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Skip to main content. Log In Sign Up. Contractive Completions of Hankel Partial Contractions. Helton R eceived December 20, A Hankel partial contraction oteyaz a Hankel matrix such that not all of its entries are elenaa, but in which every well-defined submatrix is a contraction.
We address the problem of whether a Hankel partial contraction in which the upper left triangle is known can be completed to a contraction. At the same time, we obtain necessary and sufficient conditions on the given cross-diagonals in order for the matrix to be completed.
We also study the problem of extending a contractive Hankel block of size n to one of size n algebfa 1.
Q Academ ic P ress, Inc. All rights of reproduction in any form reserved. In this article, we first study the following problem. G iven a1. We say that Problem 1. Gheondea wArG x, by C. Parrott wPar x, and by Y. Yanovskaya wSYx, and it is implicit in the work of W. There is a formulation of Problem 1. R odman wJR x and by H. R eference wWoe, Theorem 7. To analyze Problem 1. We can then provide ooteyza direct proof of the relevant portion of wJRTheorem 1x.
In the positive direction, we establish a number of criteria for the existence of contractive completions, under suitable restrictions on a, b, c, d. Of particular interest are the cases with two of a, b, c, d equal to zero, for which we determine explicitly the ranges of values admitting contrac- tive completions. As a consequence, we see that even simple cases like a s 0, b s y 34c s 38d s 0, which produce partial contractions, do not admit contractive completions.
These results are presented in Altebra 3. We now turn our attention to a related problem, treated in Section 4. Given real numbers a1. As with Problem 1.
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The techniques employed in our work on Problem 1. In Section 4, after giving a detailed discussion of the case n s 2, we proceed with the analysis of the case n s 3, with particular emphasis in the situations arising when one of the given numbers equals zero. We also establish a link between Problem 1. For both Problem 1. Alyebra we explain in Section 5, the algebfa cases can be derived from the non-extremal ones by using a limit argument; we present in Section 3, nevertheless, a concrete analysis of some extremal cases.
As a matter of fact, we can subsume the study of the extremal cases in the study of Problem 1. The present work was motivated by a question of C.
We are also indebted to T. R odman, and H. Woerdeman for correspondence and insightful comments on the material herein, and to the referee for many valuable suggestions which improved the presentation.
Much of the research was done while the second and third otfyza authors spent a sabbatical leave at The University of Iowa. Most of the calculations in this paper, and some of the ideas, were first obtained through computer experiments using the software tool Mathe- matica wWolx.
As a consequence, we can now give the proofs of Lemma 1. Proof of Lemma 1. Now use the multilinearity of det.
Proof of Theorem 1. Another application of Lemma 2. The proof of Corollary 1. Proof of Corollary 1. G 0, it is clear that the graph of det P as a function of x is a downward parabola with non-negative discriminant, assuring the existence of real x-intercepts x l F x r.
Any value of x between x l and x r makes det P G 0, which in turn guarantees that T is a contrac- tion. The solution set is the closed interval determined by the roots of det P s 0, namely, y1 y a q b 2 y1 q a q b 2 xl sxr s ; 1qa y1 q a of course, y1 F x l F x r F 1.
In either case, the resulting P is positive, so H is a contraction. As an illustration of what we do in Section 5, let us show here another approach to the extremal case. Thus, for example, Hn n s H and Pn n s P. We shall see that Problem 1. Having found x, a new application of Theorem 1. Since both p12 and p13 must then be zero, we obtain yab y bc y cx s 0 y ac y bx y cy s 0.
If c s 0, then a2 q b 2 s 1 and ab s 0, and we can easily find the values of x and y. It is worth mentioning that the roles of a and c in the previous discussion are interchangeable; this will be helpful in Section 4. We must find x such that H remains a partial contraction, so we need the maximal completely determined submatrices H24 and H33 to be contractions.
The crux of the matter lies in that the values of x solving each problem must be compatible, i. Our next task is to obtain a useful criterion to determine when two given parabolas are simultaneously non-negative. The following result is quite elementary, but extremely useful. The graphs of f and g are identical. The graphs of f and g intersect in a unique simple point x 0and two subcases arise: Cases of intersection of two downward parabolas. The graphs of f and g intersect in two simple points x 1 and x 2and two subcases arise: It follows from Theorem 3.
This is what we proceed to do now for a number of important instances. But first let us summarize conceptually our algorithm for solving Problem 1. We then conclude that Problem 1. Two Variables Equal to Zero. Of the six possible cases, five are straightforward. In order to know if the deriva- tives have the same sign at x 1we first evaluate them at x 1we then multiply these two values, and we finally analyze the sign of the product: H44 is a partial contraction and the signs of the derivatives are different in the shaded regions.
Drawing the solutions for c, we find that each of the regions in Fig. We begin with the case n s 2. The Case n s 2. Even in this simple case, Problem 1. Our ploy for solving Problem 1. Using the results of Section 2. It follows that Problem 1. The shaded region in Figure 7 shows the places where Problem 1. The Case n s 3. Three submatrices are completely determined: H13H13and H One can attempt to solve this case by imitating the method used in Section 4.
Unlike the situation there, however, we do not have at our disposal explicit formulas for the solution interval w x lx r x of Problem 1.
We are forced, therefore, to proceed differently. It follows that the parabolas do indeed meet. In terms of determining whether a solution to Problem 1. Case c s 0. Therefore, Prob- lem 1. Case a s 0. We summarize our findings as follows. We conclude this section with an observation linking Problems 1. Figure 1 with a replaced by c.
The extremal cases considered in Section 2 always provided unique solutions for Problem 1. The next result shows that this was no accident. Since 1 y r 1 r U1 s 0, and since P G 0 by hypothe- sis, we must necessarily have yr 1 r U2 s??? Now, for i s 1.