A Physical Interpretation of Tight Frames

A Physical Interpretation of Tight Frames - A Physical...

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A Physical Interpretation of Tight Frames P. G. Casazza 1 ,M .F ickus 2 ,J .Kovaˇ cevi´ c 3 , M. T. Leon 4 , and J. C. Tremain 5 1 Mathematics Department, University of Missouri, Columbia, Missouri 65211 USA, pete@math.missouri.edu 2 Department of Mathematics and Statistics, Air Force Institute of Technology, Wright-Patterson AFB, Ohio 45433 USA, Matthew.Fickus@afit.edu 3 Department of Biomedical Engineering, Carnegie Mellon University, Pittsburgh, Pennsylvania 15213, USA, jelenak@cmu.edu 4 Mathematics Department, University of Missouri, Columbia, Missouri 65211 USA, mleon@math.missouri.edu 5 Mathematics Department, University of Missouri, Columbia, Missouri 65211 USA, janet@math.missouri.edu Summary. We ±nd ±nite tight frames when the lengths of the frame elements are predetermined. In particular, we derive a “fundamental inequality” which completely characterizes those sequences which arise as the lengths of a tight frame’s elements. Furthermore, using concepts from classical physics, we show that this characteriza- tion has an intuitive physical interpretation. 1 Introduction Let H N be a fnite N -dimensional Hilbert space. A fnite sequence { f m } M m =1 o± vectors is A -tight ±or H N i± there exists A 0 such that, A k f k 2 = M X m =1 |h f,f m i| 2 , ±or all f H N .An A -tight frame is an A -tight sequence ±or which A> 0. By polarization, { f m } M m =1 is A -tight ±or H N i± and only i±, Af = M X m =1 h m i f m , ±or all f H N . Clearly, any orthonormal basis is a 1-tight ±rame. However, the converse is ±alse. For example, the vertices o± a tetrahedron, appropriately centered and scaled, ±orm a 1-tight ±rame o± ±our elements ±or R 3 .Moreover , while the elements o± an orthonormal basis are o± unit length a priori , there are no explicit assumptions made about the lengths o± a tight ±rame’s elements. This raises the question,
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2 P. G. Casazza, M. Fickus, J. Kovaˇ cevi´ c, M. T. Leon, and J. C. Tremain Given positive integers M and N , for what sequences of nonnegative numbers { a m } M m =1 do there exist tight frames { f m } M m =1 for H N , such that k f m k = a m for all m ? The answer to this question is the subject of this chapter. To be more precise, we Frst derive a necessary condition upon the lengths. Proposition 1. If { f m } M m =1 is A -tight for H N then, max m =1 ,...,M k f m k 2 A = 1 N M X m =1 k f m k 2 . (1) Proof. ±irst note that for any m =1 ,...,M , k f m k 4 = |h f m ,f m i| 2 M X m 0 =1 |h f m m 0 i| 2 = A k f m k 2 . Thus, k f m k 2 A for all m , yielding the inequality in (1). ±or the equality, let { e n } N n =1 be an orthonormal basis for H N . By Parseval’s identity, 1 N M X m =1 k f m k 2 = 1 N M X m =1 N X n =1 |h f m ,e n i| 2 = 1 N N X n =1 A k e n k 2 = A. ut Thus, the lengths { a m } M m =1 of a tight frame of M elements for an N - dimensional space must satisfy the fundamental inequality , max m =1 ,...,M a 2 m 1 N M X m =1 a 2 m . (2) Remarkably, this easily-found necessary condition will prove sufficient as well. That is, we shall show that for any sequence { a m } M m =1 which satisFes (2), and for any N -dimensional Hilbert space H N , there exists a tight frame { f m } M m =1 for H N
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A Physical Interpretation of Tight Frames - A Physical...

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