Partitions of finite frames

Transcription

1 Rowan University Rowan Digital Works Theses and Dissertations Partitions of finite fraes Jaes Michael Rosado Rowan University, Follow this and additional works at: Part of the Matheatics Coons Recoended Citation Rosado, Jaes Michael, "Partitions of finite fraes" (06). Theses and Dissertations This Thesis is brought to you for free and open access by Rowan Digital Works. It has been accepted for inclusion in Theses and Dissertations by an authorized adinistrator of Rowan Digital Works. For ore inforation, please contact

2 PARTITIONS OF FINITE FRAMES by Jaes M. Rosado A Thesis Subitted to the Departent of Matheatics College of Coputer Science and Matheatics In partial fulfillent of the requireent For the degree of Master of Arts in Matheatics at Rowan University May 0, 06 Thesis Chair: Hieu Nguyen, Ph.D.

3 06 Jaes M. Rosado

4 Dedications I dedicate this thesis to y sister Victoria and y parents Maria and Victor.

5 Acknowledgeents I would first like to thank y thesis advisor Dr. Hieu Nguyen of the Matheatics Departent at Rowan University. Dr. Nguyen s wealth of knowledge in atheatics has helped e understand the intricacies of this project and has broadened y knowledge in atheatics. Our regularly scheduled eetings not only developed our project, but I also becae ore passionate about atheat-ics. It was during these eetings that I observed Dr. Nguyen s enthusias for atheatics and as a result y zeal for atheatics becae ore profound. I would also like to express y sincere gratitude to Dr. Ronald Czochor and Dr. Thoas Osler for being on the thesis coittee. I had the pleasure of having both professors for graduate classes and it was their rigor and passion for athe-atics that was ipressed upon on e the ost. I a also very grateful for all y graduate professors in the Math Departent; they have all reinforced y passion for atheatics and have supported e to continue y career in atheatics. I a very fortunate to have had such an in-credibly knowledgeable group of professors for all y graduate studies. Without their guidance and teachings I would not have been able to coe this far in atheatics. iv

6 Abstract Jaes M. Rosado PARTITIONS OF FINITE FRAMES Hieu Nguyen, Ph.D. Master of Arts in Matheatics An open question stated by Marcus, Spielan, and Srivastava [0] asks "whether one can design an efficient algorith to find the partitions guaranteed by Corollary.5." This corollary states that given a set of vectors in C whose outer products su to the identity there exists a partition of these vectors such that nors of the outer-product sus of each subset satisfy an inequality bound. Here particular types of vector sets called finite fraes are analyzed and constructed to satisfy the inequality described in Corollary.5. In this thesis, rigorous proofs and forula-tions of outer-product nors are utilized to find these partitions and to identify constraints on the finite fraes in order to satisfy Corollary.5. v

7 Table of Contents Acknowledgeents iv Abstract v List of Figures List of Tables Chapter : Introduction The Kadison-Singer Conjecture and Fraes Preliinaries What are Finite Fraes? Standard Unit Vectors Orthonoral Bases Chapter : Partitions of Fraes Equinor-Equiangular Fraes Partitions of Equinor-Equiangular Fraes The Discriinant Proble Two-Partitions: S = S Two-Partitions: S S Alost Even Split Partition Chapter 3: Grassannian Fraes Grassannian Fraes Constructing Optial Grassannian Fraes Grassanian Fraes and the Hypothesis of Corollary Inner Products of Grassannian Frae Vectors Nors of Outer Products of Grassannian Frae Vectors ( Vectors) Case n = viii ix vi

8 Table of Contents (Continued) An Analytic Approach for n = Case n = Case n = Case n = Conclusion References Appendix A: Matrix and Vector Nor Proofs Appendix B: General Forulas vii

9 List of Figures Figure Page Figure. Mercedes-Benz Frae Figure. = 0000 frae vectors viii

10 List of Tables Table Page Table. Coparison of frae atrix nors to Corollary Table. Max Subset Size (n ax ) Until Failure Table 3. Subset size and the corresponding nors for n = 4 case Table 4. Subset size and the corresponding nors for n = 8 case Table 5. Subset size and the corresponding nors for n = 6 case Table 6. Corollary bounds for different r s (n = 6 case) Table. Max Sized Partition, Corollary Bounds, and Nor Table 8. Nors for partitions P, P, P (Consecutive) Table 9. Nors for partitions P, P, P 3 (Non-Consecutive) ix

11 Chapter Introduction The Kadison-Singer Conjecture and Fraes In 959 the Kadison-Singer proble was posed by Richard Kadison and Isadore Singer after exploring Dirac s research in quantu echanics in the 940s. The proble is stated below [4], Kadison-Singer Proble (KS) Does every pure state on the (abelian) von Neuann algebra D of bounded diagonal operators on the Hilbert space l have a unique extension to a (pure) state on B(l ), the von Neuann algebra of all bounded linear operators on l? Let us give a run down of the terinology stated by the Kadison-Singer proble (KS). A state is a linear functional that aps a vector in a Hilbert space H, which could be in R or C, to its scalars in that given Hilbert space H. As an exaple, suppose you have a vector x R n, and each coponent x i R; therefore, our doain is Hilbert space the R n. A linear functional acting on this vector could be described as f(x) = α x +α x + +α n x n, where α, α,... are arbitrary real nubers. Notice that this is the apping defined by f : R n n R where {x k } k = c, a scalar. Next one can think of the KS proble as utilizing atrices or ore generally "algebras." With the KS proble we are exaining diagonal operators as our linear functionals, which are appings between spaces with respect to diagonal eleents of atrices. As it turns out there are any states, linear functionals, and

12 all together they for a set of states that act on these algebras. The pure states are the extreal eleents of this set of states. The KS proble probes whether the said diagonal operators on our Hilbert space H all have corresponding extensions to linear operators in our sae given space H. An extension on a state is a linear functional that "reduces" to another linear functional acting on the sae space. For exaple, suppose you have a function g(a, b, c, d) = f(a, d) a, b, c, d R, then g is the extension of f over the given Hilbert space R. This is so because even though g contains ore input values than f it still reduces to the sae output values that f encopasses. Therefore, we have extended the doain of f to the doain of g, but the two have the sae range of values. Originally, Kadison conjectured the solution to the KS proble was negative; however, in 03 Marcus, Spielan, and Srivastava provided a positive solution [0] to the KS proble. Their solution was found by reforulating the KS proble into equivalent fors; for instance, Weaver deonstrated in [5] that the KS proble can be thought of as a cobinatorial proble which also addresses vectors in C n and finite atrices in coplex Hilbert s paces. The KS proble is also equivalent to the paving conjecture proposed by Anderson [] which relates linear operators to diagonal operators in a coplex Hilbert space. In [0], the authors utilized these equivalent fors of the KS proble and analysis of the largest roots to ixed characteristic polynoials of a set of atrices to prove the paving conjecture and Weaver s conjecture; hence, this resulted in a positive solution of the KS proble. In the proof that Marcus, Spielan, and Srivastava provided, they state Corollary.5, which describes an upper bound to the nors of special atrices that are

13 fored fro special partitionings (groupings) of vectors contained in a collection. One of the open questions posed by the authors at the end of their paper [0] is whether one can forulate an efficient algorith that partitions a given collection of vectors such that it satisfies Corollary.5. This thesis explores a special type of vector collection known as fraes. We show that these fraes eet the hypothesis of the corollary and explore ways of finding partitions of these fraes to satisfy the conclusion of Corollary.5. The construction of fraes has its origins in haronic and functional analysis, operator theory, linear algebra, and atrix theory. They were a result of the liited capabilities of regular bases being utilized in signal processing and sapling theory. Originally, transission signals were decoposed and expanded using the Fourier transfor, but this left out or hid soe iportant coponents contained in the signal. Gabor rectified this by ebedding ore and redundant signal inforation into the transission [9]. This extra inforation is unique to fraes and Gabor s efforts arked the early steps in the construction of fraes. Unlike orthonoral bases which have a liited nuber of vectors and no redundancy, a frae can carry ore inforation about the original signal being transitted. The foral definition of a frae was forulated by Duffin and Schaeffer in 95 [], and they utilized to copute the coefficients of linear cobinations of vectors that were linearly dependent in a given Hilbert space. A frae in essence can be thought of as a relaxation of a basis, such that linear independence is not a requireent. Thus we can have repeated eleents in our frae, but the set of vectors in the frae ust still span the Hilbert space that we are working in. In addition to these qualities, a frae has what are known as frae bounds, which restrict the 3

14 axiu and iniu sizes of sus of inner products of a given vector with our frae vectors. As a result of having fraes, the challenges of reconstructing a signal with inial loss of inforation was reedied. This has to do with the fact that fraes are not as restrictive in their features copared to regular bases. The first part of this chapter was dedicated to giving a brief history of the Kadison-Singer proble and its iplications to finding partitions of finite fraes. In the rest of this chapter we will delve into the preliinary theores involving the KS proble and how finite fraes can be associated with Corollary.5. The second chapter explores a class of finite fraes that are equinoral and equiangular. Moreover, we shall prove various ways of partitioning the frae given certain frae characteristics. In the third chapter we will exaine Grassannian fraes and show that the size of the Grassannian frae affects whether the frae subsets will eet the bound described by Corollary.5. Throughout this thesis we will designate partitions and the corresponding subsets that eet the corollary bound as being valid. Our research has yielded the following results and questions that still need to be addressed. Soe results: We have found ways to partition equinor-equiangular fraes using algebraic and nuerical analysis. We have shown nuerically that for sall order Grassannian fraes any partition will be valid. Soe open probles that are still being analyzed are: 4

15 How do we find the eigenvalues for Heritian atrices generated by outer products in C n n where n is large? How does the choice of the frae vectors affect the corresponding nors of the outer products, particularly for higher-order Grassannian fraes? Once we obtain forulas for the nors of partial fraes, then can we find efficient algoriths to partition these fraes? Preliinaries Let us begin by stating Corollary.5 fro [0]: Theore. (MSS Corollary.5). Let r be a positive integer and let u,..., u C d be vectors such that u i u i = I (.) i= and u i δ for all i. Then there exists a partition {S,..., S r } of [] = {... } such that ( u i u i + δ) (.) i S r j for j =,..., r. As a side note, the calculation uu in C d d and uu T in R d d is known as the outer product or tensor product of two vectors and is used extensively in this thesis. In essence, given a set of d diensional vectors in the coplex doain with square agnitudes less than or equal to a given constant δ, we can always group the vectors into subsets such that the nor of the outer-product su never exceeds a 5

16 certain value (upper bound). The nor that we will calculate is defined by Meyer []: Definition. (Induced Matrix Nors). A vector nor that is defined on C n, induces a atrix nor on C n n by setting A = ax Ax, x = where A C n n and x C n. As a result of this definition we need to find the spectral values, the λ s, of our atrices by calculating the solutions to det(a λi) = 0. The axiu of these spectral values is our nor; therefore A = ax{λ, λ,...}. What akes the Corollary interesting is that the upper bound only depends on the nuber of "subsets" and the constant δ. The following conjecture [0] (KS ) precedes Corollary.5: Theore. (Conjecture. (KS )). There exist universal constants η and θ > 0 so that the following holds. Let w,..., w C d satisfy w i for all i and suppose u, w i = η i= for every unit vector u C d. Then there exists a partition S, S of [] so that 6

17 u, w i η θ. (.3) i S j Therefore, the Conjecture. is a special case of Corollary.5. Here we are finding a "-subset partition" of a set of vectors in C d, this eans [] = S S. Notice that the conclusion of Conjecture. is reiniscent of a what is known as a finite frae. Below is the definition as described in [5, 6, 8]. Definition.. A frae for a Hilbert space H is a sequence of vectors {f i } H for which there exist constants 0 < A B < such that, for every x H, A x i x, f i B x, (.4) where the values A and B are the frae bounds. Notice that if we express u = x x, then (.3) can be rewritten as follows: x, w i η θ = x, w x i (η θ) x. (.5) i S j i S j By coparing (.4) and (.5) we have that the upper frae bound is given by B = η θ and the frae vectors becoe f i = w i. A natural question that arises is whether Corollary.5 holds for fraes given that they are applicable to Conjecture.. Let us exaine (.) ore closely by taking the nor of the su of outer

18 products and utilizing the Triangle Inequality for atrices to get u i u i u i u i. i= But recall fro the hypothesis of Corollary.5 that u i δ. We cobine this with the identity uu = u (A.4) to get i= u i u i u i u i i= i= = u i i= δ. (.6) We then ultiply (.6) by x to get u i x = i= ( u i x ) i= δ x. (.) By the CBS inequality we get x, u i i= x u i δ x. (.8) i= So clearly based on the hypothesis of Corollary.5 we can achieve an upper frae bound for the vectors u,..., u. In fact, we can apply the sae steps used to find (.8) on the subsets of the partitions described by the conclusion of Corollary 8

19 .5 to get i S j x, u i S j δ x, (.9) where = S j is the cardinality of the set S j. It appears that utilizing finite fraes will be appropriate in deterining an algorith for partitioning a frae because (.9) is analogous to the definition of a frae with upper bound B = S j δ. What are Finite Fraes? To describe what a frae is we recall the notion of a vector basis. As described by Meyer [], a basis is a linearly independent spanning set of vectors for a vector space V. In other words, a basis is a collection of vectors that can be used to generate all the other vectors through linear cobinations. Moreover, the vectors in this collection are independent of each other. This idea of independence eans that one vector in our collection cannot be fored fro any cobination of the other vectors in the collection. This definition lends it self to soe nice properties involving bases, but at the sae tie akes bases too restrictive when applying the to signal processing and reconstruction probles. We shall walk through two nice exaples of bases: The standard unit vectors in R n The orthonoral basis in a space V 9

20 Standard unit vectors. As an exaple, the standard unit vectors in R are e =, e = 0. 0 Notice that any two values x, y R can be utilized as coefficients for a linear cobination of e and e. If one would like to for a new vector v with coponents x and y, the we siply calculate v = x = xe + ye. y Suppose we let x = and y =, then our exaple vector v is v = e + e =. Since we can choose any x, y R to create a vector v R, then this eans that our two vectors e and e span the real plane R. Also note that e and e are independent of each other. Meaning that the solution to the equation c e + c e = 0 is {c, c } = {0, 0}; therefore, it is ipossible for one vector to be expressed in ters of another vector without triviality. Hence, the set S = {e, e } is a basis in R. In fact, we can generalize this to the set of standard unit vectors S = {e, e,..., e n } e i R n and this also fors a basis in R n. Notice that these vectors do not repeat 0

21 in S, nor do the vectors depend on each other. Orthonoral bases. Another failiar type of basis is the orthonoral basis for a vector space V with an inner product,. The definition is siilar to that of a general basis except we have the following two restrictions []: Each vector has unit length (hence "noral"). The set of vectors for an orthogonal set. In essence, suppose you have a set B = {u, u,..., u n } where each u i is fro an n-diensional vector space V. Then we say that B is an orthonoral basis if for all u i, u j B i = j u i, u j = 0 i = j. Keep in ind you ay loosely think of "orthogonal" to ean perpendicular. Since all the vectors are orthogonal to each other, then no one vector in B can be expressed as the linear cobination of the other reaining vectors. Therefore, the only solution set to the coefficients satisfying the equation c j u j = 0 j is {c, c,..., c n } = {0, 0,..., 0}. This eans the vectors in B are linearly independent and for an orthonoral basis for V. Notice that the nuber of vectors described by the definition of a basis is liited to the nuber of diensions of the vector space V. For instance, the standard

22 orthonoral basis in R n will have exactly n vectors. To suarize, a frae can be thought of as a "relaxed" for of a basis. One of the key coponents is that a frae ay contain repeated vectors and/or vectors that are not linearly independent. The second is that it still spans the vector space. For exaple, let F = {e, e, e, e } which contains linear cobinations of the standard basis in R and still spans R. If we were to utilize these previously discussed bases to transit a signal, the vectors would encounter natural perturbations during transission. Soe of these changes can be attributed to noise, erasures, or other natural phenoena that utate the receiving signal [5]. At any rate with standard unit vectors or orthonoral bases we have a liited nuber of vectors to choose fro and there is no guarantee that all of the extracted signal using these vectors will survive the actual transission process. Fraes reedy this proble by providing redundancy to a transitted signal.

23 Chapter Partitions of Fraes Equinor-Equiangular Fraes Equinor and equiangular fraes in R can be thought of as vectors that have the sae angle between the. We will define this frae in the following way: F = {f j } j=0 = cos( πj ). (.) sin( πj Notice that the nor for every vector f j equals consecutive vectors is always π ) j=0 and the angle between two radians. As an exaple if we let = 3, then our frae would be { 3 F = {f 0, f, f } =, 6, } 6. 0 Figure is the Mercedes-Benz Frae placed in the coordinate plane, called so because it is siilar to the logo of the auto anufacturer Mercedes-Benz. 3

24 Figure. Mercedes-Benz Frae, = 3 4

25 Let us find the frae bounds of (.). To do this we will exaine the suation found in (.4). Let x = {α, β} where α, β R. Then x, f j = ( ) ( ) πj πj α cos + β sin j=0 j=0 = + αβ (α j=0 j=0 ( ) πj cos ( πj cos ) sin ( πj ) + β j=0 ( )) πj sin. (.) If we apply the trigonoetric identities for double angles, (B.3), and (B.4) on the three suations in (.), then we have j=0 ( ) πj cos = ( ( )) 4πj + cos j=0 = + ( ) 4πj cos j=0 =, (.3) j=0 ( πj cos ) sin ( ) πj = j=0 ( ) 4πj sin = 0, (.4) 5

26 and j=0 ( ) πj sin = ( ( )) 4πj cos j=0 = ( ) 4πj cos j=0 =. (.5) Now substitute (.3), (.4), and (.5) back into (.) to get x, f j = ( α + αβ 0 + β ) j=0 = α + β = x. (.6) Interestingly, if we copare the result (.6) to the frae definition (.4) the bounds of the frae are equal, naely A = B =. When A = B the frae is known as a tight frae. If both frae bounds equal, then it is called a Parseval frae. We now show that the frae described by (.) satisfies the hypothesis of MSS 6

27 Corollary.5: j=0 f j f T j = j=0 = j=0 cos ( πj = 0 0 = I. cos ( ) πj ) ( sin πj ) + cos ( ) 4πj sin ( ) 4πj cos ( πj ) sin ( πj sin ( ) πj sin ( ) 4πj cos ( 4πj ) ) It suffices to now find pa rtitions fo r th ese eq uinor-equiangular fr aes guaranteed by MSS Corollary.5, which we investigate next. Partitions of Equinor-Equiangular Fraes In this section we will exaine the partitions of equinor-equiangular fraes. Our analysis will involve partitioning the frae F into two subsets, S and S, where S + S =. We then copare the nor of the su of the outer prod-ucts to the bound in MSS Corollary.5 [0]. First we shall establish a relationship between the vectors corresponding to the subsets S and S to the discriinant cal-culated fro the nor of a atrix. Second we will exaine algoriths for three special cases of partitions: S = S

28 S S S = S + The discriinant proble. We will now begin exaining the nor of the atrix fored fro the su of outer products of vectors corresponding to a subset S k. Recall that [] = S S. Our goal in this section is to show the existence of a relationship between the vectors in a subset to the discriinant corresponding to the nor of the outer product su. First we will derive the su of outer products when S k = S k then we will extend it to the case when S >. Outer Product Su for Vectors in a Subset. Suppose we let S = {a, a } where a, a []. If we utilize the trigonoetric double angle identities and copute the su of the outer products of frae vectors corresponding to the eleents in 8

29 our subset S we have F S = j S f j f T j = cos ( ) πj [ j S sin ( ) πj cos ( πj ) ( sin πj ) ] = j S cos ( ) πj cos ( πj ) sin ( πj ) cos ( πj ) ( sin πj ) ) sin ( πj = + cos ( 4πa sin ( 4πa ) ( + cos 4πa ) ) ( + sin 4πa ) sin ( 4πa cos ( 4πa ) ( + sin 4πa ) ) cos ( 4πa ). Notice that the atrix F S is Heritian. Outer Product Su for n Vectors in a Subset. Next we can generalize this to the case where S = {a, a,..., a n } and S = n is any positive nuber less than. Fix S = p. We have F S = n + j S cos ( 4πj j S sin ( ) 4πj ) j S sin ( ) 4πj n j S cos ( 4πj ) (.) 9

30 and F S = p + j S cos ( 4πj j S sin ( ) 4πj ) j S sin ( ) 4πj p j S cos ( 4πj ). (.8) Therefore, (.) and (.8) give the generalized outer product sus of the frae vectors extracted by the subsets S and S. Our goal is to find the nor (.) and copare it to the bound described by (.). Since we are dealing with two subsets of F we have r = and the largest odulus of each frae vector f j is so that δ =. If suffices to deterine S so that (.) holds: F S ( + ). (.9) Let us deterine the nor of the atrix F S. To siplify our notation, let Ψ = j S cos ( ) 4πj and Ω = j S sin ( ) 4πj. Therefore, (.) becoes F S = n + Ψ Ω Ω. n Ψ 0

31 Then n + Ψ λ F S λi = Ω Ω n Ψ λ = n + Ψ λ Ω Ω, (.0) n Ψ λ Which has characteristic equation n + Ψ λ Ω Ω n Ψ λ = 0. This yields the corresponding quadratic equation in λ: (n λ) Ψ Ω = n nλ + λ (Ψ + Ω ) = λ nλ + n (Ψ + Ω ) = 0. Denote λ, λ to be the solutions of this equation. It follows that F S = ax{λ, λ } = n + Ψ + Ω. (.) By the sae analysis, we obtain F S = p + Ψ + Ω. (.)

32 It appears that the solution to our nor depends on the discriinant in (.) and (.): Ψ + Ω = ( j S cos ( ) ) ( 4πj ( ) ) 4πj + sin. (.3) j S In the next several sections we will exaine how to siplify (.3) for special cases of partitions. Two-Partitions: S = S. Suppose our frae has the following restrictions: The frae F contains an even nuber of vectors, F = q where q Z +. We partition [] into two equal subsets such that [] = S S, where S = {,, 3,..., } and S = { +, +,..., } S = S = n =. The eleents in S and S correspond to consecutive vectors in our frae. Essentially, S corresponds to the first half of F and S corresponds to the second half of F. We are going to calculate the nor of the tensor product sus corresponding to the subsets S and S. With these restrictions described above (.3) becoes / ( ) / Ψ + Ω = 4πj ( ) cos + 4πj sin. (.4) j= j=

33 What are these suations? They do not appear trivial. Let us closely exaine (B.) by replacing and y with and 4π, respectively. Then / ( ) / 4πj sin = j= j=0 sin ( ) 4πj ( = sin 4 4π ) ( ( + ) sin 4π ) 4 ( ) ( + )π = sin π sin sin ( π ) sin ( π = 0. (.5) ) We can repeat the analysis on the cosine ter utilizing (B.) to get / ( ) / 4πj cos = + j=0 j= j= cos ( ) 4πj ( = cos 4 4π ) ( ( + ) sin 4π ) 4 ( = sin π + π ) ( = sin π cos sin ( π ) ( ) π + cos π sin sin ( π ( )) π ) sin ( π ) = / ( ) 4πj cos = 0. (.6) 3

34 We substitute (.5) and (.6) back into (.4) to get that Ψ + Ω = 0. Therefore the nor described in (.) becoes F S = n. Recall one of the restrictions is that n = ; therefore, if we substitute this into the line above then we validate (.9) which is ( F S = + ). Therefore, a subset fored by the first / frae vectors is valid for (.). But what about the second subset S which represents the second half of the frae? Since we are considering S = { +, +,..., }, our discriinant (.3) becoes Ψ + Ω = j=/+ cos ( ) 4πj + j=/+ sin ( ) 4πj. (.) We arrive at a siilar question, what are these suations? If we exaine the cosine suation in (.) we have j=/+ cos ( ) 4πj = ( ) / 4πj cos j= j= cos ( ) 4πj. (.8) 4

35 Notice that the two suations on the right hand side of (.8) were already calculated in (.6) and (B.4); hence, j=/+ cos ( ) 4πj = 0. (.9) We can follow the sae analysis for the sine suation in (.) and arrive at, j=/+ sin ( ) 4πj = 0. (.0) Then we substitute (.9) and (.0) back into (.) to get the sae result for the discriinant Ψ + Ω = 0. Therefore, the nor corresponding to the second subset S will also be, and we have ( F S = F S = + ). Fro here we can conclude that given an even nuber of vectors that for an equinor-equiangular frae F then we can always partition the frae into two subsets with equal cardinality. The subsets are such that they correspond to consecutive vectors in our frae F and the set {S, S } is a valid partition for (.). Two-Partitions: S S. Let us exaine another partition with the following qualities: The cardinalities of S and S are not equal; therefore, S = n and S = p with n + p = and n p. 5

36 We will consider the "nice" case when the indices in S and S are consecutive, S = {,, 3,..., n} and S = {n +, n +, n + 3,...,, }. There are any nuber of vectors in our frae F, where F =. As in the previous section, the proble reduces to deterining the value of the discriinant (.3). The following two subsections are dedicated to finding closed for equations describing the nors of the atrices corresponding to the outer products of frae vectors. The first subsection analyzes S and the second analyzes S. The Subset S. Our goal is to copute the discriinant corresponding to the subset S. Thus we have ( n Ψ + Ω = cos j= ( ) ) ( n 4πj + sin j= We begin by using the series (B.) and (B.) to get n ( ) 4πj sin = sin j= ( π(n ) ) sin ( ) πn ( ) ) 4πj. (.) sin ( π ) (.) and n ( ) 4πj cos = cos j= ( π(n ) ) sin ( ) πn sin ( π ). (.3) 6

37 We substitute (.) and (.3) back into (.) to get ( ) πn Ψ + Ω = sin ( ) πn = sin sin ( π sin ( π ) ( ( ) ( )) π(n ) π(n ) cos + sin ). (.4) We substitute this result into the nor described by (.) and we have F S = n + sin ( ) πn sin ( ). (.5) π Interestingly, this nor depends only on the nuber of vectors n in the subset S. The Subset S. We still need to exaine the second subset S = {n+, n+,..., }. Our discriinant fro (.3) becoes Ψ + Ω = ( j=n+ cos ( ) ) ( 4πj ( ) ) 4πj + sin. (.6) j=n+

38 This is not as nice as before, but we ay utilize a few results involving trigonoetric suations. If we utilize (.) and (B.3) we obtain ( ) 4πj sin = j= n ( ) 4πj sin + j= ( π(n ) 0 = sin j=n+ sin j=n+ ) sin sin ( ) πn ( ) 4πj = sin ( ) 4πj sin ( π ( π(n ) ) + j=n+ ) sin sin ( ) πn ( ) 4πj sin ( π ). (.) If we follow the sae analysis on the cosine ter of (.6), then we get j=n+ cos ( ) 4πj = cos ( π(n ) ) sin ( ) πn sin ( π ). (.8) Substitute (.8) and (.) into (.6) and we arrive at the sae result for the subset S, the discriinant (.4) becoes With p = n, (.) becoes ( ) πn Ψ + Ω = sin F S = n sin ( π ). + sin ( πn ) sin ( ). (.9) π 8

39 This nor ay also be described in ters F S : F S = n + F S = p n + F S. (.30) If we let n p, then F S F S. Notice that equality occurs if we have an even nuber of vectors in our frae, and we would arrive at the sae conclusion fro the section on the S = S partition. We have thus found closed for equations for the nors of the partitioned frae vector sets S and S. However, we need to explore how do the nors corresponding to these subsets behave in relation to the conclusion of the corollary (.)? In the subsequent sections we will analyze the behavior of our atrix nor equations corresponding to the subsets S and S. Fro there we shall identify soe ways in deterining valid partitions of equinor-equiangular fraes for the case where S S. Matrix Nor Analysis. In this section we will analyze the behavior of the nor equations derived in the previous two sections. Since we are only foring two subsets of the frae and they are not equal in size, this results in one subset containing ore than half the frae eleents and the other contains less than half. Hence, with out loss of generality we will let S = n > and as a consequence S = p < n. Our focus will be on analyzing the nors corresponding to the subset S because if n > p then F S > F S. If we can find a way to obtain the subset S, then S is siply its copleent. F 9

40 In Table. there are four figures, each figure is showing how the nor (corresponding to the subset S ) copares to the corollary bound for various cardinalities of S. The horizontal axis represents the nuber of vectors in the subset, S = n, and the vertical axis is the value of the nor. The horizontal dashed line (which is constant since it only depends on ) is the corollary bound fro (.) and the curve is the function described by (.5). Notice in figures.a a nd.b n o atter t he s ize o f t he s ubset S, t he nor always reains less than the bound described by the corollary (.). On the other hand, in figures.c and.d the nors fail when the size of S is sufficiently large. Thus it suffices to find the largest value for n where the nor satisfies the corollary bound? I will refer to this special value as n ax and it will be the size of the largest S subset of an vector frae just before we fail the corollary bound. Visually, in the figures n ax is the intersection of the horizontal line and the curve, assuing they intersect at all. 30

41 Table Coparison of frae atrix nors to Corollary.5.0 Partition Size vs. Nor.0 Partition Size vs. Nor.5.5 FS.0 FS S n S n (A) = 0 frae vectors (B) = 0 frae vectors.0 Partition Size vs. Nor.0 Partition Size vs. Nor.5.5 FS.0 FS S n S n (C) = 30 frae vectors (D) = 40 frae vectors 3

42 Below is another figure deonstrating an extree case. Notice that for such a large value it appears n ax is roughly half of..0 Partition Size vs. Nor.5 FS S =n Figure. = 0000 frae vectors. Table is a table of values for various sizes of finite fraes,, and n ax. For instance, if we exaine the third row when = 40 we have a subset containing 3 vectors and it is valid with the corollary, but as soon as we go beyond to 38 vectors in the subset then we fail the corollary. Hence, n ax = 3 for a frae consisting of 40 vectors. 3

43 Table Max Subset Size (n ax ) Until Failure F = n ax F S Cor.-.5 n ax /

44 Interestingly, when there are 3 or fewer vectors contained in the frae any size of S will be valid. Once we go beyond 3 vectors in a frae then we need to solve an equation in order to find n ax. Recall that for extreely large it appears that the largest possible subset size approaches n ax = /. This is easily shown by taking the liit of (.): li ( + ) =. Next, we solve the equation F S = ( + ). This equation can be written in ters of the equation for F S (.5) to get x + sin ( πx ) ( sin ( ) = π + The solution x is related to n ax in the following way ). (.3) n ax = x. Equation (.3) is not eleentary, and we use Matheatica to solve this equation for x, and therefore n ax, for various values of in Table. In conclusion, a valid partition {S, S } of [], where S = S, can always be fored if S n ax. 34

45 Alost even split partition. Let us exaine another partition, this tie we have the following criteria: F is odd or F = q + where q Z +. Let S = n = + and S = p =. We ake the above substitutions into (.5) to get F S = + = + + ( ) sin π(+) + sin ( ) π ( sin π + π ) sin ( π ) cos ( π = + cos ( ). π Next we need to deterine when this quantity less than the bound (.). Let us exaine the following inequality proble: ) + cos ( ) π + + cos ( ) + π cos ( ) + π cos ( ) + 3 π. (.3) 35

46 The inequality (.3) holds for all positive integers.we can follow this sae procedure if we substitute n = into (.9). Then (.3) becoes cos ( π ) + 5 and we arrive at the sae conclusion as before. Therefore, given a frae F with an odd nuber of vectors we can always a for a valid partition {S, S } of [] such that S = + and S =. 36

47 Chapter 3 Grassannian Fraes Grassannian Fraes Grassanian fraes are a class of fraes that satisfy a condition involving axial frae correlation. The defintion provided by Stroher and Heath [4] of axial frae correlation is the following: Definition 3.. For a given unit nor frae {f j } n j= in E we define the axial frae correlation M ( {f j } n j=) by M ( {f j } n j=) = ax j,k,j k { f j, f k }. (3.) As an exaple suppose we have a unit frae with three vectors {v, v, v 3 } R. We can think of the axial correlation of this set of vectors as the largest agnitude of their dot products. Suppose, denotes the dot product of two vectors and that v, v = 0.8, v, v 3 =, and v, v 3 = 0.5. Then the axial frae correlation M ({v, v, v 3 }) =. As a side note if the set of vectors are orthonoral then the axial correlation is 0. The Grassannian frae is defined as follows [4]: Definition 3.. A sequence of vectors {u j } n j= in E is called a Grassannian frae if it is the solution to in{m ( {f j } n j=) } (3.) 3

48 where the iniu is taken over all unit nor fraes {f j } n j= in E. In other words, if we have a set of finite fraes, we find th e s allest of th e axial frae correlations. Then a frae that has a axial frae equal to this is a Grassannian frae. For our purposes we will be working in C. In the subsequent sections we will go through the construction of Grassannian fraes using skew-syetric conference atrices. Then we will deonstrate that the optial Grassannian fraes eet the hypothesis of Corollary.5. We will also exaine the inner products of Grassannian frae vectors. The last section will be dedicated to exaining the partitions for Grassannian fraes for various frae sizes. Constructing Optial Grassannian Fraes We will explore optial Grassannian fraes and this occurs when n =. There-fore, the iniu of the axial correlations is in{m ( {f j } n j= ) } = n (n ) =. Stroher and Heath give a construction of optial Grassannian fraes [4] using a recursive algorith. atrix We first begin with a skew-syetric conference C =

49 and define higher order ones by the recurrence C = C C I. C + I C Then we define R := R = αc + I, (3.3) where α = i and n =. Notice that this algorith will generate conference atrices that are sizes of, 4, 8, 6, and so on, which are powers of. As an exaple if we let =, then n = 4 and we will have the 4 4 conference atrix 0 C 4 = C = C C I 0 =. C + I C 0 0 Therefore, α = i 3 and our R-atrix is R = i 3 i 3 i i 3 3 i 3 i 3 i 3 i 3 i 3 i 3 i 3 i 3. Next we take our R-atrix and copute the spectral factorization R = WΛW, (3.4) 39

50 where W is a unitary atrix with coluns that are the eigen-vectors of R and Λ is a n n diagonal atrix with diagonal entries being the eigenvalues of the R-atrix. For this thesis we are going to analyze optial Grassannian fraes and let n =. Fro Stroher and Heath s construction it is known that the eigenvalues are 0 and, each with ultiplicity. Thus Λ is a diagonal atrix that has s in the left half and 0 s in the right half. Going back to our exaple when n = 4 we have Λ = Next we define the frae vectors f k to be f k = n {W k,l} l= = {W k,l } l=. (3.5) Our frae F is given by F = {f k } n k=. (3.6) In essence, we take the W-atrix (with eigen-vectors as its coluns) and isolate the first coluns corresponding to the eigenvalue λ =. We then take the rows of these first coluns and ultiply the by the coefficient. These rows 40

51 constitute the frae vectors f k. Notice that, ω T R = ω ω ω ω T n ω T n = ω ω + ω ω + + ω ω = ω k ωk. (3.) k= This verifies that our frae need only constitute vectors fro the left half of W. Notice that a bit of diension analysis verifies (3.), naely [n n] = [n n][n n][n n]. Grassanian Fraes and the Hypothesis of Corollary.5 In this section we will prove that Grassanian fraes eet the hypothesis of Corollary.5 if we oit the coefficient fro our original definition (3.5). Let ω k C n be the eigen-vectors of R and φ j C be the rows of W, where W consists of the vectors ω k as coluns: φ T W = ω ω ω = φ T. φ T n 4

52 and (3.) becoes R = WW. (3.8) Again with soe diension analysis we have [n n] = [n ][ n]. Recall that W was coputed fro the spectral factorization of R, and W is unitary. This eans WW = W W = I and the coluns and rows of W for an orthonoral basis. Since W consists of the first coluns of W, then the coluns of W are orthonoral. If we utilize the standard orthonoral vectors in n-space we can re-express our frae vectors fro (3.5) as f k = W k = W T e k = φ k. Hence f = W T e = φ f = W T e = φ. f n = W T e n = φ n. 4

53 If we now attept to verify the hypothesis of the corollary (.), then we get n n ( ) f k fk = W T e k W T e k k= k= n = W T e k e T k W k= ) ( n = W T e k e T k W k= = W T I n n W ω T ω T = ω ω ω ω T ω T ω ω T ω 0 0 = ω T ω ωω T = I. 43

54 Therefore, our frae does not quite satisfy the hypothesis (.), but this can be easily rectified by rescaling each f k by a factor of : f k = {W k,l } l= = W k = W T e k = φ k F = {f k } n k=. (3.9) If we now repeat the sae analysis as above we get n fk f k = I. k= Hence our redefined frae (3.9) satisfies the hypothesis (.) in Corollary.5. Inner Products of Grassannian Frae Vectors In this section we will analyze the inner products of the vectors that are fored in W and W. Recall that since we have the spectral decoposition of R = WΛW = WW, then the coluns of W and W constitute an orthonoral basis, i.e., for all k, j [n], j = k; ω j, ω k = 0 j k. (3.0) So clearly the coluns of W are noral and orthogonal, but we are interested in the rows of W because this constitutes our frae F which we defined in (3.9). Notice that we can express each row of W using our standard orthonoral vectors φ k = W T e k. (3.) 44

55 We will now derive the nor of the vector φ k : φ k = φ kφ k = ( W T e k ) W T e k = e T k WW T e k = et k R T e k =. Therefore φ k =. (3.) More generally the inner product of φ k, φ j is given by φ kφ j = ( W T e k ) W T e j = e T k WW T e j = et k R T e j. This extracts an entry that is off the diagonal of R. Hence, φ kφ j = ±i n. (3.3) The cobined results of (3.3) and (3.) and (3.9) yield j = k; f j, f k = φ j, φ k = ±i j k. n (3.4) 45

56 Recall that the vectors φ k are the rows of W, which are our frae vectors f k. More iportantly notice that our frae vectors are not orthogonal. This eans our calculation for the nor of the outer product su will be ore involved. Nors of Outer Products of Grassannian Frae Vectors ( Vectors) Now we will attept to establish a closed forula for the nor of the outer prod-uct of the su of frae vectors. First, if we utilize the result fro (A.4) and (3.4) we have f k fk = f k =. (3.5) Hence the nor of the outer product of one frae vector has a nice closed for. Next we calculate the nor of the su of two frae vectors: ( ) F = f k fk + f j fj = W T e k W T ( ) e k + W T e j W T e j = W T e k e T k W + W T e j e T j W = W ( ) T e k e T k + e j e T j W. (3.6) Observe that F is Heritian, i.e., F = F. If we want to find F, then we need to copute Fx where x is any unit vector. Hence, we utilize the results of (A.8) to get Fx = x ( f kf k + f k f k f j f j + f j f j f k f k + f jf j ) x. Reeber that fk f j is equal to the off diagonal entries of R and R is Heritian. Therefore f k f j = f j f k = ρ, 46

57 where ρ = i. We then have n Fx = x ( f kf k + ρ ( f k f j f j f k ) ) + f jfj x. (3.) Observe that f k f k = W T e k e T k W f k f j = W T e k e T j W. As a result, equation (3.) becoes Fx = x W T ( e ke T k + ρ ( e k e T j e j e T k ) ) + e je T j Wx. (3.8) While the equation above appears nice, it is not trivial to use Lagrange ultipliers to solve for λ ax. We will exaine soe particular exaples for n = 4, 8, 6, and 3 and utilize nuerical analysis to deterine what partitions are valid with respect to the Corollary.5. Case n = 4. Let us exaine the case when n = 4. This has the iplication that there will be 4 frae vectors in C. Below is the corresponding R atrix: R = i 3 i 3 i i 3 3 i 3 i 3 i 3 i 3 i 3 i 3 i 3 i 3. (3.9) 4

58 An orthonoral basis for W can be chosen so that W = 6 6 ( 3i 3 ) ( 3i + 3 ) i 6 6 ( ) i 6 6 3i 3 i ( ) 3i i i 3 6 i 6, (3.0) where the coluns of W are the eigenvectors of R. We can then for our frae using (3.9) by siply extracting the first two coluns of (3.0): W = 6 6 ( 3i 3 ) i 6 ( 3i + 3 ) i i 6. (3.) Each row of W is a frae vector f k ; hence there are 4 vectors in C : F = {f k } 4 k= ( ) 6 3i 3 =, i 6 6 ( ) 3i + 3 0,, i 6 3 i 6. (3.) The nor of each frae vector is. Let us now describe our goal: The four frae vectors {f, f, f 3, f 4 } correspond to the four eleents [4] = {,, 3, 4}. 48

59 We need to find a partition {S,..., S r } of [4] such that S S r = [4], and the nors of the su of outer products of frae vectors corresponding to each subset S k is satisfies bound (.). Let us exaine soe partitions. Recall the atrix F which is the su of the outer products of vectors corresponding to a subset S k. Table 3 Subset size and the corresponding nors for n = 4 case. S k = { } F {} 0.5 {} 0.5 {3} 0.5 {4} 0.5 {, } {, 3} {, 4} {, 3} {, 4} {3, 4} {,, 3} {,, 4} {, 3, 4} {, 3, 4} {,, 3, 4} 49

60 Fro Table 3 it appears that the nor of F takes on only certain values depending only on the size of S k. It is clear fro Corollary.5 that r n = 4. (3.3) Recall that δ =. Thus we can anipulate (3.3) to get r 4 r r + r + + ( + ) ( r + ) ( + ) 3 + ( r + ) = (3.4) This shows that the corollary bound is greater than unity for all r. If we copare this to Table 3. then any subset corresponding to our frae F will satisfy (.) for the case n = 4. An analytic approach for n = 4. We copute analytically the nor of F for a subset S = {, } corresponding to the two vectors f, f. Recall that each frae 50

61 vector f k is obtained fro a row of W. Therefore f k = w k, w k, Suppose we rotate f by a unitary atrix U to obtain f = Uf = {a, 0} T so that f = aa = f =. Then f = Uf = {b, c} T and f = bb + cc = f =. But notice that f, f = {a, 0} {b, c} = ab = f, f, where b = f, f a b = f, f a We can calculate bb and cc to get bb = f, f f, f aa = f, f f = 6 and cc = bb = 3. 5

62 Thus F = f f + f f = aa + bb bc bc. cc To find the nor of the above atrix we solve the characteristic equation (aa + bb λ)(cc λ) bbcc = 0. Interestingly, we know all of the quantities and can therefore solve for λ. Hence, ( ) ( ) 3 λ 3 λ 8 = 0. Which becoes, λ λ + 6 = 0 The solutions to this equation are {λ, λ } = { 6 is equal to the largest eigenvalue; hence, F = ax{λ, λ } = 6 ( ) ( ) 3 3, }. The nor ( 3 + ) 3 = This nor is exactly the value fro Table 3.. Suppose we consider the case where S = {,, 3} where a third vector f 3 is added. We find the corresponding vector f 3 = Uf 3 = {d, e} T. If we repeat a siilar analysis as we did for f we get that dd = and ee =. Our su of outer 6 5

63 products is F = f f + f f + f 3 f 3 = aa + bb + dd bc + de bc + de. cc + ee The corresponding characteristic equation becoes ( ) ( ) 5 3 λ 6 λ 8 = 0, which siplifies to λ 3 λ + = 0. The solutions to this equation are {λ, λ } = {, }. The nor is equal to the largest eigenvalue; hence, F = ax{λ, λ } =. This nor is exactly the value fro Table

64 Case n = 8. Let us consider n = 8. This will yield eight f k vectors in C 4, where the R- atrix is given by R = i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i (3.5) The corresponding W-atrix is W = i(i+ ) 4 i(i+ ) 4 i 4 3+i 4 i i(i+ ) 3i i 4 3 +i +i i i i 4 i 4 i 4 i i( i+ ) i i(i+ ) i 4 i 4 i 4 (3.6) Each row in (3.6) is one frae vector f k. Thus our frae F = {f,..., f 8 }. We now consider the proble of finding a partition {S,..., S r } of [8] such that [8] = {,..., 8} = S S r, where r is between and 8. In order to exaine 54

65 the nors corresponding to the subsets of the partition, we need to calculate the nors of all the different possible cobinations of the vectors. There are n = 8 = 55 possible cobinations of picking a set of vectors out of F. Table 4 shows the corresponding nors for the given cobinations of vectors. As before, we observe that the nors only take on specific values. Table 4 Subset size and the corresponding nors for n = 8 case S k F or Let us perfor the sae analysis on the nuber of subsets r as we did in the n = 4 case to estiate the bound in Corollary.5: r 8 r r 4 + r ( + ) ( r + ) ( 3 ) 4 ( + ) ( r + ) 9 8 =.5 (3.) Notice that our corollary bound is again always greater than unity; hence, any 55

66 partition of F will satisfy the corollary.5. Interestingly, our bound has decreased fro for the n = 4 case to.5 for n = 8 case. This is to be expected because as r increases the corollary bound will approach 0.5. We need to exaine larger fraes to see if this continues to hold and we need to find a relationship between n and r. Case n = 6. If we repeat the sae procedure for the case of n = 6 we have that the entries of the 6 6 R-atrix is given by j = k; R j,k = ±i 5 j k. (3.8) The corresponding W-atrix with rows that are the frae vectors f k is given by 56

67 W = i i ( 5i + 5 ) i i i 05 i 05 5+i ( ) 5i i ( 5i + 5 ) i i(i+ 5) 0 5+3i ( + i 5 ) 5 i 5 +3i i ( ) i 5 5 i i i i i 3 5 i(i+ 5) 6 0 i(i+ 5) 6 0 i i i i i i i 5 30 i(i+ 5) 05 +i i 5 05 i i i i i 5 05 i 5 05 i 30 i 30 i 30 i 30 i 30 i 30 i i i i 3 5 i( i+ 5) i i 5 05 i 30 i i 05 +i i i i i i( i+ 5) i 4 5 i i(i+ 5) 05 i( i ) 3 0 i 3 5 i 3 5 i 5 i( i+ 5) 6 0 +i i(3i+ 5) i( i+ 5) 05 i(i+ 5) i i i 5 30 i( i+ 5) 05 i(i+ 5) 05 i( i+ 5) 05 i i(i+ 5) i 30 i 30 i 30 i 30 i 30 i 30 i 30 i 30 5

68 There are 6 possible subsets of F. Table 5 shows the nors corresponding to a subset S k with cardinality k. Table 5 Subset size and the corresponding nors for n = 6 case. S k F or or or or or or or or or... or or or or or or or 9 Two noticeable observations for n = 6: As we increase the nuber of eleents in a subset the nuber of values, not equal to, that the corresponding nor takes on increases. For instance, a subset with 4 eleents, S k = 4, corresponded with two possible nors not equal to. If a subset contained 6 eleents, S k = 6, this corresponded to five possible nors not equal to. If the cardinality of a subset is greater than 9, the corresponding nor is always. Nine is one ore than half the cardinality of F. 58

69 If we analyze the range of values on r we have, r 6 r 4 r 4 + r r 4 ( + ) ( r + ) ( + ) 4 = = Notice that the corollary bound falls below unity; therefore, the lowest that our corollary bound ay achieve occurs when r = 6 and the bound is The question that reains is whether there exists a subset S k with corresponding nor that is greater than the corollary bound. Keep in ind that the size of the largest subset will dictate the nuber of r partitions that can be fored fro [6] = {,..., 6}. For instance, suppose S = 8. There are only eight reaining vectors left in our frae to distribute aong the other subsets; therefore, the largest nuber of subsets that can be fored, with one subset of size equal to eight, is r = 9. This eans the set of cardinalities for the nine subsets is { S k } 9 k= = {8,,,,,,,, }. In Table 6, which shows the corollary bounds for all the possible r subsets, we see that that the bound for r = 9 is.085. Since all the corresponding nors in Table 5 fall below.085, this shows that {S k } 9 k= 59