r and D r is contained in Dom(C α,β ).Moreover, we haverC αr ,βr = C α,β Υ −1 , O Cr,Ar = rΥO C,A and T Ωr = ΥT Ω Υ −1 .Furthermore, in case T Ω is invertible, then T Ωr is invertible as well and T −1 Ωr = ΥT −Proof.Since the realization (1.2) of Ω is minimal, it follows from the definition of r 0 that A r = rA is still stable.Since r > 1 and α is semi-stable, α r = r −1 α is stable.This implies the boundedness of O C,A , O Cr,Ar and C αr ,βr , as well as the fact that O C,A maps into D r .The identity O Cr,Ar = rΥO C,A is straightforward from the definitions.To show that D r is in the domain of C α,β and that rC αr ,βr= C α,β Υ −1 holds on H p m , let f (z) = ∞ k=0 f k z k ∈ H p m , then f is integrable in T and so is the rational matrix function (zI − α r ) −1 β r = ∞ k=0 α k r β r z k .It follows from Lemma 1.5 on page 81 of[17] thatC αr ,βr f = ∞ k=0 α k r β r f k = ∞ k=0 r −k−1 α k βf k = r −1 ∞ k=0 α k βr −k f k = r −1 C α,β f 1/r = r −1 C α,β Υ −1 f.The above computation shows in particular that Υ −1 f is in Dom(C α,β ) for each f ∈ H p m so that D r ⊂ Dom(C α,β ).The relation between T Ω and T Ωr in (3.3) implies that ΥT Ω | Dr = T Ωr Υ.Since the range of Υ −1 is equal to D r and ΥΥ −1 = I, we have T Ωr = ΥT Ω Υ −1 , as claimed.Assume T Ω is invertible.By Theorem 1.2, also T Ωr is invertible.By Proposition 3.1, T Ω maps D r into D r and T −1