Uniform Tight Frames with Erasures - Semantic Scholar

Uniform Tight Frames with Erasures Peter G. Casazza

Department of Mathematics, University of Missouri-Columbia Columbia, MO 65211 E-mail: [email protected]

and Jelena Kovacevic

Mathematics of Communication Research, Bell Labs, Lucent Technologies, 600 Mountain Avenue, Murray Hill, NJ 07974 USA E-mail: [email protected]

Uniform tight frames have been shown to be useful for robust data transmission. The losses in the network are modeled as erasures of transmitted frame coecients. We give the rst systematic study of the general class of uniform tight frames and their properties. We search for ecient constructions of such frames. We show that the only uniform tight frames with the group structure and one or two generators are the generalized harmonic frames. Finally, we give a complete classication of frames in terms of their robustness to erasures.

1. INTRODUCTION

Frames are redundant sets of vectors in a Hilbert space which yield one natural representation for each vector in the space, but which may have innitely many dierent representations for a given vector 4, 5, 6, 7, 8, 9, 10, 11, 16, 18, 20, 23]. Frames have been used in signal processing because of their resilience to additive noise 10], resilience to quantization 14], as well as their numerical stability of reconstruction 10], and greater freedom to capture signal characteristics 2, 3]. Recently, several new applications for (uniform tight) frames have been developed. The rst, developed by Goyal, Kovacevic and Vetterli 15, 25, 24, 26], uses the redundancy of a frame to mitigate the eect of losses in packet-based communication systems. Modern communication networks transport packets of data from a Some of the results in this paper were reported at the SPIE Conf. on Wavelet Appl. in Signal and Image Proc. (San Diego, CA, July 2001). The rst author was supported by NSF DMS 0102686. 1

2

CASAZZA AND KOVAC EVIC

source to a recipient. These packets are sequences of information bits of a certain length surrounded by error-control, addressing, and timing information that assure that the packet is delivered without errors. This is accomplished by not delivering the packet if it contains errors. Failures here are due primarily to buer overows at intermediate nodes in the network. So to most users, the behavior of a packet network is not characterized by random loss, but by unpredictable transport time. This is due to a protocol, invisible to the user, that retransmits lost packets. Retransmission of packets takes much longer than the original transmission. In many applications, retransmission of lost packets is not feasible and the potential for large delay is unacceptable. If a lost packet is independent of the other transmitted data, then the information is truly lost to the receiver. However, if there are dependencies between transmitted packets, one could have partial or complete recovery despite losses. This leads us naturally to use frames for encoding. The question, however, is: What are the best frames for this purpose? With an additive noise model for quantization, in 25], the authors show that a uniform frame minimizes mean-squared error if and only if it is tight. So it is this class of frames - the uniform normalized tight frames - which we seek to identify and study. Another recent important application of uniform normalized tight frames is in multiple-antenna code design 17]. Much theoretical work has been done to show that communication systems which employ multiple antennas can have very high channel capacities 13, 21]. These methods rely on the assumption that the receiver knows the complex-valued Rayleigh fading coecients. To remove this assumption, in 19] new classes of unitary space-time signals are proposed. If we have N transmitting antennas and we transmit in blocks of M time samples (over which the fading coecients are approximately pconstant), then a constellation of K unitary space-time signals is a (weighted by M ) collection of M N complex matrices fk g for which k k = I . The nth column of any k contains the signal transmitted on antenna n as a function of time. The only structure required in general is the time-orthogonality of the signals. Originally it was believed that designing such constellations was a too cumbersome and dicult optimization problem for practice. However, in 19], it was shown that constellations arising in a \systematic" fashion can be done with relatively little eort. Systematic here means that we need to design high-rate space-time constellations with low encoding and decoding complexity. It is known that full transmitter diversity (that is, where the constellation is a set of unitary matrices whose dierences have nonzero determinant) is a desirable property for good performance. In a tour-de-force, in 17], the authors used xed-point-free groups and their representations to design high-rate constellations with full diversity. Moreover, they classied all full-diversity constellations that form a group, for all rates and numbers of transmitting antennas. For these applications, and a host of other applications in signal processing, it has become important that we understand the class of uniform normalized tight frames. In this paper, we make the rst systematic study of such frames. In Section 2 we review basic notions on frames. In particular, we state the Naimark's Theorem 1] which has been rediscovered several times in recent years 16, 12] although it has been used for several decades in operator theory. We give examples of uniform

UNIFORM TIGHT FRAMES WITH ERASURES

3

normalized tight frames such as harmonic and Gabor frames. For harmonic frames, we dene a more general class { general harmonic frames (GHF) { and study when harmonic frames are equivalent to each other and general harmonic frames. In Proposition 2.4, we give a simple equivalence condition on general harmonic frames which states that the inner product between any two frame vectors 'i and 'j in a GHF is equal to the inner product between 'i+1 and 'j+1 . Section 3 concentrates on uniform tight frames (UTF). We start with a review in Section 3.1 and proceed with several ways of classifying UTFs including providing a correspondence between subspaces of the original space and the UTFs, obtaining UTFs as alternate dual frames for a given frame as well as nding UTFs through frames equivalent to them. In Section 4, we shift our attention to UTFs with group structure. We show that ;1 the UNTFs generated by the family fU k '0 gM k=0 (where U is unitary and '0 2 H ) are precisely the GHFs. We then extend our discussion to UNTFs generated by fU k V j '0 g and higher numbers of generators. Finally, Section 5 introduces erasures modeled as losses of transform coecients hf 'i i, where f is the signal to be transmitted and f'i gi2I is a set of frames vectors corresponding to erased transform coecients. We give a complete classication of frames with respect to their robustness to erasures. We study when we can obtain frames robust to a certain number of erasures as a projection from another frame with a dierent number of erasures.

2. FRAME REVIEW A set of vectors = f'i gi2I in a Hilbert space H , is called a frame if 0 < Akxk2

X i2I

jhx 'i ij2 < B kxk2 < +1 x 6= 0

(1)

where I is the index set and the constants A B are called frame bounds. Although many of our results hold in more general settings, in this paper, we concentrate mostly on the N -dimensional real or complex Hilbert spaces RN and C N (which we denote H N ) with the usual Euclidean inner product. When results generalize to the innite-dimensional setting, we will point it out. When A = B the frame is tight (TF). If A = B = 1, the frame is normalized tight (NTF). A frame is uniform (UF) if all its elements have the same norm c,1 k'i k = c. For a uniform tight frame (UTF), the frame bound A gives the redundancy ratio. A UNTF of norm-1 vectors is an orthonormal basis (ONB). The analysis frame operator2 F maps the Hilbert space H into `2 (I ) (Fx)i = h'i xi

(2)

Often, uniform is used to mean c = 1 here, however, we use the more general denition. is sometimes called just frame operator. Here, we use analysis frame operator for F , synthesis frame operator for F and frame operator for F F . 1 2F

4


for i 2 I . When H = H N , the analysis frame operator is an M N matrix whose rows are the transposed frame vectors 'i :

F =

0 ' ::: ' 1 B@ ...11 : : : 1...N CA :

(3)

'M 1 : : : 'MN We say that two frames f'i gi2I and fi gi2I for H are equivalent if there is an invertible operator L on H for which L'i = i for all i 2 I , and they are unitarily equivalent if L can be chosen to be a unitary operator. A direct calculation shows that fS ;1=2 'i g is a normalized tight frame for any frame f'i g. Thus, every frame is equivalent to a normalized tight frame. If f'i gi2I is a frame with frame bounds A B , and if P is an orthogonal projection on H , then by (1) we have that fP'i gi2I is a frame for P H with frame bounds A and B . In particular, if f'i gi2I is a

normalized tight frame for H (for example, if it is an orthonormal basis for H ), then fP'i gi2I is a normalized tight frame for P H . The following theorem tells us that every normalized tight frame can be realized as a projection of an orthonormal basis from a larger space. It serves as a converse to the observation above that orthogonal projections of normalized tight frames produce normalized tight frames for their span. Theorem 2.1 (Naimark 1], Han & Larson 16]). 3 A family f'i gi2I in a Hilbert space H is a normalized tight frame for H if and only if there is a larger Hilbert space H K and an orthonormal basis fei gi2I for K so that the orthogonal projection P of K onto H satises: Pei = 'i , for all i 2 I . Let us now go through certain important frame notions. Using the analysis frame operator F , (1) can be rewritten as

AI F F BI:

(4)

We call S = F F the frame operator. It follows that S is invertible (Lemma 3.2.2 in 10]), and furthermore

B ;1 I S ;1 A;1 I:

(5)

It follows that f'i gi2I is a normalized tight frame if and only if F is an isometry. Also, S is a positive self-adjoint invertible operator on H and S = AI if and only if the frame is tight. In nite dimensions, the canonical dual frame of is a frame ;1 M dened as ~ = f'~i gM i=1 = fS 'i gi=1 , where

'~i = S ;1 'i

(6)

3 This theorem has been rediscovered by several people in recent years: The rst author rst heard it from I. Daubechies in the mid-90's. Han and Larson rediscovered it in 16] they came up with the idea that a frame could be obtained by compressing a basis in a larger space and that the process is reversible (the statement in this paper is due to Han and Larson). Finally, it was pointed out to the rst author by E. S oljanin 22] that this is, in fact, the Naimark's theorem, which has been widely known in operator theory and has been used in quantum theory.

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for k = 1 : : : M . Now,

f =

X i2I

hf S ;1 'i i'i for all f 2 H:

So the canonical dual frame can be used to reconstruct the elements of H from the frame. However, there may be other sequences in H which give reconstruction. This formula points out both the strengths and weaknesses of frames. First, we see that every element f 2 H has at least one natural series representation in terms of the frame elements. Also, this element may have innitely many other representations. However, in order to nd this natural representation of f , we need to invert the frame operator, which may be dicult or even impossible in practice. The best frames then are clearly the tight frames since in this case the frame operator becomes a multiple of the identity. Noting that '~i = 'i S ;1 and stacking '~1 , '~2 , : : :, '~M in a matrix, the frame operator associated with ~ is

F~ = FS ;1 :

(7)

Since F~ F~ = S ;1 , (5) shows that B ;1 and A;1 are frame bounds for ~ . Another important concept is that of a pseudo-inverse F y . It is the analysis frame operator associated with the dual frame,

F y = F~ :

(8)

Note that for any matrix F with rows 'i

S = F F =

M X i=1

'i 'i :

(9)

This identity will prove to be useful in many proofs.

2.1. The Role of Eigenvalues

The product S = F F will appear everywhere and its eigenstructure will play an important role. Denote by k 's the eigenvalues of S = F F . We now summarize the important eigenvalue properties. General Frame. For any frame in H N , the sum of the eigenvalues of S = F F , equals the sum of the lengths of the frame vectors: N X k=1

k =

M X i=1

k'i k2 :

(10)

Uniform Frame. For a uniform frame, that is, when k'i k = c, i = 1 : : : M , N X k=1

k =

M X i=1

k'i k2 = M c2 :

(11)


6

Tight Frame. Since tightness means A = B , for a TF, we have from (1) M X i=1

jhf 'i ij2 = Akf k2

(12)

for all f 2 H N . Moreover, according to (5), a frame is a TF if and only if

F F = A IN :

(13)

Thus, for a TF, all the eigenvalues of the frame operator S = F F are equal to A. Then, using (10), the sum of the eigenvalues of S = F F is as follows: N X

N A =

k=1

k =

M X i=1

k'i k2 :

(14)

Normalized Tight Frame. If a frame is an NTF, that is, A = B = 1, then M X i=1

jhf 'i ij2 = kf k2

(15)

for all f 2 H N . In operator notation, a frame is an NTF if and only if

S = F F = IN :

(16)

For an NTF, all the eigenvalues of S = F F are equal to 1. Then, using (10), the sum of the eigenvalues of S = F F is as follows:

N =

N X k=1

k =

M X i=1

k'i k2 :

(17)

Uniform Tight Frame. From (11) and (14), we see that

N A =

N X k=1

k =

M X i=1

k'i k2 = M c2 :

(18)

Then, from (12) and (18), M X

jhf 'i ij2 = M c2 kf k2 N i=1

(19)

2 A = M N c :

(20)

for all f 2 H N . The redundancy ratio is then

Since S = F F = (M=N )I , the following is obvious: M X

j'ik j2 = M N: i=1

(21)


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Uniform Normalized Tight Frame. If a frame is a UNTF, that is, we also ask for A = B = 1, and if the frame vectors have norm c, then

N =

N X k=1

k =

M X i=1

k'i k2 = c2 M:

Thus, a UNTF with norm c = 1 is an orthonormal basis.

2.2. Examples of Uniform Normalized Tight Frames

We start with a simple example of a frame" 3 vectors in 2 dimension. The particular frame we examine is termed Mercedes-Benz (MB) frame (for obvious reasons, just draw the vectors). The UNTF version of it is given by the analysis frame operator 1 r 0 p0 1 2 F = 3 @ ;p 3=2 ;1=2 A : (22) 3=2 ;1=2 This is obviously a UNTF since (2=3)F F = I . We could make this frame just a UTF with norm 1 (as in 25]) by having the analysis frame operator be simply F . We now review two general classes of uniform tight frames which are commonly used. The (general) harmonic frames and the tight Gabor frames. More general classes of uniform tight frames are the full-diversity constellations that form a group given in 17].

2.3. Harmonic Frames

Harmonic tight frames (HTF) are obtained by keeping the rst N coordinates of an M M discrete Fourier transform basis as in Naimark's Theorem 2.1. They have been proven to be useful in applications 19]. An HTF is given by:

'k = (w1k w2k wNk )

(23)

for k = 0 : : : M ; 1, where wi are distinct M th roots of unity. This is a TF { F F = MI . A more general denition of the harmonic frame (general harmonic frame) is as follows: 1 N Definition 2.1. Fix M N , jcj = 1 and fbi gN i=1 with jbi j = pM . Let fci gi=1 be distinct M th roots of c, and for 0 k M ; 1, let

'k = (ck1 b1 ck2 b2 : : : ckN bN ):

;1 Then f'k gM k=0 is a uniform normalized tight frame for H N called a general harmonic frame (GHF). Note the dierence between the HTFs and GHFs" the former ones are only tight, while the latter ones are uniform, normalized and tight. We will now examine the general harmonic frames in detail. Our goal is to determine when two such frames are unitarily equivalent. We start by showing that the harmonic frames (not GHFs) are unique up to a permutation of the orthonormal basis (that is, if and only if their columns are permutations of each other).


8

;1 ;1 Proposition 2.1. Let f'k gM and fk gM be harmonic frames on H N . k =0 k =0 ;1 M ;1 if and only if there is a permutation of Then f'k gM is equivalent to f g k k=0 k=0 f1 2 : : : N g so that 'kj = k(j) , for all 0 k M ; 1 and all 1 j N . Hence,

two harmonic frames are equivalent if and only if they are unitarily equivalent.

Proof (Proposition 2.1). Let fej gNj=1 be the natural orthonormal basis for H N and

'k = (w1k w2k wNk ) = and

k = (v1k v2k vNk ) =

N X

wjk ej

j =1 N X j =1

vjk ej

where wj vj are sets of distinct M th roots of unity. If

fwj j1 j N g 6= fvj j1 j N g then, without loss of generality, we may assume that v1 2= fwj j1 j N g. Therefore, MX ;1 k=0

v k1 'k =

On the other hand,

N MX ;1 X j =1

MX ;1 k=0

k=0

( v 1 w j )k

vk1 k (1) =

!

MX ;1 k=0

ej =

N X j =1

0 ej = 0:

jv1k j2 = M:

;1 M ;1 It follows that f'k gM k=0 is not equivalent to fk gk=0 . The other direction is immediate. We now show that every general harmonic frame is unitarily equivalent to a simple variation of a harmonic frame. Proposition 2.2. Every general harmonic frame is unitarily equivalent to a ;1 M ;1 frame of the form fck k gM k=0 , where jcj = 1 and fk gk=0 is a harmonic frame.

p

Proof (Proposition 2.2). Let M N , jcj = 1 and jbk j = 1= M , for all 1 k N . Let fcj gNj=1 be distinct M th roots of c and let the GHF be

'k = (ck1 b1 ck2 b2 : : : ckN bN ) for all 0 k M ; 1. If c = ei , let d = ei=M . Then there exist distinct M th roots of unity w1 w2 : : : wN with cj = ei=M wj = dwj . For all sets of complex ;1 numbers fak gM k=0 we have:

k

MX ;1 k=0

ak 'k k2 =

N MX ;1 X

N MX ;1 X

ak ckj bj j2 = M1 j ak ckj j2 = j =1 k=0 j =1 k=0

j


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N MX ;1 M ;1 1 X ik=M wk j2 = 1 k X ak dk k k2 j a e k j M j=1 k=0 M k=0 ;1 where k = (w1k w2k : : : wNk ). Thus, the GHF f'k gM k=0 is unitarily equivalent to M ; 1 M ; 1 1 p fk g k=0 , with fk gk=0 an HTF. M Finally, we show that the frames given in Proposition 2.2 are not unitarily equivalent to each other or to harmonic frames except in the trivial case. ;1 ;1 Proposition 2.3. Let f'k gM and fk gM be harmonic frames and let jcj = k =0 k =0 M ;1 M ;1 k 1. Then fc 'k gk=0 is equivalent to fk gk=0 if and only if c is an M th root of unity and there is a permutation of f1 2 : : : N g so that 'kj = k(j) , for all

0 k M ; 1 and all 1 j N . In particular, a general harmonic frame is equivalent to a harmonic tight frame if and only if it equals a harmonic tight frame.

P

Proof (Proposition 2.3). Let 'k = Nj=1 wjk ej , for all 0 k M ; 1. If c = wj;1 , P ;1 = 0, if fck ' g is equivalent to for some j , we are done. Otherwise, since M k k=0 k P fk g then also Mk=0;1 ck 'k = 0. Hence, for all 1 j N we have MX ;1 k=0

ck wjk =

MX ;1

j )M = 0: (cwj )k = 1 ;1 ;(cw cwj k=0

Hence, (cwj )M = cM wjM = cM = 1. The proposition now follows from Proposition 2.1. We now have immediately, ;1 M ;1 be harmonic frames and let jcj = Corollary 2.1. Let f'k gM k=0 and fk gMk=0 M ;1 i i jdj = 1. The frames fc 'k gk=0 and fd k gk=0;1 are equivalent if and only if c = d and there is a permutation of f1 2 : : : N g so that 'kj = k(j) , for all 0 k

M ; 1.

The next proposition gives a classication of GHFs. Proposition 2.4. A family

0 i j M ; 1 we have

f'i gMi=0;1 in

HN

is a GHF if and only if for all

h'i 'j i = h'i+1 'j+1 i where 'M = '0 . ;1 Proof. If f'i gM i=0 in H N is a generalized harmonic frame then there is a unitary operator U on H N so that U i '0 = 'i , for all 0 i M ; 1. Hence, for all 0 i j M ; 1 we have

h'i+1 'j+1 i = hU i+1 '0 U j+1 '0 i = hU i '0 U j '0 i = h'i 'j i:


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;1 Conversely, given our equality, for any sequence of scalars fai gM i=0 ,

k

MX ;1 i=0

MX ;1

ai 'i+1 k2 = h =

i=0 MX ;1

i=0

MX ;1 i=0

ai 'i+1 i

ai aj h'i+1 'j+1 i =

ij =0 MX ;1

= h

ai 'i+1

ai ' i

MX ;1 i=0

MX ;1

ai aj h'i 'j i

ij =0 MX ;1

ai 'i i = k

i=0

ai 'i k2 :

It follows that U'i = 'i+1 is a unitary operator and U i '0 = 'i , for all 0 i M ; 1. As we will see in the next section, UTFs with M = N + 1 are all unitarily equivalent. Thus, as a direct consequence of Theorem 3.3, any UTF with M = N +1 is unitarily equivalent to the HTF with M = N + 1. This is a very useful result since we have HTFs for any N and M " thus, for M = N + 1, we always have an expression for all UTFs. For example, this means that the MB frame we introduced earlier is equivalent to the HTF with M = 3 and N = 2. Another interesting property of an HTF is that it is the only UNTF such that its elements are generated by a group of unitary operators with one generator, as i M we will see in Section 4. That is, = f'i gM i=1 = fU '0 gi=1 , where U is a unitary operator. Moreover, HTFs have a very convenient property when it comes to erasures. We can erase any e (M ; N ) elements from the original frame and what is left is still a frame (Theorem 4.2 from 25]). We provide a complete classication of frames in terms of their robustness to erasures in Section 5.

2.4. Gabor or Weyl-Heisenberg Frames

For the other general class of frames, we introduce two special operators on L2 (R). Fix 0 < a b and for f 2 L2(R) dene translation by a as

Taf (t) = f (t ; a) and modulation by b as

Eb f (t) = e2imbt f (t): Now, x g 2 L2 (R). If fEmb Tna ggmn2Z is a frame for L2 (R), we call it a Gabor

frame (or a Weyl-Heisenberg frame). It is clear that in this class the frames are uniform. Also, since the frame operator S for a Gabor frame fEmb Tnaggmn2Z must commute with translation and modulation, each Gabor frame is equivalent to the (uniform) normalized tight Gabor frame fEmb Tna S ;1=2 ggmn2Z. For an introduction to Gabor frames we refer the reader to 7] and 18].


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3. UNIFORM TIGHT FRAMES 3.1. What is Known about UTFs?

As we have mentioned in the introduction, UNTFs have become popular in applications. In particular, in 25], the authors attack the problem of robust transmission over the Internet by using frames. Being redundant sets of vectors, they provide robustness to losses, which are modeled as erasures of certain frame coecients. It is further shown in 25] that, assuming a particular quantization model, a uniform frame with quantized coecients, minimizes mean-squared error if and only if it is tight. Moreover, the same is true when we consider one erasure and look at both the average- and worst-case MSE. As a result, our aim is to construct useful families of frames for such applications. To that end, it is important to classify uniform normalized tight frames in order to facilitate the search for useful families. Our notion of usefulness includes any families which would be computationally ecient" for example, those that can be obtained from one or more generating vectors or those with a simple structure, for example, such that any of the frame operators could be expressed as a product of sparse matrices. Finally, we are interested in the robustness of our frames to coecient (frame vector) erasures. These and other issues will be explored in the rest of the paper. 3.1.1. Construction of UTFs There is a general well known method for getting nite UNTFs. We state this result here in its standard form and will give a new proof of the result in Section 5. In Section 5 we will also extend this result to uniform frames. Theorem 3.1. There is a unique way to get UNTFs with M elements in H N . Take any orthonormal set f k gNk=1 in H M which has the property N X k=1

j ki j2 = c for all i:

Thinking of the k as row vectors, switch to the M column vectors and divide by pc. This family is a uniform tight frame for H M with M elements, and all UNTFs for H N with M elements are obtained in this way. Corollary 3.1. If

f'i gMi=1 is a uniform normalized tight frame for H N then k'i k2 = M N:

There is a detailed discussion concerning the uniform normalized tight frames for

R2 in 25]. In 17], there is a deep classication of groups of unitary operators which

generate uniform normalized tight frames. The simplest case of this is the harmonic frames. In Section 4 we do not assume that we have a \group" of unitaries, but instead conclude that our family of unitaries must be a group. The picture becomes much more complicated if the uniform normalized tight frame is generated by a group of unitaries with more than one generator (see 17])


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or worse, if the uniform tight frame comes from a subset of the elements of such a group. 3.1.2. Classication of UNTFs Although we would like to classify all uniform tight frames, especially those which can be obtained by reasonable algorithms, this is an impossible task in general. The following theorem shows that every nite family of norm-1 vectors in a Hilbert space can be extended to become a uniform tight frame. Theorem 3.2. If f'i gM i=1 is a family of norm-1 vectors in a Hilbert space H , then there is a uniform tight frame for H which contains the family f'i gM i=1 . Proof (Theorem 3.2). For each 1 i M choose an orthonormal basis feij gj2J for H which contains the vector 'i . Now the family feij gM i=1j 2J is made up of norm-1 vectors and for any f 2 H we have M X X i=1 j 2J

jhf eij ij2 =

M X i=1

kf k2 = M kf k2:

In the above construction we get as a tight frame bound the number of elements

M in the family f'i gM i=1 . In general, this is the best possible. For example, just let 'i = 'j be norm-1 vectors for all 1 i j M . Then M X j =1

jh'i 'j ij2 = M:

Hence, any tight frame containing the family f'i g has the tight frame bound at least M . There is another general class of uniform tight frames (see 25]). The result below tells us that all UNTFs with M = N + 1 are unitarily equivalent. It is a direct consequence of Theorem 2.6 from 25] where it is stated for UTFs with norm 1. +1 Theorem 3.3 (Goyal, Kovacevic and Kelner 25]). A family f'i gN i=1 is a uniN +1 form normalized tight frame for H N if and only if f'i gi=1 is unitarily equivalent to the frame fPei gNi=1+1 where fei gNi=1+1 is an orthonormal basis for H N +1 and P is the orthogonal projection of H N +1 onto P the orthogonal complement of the onedimensional subspace of H N +1 spanned by Ni=1+1 ei . Since there are HTFs with N + 1-elements in H N , it follows that every UNTF for H N with N + 1-elements is unitarily equivalent to a HTF.

3.2. Normalized and Uniform Tight Frames and Subspaces of the Hilbert Space

Here, we begin to classify NTFs and UNTFs by providing a correspondence with subspaces of the original Hilbert space. We give two results, one for NTFs and another for UNTFs with the correspondence being one-to-one. The material in this section grew out of conversations between the rst author and V. Paulsen.

13


As we saw in Theorem 2.1, there is a unique way to get normalized tight frames in H N with M elements. Namely, we take an orthonormal basis fei gM i=1 for H M and take the orthogonal projection PHN of H M onto H N . Then fPHN ei gM i=1 is a normalized tight frame for H N with M elements. In particular, there is a natural one-to-one correspondence between the normalized tight frames for H N with M elements and the orthonormal bases for H M . Then, the uniform normalized tight frames for H N are the ones for which kPeik = kPej k, for all 1 i j M . We use this to exhibit a natural correspondence between these families and certain subspaces of H M . Here we treat two frames as the same if they are unitarily equivalent. Theorem 3.4. Let P be a rank-N orthogonal projection on H M and let fei gM i=1 be an orthonormal basis for H M . There is a natural one-to-one correspondence between the normalized tight frames for P H M with M elements and the family of all N -dimensional subspaces of H M .

Proof (Theorem 3.4). If f'i gM i=1 is any normalized tight frame for P H M , then by Naimark's Theorem, there is an orthonormal basis fei gM i=1 for H M so that Pei = 'i . Dene a unitary operator U on H M by Uei = ei . Now, 'i = PUei which is unitarily equivalent to 'i = U PUei . So we will associate our normalized tight frame f'i g with the subspace U PU H M . Now we need to check that this correspondence is one-to-one. Let fi g be another normalized tight frame for P H M which is associated with the same subspace of H M , namely U PU H M . Then there is an orthonormal basis fei g and a unitary operator V ei = ei with V PV = U PU and PV ei = ei . Hence, V PV ei = V i = U PUei = U 'i . This implies that the normalized tight frames f'i g and fi g are unitarily equivalent (and hence the same). Finally, we need to see that this correspondence covers all subspaces. If W is any subspace of H M of dimension N , we can dene a unitary operator U on H M so that UW = P H M . Then U PU = PW while fPUei gM i=1 is a normalized tight frame for P H M (which under our association corresponds to W ). One of the consequences of the above result is that if we have one normalized tight frame for H N , then all the others can be obtained from it by this process. We now turn our attention to uniform frames: 0

0

0

0

00

00

00

Theorem 3.5. Fix an orthonormal basis fei gM i=1 for H M so that if P is the M orthogonal projection of H M onto H N then fPeigi=1 is a uniform normalized tight frame for H N . Then there is a natural one-to-one correspondence between the uniform normalized tight frames for H N with M elements and the subspaces W of H M for which kPW ei k2 = M=N , for all 1 i M .

Proof (Theorem 3.5). We already have our classication of the normalized tight frames in terms of all subspaces. Now we need to see which of these subspaces correspond to the uniform normalized tight frames. Suppose U is any unitary operator on H so that fPUei g is a uniform normalized tight frame for P H M . Then this frame is associated to the subspace W = U P H M and PW = U PU . Hence, PUei = UPW ei , for all i. But, fPUeig is a uniform normalized tight frame for

14


P H M and so kPUei k = N=M , for all i. Hence, N = kUP e k = kP e k: kPUeik = M W i W i

So our association between normalized tight frames and subspaces given in Theorem 3.4 identies the subspaces W of H M for which kPW ei k = N=M , for all i.

3.3. Uniform Dual Frames

Another method for classifying UTFs uses a back door: get UTFs as alternate dual frames for a given frame. To do this, we need the denition of alternate dual frames: Definition 3.1. Let f'i gi2I be a frame for a Hilbert space H . A family fi gi2I is called an alternate dual frame for f'i gi2I if

f =

X i2I

hf i i'i for all f 2 H :

In Denition 3.1, if we let F1 (respectively, F2 ) be the analysis frame operators for f'i g (respectively fi g), then we see that F2 F1 = I . There are many alternate dual frames for a given frame. In fact, a frame has a unique alternate dual frame (the canonical dual frame) if and only if it is a Riesz basis 16]. Moreover, no two distinct alternate dual frames for a given frame are equivalent 16]. For a normalized tight frame, its canonical dual frame is the frame itself since F~ = FS ;1 = F I = F . Moreover, Proposition 3.1. If f'i gi2I is a normalized tight frame for H N , then the only normalized tight alternate dual frame for f'i gi2I is f'i gi2I itself. Proof (Proposition 3.1). Let F1 be the analysis frame operator for f'i gi2I . Let fi gi2I be any normalized tight alternate dual frame for f'i gi2I with analysis frame operator F2 . It follows that F1 F2 are isometries. Now,

F2 (F1 ; F2 ) = F2 F1 ; F2 F2 = I ; I = 0: Hence, for every f g 2 H we have

hF2 g (F1 ; F2 )f i = hg F2 (F1 ; F2 )f i = hg 0i = 0: Since f'i gi2I is an NTF, for every f 2 H we have kf k2 =

M X i=1

jhf 'i ij2 = kF1 f k2 :

(24)

We can further expand this as

kF1 f k2 = k(F1 F2 )f k2 = kF2 f k2 + k(F1 ; F2 )f k2 + h|F2 (F1 ;{zF2 )f f }i + h|F2 (F1 ;{zF2 )f f i} : 0

0


15

However, since F2 corresponds to an NTF as well,

kF2 f k2 + k(F1 ; F2 )f k2 = kf k2 + k(F1 ; F2 )f k2 :

(25)

Equating (24) and (25), we get

k(F1 ; F2 )f k2 = 0: Hence, F1 = F2 and so 'i = i , for all i 2 I . If we ask for the dual frame just to be tight, but not normalized, then: Proposition 3.2. If f'i gM i=1 is a normalized tight frame for H N and M < 2N , M then the only tight dual frame for f'i gM i=1 is f'i gi=1 itself. If M 2N then there are innitely many (nonequivalent) tight alternate dual frames for f'i gM i=1 . Proof (Proposition 3.2). With the notation of the proof of Proposition 3.1, the only change is that, since the dual frame is only tight and not normalized tight, there is a frame bound A > 0 so that

kF2 k2 = Akf k2 for all f 2 H : Hence, as in the proof of Proposition 3.1 we have

kf k2 = kF1 f k2 = kF2 f k2 + k(F1 ; F2 )f k2 = Akf k2 + k(F1 ; F2 k2 : Hence, k(F1 ; F2 )f k2 = (1 ; A)kf k2 for all f 2 H . Hence, either F1 ; F2 = 0 and A = 1 which gives uspthe NTF as in the previous proposition, (and so 'i = i for all 1 i M ) or (1= 1 ; A)(F2 ; F1 ) is an isometry on H N . In the latter case it follows that dim (F1 H N )? N , and so M 2N . Thus, if M < 2N , the only tight

dual frame is the frame itself. On the other hand, if M 2N given any F : H N ! `M 2 which is a constant times an isometry, and with F1 H N ? F H N , F1 + F denes a tight frame which is an alternate dual frame for f'i g. However, since it is really the uniform case we are interested in, Proposition 3.3. If f'i gM i=1 is a uniform normalized tight frame for H N and M = 2N , then there are innitely many uniform tight alternate dual frames for N. f'i g2i=1 Proof (Proposition 3.3). Again we use the notation of the proof of Proposition 3.1. Choose any F2 : H N ! `M 2 which is a constant multiple of an isometry and with F2 H N = (F1 H N )? . Then F1 + F2 denes a tight alternate dual frame for f'i gM i=1 , say i = (F1 + F2 )ei , for all 1 i M . We just need to check that this frame is uniform. Let P : `M 2 ! F1 H N be the orthogonal projection, so that Pei = F1 'i , for all 1 i M . It follows that kPeik2 = N=M and k(I ; P )ei k2 = 1 ; N=M , for all 1 i M . Now,

i = F1 ei + F2 ei = 'i + F2 (I ; P )ei :


16

Also, there is an a > 0 so that for every 1 i M we have

N + ak(I ; P )e k2 ki k2 = ki k2 + kF2 (I ; P )ei k2 = M i

N + a 1 ; N = a + (1 ; a) N : =M M M Hence, fi gM i=1 is a uniform tight frame. It can be shown that the results of this section actually classify when NTFs or UNTFs have tight (respectively uniform tight) alternate dual frames, that is, all tight alternate dual frames for f'i g are obtained by the methods of this section. We do not know in general which frames (not tight) have tight (respectively uniform tight) alternate dual frames.

3.4. Frames Equivalent to UTFs

Another approach to the classication of UTFs looks into families of frames obtained from a UTF by equivalence relations. For example, we know that every frame is equivalent to a normalized tight frame. That is, given any frame f'i gi2I with the frame operator S , the frame fS ;1=2'i gi2I is a normalized tight frame which is equivalent to f'i gi2I . Therefore, it is natural to try to nd ways to turn frames into uniform normalized tight frames. As it turns out, this is not possible in most cases: Theorem 3.6. If a frame f'i gi2I with the frame operator S is equivalent to a UTF, then fS ;1=2 'i gi2I is a UNTF. In particular, a tight frame which is not uniform cannot be equivalent to any UNTF. Proof (Theorem 3.6). It is known that fS ;1=2'k g is a normalized tight frame which is equivalent to f'k g. So if f'k g is equivalent to a uniform tight frame, say fk g, and kk k = c, for all k, then fS ;1=2'k g is equivalent to fk g. That is, there is an invertible operatorpT on H so that TS ;1=2'k = k . Now we show that T= A is a unitary operator. Let A be the tight frame constant of fTS ;1=2'k g. Then for all f 2 H , the tightness of fTS ;1=2'k g implies

Akf k2 =

X k

jhf TS ;1=2'k ij2 =

X k

jhT f S ;1=2 'k ij2 :

(26)

However, since fS ;1=2 'k g is an NTF, then this leads us to

X k

jhT f S ;1=2'k ij2 = kT f k2:

Equating (26) and (27), we get that

p

Akf k2 = kT f k2

that is, T= A is unitary. Hence, the frame

S ;1=2 'k = T ;1 k

(27)


17

is a UNTF since

kT ;1k k = p1 kk k = p1 c A

A

for all k = 1 : : : M:

4. UNIFORM TIGHT FRAMES WITH GROUP STRUCTURE

In this section, we examine frames which are generated by a group of unitary operators applied to a xed vector in the Hilbert space. These classes are especially important in applications. The traditional use of one of these classes is in the Gabor frames used in signal/image processing. These are groups with two generators. There is also a class of wavelet frames which has two generators 9]. Recently, several new applications have arisen. In 25], the authors proposed using the redundancy of frames to mitigate the losses in packet-based communication systems such as the Internet. In 17], frames are used for multiple antenna coding/decoding (see the discussion Section 4.3 for further explanation). In 22, 12], connections between quantum mechanics and tight frames are established. Although each of these applications requires a dierent class of UNTFs, they all have one important common constraint. Namely, calculations for the frame must be easily implementable on the computer. Since UNTFs generated by a group of unitary operators satisfy this constraint, this class is one of the most important to understand. ;1 In this section we will show that the UNTFs generated by the family fU k '0 gM k=0 (where U is unitary and '0 2 H ) are precisely the GHFs. We then extend our discussion to UNTFs generated by fU k V j '0 g and higher numbers of generators. In the discussion Section 4.3 we will relate the results of this section to the literature.

4.1. UTFs with a Single Generator

;1 Here, we give a complete classication of UNTFs of the form fU k '0 gM k=0 where ;1 U is a unitary operator on H N . We will, in fact, nd that in this case fU k gM k=0 must be a group. Moreover, we will see the the GHFs will be that special class. Since the proofs of the following results are long, the reader can nd them in the Appendix. Our rst result classies GHFs as those frames generated by powers of a special class of unitary operators applied to a xed vector in H . ;1 Proposition 4.1. f'k gM k=0 is a general harmonic frame for H N if and only if N , an orthonormal basis fei gN for H N there is a vector '0 2 H N with k'0 k2 = M i=1 and a unitary operator U on H N with Uei = ci ei , with fci gNi=1 distinct M th roots of some jcj = 1 so that 'k = U k '0 , for all 0 k M ; 1.

Proof. See Appendix A.1. We now show that the only UNTFs with group structure and a single generator are GHFs. This result is an excellent result and a disappointment at the same time. On the one hand, the fact that GHFs are the only ones with such a group structure proves once more why they are so universally used. It also spares us the trouble of looking for other such UNTFs. On the other hand, we cannot nd any other useful families via this route as we were hoping for. Another important property of


18

GHFs (Theorem 4.2 from 25]) is that GHFs are robust to the maximum number of possible erasures, that is, a GHF with M elements in H N is robust to e erasures for any e (M ; N ). Our next theorem completes the classication of GHFs as being those UNTFs of ;1 the form fU k '0 gM k=0 where U isM ;a 1unitary operator on H . As a consequence, we discover that, in this case, fU k gk=0 is a group. Theorem 4.1. Let U be a unitary operator on H N , '0 2 H N and assume fU k '0 gMk=0;1 is a UNTF for H N . Then U M = cI for some jcj = 1 and fU k '0 gMk=0;1 is a general harmonic frame. That is, the GHFs are the only UNTFs generated by a group of unitary operators with a single generator. Proof. See Appendix A.2.

4.2. UTFs with Two or More Generators

Having completely characterized UTFs which come from a group of unitary operators with one generator, we now turn our attention to those with more than one. The fundamental examples of frames with two generators are the Gabor frames fEmb Tma ggmn2Z (or, in our case, the nite discrete Gabor frames). Each of these frames is equivalent to the UNTF Gabor frame fEmb Tna S ;1 ggmn2Z where S is the frame operator for fEmb Tma ggmn2Z. ;1M ;1 j M ;1 We will classify the UNTFs of the form fU i V j '0 gLi=0 j =0 where fV '0 gj =0 is a UNTF for its span. In the process of proving these results, we introduce a general method for using special families of nite-rank projections to produce frames for a space. If we associate a frame f'i gM i=1 with the family of rank-1 orthogonal projections Pi taking H onto span 'i , then we can view the results of this section as a generalization of the very notation of a frame to the case where the Pi have arbitrary rank. Our rst result states that given a GHF generated by a unitary operator V , a unitary operator U and an integer M , we can always nd a corresponding UNTF generated by U V . Proposition 4.2. Let V be a unitary operator on H N , let m 2 N and 0 2 H ;1 so that fV i 0 gm U on H N i=0 is a UNTF for H N . Then for every unitary operator ;1m;1 and every M 2 N , there is a vector '0 2 H N such that fU k V j '0 gM k=0j =0 is a uniform, normalized tight frame for H N .

p

Proof (Proposition 4.2). Let '0 = 0 = M . Now, for every f 2 H N we have MX ;1 m X;1 k=0 j =0

jhf U k V j '0 ij2 =

MX ;1 m X;1 k=0 j =0

jhU

;k

MX ;1 m X;1 ;k j 2 1 j 2 f V '0 ij = M jhU f V 0 ij : k=0 j =0

;1 Since fV j 0 gm j =0 is a UNTF, the previous sum equals ;1 MX ;1 1 MX ;k f k2 = 1 2 2 k U M k=0 M k=0 kf k = kf k :


19

It is easily seen that an invertible operator on H maps a frame for H to another frame for H . We next observe that there is a natural relationship between the frame operators of these equivalent frames. Proposition 4.3. If f'k g is a frame for H with the frame operator S , and T is an invertible operator on H , then TST is the frame operator for the frame fT'k g. In particular, if f'k g is a UNTF then every frame unitarily equivalent to f'k g is also a UNTF. Proof (Proposition 4.3). Let L be the frame operator for fT'k g. For any f 2 H we have:

Lf =

X k

hf T'k iT'k = T

X k

!

hT f 'k i'k = T S (T )f ]:

As a consequence of Proposition 4.3, we can apply a unitary operator to any UNTF and get a new UNTF. If this unitary operator is part of the generating set for our UNTF, then our new UNTF has a special form as given in the next proposition. Proposition 4.4. If U is unitary and fU k V j '0 g is a UNTF for H N , then for every M 2 Z and for every f 2 H N we have,

f =

X kj

hf U k+M V j '0 iU k+M V j '0

that is, fU k+M V j '0 gkj is a UNTF as well. Proof (Proposition 4.4). Since U M is unitary and fU k V j '0 gkj is a UNTF for H N , it follows that fU M U k V j '0 gkj is also a UNTF for H N . To explain where we are going from here, let us revisit the notion of a UNTF and look at it from a slightly dierent point of view. Suppose f'i gi2I is a UNTF for a (nite or innite-dimensional) Hilbert space H with k'i k = c, for all i 2 I . For each i 2 I , let Pi be the orthogonal projection of H onto the one-dimensional subspace of H spanned by 'i . Then, for all f 2 H

X i2I

kPi f k2 = =

X i2I X i2I

khf 'i i'i k2 jhf 'i ij2 k'i k2 = c2

X i2I

jhf 'i ij2 = c2 kf k2:

Conversely, let fPi gi2I be rank-1 orthogonal projections on H satisfying

X i2I

kPi f k2 = akf k2 for all f 2 H :


20

Then f'i gi2I is a UNTF for H where 'i 2 Pi H and k'i k2 = 1=a. That is, if f 2 H then X X X jhf 'i ij2 = jhPi f 'i ij2 kPi f k2 k'i k2 i2I

= a1

X i2I

i2I

i2I

kPi f k2 = a1 akf k2 = kf k2:

It follows that the condition on f'i g which guarantees that we have a UNTF is really a condition on the rank-1 orthogonal projections of H onto the span of 'i . It is this which we now generalize to nd our next general class of UNTFs. Since this material is also new for general frames, we start here. Proposition 4.5. Let fPi gi2I be a family of projections (of any rank) on a (nite or innite-dimensional) Hilbert space H and assume there are constants 0 < A B satisfying,

Akf k2

X i2I

kPi f k2 B kf k2 for all f 2 H :

i Let f'ij gM j =1 be a frame for Pi H with frame bounds Ai Bi , for all i 2 I . Then M i f'ij gj=1i2I has frame bounds,

A infi2I Ai B supi2I Bi : Proof. For any f 2 H we have M X i

j =1i2I

jhf 'ij ij2 = =

M XX i

i2I j =1 Mi XX i2I i=1

jhf 'ij ij2 jhPi f 'ij ij2

supi2I Bi

X i2I

kPi f k2

X i2I

Bi kPi f k2

B (supi2I Bi ) kf k2:

The lower frame bound is computed similarly. The cases of interest to us are the next two corollaries: the rst for NTFs and the second for UNTFs. Corollary 4.1. Let

satisfying:

fPi gi2I be a family of projections on a Hilbert space

X i2I

H

kPi f k2 = a2 kf k2 for all f 2 H :

Mi 1 i Then a 1, and if f'ij gM j =1 is an NTF for Pi H , for all i 2 I , then f a 'ij gj =1i2I is an NTF for H .


21

In the next corollary we assume that the projections Pi all have the same rank and the frames for Pi H all have the same number of elements. This assumption is necessary to guarantee that our frame is uniform. That is, as we observed in Section 2.1, if dim Pi H = N and f'ij gj2J is a UNTF for Pi H then

k'ij k2 = jNJ j : Corollary 4.2. Let fPi gi2I be a sequence of projections (all with the same rank) on a Hilbert space H satisfying:

X i2I


Then a 1, and if f'ij gj2J is a UNTF for Pi H , for all i 2 I , then f a1 'ij gj2Ji2I is a UNTF for H . We are now ready to consider UNTFs with two generators. ;1 Theorem 4.2. Let U be a unitary operator on H , and let fV j 0 gM j =0 be a UTF (with tight frame bound 1=a) for its closed linear span in H . Let P0 be the orthogonal ;1M ;1 projection of H onto this span. Assume that fU i V j '0 gLi=0 j =0 is a UNTF for H . i If Pi is the orthogonal projection of H onto the span of U P0 H , for all 1 i L ; 1, then LX ;1 i=0

kPi f k2 = akf k2 for all f 2 H :

Proof. Note that for all 1 i K ; 1 we have Pi = U i P0 U ;i . Now, for all f 2 H we compute, LX ;1 i=0

kPi f k2 = =

LX ;1 i=0 LX ;1

kU i P0 U ;i f k2 kP0 U ;i f k2 =

i=0 KX ;1 M X;1

= a

i=0 j =0

LX ;1 M X;1 i=0

a

j =0

jhU ;i f V j '0 ij2

jhf U i V j '0 ij2 = akf k2:

The next theorem is the converse of the above. ;1 Theorem 4.3. Let fPi gLi=0 be a family of orthogonal projections on H satisfying LX ;1 i=0


Assume that U is a unitary operator on H so that U i P0 H = Pi H . Let '0 2 P0 H ;1 and let V be a unitary operator on P0 H such that fV j '0 gM j =0 is a UNTF for P0 H . L ; 1 M ; 1 Then f a1 U i V j '0 gi=0j=0 is a UNTF for H .


22

Proof. For all f 2 H we have, LX ;1 M X;1

LX ;1 M X;1

i=0

i=0 j =0

jhf a1 U i V j '0 ij2 = a12 j =0

jhU ;i f V j '0 ij2 =

;1 ;1 L;1 1 LX 1 LX ;i 2 i P0 U ;i f k2 = 1 X kPi f k2 = kf k2: k P U f k = k U 0 a2 a2 a2

i=0

i=0

Some remarks are in order:

i=0

4.3. Discussion

;1 1. The assumptions in Theorems 4.2 and 4.3 that fV j '0 gM j =0 is a UNTF for its span is sucient for the conclusion of the theorems but not necessary. For example, it is known to Gabor frame specialists that for ab < 1, there are Gabor frames fEmb Tna ggmn2Z for which neither of fEmb Tnaggm2Z nor fEmb Tnaggn2Z is a frame for its closed linear span. The equivalent UNTF Gabor frame fEmb Tna S ;1=2 ggmn2Z fails the hypotheses of Theorems 4.2 and 4.3 but satises the conclusion. 2. The results of Section 4.2 generalize to three or more generators. For example, for three generators, Theorem 4.3 becomes: ;1M ;1 Theorem 4.4. Let U V be unitary operators with `fU i V j '0 gLi=0 j =0 a UNTF for its span. Let P0 be the projection of the Hilbert space H onto this span. Let W be

a unitary operator on H so that KX ;1 k=0

kW k P0 W ;k f k2 = a2 kf k2 for all f 2 H :

;1L;1M ;1 Then f a1 W k U i V j '0 gKk=0 i=0j =0 is a UNTF for H .

3. It would be very interesting to classify when a nite group of unitaries G generates a UNTF for H of the form fU'0 gU 2G. Since nite Abelian groups are isomorphic to a direct product of cyclic groups, the results of this section give sucient conditions for fU'0 gU 2G to generate a UNTF for H . Frames generated by Abelian groups of unitaries were studied in 4] where they are called geometrically uniform frames (GU frames). It is shown there that the canonical dual frame of a GU frame is also a GU frame and that the equivalent NTF frame is also GU. Since GU frames have strong symmetry properties, they are particularly useful in applications. In 4], it is further shown that the frame bounds resulting from removing a single vector of a GU frame are the same regardless of the particular vector removed. This result for HTFs was observed in 25]. 4. Frames generated by possibly noncommutative groups of unitaries are important in multiple antenna coding and decoding 17]. Here, one needs classes of

23


unitary space-time signals (that is, frames generated by classes of unitary operators) called constellations. It is also desirable in this setting to have full transmitter diversity, meaning that det (I ; U ) 6= 0 for all I 6= U 2 G: Groups with this property are called xed-point free groups. In a tour-de-force, the authors in 17] classify all full-diversity constellations that form a group, for all rates and numbers of transmitting antennas. Along the way they correct some errors in the classication theory of xed point free groups.

5. CLASSIFICATION OF UNIFORM TIGHT FRAMES WITH ERASURES

As mentioned in the introduction, one of the main applications that motivated us to examine uniform tight frames is that of robust data transmission. This means that at some point, the system experiences losses. These losses are modeled as erasures of transmitted frame coecients. Since at the receiver side, this looks like the original frame was the one without vectors corresponding to erased coecients, we examine the structure of our frames after losses. The rst question is whether after erasures what we have is still a frame? For the MB frame, if we erase any one element, the remaining two are enough for reconstruction. However, if we erase any two, we have lost one subspace and reconstruction is not possible. We can provide more robustness by adding more vectors. For example, take a uniform tight frame in R2 made up of two orthonormal bases (1 0) (0 1) (;1 0) (0 ;1). This frame is robust to any one erasure" what is left contains at least one of the two bases. However, if two elements are erased, the situation is not that clear anymore. We could erase coecients corresponding to one of the bases and still have a basis able to reconstruct. If, on the other hand, we lose coecients corresponding to two collinear vectors, we are stuck" we have lost one entire subspace and cannot reconstruct. We could, however, take another UNTF with M = 4 vectors in R2 { the HTF with the analysis frame operator

F =

0 1 0 1 p p BB 2=2 2=2 CC : @ p0 p 1 A ; 2=2

(28)

2=2

This frame is also made up of two orthonormal bases. However, this frame is robust to any two erasures. It is thus more robust than the MB frame" we pay for this added benet with more redundancy (more vectors). It is also more robust than the previous frame with 4 vectors. These simple examples demonstrate the types of questions we will be asking in this section. One might think that starting with a tight frame might provide some resilience to losses (as opposed to starting with a general frame). As we have seen in our example, this is not the case. Thus, we are searching for frames robust to a certain number of erasures. That is, frames which remain frames after erasures. There are frames on H N with M elements which are robust to M ; N erasures,


24

namely the HTFs (Theorem 4.2 from 25]). Such frames are the best. This is clearly optimal since any more erasures would not leave enough elements to span H N . In this section, we will characterize frames based on their robustness to erasures.

5.1. UTFs and UNTFs Robust to One Erasure

In this section we will consider UNTFs which are robust to one erasure. This contains the basic idea for the general case. We start with the notion of a frame robust to k erasures, or a k-robust frame. Definition 5.1. A frame f'i gM i=1 is said to be robust to k erasures if f'i gi2I c is still a frame, for I any index set of k erasures, I f1 2 : : :M g and jI j = k. The following property tells us we do not destroy the robustness of the frame by projection. This observation lead us to the idea that we could classify frames by starting from a large space and \step down"using projections. For example, we could start with a frame robust to one erasure and step down to frames (hopefully) robust to two erasures, then once more to frames robust to three erasures and so on. Although this will not be the case, the idea of stepping down lead us to the results in this section. Note that it is clear from the denition of a frame that if we apply an orthogonal projection P on H to a frame we will get a frame for P H with the same frame bounds. Proposition 5.1. Let f'i gM i=1 be a frame in H N robust to k erasures and let P be an orthogonal projection on H N . Then fP'i gM i=1 is a frame for P H N robust to

k erasures.

Proof. Let I be the index set of erasures, I f1 2 : : : M g with jI j = k. Now, f'i gi2I spans H N and so fP'i gi2I is a frame for P H N with the same frame c

c

bounds. The main ingredient for classifying frames robust to one erasure is contained in the next proposition. Proposition 5.2. Let

equivalent:

f'i gMi=1 be a set of vectors in

HN.

The following are

1.f'i gM i=1 is a frame robust to one erasure. 2.There are nonzero scalars ai 6= 0, for 1 i M so that M X i=1

ai 'i = 0:

Proof. (1) ) (2): Choose I f1 2 : : :M g maximal for which there are nonzero

ai 's, i 2 I and

g =

X i 2I

ai 'i = 0:


25

We claim that jI j = M . We proceed by contradiction. If I 6= f1 2 : : : M g, choose m 2 I c . Since f'i gM i=1 is robust to one erasure, there are scalars fbi g, not all zero, so that if 'm is erased, it can be recovered from the rest as

'm = or

h = 'm ;

X i2I

X

i6=m

bi 'i ;

bi 'i

X m6=i2I c

bi 'i = 0:

We have two cases: Case I: Assume that bi = 0 for all i 2 I . P Then, h = 'm ; m6=i2I c bi 'i = 0: In this case, g + h = 0 and has nonzero coecients on every 'i , i 2 I , plus a nonzero coecient on 'm contradicting the maximality of I . Case II: At least one bi 6= 0 for some i 2 I . Since ai 6= 0 for all i 2 I , we can choose an > 0 so that

6= ai =bi for all i 2 I: Now, g + h = 0 and has nonzero coordinates on 'i , for all i 2 I , as well as for a coordinate on 'm , again contradicting the maximality of I . (2) ) (1): Assume ai 6= 0, for all 1 i M and

X i2I

ai 'i = 0:

Then for each m 2 I we have:

'm = ;

X ai am 'i

m6=i2I

that is, any vector lost can be recovered using the rest and so f'i gM i=1 is robust to the erasure of 'm , for an arbitrary m 2 I . Note that to construct a frame guaranteed not to be robust to one erasure, it is enough to put one vector, say 'M , orthogonal to the span of the rest. Erasing that vector destroys frame property. Thus, we cannot nd a nonzero coecient aM P the such that M a ' = 0. i i i=1 For example, we know that the MB frame f'i g3i=1 is robust to onePerasure (Theorem 4.1 from 25]). Hence, we can nd a1 = a2 = a3 = 1 such that 3i=1 ai 'i = 0. The next results will characterize frames robust to one erasure. Since we mentioned the idea of stepping down, we will start from an orthonormal basis fei gM i=1 for H M and project it using a projection operator P (or I ; P ). We thus look into a few facts connecting P and (I ; P ). 1. If P is an orthogonal projection, then (I ; P ) is an orthogonal projection as well.


26

2. The subspaces P and (I ; P ) project onto are orthogonal. For any f 2 H ,

h(I ; P )f Pf i = hP (I ; P )f f i = h(P ; P )f f i = 0: M 3. f(I ; P )ei gM i=1 is uniform if and only if fPei gi=1 is uniform. To see this we compute, for all i j

h(I ; P )ei (I ; P )ei i h(I ; P )(I ; P )ei ei i h(I ; P )ei ei i 1 ; hPei ei i hP Pei ei i hPei Pei i

= = = = = =

h(I ; P )ej (I ; P )ej i h(I ; P )(I ; P )ej ej i h(I ; P )ej ej i 1 ; hPej ej i hP Pej ej i hPej Pej i:

Here we used the fact that P is an orthogonal projection and thus P = P 2 = P P . M 4. Both fPei gM i=1 and f(I ; P )ei gi=1 are NTFs.

M By Theorem 2.1, all NTFs are of the form fPei gM i=1 , where fei gi=1 is an orthonormal basis for H and P is an orthogonal projection on H . The UNTFs form a subclass of these frames. In the rest of this paper we will identify this subclass. So we will be working with frames of the form above. We will discover that the subspace P H determines when the frame is a UNTF. From our earlier discussion, if P is an orthogonal projection on H , we may work either with fPei g or f(I ; P )eig to classify UNTFs. We will freely switch between these classes because each of them has certain advantages for specic results. The class fPei g works well for classications when the dimension of P H is \large" relative to the dimension of H . On the other hand, f(I ; P )ei g is easier to work with when the dimension of P H is \small" relative to the dimension of H . Also, as we will see, important information about fPei g is often contained in (I ; P )H (and vice-versa). So we will need to bring both of these into our discussion. Now we give the general classication for frames robust to one erasure. Corollary 5.1. Let

fei gMi=1 be an orthonormal basis for H M and let P be an

orthogonal projection on H M . The following are equivalent:

1.f(I ; P )ei gM i=1 is a frame robust to one erasure. 2.There are nonzero scalars ai 6= 0, for 1 i M so that M X i=1

ai (I ; P )ei = 0:

3.There is an f 2 span1iM Pei so that

hf ei i 6= 0 for all 1 i M:

27


Proof. The equivalence of (1) and (2) comes from Proposition 5.2. Let us check the equivalence of (2) and (3). (2) ) (3): Given (2), choose f to be

f =

M X i=1

ai ei :

P a e = PM a Pe and thus By our assumption, M i=1 i i i=1 i i f =

M X i=1

ai Pei 2 P H M :

Moreover, hf ei i = ai 6= 0, for all 1 i M . (3) ) (2): Choose f 2 P H M as in (3) and call ai = hf ei i 6= 0, for all 1 i M . Now M X i=1

ai (I ; P )ei =

M X

M X

M X

i=1

i=1

i=1

hf ei i(I ; P )ei =

hf ei iei ;

hPf ei iei = 0

since f 2 P H M , and thus Pf = f . Corollary 5.1 classies frames which are robust to one erasure. We now extend this to a classication of UNTFs robust to one erasure. Corollary 5.2. Let fei gM i=1 be an orthonormal basis for H M and let P be an orthogonal projection on H M . The following are equivalent: 1.f(I ; P )ei gM robust to one erasure. i=1 is Pa frame 2.There is a g = M a e with nonzero scalars ai 6= 0, for all 1 i M such i=1 i i that Pf = hf gig, for all f 2 H M , and kgk = 1. Moreover, if P is rank-1, f(I ; P )ei gM i=1 is a uniform frame if and only if jai j = p 1= M , for all 1 i M . Proof. (1) , (2) is immediate from Corollary 5.1. For the moreover part, we know that f(I ; P )ei gM i=1 is uniform if and only if fPei gMi=1 is uniform. Also, since P is rank-1, for all 1 i M and using the expression expression Pf = hf gig in Part 2: with f = ei we have,

Pei = ai g: Hence, kPei k = kPej k is true if and only if jai j = jaj j, for all 1 i j M . M That is, fPei gM i=1 is uniform (and hence f(I ; P )ei gi=1 is uniform)pif and only if jai j = jaj j, for all 1 i j M . Finally, kPei k = jai jkgk = jai j = 1= M .

5.2. UTFs Robust to More than One Erasure

We now try to apply the same ideas from the previous section to classify frames robust to more than one erasure. The classication of UNTFs with k erasures in


28

this section is useful if we have M vectors in an N -dimensional space and M ; N is \small". In the next section we will give another classication of this family which works best when M ; N is \large". In what follows, I f1 2 : : : M g will denote the erasure index set. Proposition 5.3. Let fei gM i=1 be an orthonormal basis for H M . Let P be an orthogonal projection of H M onto an L-dimensional subspace H L . Fix I f1 2 : : : M g with jI j = k L and let K = spani2I c ei . If f'j gLj=1 is any orthonormal basis for H L and we have the following L M matrix

A = (h'j ei i)jL=1iM=1 then dim ker (I ; P )jK ] = L ; row rank of the k columns of A indexed by I ]: Proof. We have

ker (I ; P )jK = ff 2 K j(I ; P )f = 0g = ker (I ; P )] \ K : L Now, apply row reduction (relative to fei gM i=1 ) to f'i gi=1 using the elements of the columns in I . This gives a linearly independent set fgj gsj=1 fgj gj=Ls+1 spanning K with s = row rank of (h'j ei i)jL=1i2I gj 2 K for s + 1 j L and

(span1js gj ) \ K = 0: Hence, Therefore,

ker (I ; P ) \ K = spans+1jL gj :

dim (ker (I ; P )) \ K = dim (ker (I ; P )jK = L ; (s + 1) + 1 = L ; s: We are looking for the NTFs which are robust to k erasures. Proposition 5.3 actually yields a stronger result. Namely, it gives necessary and sucient conditions for fPei g to be robust to one particular choice of k erasures. We state this stronger result in two dierent forms in the next two propositions. Proposition 5.4. With the notation of Proposition 5.3, the following are equivalent: 1.f(I ; P )ei gM i=1 is robust to the erasure of the elements f(I ; P )ei gi2I . 2.We have rank (I ; P )jK = M ; L. 3.We have dim ker (I ; P )jK ] = L ; k.


29

Proof. (1) , (2): f(I ; P )ei gM i=1 is robust to the erasure of the elements f(I ; P )ei gi2I if and only if (I ; P )jK is full rank, which must be rank (I ; P ) = M ; L. (2) , (3): By Proposition 5.3,

dim (ker (I ; P )jK = L ; row rank of the k columns of A indexed by I ] L ; k: On the other hand, (I ; P )jK is full rank if and only if (I ; P )(I ; Pi ) is full rank, where Pi is the orthogonal projection of H M onto spani2I ei . Hence, dim (ker (I ; P )(I ; Pi )) L: But, ei 2 ker (I ; Pi ), for all i 2 I . Hence, dim (ker (I ; P )(I ; Pi ) = dim (ker (I ; P ) + k L: Therefore,

dim (ker (I ; P ) L ; k:

Proposition 5.4 gives precise information when a frame is robust to the erasure of one particular choice of k elements in terms of the coecients of f'i g relative to the unit vector basis of H L . The following corollary is a statement for a xed choice of erasures as well. Corollary 5.3. With the notation of Proposition 5.3, the following are equivalent: 1.f(I ; P )ei gM i=1 is robust to the erasure of the elements f(I ; P )ei gi2I . 2.The row rank of (h'j ei i)Lj=1i2I = k. Applying the above result to every choice of erasures results in Theorem 5.1 and a classication of when our frame is robust to any k erasures. We are dealing here with general frames. We will deal with uniform frames right afterwards. Theorem 5.1. With the notation of Proposition 5.3, the following are equivalent: 1.f(I ; P )ei gM i=1 is robust to k erasures. 2.For every I f1 2 : : :M g with jI j = k, the row rank of A = (h'j ei i)Lj=1i2I equals k. Note the crucial role played by erased elements in the theorem, since in the statement, the indices are from I . Next we see what it takes for such frames to be uniform. Here, we call the \angle" between two vectors the inner product when this is really the cosine of the angle. Note that this result classies when f(I ; P )ei g is a UNTF without any reference to erasures. Theorem 5.2. With the notation of Proposition 5.3, the following are equivalent:


30

M 1.fPeigM i=1 is uniform (and hence f(I ; P )ei gi=1 is uniform). 2.For every 1 i M we have,

(A A)ii =

L X

jh'j ei ij2 = ML : j =1

That is, the columns of A = f'j gLj=1 all have the same square sums.

Moreover, the angle between (I ; P )ei and (I ; P )ej is given by the inner product of the ith and j th columns of f'j gLj=1 . M Proof. (1) , (2): We know that f(I ;P )ei gM i=1 is uniform if and only if fPei gi=1 which, in turn, is uniform if and only if the diagonal elements of the matrix for P relative to fei gM i=1 have constant modulus. We have,

Pei =

L X

hei 'n i'n =

n=1

M X L X

m=1 n=1

!

hei 'n ih'n emi em :

The diagonal element of Pei is: L X

hei 'n ih'n ei i =

n=1

L X n=1

jh'n ei ij2 :

So fPei gM i=1 is a uniform tight frame if and only if for all 1 i j M we have L X n=1

jh'n ei ij2 =

L X n=1

jh'n ej ij2 :

For the moreover part, we compute for 1 i 6= j M :

h(I ; P )ei (I ; P )ej i = hei ej i ; hPei ej i ; hei Pej i + hPei Pej i = hPei Pej i = hei Pej i

hei

L X

hej 'n i'n i = hei

n=1

M X L X

m=1 n=1

!

hej 'n ih'n em i em i =

L X n=1

hej 'n ih'n ei i:

The right-hand side of the above equality is precisely the inner product of the ith and j th columns of f'j gLj=1 . The following consequence of the above is surprising at rst. It says that we can almost never get (in the real case) frames robust to k erasures by stepping down from frames robust to one erasure, then to frames robust to two erasures etc. Corollary 5.4. In the real case, if M 3, there do not exist rank-1 orthogonal projections P1 P2 with P1 P2 = 0 so that f(I ; P1 )ei gM i=1 is uniform and f(I ; M P2 )(I ; P1 )ei gi=1 is uniform and robust to two erasures.


31

Proof. Assume such P1 P2 exist. Since f(I ; P1 )ei gM i=1 is uniform, by TheP M orem 5.2, there is a vector '1 = i=1 ai ei 2 H M with P1 f = hP f '1 i'1 , for all f 2 H M and jai j = jaj j, for all 1 i j M . Now choose '2 = M i=1 bi ei 2 H M so the P2 f = hf '2 i'2 , for all f 2 H M . Since P1 P2 are rank-1, P1 P2 = 0 implies P2 P1 = 0. That is, h'1 '2 i = 0. Let P = P1 + P2 . Then (I ; P2 )(I ; P1 ) = I ; P . Since f(I ; P )gM i=1 is uniform, by Theorem 5.2 we have

a2i + b2i = a2j + b2j for all a i j M: Hence, b2i = b2j , for all 1 i j M . Since k'1 k = k'2 k = 1, it follows that

a2i = b2i = M1 for all 1 i j M: Since f(I ; P )ei gM i=1 is robust to two erasures, by Theorem 5.2

ai bj ; aj bi 6= 0 for all 1 i j M:

(29)

However, for all i j we have ai = aj = bi . Hence, there is some i j with either ai = aj and bi = bj or ai = ;aj and bi = ;bj . In either case (29) fails. Corollary 5.4 is heavily dependent on having a real Hilbert space. In the complex case, we will show in Corollary 5.6 that the GHFs have the property that we can step down to the frame by successively applying rank-1 projections to the orthonormal basis. Corollary 5.4 shows that in general we cannot step down one dimension at a time to construct frames robust to k erasures. It would be interesting to extend this to a classication of when we can step down from k1 erasures to k2 erasures and so on. One reason is that this has implications for entangled states in quantum detection theory 12, 22]. This raises the question of whether we can step down one dimension at a time as in Corollary 5.4 if we only ask for each level to be uniform. The answer is no as the next example shows. Example 5.1. Let fei g4i=1 be an orthonormal basis for H 4 and let

'1 = p1 e1 + 12 e3 + 12 e4 2

and

'2 = p1 e2 + 21 e3 + ;21 e4 :

2 Then h'1 '2 i = 0, and if P is the orthogonal projection of H 4 onto the span of f'1 '2 g, then by Theorem 5.2, f(I ; P )ei g4i=1 is a uniform normalized tight frame. However, there does not exist a rank-1 orthogonal projection P1 of H 4 into range of P so that f(I ; P1 )ei g4i=1 is a uniform frame.

32


Proof. By Theorem 5.2, in order for such a P1 to exist, there must exist a vector P 4 in P H 4 of the form i=1 bi ei with jbi j = jbj j, for all 1 i j 4. For any a1 a2 ,

a1 '1 + a2 '2 = pa1 e1 + pa2 e2 + a1 +2 a2 e3 + a1 ;2 a2 e4 : 2

2

If (ja1 + a2 j)=2 = (ja1 ; a2 j)=2, then one of ja1 j ja2 j equals 0 so ja1 j 6= ja2 j. The next question is whether we can step down one dimension at a time as in Corollary 5.4 if all we want is for our frame to be robust to j erasures at the j th level but are willing to give up uniformity at each level. Surprisingly, the answer here is yes. Proposition 5.5. Let P be an orthogonal projection of H M onto an L-dimensional M subspace H L . Let fei gM i=1 be an orthonormal basis for H M . If f(I ; P )ei gi=1 is robust to k erasures (k L), then there are mutually orthogonal rank-1 projections fPi gLi=1 taking H M into H L so that f(I ; Pj )(I ; Pj;1 ) (I ; P1 )ei gMi=1 is robust to j erasures, for all 1 j k, and robust to k erasures for all k j L. Proof. See Appendix A.3.

5.3. UTFs Robust to More than One Erasure: An Alternative Approach

In the previous sections we used the orthogonal projection P on H M to classify when f(I ; P )ei gM i=1 is robust to k erasures and when it is uniform. However, if the dimension of P H M is large (that is, close to the dimension of H M ) these results become dicult to implement since they require knowledge about the orthonormal bases for the large-dimensional space P H M . In this case, f(I ; P )ei gM i=1 is a large number of vectors in a small-dimensional space (I ; P )H M . So it is easier in this case to work directly with (I ; P )H and f(I ; P )ei g. Or, equivalently, with P H and fPei g, where dim P H is small. M Proposition 5.6. Let f'i gN i=1 be an orthonormal sequence in H M , let fei gi=1 be an orthonormal basis for H M and let P be the orthogonal projection of H M onto the span of f'i gNi=1 . Then the normalized tight frame fPeigM i=1 is unitarily equivalent to fgigM where for 1 j M we have i=1

gj =

N X i=1

h'i ej iei :

N That is, fPei gM i=1 is the frame obtained by turning the columns of f'i gi=1 into row vectors in H N .

Proof. For any 1 i M we have:

Pei =

L X n=1

hei 'n i'n :


33

Since f'n gLn=1 is an orthonormal sequence, the operator T'n = en, for 1 n L is a unitary operator which takes Pei to fgj gM j =1 where for 1 j M and we have

gj =

N X i=1

h'i ej iei :

Corollary 5.5. Given the conditions in Proposition 5.6 we have: 1.fgj gM j =1 is a uniform frame for H N if and only if N X

N for all 1 j M: jh'I ej ij2 = M i=1

2.The following are equivalent: (a) fgj gM j =1 is robust to the erasure of fgj gj 2I for some I f1 2 : : :M g. (b) We have that (h'i ej i)Ni=1j2I c has row rank N . Hence, fgj gM j =1 is robust to any choice of k erasures, k M ; N , if and only if every set of M ; k columns of the matrix N M A = (h'i ej i)i=1 j =1

has row rank N . Finally, the angle between gi and gj is given by the inner product of the ith and j th columns of A.

It would be interesting and useful to classify the GHFs in the format of this section. That is, precisely when is f(I ; P )ei gM i=1 a GHF? One important property of GHFs is contained in the following result which says they have the step-down property of Section 5.1. Corollary 5.6. Let P be an orthogonal projection of H M onto an L-dimensional ;1 M ;1 subspace H L and let fei gM i=0 be an orthonormal basis for H M . If f(I ; P )ei gi=0 L is a GHF then there is an orthogonal sequence of rank-1 projections fPi gi=1 of H M into H L so that for all 1 j L and for all permutations of f1 2 : : : Lg we have that jY ;1

f

;1 (I ; P(m) )ei gM i=0

m=0

is a GHF and hence is a UNTF which is robust to j erasures. Proof. We will do the proof for HTFs since the GHF case requires only notational changes but obscures the basic ideas of the proof. By Proposition 5.6, there


34

;1 is a unique way to get HTFs. Namely, let fei gM be the natural orthonormal basis i =0 ;1 for H M . Let fwi gM be distinct M th roots of unity and consider the orthonormal i =0 ;1 basis f'i gM i=0 for H M given by:

'i =

MX ;1 j =0

wij ej :

Without loss of generality, we may as well assume that H L = span f'i gLi=0 . ;1 Now, turning the row vectors of f'i gM i=L into column vectors gives a HTF for M ; L H . Moreover, again by Proposition 5.6, this HTF is unitarily equivalent to f(I ; P )ei gMi=0;1 where I ; P is the orthogonal projection onto (H L )? . Now, let Pi be the orthogonal rank-1 projection of H M onto span 'i for 0 i L ; 1. Fix a permutation of f1 2 : : : Lg and let Ij = f(0) (1) : : : (j ; 1)g. Then, again by Proposition 5.6, jY ;1

f

;1 (I ; P(m) )ei gM i=0

m=0

is unitarily equivalent to the HTF obtained by turning the row vectors of f'i gi=2Ij into column vectors. It is possible that the property in Corollary 5.6 characterizes GHFs, but we do not have a proof for this. We can show that the use of permutations is necessary in Proposition 5.6. That is, there are UNTFs which have the step-down property for erasures and for uniformness for one xed ordering of the rank-1 projections Pi while failing to be equivalent to a GHF. This is the point of the next example. Example 5.2. In C 4 , let

'1 =

4 X i=1

ei

'2 =

4 X i=1

wi ei

'3 = e1 + e2 ; e3 ; e4

where w1 = 1, w2 = ;1, w3 = i and w4 = ;i. Let fPi g3i=1 be the rank-1 orthogonal projections of C 4 onto the span of 'i . Then it follows easily from our results that: 1. f(I ; P1 )ei g4i=1 is a UNTF which is robust to one erasure. 2. f(I ; P2 )(I ; P1 )ei g4i=1 is a UNTF which is robust to 2 erasures. 3. f(I ; P3 )(I ; P2 )(I ; P1 )ei g4i=1 is a UNTF which is robust to 3 erasures, but is not a harmonic frame since f(I ; P3 )(I ; P1 )ei g4i=1 is a UNTF which is not robust to 2 erasures. Example 5.3. If 'i = (1 wi ) in C 2 for 1 i M , where jwi jP = 1 and the fwi gMi=1 are distinct, then f'i gMi=1 is a UNTF for C 2 if and only if Mi=1 wi = 0. In this case, if Pi is an orthogonal rank-1 projection onto the span 'i , then f(I ; M Pj )ei gM i=1 is a UNTF which is robust to one erasure while f(I ; Pj )(I ; Pk )ei gi=1

is a UNTF which is robust to 2 erasures for all 1 j 6= k M .

35


Proof. This family is certainly uniform. The assumption tees that the vectors

PM w = 0 guarani=1 i

(1 1 1 : : : 1) and (w1 w2 : : : wM ) 2 are orthogonal in H M (and conversely). Hence, f'i gM i=1 is a UNTF for C by Proposition 5.6. Example 5.4. In general, the frames constructed in Example 5.3 are not equivalent to harmonic frames even after a permutation. ;1 Proof. The reason is that f'j gM j =0 is a GHF implies that

h'j 'j+1 i = h'j+1 'j+2 i for all 0 j M ; 2 (see Proposition 2.4). This would imply in Example 5.3 that

h'j 'j+1 i = 1 + wj wj+1 = 1 + wj+1 wj+2 = h'j+1 'j+2 i:

p

w1 = 1, w2 = ;1=2 + 3=4i and w2 = Hence, wpj wj+1 = wj+1 wj+2 . Now, let P ;1=2 ; 3=4i. Then jwj j = 1 and 3j=1 wj = 0. Hence, by Example 5.3, f(1 wj )g3j=1 is a UNTF for C 2 . However, for any permutation of f1 2 3g, we do not have w(j) w(j+1) = w(j+1) w(j+2) , for all 0 j 2 (with w4 = w1 ). It is not hard to see that in the case of R2 , the condition in Corollary 5.6 is sucient.

APPENDIX: PROOFS A.1. PROOF OF PROPOSITION 4.1 First note that if fci gNi=1 are distinct M th roots of c with jcj = 1, then MX ;1 k=0

cki = 0

MX ;1 k=0

jcki j2 =

M

MX ;1 k

=0

(ci cl )k = 0

(A.1)

i6=l

and thus MX ;1 k=0

(ci cl )k = M i;l :

(A.2)

We now check the necessity of our condition for a frame to be a general harmonic frame. If f'k gM k=1 is a general harmonic frame for H N then by the denition:

'k = (ck1 b1 ck2 b2 ckN bN ) =

N X i=1

cki bi ei


36

p

where feigNi=1 is the natural unit vector orthonormal basis of H N , jbi j = 1= M , and fci gNi=1 are distinct M th roots of c with jcj = 1. Now let

'0 =

n X i=1

bi ei :

Dene a unitary operator U on H N by Uei = ci ei . Then, for all 0 k M ; 1 we have U k ei = cki ei , and thus, U k '0 = 'k for all 0 k M ; 1. So our frame has the form given in the proposition. The suciency of the condition is checked similarly. If we assume that f'k g is of the form described in the proposition, then:

k'k k = kU k '0 k =

k'0 k =

r

N M

since U is a unitary operator. So fP 'k g is a uniform frame. To see that f'k g is a normalized tight frame, we let f = Ni=1 ai ei 2 H N and compute: MX ;1 k=0

jhf U k '0 ij2 = =

MX ;1 X N

j

k=0 i=1 MX ;1 X N k=0 i=1 N X

ai cki bi j2 =

jai cki bi j2 +

X m6=`

am b`

MX ;1

(cm c` )k

k=0 N X = M jai j2 jbi j2 + 0 = M jai j2 j p1 j2 = M i=1 i=1

kf k2

where in the next to last equality we used the fact that the fci g is a family of distinct M th roots of c (see (A.2)). Hence, f'k g is a UNTF. Thus, using (15) and writing f = ej , we see that

kej k2 = 1 = p

M X k=1

jhej U k '0 ij2 =

M X k=1

jhej

N X i=1

cki bi ei ij2 =

M X k=1

jckj bj j2 = M jbj j2 :

So jbj j = 1= M , for all j 0 j N ; 1. This further means that

p

'0 = (b1 b2 : : : bN )

with jbj j = 1= M and since 'k = U k '0 ,

'k = (ck1 b1 ck2 b2 : : : ckN bN ):

A.2. PROOF OF THEOREM 4.1

We prove the theorem in steps.

Step I: We rst prove that there is a constant jcj = 1 so that U M = cI . This will ;1 show that the operators fU i gM i=0 , in fact, form a . Since the frame is an NTF, for

37


every f 2 H N we have,

f = Hence,

Uf = =

MX ;1 k=0 MX ;1 k=0

= U

MX ;1 k=0

hf U k '0 iU k '0 :

hUf U k '0 iU k '0 = hf U k;1 '0 iU k '0

MX;1 k=0

since U = U ;1

hf U k;1 '0 iU k;1 '0

!

= U

MX;2

k=;1

hf U k '0 iU k '0

!

: (A.3)

Since U is one-to-one,

f =

MX ;2 k=;1

hf U k '0 iU k '0 =

MX ;1 k=0

hf U k '0 iU k '0 :

(A.4)

From (A.3) and (A.4), it follows that,

hf U M ;1 '0 iU M ;1 '0 = hf U ;1 '0 iU ;1 '0 : Applying U we have

hf U M ;1 '0 iU M '0 = hf U ;1 '0 iU 0 '0 = hf U ;1 '0 i'0 :

(A.5)

Hence, there is a c 2 C so that U M '0 = c'0 . Replacing f by U ;1 f in (A.5) gives:

hU ;1 f U M ;1 '0 iU M '0 = hf U M '0 iU M '0 = hf c'0 ic'0 = jcj2 hf '0 i'0 = hU ;1 f U ;1 '0 i'0 = hf '0 i'0 : So jcj2 = 1. Also, for all 0 k M ; 1,

U M U k '0 = U k U M '0 = U k c'0 = cU k '0 :

P

Since fU k '0 g spans H N , that is, for any f 2 H N f = k hf U k '0 iU k '0 , it follows that U M = cI . This completes Step I. Step II: We want to prove: 1. U is diagonalizable with respect to an orthonormal basis feigNi=1 with diagonal elements fci gNi=1 with ci an M th root of c. q P 2. '0 = Ni=1 bi ei and jbi j = M1 , for all 1 i N . These two steps give us frame elements as in Denition 2.1.


38

Since U is unitary (and hence normal), and our space is nite dimensional, it follows that U is diagonalizable with respect to an orthonormal basis fei gNi=1 with diagonal elements fci gNi=1 . Writing '0 as in (2), we have that

U k '0 =

N X i=1

cki bi ei :

Since U M = cI , it follows that each ci is an M th root of c. Also, since the frame is an NTF, for all 1 j N we have 1 =

M X k=1

jhej U k '0 ij2 =

p

M X k=1

jhej

N X i=1

cki bi ei ij2 =

N X k=1

jckj bj j2 = M jbj j2 :

So jbj j = 1= M . This completes Step II. Step III: We nally prove that the cj are distinct M th roots of c. This will complete the proof. Fix 1 m 6= ` M . Since the columns of a normalized tight frame are orthogonal vectors, that is, F F = I , we have for columns m `, 0 = Hence,

MX ;1 k=0

bm ckm b` ck` = bm b`

MX ;1 k=0

(cm c` )k :

MX ;1 k=0

(cm c` )k = 0:

Thus, cm and c` are distinct M th roots of c. This completes the proof of Step III and the theorem.

A.3. PROOF OF PROPOSITION 5.5

For all 1 j L, let

gj =

M X i=1

hgj ei iei

be an orthonormal basis for H L . We will construct the projections fPj gLj=1 by nite induction. We start with P1 . Since f(I ; P )ei gM i=1 is robust to 1 erasure, by Theorem 5.1, for every 1 i M there exists a 1 j L so that hgj ei i 6= 0. Choose scalars a1 a2 : : : aL so that for any 1 j L, if hgj ei i 6= 0 for some 1 i M , then

jaj hgj ei ij 2 It follows easily that

'1 =

j ;1 X k=1

jak hgk ei ij:

PL a g j =1 j j P k L agk j =1 j j

39


then h'1 ei i 6= 0, for all 1 i M . Now, let P1 be the orthogonal projection onto the span f'1 g, then f(I ; P1 )ei gM i=1 is robust to 1 erasure by Corollary 5.1. Now assume we have found orthonormal vectors f'j gnj=1 , n k ; 1 so that the rank-1 projections Pj onto span f'j g satisfy the proposition and span f'j gnj=1 = span fgj gnj=1 . We will construct 'n+1 and Pn+1 . Let fgj gLj=1 be an orthonormal basis for the orthogonal complement of span f'j gnj=1 in H L . Fix 1 i1 < i2 < : : : < in+1 M . Since

f(I ; Pn )(I ; Pn;1 ) (I ; P1 )ei gMi=1 is robust to n erasures, if we row reduce f'j gnj=1 using the columns fi1 i2 : : : in g (and switching rows if necessary) we obtain vectors fhj gnj=1 with span fhj gnj=1 = span f'j gnj=1 and for every 1 j n we have,

hhj ei i = j` : l

Now we verify the following claim: Claim: For all but nitely many 0 < x < 1, if L X

hn+1 = then the row rank of

j =n+1

xj gj

(hhj ei` i)nj`+1=1

is n + 1. Proof of Claim: Fix 0 < x < 1. We row reduce this matrix by taking, for all 1 ` M,

;h

L X

j =n+1

xj gj ei` ih` + hn+1 :

We then arrive at a matrix with 1s in the j th row and the ij th column for 1 j n, zeroes otherwise, zeroes in the (n + 1)st row for columns ij for all 1 j n, and in the (n + 1 n + 1)st position we have:

hhn+1 ei +1 i ; n

n X L X

h

`=1 j =n+1

xj gj ei` ihh` ei` i:

If this number is nonzero, then our matrix is of rank n + 1. Now we check what it takes for this to be nonzero.

hhn+1 ei +1 i ; n

=

L X j =n+1

n X L X

h

`=1 j =n+1

xj hgj ein+1 i ;

L X j =n+1

xj gj ei` ihh` ei` i

xj

n X j =1

hgj ei ihh` ei i `

`


40 =

L X j =n+1

xj

"

hgj ei +1 i ; n

By the hypotheses on the proposition,

n X `=1

#

hgj ei ihh` ei i : `

`

+1 (hgj ei` i)jL=1`n=1

has rank n + 1. It follows that there is a n + 1 j L so that

hgj ei +1 i ; n

n X `=1

hgj ei ihh` ei i 6= 0: `

`

Hence, the number of x's with

hhn+1 ei +1 i ; n

n X L X

h

`=1 j =n+1

xj gj ei` ihh` ei` i = 0

is nite. This establishes the claim. Applying the claim to all choices of 1 i1 < i2 < in+1 M , we see that for all but nitely many 0 < x < 1, for all 1 i1 < i2 < < in+1 L, the rank of n+1 (hhj ei` i)jM =1`=1

is n + 1. Let

'n+1 = khhn+1 k n+1

and let Pn+1 be the orthogonal projection of H M onto span f'n+1 g. Now,

f(I ; Pn+1 )(I ; Pn ) (I ; P1 )ei iMi=1 is robust to (n + 1) erasures by Theorem 5.1. This completes the proof of the proposition.

ACKNOWLEDGMENT

The rst author would like to thank V. Paulsen for fruitful discussions.

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