equal-norm tight frames with erasures

Uniform Tight Frames with Erasures

Peter G� Casazza

Department of Mathematics� University of Missouri�Columbia

Columbia� MO ��

E�mail� pete�math�missouri�edu

and

Jelena Kova�cevi�c

Mathematics of Communication Research� Bell Labs� Lucent Technologies�

�� Mountain Avenue� Murray Hill� NJ �� USA

E�mail� jelena�bell�labs�com

Uniform tight frames have been shown to be useful for robust data trans�

mission� The losses in the network are modeled as erasures of transmitted

frame coe�cients� We give the �rst systematic study of the general class

of uniform tight frames and their properties� We search for e�cient con�

structions 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 classi�cation of frames in terms of their

robustness to erasures�

�� 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 in�nitely many

di�erent representations for a given vector ��

Frames have been used in signal processing because of their resilience to additive

noise � �� resilience to quantization � �� as well as their numerical stability of

reconstruction � �� and greater freedom to capture signal characteristics �� Re�

cently� several new applications for �uniform tight� frames have been developed�

The �rst� developed by Goyal� Kova�cevi�c and Vetterli � �� uses the

redundancy of a frame to mitigate the e�ect of losses in packet�based communica�

tion 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 Signaland Image Proc� �San Diego� CA� July �� The �rst author was supported by NSF DMS��

�

� CASAZZA AND KOVACEVI�C

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 bu�er over�ows

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� Re�

transmission 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 �� 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 � �� Much theoretical work has been done to show

that communication systems which employ multiple antennas can have very high

channel capacities � �� These methods rely on the assumption that the receiver

knows the complex�valued Rayleigh fading coe�cients� To remove this assumption�

in � �� new classes of unitary space�time signals are proposed� If we have N trans�

mitting antennas and we transmit in blocks of M time samples �over which the

fading coe�cients are approximately constant�� then a constellation of K unitary

space�time signals is a �weighted bypM� collection of M � N complex matrices

f�kg for which ��k�k � I � The nth column of any �k contains the signal trans�

mitted 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 cumber�

some and di�cult optimization problem for practice� However� in � �� it was shown

that constellations arising in a �systematic� fashion can be done with relatively

little e�ort� 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 di�erences have nonzero determinant� is a desirable property for good per�

formance� In a tour�de�force� in � �� the authors used �xed�point�free groups and

their representations to design high�rate constellations with full diversity� More�

over� they classi�ed 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 � we

review basic notions on frames� In particular� we state the Naimark�s Theorem

� � which has been rediscovered several times in recent years � � �� although it

has been used for several decades in operator theory� We give examples of uniform

UNIFORM TIGHT FRAMES WITH ERASURES �

normalized tight frames such as harmonic and Gabor frames� For harmonic frames�

we de�ne a more general class � general harmonic frames �GHF� � and study when

harmonic frames are equivalent to each other and general harmonic frames� In

Proposition �� 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�� and �j�� Section � concentrates

on uniform tight frames �UTF�� We start with a review in Section �� 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 �� we shift our attention to UTFs with group structure� We show that

the UNTFs generated by the family fUk��gM��k�� where U is unitary and ��

H � are precisely the GHFs� We then extend our discussion to UNTFs generated

by fUkV j��g and higher numbers of generators� Finally� Section � introduceserasures modeled as losses of transform coe�cients hf� �ii� where f is the signalto be transmitted and f�igi�I is a set of frames vectors corresponding to erasedtransform coe�cients� We give a complete classi�cation 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 di�erent

number of erasures�

�� FRAME REVIEW

A set of vectors � � f�igi�I in a Hilbert space H � is called a frame if

� � Akxk� �Xi�I

jhx� �iij� � Bkxk� � �� x ��

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 CN �which

we denote HN � with the usual Euclidean inner product� When results generalize

to the in�nite�dimensional setting� we will point it out�

When A � B the frame is tight �TF�� If A � B � � the frame is normalized tight

�NTF�� A frame is uniform �UF� if all its elements have the same norm c�� k�ik � c�

For a uniform tight frame �UTF�� the frame bound A gives the redundancy ratio�

A UNTF of norm� vectors is an orthonormal basis �ONB��

The analysis frame operator� F maps the Hilbert space H into ��I�

�Fx�i � h�i� xi� ��

�Often� uniform is used to mean c � � here� however� we use the more general de�nition��F 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 �


for i � I � When H � HN � the analysis frame operator is an M �N matrix whose

rows are the transposed frame vectors ��i �

F �

�B�

�� N��

��

��M� � � � ��MN

�CA � ��

We say that two frames f�igi�I and f�igi�I for H are equivalent if there is an

invertible operator L on H for which L�i � �i for all i � I � and they are unitarily

equivalent if L can be chosen to be a unitary operator� A direct calculation shows

that fS��ig is a normalized tight frame for any frame f�ig� Thus� every frameis equivalent to a normalized tight frame� If f�igi�I is a frame with frame boundsA�B� and if P is an orthogonal projection on H � then by � � we have that fP�igi�Iis a frame for PH with frame bounds A and B� In particular� if f�igi�I is anormalized tight frame for H �for example� if it is an orthonormal basis for H ��

then fP�igi�I is a normalized tight frame for PH �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 �� Naimark � �� Han Larson � �� A family f�igi�I 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 feigi�I for K so that the orthogonal projection P

of K onto H satis�es� Pei � �i� for all i � I�

Let us now go through certain important frame notions� Using the analysis frame

operator F � � � can be rewritten as

AI � F �F � BI� ��

We call S � F �F the frame operator� It follows that S is invertible �Lemma ��

in � �� and furthermore

B��I � S�� A��I� ��

It follows that f�igi�I 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

de�ned as !� � f !�igMi�� fS��igMi�� where

!�i � S��i� ��

�This theorem has been rediscovered by several people in recent years� The �rst author �rstheard it from I� Daubechies in the mid��s� Han and Larson rediscovered it in �� they came upwith the idea that a frame could be obtained by compressing a basis in a larger space and thatthe process is reversible �the statement in this paper is due to Han and Larson�� Finally� it waspointed out to the �rst author by E� Soljanin �� that this is� in fact� the Naimark�s theorem�which has been widely known in operator theory and has been used in quantum theory�


for k � � � � � �M � Now�

f �Xi�Ihf� S��ii�i� for all f � 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 � H has at least one natural series representation in

terms of the frame elements� Also� this element may have in�nitely many other

representations� However� in order to �nd this natural representation of f � we

need to invert the frame operator� which may be di�cult 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 � ��iS�� and stacking !�� !�

�� !�

�M in a matrix� the frame

operator associated with !� is

!F � FS��

Since !F � !F � S�� shows that B�� and A�� 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 ��

Note that for any matrix F with rows ��i

S � F �F �MXi��

�i��i � ��

This identity will prove to be useful in many proofs�

�� The Role of Eigenvalues

The product S � F �F will appear everywhere and its eigenstructure will play animportant role� Denote by �k �s the eigenvalues of S � F �F � We now summarizethe important eigenvalue properties�

General Frame� For any frame in HN � the sum of the eigenvalues of S � F �F �equals the sum of the lengths of the frame vectors�

NXk��

�k �

MXi��

k�ik��

Uniform Frame� For a uniform frame� that is� when k�ik � c� i � � � � � �M �

NXk��

�k �

MXi��

k�ik� � M � c��

CASAZZA AND KOVACEVI�C

Tight Frame� Since tightness means A � B� for a TF� we have from � �

MXi��

jhf� �iij� � Akfk��

for all f � HN � Moreover� according to �� a frame is a TF if and only if

F �F � A � IN � � ��

Thus� for a TF� all the eigenvalues of the frame operator S � F �F are equal to A�Then� using � �� the sum of the eigenvalues of S � F �F is as follows�

N � A �NXk��

�k �MXi��

k�ik��

Normalized Tight Frame� If a frame is an NTF� that is� A � B � � then

MXi��

jhf� �iij� � kfk��

for all f � HN � In operator notation� a frame is an NTF if and only if

S � F �F � IN � � �

For an NTF� all the eigenvalues of S � F �F are equal to �Then� using � �� the sum of the eigenvalues of S � F �F is as follows�

N �

NXk��

�k �

MXi��

k�ik��

Uniform Tight Frame� From � � and � �� we see that

N �A �

NXk��

�k �

MXi��

k�ik� � M � c��

Then� from � �� and � ��

MXi��

jhf� �iij� � M

Nc�kfk��

for all f � HN � The redundancy ratio is then

A �M

N� c��

Since S � F �F � �M�N�I � the following is obvious�

MXi��

j�ik j� � M

N� ��

UNIFORM TIGHT FRAMES WITH ERASURES

Uniform Normalized Tight Frame� If a frame is a UNTF� that is� we also ask

for A � B � � and if the frame vectors have norm c� then

N �

NXk��

�k �

MXi��

k�ik� � c�M�

Thus� a UNTF with norm c � is an orthonormal basis�

�� Examples of Uniform Normalized Tight Frames

We start with a simple example of a frame" � vectors in � 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

F �

r�

�

��

�p�� p��

�A � ��

This is obviously a UNTF since ��F �F � I � We could make this frame just a

UTF with norm �as in �� 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 � ��

�� 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 �� They

have been proven to be useful in applications � ��

An HTF is given by�

�k � �wk� � w

k� � � � � wk

N ��

for k � �� M � � where wi are distinct Mth roots of unity� This is a TF �F �F � MI � A more general de�nition of the harmonic frame �general harmonic

frame� is as follows�

Definition �� Fix M � N � jcj � and fbigNi�� with jbij � �pM� Let fcigNi��

be distinct Mth roots of c� and for � � k �M � � let

�k � �ck�b�� ck�b�� c

kNbN ��

Then f�kgM��k�� is a uniform normalized tight frame for HN called a general har�

monic frame �GHF��

Note the di�erence 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��


Proposition �� Let f�kgM��k�� and f�kgM��

k�� be harmonic frames on HN �

Then f�kgM��k�� is equivalent to f�kgM��

k�� if and only if there is a permutation � of

f � �� Ng so that �kj � �k��j�� for all � � k �M� and all � j � N � Hence�

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

Proof �Proposition �� Let fejgNj�� be the natural orthonormal basis for HNand

�k � �wk� � w

k� � � � � wk

N � �

NXj��

wkj ej �

and

�k � �vk� � vk� � � � � � vkN � �

NXj��

vkj ej �

where wj � vj are sets of distinct Mth roots of unity� If

fwj j � j � Ng �� fvj j � j � Ng�

then� without loss of generality� we may assume that v� �� fwj j � j � Ng�Therefore�

M��Xk��

vk��k �

NXj��

�M��Xk��

�v�wj�k

�ej �

NXj��

� � ej � ��

On the other hand�

M��Xk��

vk��k� � �

M��Xk��

jvk� j� � M�

It follows that f�kgM��k�� is not equivalent to f�kgM��

k�� The other direction is imme�

diate�

We now show that every general harmonic frame is unitarily equivalent to a

simple variation of a harmonic frame�

Proposition �� Every general harmonic frame is unitarily equivalent to a

frame of the form fck�kgM��k�� where jcj � and f�kgM��

k�� is a harmonic frame�

Proof �Proposition �� Let M � N � jcj � and jbkj � �pM � for all � k �

N � Let fcjgNj�� be distinct Mth roots of c and let the GHF be

�k � �ck�b�� ck�b�� c

kNbN ��

for all � � k � M � � If c � ei�� let d � ei��M � Then there exist distinct Mth

roots of unity w�� w� � � � � wN with cj � ei��Mwj � dwj � For all sets of complex

numbers fakgM��k�� we have�

kM��Xk��

ak�kk� �NXj��

jM��Xk��

akckj bj j� �

M

NXj��

jM��Xk��

akckj j� �


M

NXj��

jM��Xk��

akeik��Mwk

j j� �

MkM��Xk��

akdk�kk��

where �k � �wk� � w

k� � � � � � w

kN �� Thus� the GHF f�kgM��

k�� is unitarily equivalent to�pMf�kgM��

k�� with f�kgM��k�� an HTF�

Finally� we show that the frames given in Proposition �� are not unitarily equiv�

alent to each other or to harmonic frames except in the trivial case�

Proposition �� Let f�kgM��k�� and f�kgM��

k�� be harmonic frames and let jcj � � Then fck�kgM��

k�� is equivalent to f�kgM��k�� if and only if c is an M th root of

unity and there is a permutation � of f � �� Ng so that �kj � �k��j�� for all

� � k � M � and all � 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�

Proof �Proposition �� Let �k �PN

j�� wkj ej � for all � � k �M� � If c � w��j �

for some j� we are done� Otherwise� sincePM��

k�� k � �� if fck�kg is equivalent tof�kg then also

PM��k�� ck�k � �� Hence� for all � j � N we have

M��Xk��

ckwkj �

M��Xk��

�cwj�k �

� �cwj�M � cwj

� ��

Hence� �cwj�M � cMwM

j � cM � � The proposition now follows from Proposi�

tion ��

We now have immediately�

Corollary �� Let f�kgM��k�� and f�kgM��

k�� be harmonic frames and let jcj �jdj � � The frames fci�kgM��

k�� and fdi�kgM��k�� are equivalent if and only if c � d

and there is a permutation � of f � �� Ng so that �kj � �k��j�� for all � � k �M � �The next proposition gives a classi�cation of GHFs�

Proposition �� A family f�igM��i�� in HN is a GHF if and only if for all

� � i� j �M � we have

h�i� �ji � h�i�� j��i�

where �M � ��

Proof� If f�igM��i�� in HN is a generalized harmonic frame then there is a unitary

operator U on HN so that U i�� i� for all � � i � M � � Hence� for all� � i� j �M � we have

h�i�� j��i � hU i�� Uj��i � hU i�� U

j��i � h�i� �ji�


Conversely� given our equality� for any sequence of scalars faigM��i��

kM��Xi��

ai�i��k� � hM��Xi��

ai�i��M��Xi��

ai�i��i

�

M��Xi�j��

aiajh�i�� j��i �M��Xi�j��

aiajh�i� �ji

� hM��Xi��

ai�i�

M��Xi��

ai�ii � kM��Xi��

ai�ik��

It follows that U�i � �i�� is a unitary operator and Ui�� i� for all � � i �M�

�

As we will see in the next section� UTFs with M � N � are all unitarily

equivalent� Thus� as a direct consequence of Theorem �� any UTF withM � N�

is unitarily equivalent to the HTF with M � N � � This is a very useful result

since we have HTFs for any N and M " thus� for M � N � � 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 � � and N � ��

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

we will see in Section �� That is� � � f�igMi�� fU i��gMi�� 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 �� from �� We provide a complete classi�cation of frames in

terms of their robustness to erasures in Section ��

�� Gabor or Weyl�Heisenberg Frames

For the other general class of frames� we introduce two special operators on L��R��

Fix � � a� b and for f � L��R� de�ne translation by a as

Taf�t� � f�t� a��

and modulation by b as

Ebf�t� � e��imbtf�t��

Now� �x g � L��R�� If fEmbTnaggm�n�Z is a frame for L��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 fEmbTnaggm�n�Zmust commute with translation and modulation� each Gabor frame is equivalent

to the �uniform� normalized tight Gabor frame fEmbTnaS��ggm�n�Z� For an

introduction to Gabor frames we refer the reader to �� and � ��


�� UNIFORM TIGHT FRAMES

�� What is Known about UTFs�

As we have mentioned in the introduction� UNTFs have become popular in appli�

cations� In particular� in �� 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 coe�cients� It

is further shown in �� that� assuming a particular quantization model� a uniform

frame with quantized coe�cients� 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

e�cient" 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 coe�cient �frame

vector� erasures� These and other issues will be explored in the rest of the paper�

�� 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 ��

In Section � we will also extend this result to uniform frames�

Theorem �� There is a unique way to get UNTFs with M elements in HN �

Take any orthonormal set fkgNk�� in HM which has the property

NXk��

jkij� � c� for all i�

Thinking of the k as row vectors� switch to the M column vectors and divide bypc� This family is a uniform tight frame for HM with M elements� and all UNTFs

for HN with M elements are obtained in this way�

Corollary �� If f�igMi�� is a uniform normalized tight frame for HN then

k�ik� � M

N�

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

R� in �� In � �� there is a deep classi�cation of groups of unitary operators which

generate uniform normalized tight frames� The simplest case of this is the harmonic

frames� In Section � 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 � ��


or worse� if the uniform tight frame comes from a subset of the elements of such a

group�

�� Classi�cation 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� vectors in a Hilbert space

can be extended to become a uniform tight frame�

Theorem �� If f�igMi�� is a family of norm� vectors in a Hilbert space H �

then there is a uniform tight frame for H which contains the family f�igMi��

Proof �Theorem �� For each � i �M choose an orthonormal basis feijgj�Jfor H which contains the vector �i� Now the family feijgMi��j�J is made up ofnorm� vectors and for any f � H we have

MXi��

Xj�J

jhf� eijij� �MXi��

kfk� � Mkfk��

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

M in the family f�igMi�� In general� this is the best possible� For example� just let�i � �j be norm� vectors for all � i� j �M � Then

MXj��

jh�i� �jij� � M�

Hence� any tight frame containing the family f�ig has the tight frame bound atleast M �

There is another general class of uniform tight frames �see �� The result below

tells us that all UNTFs with M � N � are unitarily equivalent� It is a direct

consequence of Theorem �� from �� where it is stated for UTFs with norm �

Theorem �� Goyal� Kova�cevi�c and Kelner �� A family f�igN��i�� is a uni�

form normalized tight frame for HN if and only if f�igN��i�� is unitarily equivalent

to the frame fPeigN��i�� where feigN��

i�� is an orthonormal basis for HN�� and P

is the orthogonal projection of HN�� onto the orthogonal complement of the one�

dimensional subspace of HN�� spanned byPN��

i�� ei�

Since there are HTFs with N � �elements in HN � it follows that every UNTF

for HN with N � �elements is unitarily equivalent to a HTF�

�� 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�


As we saw in Theorem �� there is a unique way to get normalized tight frames

in HN with M elements� Namely� we take an orthonormal basis feigMi�� for HMand take the orthogonal projection PHN of HM onto HN � Then fPHN eigMi�� is anormalized tight frame for HN with M elements� In particular� there is a natural

one�to�one correspondence between the normalized tight frames for HN with M

elements and the orthonormal bases for HM � Then� the uniform normalized tight

frames for HN are the ones for which kPeik � kPejk� for all � i� j � M � We

use this to exhibit a natural correspondence between these families and certain

subspaces of HM � Here we treat two frames as the same if they are unitarily

equivalent�

Theorem �� Let P be a rank�N orthogonal projection on HM and let feigMi��be an orthonormal basis for HM � There is a natural one�to�one correspondence

between the normalized tight frames for PHM with M elements and the family of

all N�dimensional subspaces of HM �

Proof �Theorem �� If f�igMi�� is any normalized tight frame for PHM � then byNaimark�s Theorem� there is an orthonormal basis fe�

igMi�� for HM so that Pe�

i � �i�

De�ne a unitary operator U on HM by Uei � e�

i� Now� �i � PUei which is uni�

tarily equivalent to ��

i � U�PUei� So we will associate our normalized tight framef�ig with the subspace U�PUHM � Now we need to check that this correspon�dence is one�to�one� Let f�ig be another normalized tight frame for PHM which

is associated with the same subspace of HM � namely U�PUHM � Then there is an

orthonormal basis fe��

i g and a unitary operator V ei � e��

i with V �PV � U�PUand PV ei � e

��

i � Hence� V�PV ei � V ��i � U�PUei � U��i� This implies that

the normalized tight frames f�ig and f�ig are unitarily equivalent �and hence thesame�� Finally� we need to see that this correspondence covers all subspaces� If

W is any subspace of HM of dimension N � we can de�ne a unitary operator U

on HM so that UW � PHM � Then U�PU � PW while fPUeigMi�� is a nor�malized tight frame for PHM �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 HN � then all the others can be obtained from it by this process� We

now turn our attention to uniform frames�

Theorem �� Fix an orthonormal basis feigMi�� for HM so that if P is the

orthogonal projection of HM onto HN then fPeigMi�� is a uniform normalized tight

frame for HN � Then there is a natural one�to�one correspondence between the

uniform normalized tight frames for HN with M elements and the subspaces W of

HM for which kPW eik� �M�N � for all � i �M �

Proof �Theorem �� We already have our classi�cation 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 fPUeig is a uniform normalized tight frame for PHM � Thenthis frame is associated to the subspace W � U�PHM and PW � U�PU � Hence�PUei � UPW ei� for all i� But� fPUeig is a uniform normalized tight frame for


PHM and so kPUeik � N�M � for all i� Hence�

kPUeik � N

M� kUPW eik � kPW eik�

So our association between normalized tight frames and subspaces given in The�

orem �� identi�es the subspacesW of HM for which kPW eik � N�M � for all i�

�� 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 de�nition of alternate dual

frames�

Definition �� Let f�igi�I be a frame for a Hilbert space H � A family f�igi�Iis called an alternate dual frame for f�igi�I if

f �Xi�Ihf� �ii�i� for all f � H �

In De�nition �� if we let F� �respectively� F�� be the analysis frame operators

for f�ig �respectively f�ig�� then we see that F �� F� � 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 � �� Moreover� no two distinct alternate dual frames for a given frame are

equivalent � �� For a normalized tight frame� its canonical dual frame is the frame

itself since !F � FS�� F � I � F � Moreover�

Proposition �� If f�igi�I is a normalized tight frame for HN � then the only

normalized tight alternate dual frame for f�igi�I is f�igi�I itself�

Proof �Proposition �� Let F� be the analysis frame operator for f�igi�I �Let f�igi�I be any normalized tight alternate dual frame for f�igi�I with analysisframe operator F�� It follows that F�� F� are isometries� Now�

F �� F� � F�� F �� F� � F �� F� � I � I � ��

Hence� for every f� g � H we have

hF �� g� �F� � F��fi � hg� F �� F� � F��fi � hg� �i � ��

Since f�igi�I is an NTF� for every f � H we have

kfk� �MXi��

jhf� �iij� � kF�fk��

We can further expand this as

kF�fk� � k�F� F��fk�� kF�fk� � k�F� � F��fk� � hF �� F� � F��f� fi� �z �

�

� hF �� F� � F��f� fi�� z ��

�


However� since F� corresponds to an NTF as well�

kF�fk� � k�F� � F��fk� � kfk� � k�F� � F��fk��

Equating �� and �� we get

k�F� � F��fk� � ��

Hence� F� � F� and so �i � �i� for all i � I �

If we ask for the dual frame just to be tight� but not normalized� then�

Proposition �� If f�igMi�� is a normalized tight frame for HN and M � �N �

then the only tight dual frame for f�igMi�� is f�igMi�� itself� If M � �N then there

are in�nitely many �nonequivalent� tight alternate dual frames for f�igMi��

Proof �Proposition �� With the notation of the proof of Proposition �� the

only change is that� since the dual frame is only tight and not normalized tight�

there is a frame bound A � so that

kF�k� � Akfk�� for all f � H �

Hence� as in the proof of Proposition �� we have

kfk� � kF�fk� � kF�fk� � k�F� � F��fk� � Akfk� � k�F� � F�k��

Hence� k�F� � F��fk� � � �A�kfk� for all f � H � Hence� either F� � F� � � and

A � which gives us the NTF as in the previous proposition� �and so �i � �i for

all � i �M� or � �p �A��F� � F�� is an isometry on HN � In the latter case it

follows that dim �F�HN �� N � and so M � �N � Thus� if M � �N � the only tight

dual frame is the frame itself�

On the other hand� ifM � �N given any F � HN �M� which is a constant times

an isometry� and with F�HN � FHN � F��F de�nes a tight frame which is an alter�

nate dual frame for f�ig�However� since it is really the uniform case we are interested in�

Proposition �� If f�igMi�� is a uniform normalized tight frame for HN and

M � �N � then there are in�nitely many uniform tight alternate dual frames for

f�ig�Ni��

Proof �Proposition �� Again we use the notation of the proof of Proposi�

tion �� Choose any F� � HN �M� which is a constant multiple of an isometry

and with F�HN � �F�HN �� Then F� � F� de�nes a tight alternate dual frame

for f�igMi�� say �i � �F �� F �� ei� for all � i � M � We just need to check

that this frame is uniform� Let P � �M� F�HN be the orthogonal projection�

so that Pei � F��i� for all � i � M � It follows that kPeik� � N�M and

k�I � P �eik� � �N�M � for all � i �M � Now�

�i � F �� ei � F �� ei � �i � F �� I � P �ei�

CASAZZA AND KOVACEVI�C

Also� there is an a � so that for every � i �M we have

k�ik� � k�ik� � kF �� I � P �eik� � N

M� ak�I � P �eik�

�N

M� a

� � N

M

� a� � � a�

N

M�

Hence� f�igMi�� 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�ig are obtained by the methods of this section� Wedo not know in general which frames �not tight� have tight �respectively uniform

tight� alternate dual frames�

�� Frames Equivalent to UTFs

Another approach to the classi�cation 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�igi�Iwith the frame operator S� the frame fS��igi�I is a normalized tight framewhich is equivalent to f�igi�I � Therefore� it is natural to try to �nd ways to turnframes into uniform normalized tight frames� As it turns out� this is not possible

in most cases�

Theorem �� If a frame f�igi�I with the frame operator S is equivalent to

a UTF� then fS��igi�I is a UNTF� In particular� a tight frame which is not

uniform cannot be equivalent to any UNTF�

Proof �Theorem �� It is known that fS��kg is a normalized tight framewhich is equivalent to f�kg� So if f�kg is equivalent to a uniform tight frame� sayf�kg� and k�kk � c� for all k� then fS��kg is equivalent to f�kg� That is� thereis an invertible operator T on H so that TS��k � �k�

Now we show that T�pA is a unitary operator� Let A be the tight frame constant

of fTS��kg� Then for all f � H � the tightness of fTS��kg implies

Akfk� �Xk

jhf� TS��kij� �Xk

jhT �f� S��kij��

However� since fS��kg is an NTF� then this leads us toXk

jhT �f� S��kij� � kT �fk��

Equating �� and �� we get that

Akfk� � kT �fk��

that is� T�pA is unitary� Hence� the frame

S��k � T��k�


is a UNTF since

kT��kk � pAk�kk � p

Ac� for all k � � � � � �M�

�� 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 espe�

cially 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 genera�

tors� There is also a class of wavelet frames which has two generators �� Recently�

several new applications have arisen� In �� the authors proposed using the redun�

dancy of frames to mitigate the losses in packet�based communication systems such

as the Internet� In � �� frames are used for multiple antenna coding#decoding �see

the discussion Section �� for further explanation�� In �� connections between

quantum mechanics and tight frames are established� Although each of these ap�

plications requires a di�erent 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�

In this section we will show that the UNTFs generated by the family fUk��gM��k��

�where U is unitary and �� H � are precisely the GHFs� We then extend our

discussion to UNTFs generated by fUkV j��g and higher numbers of generators�In the discussion Section �� we will relate the results of this section to the literature�

�� UTFs with a Single Generator

Here� we give a complete classi�cation of UNTFs of the form fUk��gM��k�� where

U is a unitary operator on HN � We will� in fact� �nd that in this case fUkgM��k��

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 classi�es GHFs as those frames generated by powers of

a special class of unitary operators applied to a �xed vector in H �

Proposition �� f�kgM��k�� is a general harmonic frame for HN if and only if

there is a vector �� HN with k��k� � NM � an orthonormal basis feigNi�� for HN

and a unitary operator U on HN with Uei � ciei� with fcigNi�� distinct M th roots

of some jcj � so that �k � Uk�� for all � � k �M � �

Proof� See Appendix A� �

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


GHFs �Theorem �� from �� is that GHFs are robust to the maximum number

of possible erasures� that is� a GHF with M elements in HN is robust to e erasures

for any e � �M �N��

Our next theorem completes the classi�cation of GHFs as being those UNTFs of

the form fUk��gM��k�� where U is a unitary operator on H � As a consequence� we

discover that� in this case� fUkgM��k�� is a group�

Theorem �� Let U be a unitary operator on HN � �� HN and assume

fUk��gM��k�� is a UNTF for HN � Then UM � cI for some jcj � and fUk��gM��

k��

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��

�� UTFs with Two or More Generators

Having completely characterized UTFs which come from a group of unitary oper�

ators 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

fEmbTmaggm�n�Z �or� in our case� the �nite discrete Gabor frames�� Each of theseframes is equivalent to the UNTF Gabor frame fEmbTnaS

��ggm�n�Z where S isthe frame operator for fEmbTmaggm�n�Z�We will classify the UNTFs of the form fU iV j��gL��M��

i��j�� where fV j��gM��j�� 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�igMi�� with the family of rank� orthogonalprojections 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 �� Let V be a unitary operator on HN � let m � N and �� H

so that fV i��gm��i�� is a UNTF for HN � Then for every unitary operator U on HN

and every M � N� there is a vector �� HN such that fUkV j��gM��m��k��j�� is a

uniform� normalized tight frame for HN �

Proof �Proposition �� Let �� pM � Now� for every f � HN we have

M��Xk��

m��Xj��

jhf� UkV j��ij� �M��Xk��

m��Xj��

jhU�kf� V j��ij� �

M

M��Xk��

m��Xj��

jhU�kf� V j��ij��

Since fV j��gm��j�� is a UNTF� the previous sum equals

M

M��Xk��

kU�kfk� �

M

M��Xk��

kfk� � kfk��


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 �� If f�kg 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�kg�In particular� if f�kg is a UNTF then every frame unitarily equivalent to f�kg is

also a UNTF�

Proof �Proposition �� Let L be the frame operator for fT�kg� For any f � H

we have�

Lf �Xk

hf� T�kiT�k � T

�Xk

hT �f� �ki�k�� T �S�T ��f ��

As a consequence of Proposition �� 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 �� If U is unitary and fUkV j��g is a UNTF for HN � then for

every M � Z and for every f � HN we have�

f �Xk�j

hf� Uk�MV j��iUk�MV j��

that is� fUk�MV j��gk�j is a UNTF as well�

Proof �Proposition �� Since UM is unitary and fUkV j��gk�j is a UNTF forHN � it follows that fUMUkV j��gk�j is also a UNTF for HN �To explain where we are going from here� let us revisit the notion of a UNTF

and look at it from a slightly di�erent point of view� Suppose f�igi�I is a UNTFfor a ��nite or in�nite�dimensional� Hilbert space H with k�ik � c� for all i � I �

For each i � I � let Pi be the orthogonal projection of H onto the one�dimensional

subspace of H spanned by �i� Then� for all f � HXi�I

kPifk� �Xi�I

khf� �ii�ik�

�Xi�I

jhf� �iij�k�ik� � c�Xi�I

jhf� �iij� � c�kfk��

Conversely� let fPigi�I be rank� orthogonal projections on H satisfyingXi�I

kPifk� � akfk�� for all f � H �

�� CASAZZA AND KOVACEVI�C

Then f�igi�I is a UNTF for H where �i � PiH and k�ik� � �a� That is� if f � H

then Xi�I

jhf� �iij� �Xi�I

jhPif� �iij�Xi�I

kPifk�k�ik�

�

a

Xi�I

kPifk� �

aakfk� � kfk��

It follows that the condition on f�ig which guarantees that we have a UNTF isreally a condition on the rank� 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 �� Let fPigi�I be a family of projections �of any rank� on a

��nite or in�nite�dimensional� Hilbert space H and assume there are constants � �

A�B satisfying�

Akfk� �Xi�I

kPifk� � Bkfk�� for all f � H �

Let f�ijgMi

j�� be a frame for PiH with frame bounds Ai� Bi� for all i � I� Then

f�ijgMi

j��i�I has frame bounds�

A infi�IAi� B supi�IBi�

Proof� For any f � H we have

MiXj��i�I

jhf� �ijij� �Xi�I

MiXj��

jhf� �ijij�

�Xi�I

MiXi��

jhPif� �ijij� �Xi�I

BikPifk�

� supi�IBi

Xi�I

kPifk� � B �supi�IBi� kfk��

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 �� Let fPigi�I be a family of projections on a Hilbert space H

satisfying� Xi�I

kPifk� � a�kfk�� for all f � H �

Then a � � and if f�ijgMi

j�� is an NTF for PiH � for all i � I� then f �a�ijgMi

j��i�Iis an NTF for H �


In the next corollary we assume that the projections Pi all have the same rank

and the frames for PiH 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 �� if dim PiH � N and f�i�jgj�J is a UNTF for PiH then

k�ijk� � N

jJ j �

Corollary �� Let fPigi�I be a sequence of projections �all with the same

rank� on a Hilbert space H satisfying�Xi�I


Then a � � and if f�ijgj�J is a UNTF for PiH � for all i � I� then f �a�ijgj�J�i�Iis a UNTF for H �

We are now ready to consider UNTFs with two generators�

Theorem �� Let U be a unitary operator on H � and let fV j�gM��j�� be a UTF

�with tight frame bound �a� for its closed linear span in H � Let P� be the orthogonal

projection of H onto this span� Assume that fU iV j��gL��M��i��j�� is a UNTF for H �

If Pi is the orthogonal projection of H onto the span of U iP�H � for all � i � L� �then

L��Xi��

kPifk� � akfk�� for all f � H �

Proof� Note that for all � i � K � we have Pi � U iP�U�i� Now� for all

f � H we compute�

L��Xi��

kPifk� �L��Xi��

kU iP�U�ifk�

�L��Xi��

kP�U�ifk� �L��Xi��

aM��Xj��

jhU�if� V j��ij�

� a

K��Xi��

M��Xj��

jhf� U iV j��ij� � akfk��

The next theorem is the converse of the above�

Theorem �� Let fPigL��i�� be a family of orthogonal projections on H satisfying

L��Xi��


Assume that U is a unitary operator on H so that U iP�H � PiH � Let �� P�H

and let V be a unitary operator on P�H such that fV j��gM��j�� is a UNTF for P�H �

Then f �aU iV j��gL��M��i��j�� is a UNTF for H �


Proof� For all f � H we have�

L��Xi��

M��Xj��

jhf� aU iV j��ij� �

a�

L��Xi��

M��Xj��

jhU�if� V j��ij� �

a�

L��Xi��

kP�U�ifk� �

a�

L��Xi��

kU iP�U�ifk� �

a�

L��Xi��

kPifk� � kfk��

�� Discussion

Some remarks are in order�

� The assumptions in Theorems �� and �� that fV j��gM��j�� is a UNTF for its

span is su�cient for the conclusion of the theorems but not necessary� For exam�

ple� it is known to Gabor frame specialists that for ab � � there are Gabor frames

fEmbTnaggm�n�Z for which neither of fEmbTnaggm�Znor fEmbTnaggn�Z is a framefor its closed linear span� The equivalent UNTF Gabor frame fEmbTnaS

��ggm�n�Zfails the hypotheses of Theorems �� and �� but satis�es the conclusion�

�� The results of Section �� generalize to three or more generators� For example�

for three generators� Theorem �� becomes�

Theorem �� Let U� V be unitary operators with fU iV j��gL��M��i��j�� a UNTF for

its span� Let P� be the projection of the Hilbert space H onto this span� Let W be

a unitary operator on H so that

K��Xk��

kW kP�W�kfk� � a�kfk�� for all f � H �

Then f �aW kU iV j��gK��L��M��k��i��j�� is a UNTF for H �

�� It would be very interesting to classify when a �nite group of unitaries G

generates a UNTF for H of the form fU��gU�G� Since �nite Abelian groups areisomorphic to a direct product of cyclic groups� the results of this section give

su�cient conditions for fU��gU�G to generate a UNTF for H � Frames generatedby Abelian groups of unitaries were studied in �� 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 �� 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 ��

�� Frames generated by possibly noncommutative groups of unitaries are impor�

tant in multiple antenna coding and decoding � �� Here� one needs classes of

UNIFORM TIGHT FRAMES WITH ERASURES ��

unitary space�time signals �that is� frames generated by classes of unitary opera�

tors� called constellations� It is also desirable in this setting to have full transmitter

diversity� meaning that

det �I � U� �� for all I �� U � G�

Groups with this property are called �xed�point free groups� In a tour�de�force�

the authors in � � 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 classi�cation theory of �xed point free groups�

�� 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 coe�cients� Since at the receiver side� this looks like

the original frame was the one without vectors corresponding to erased coe�cients�

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 recon�

struction is not possible� We can provide more robustness by adding more vectors�

For example� take a uniform tight frame in R� made up of two orthonormal bases

� � �� 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 coe�cients corresponding to

one of the bases and still have a basis able to reconstruct� If� on the other hand�

we lose coe�cients 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 � � vectors in R� � the HTF with the analysis frame operator

F �

�BB�

�p��

p��

�

�p�� p��

�CCA � ��

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 bene�t with more redundancy �more vectors�� It is also more robust than

the previous frame with � 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 HN with M elements which are robust to M � N erasures�


namely the HTFs �Theorem �� from �� Such frames are the best� This is clearly

optimal since any more erasures would not leave enough elements to span HN � In

this section� we will characterize frames based on their robustness to erasures�

�� 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 �� A frame f�igMi�� is said to be robust to k erasures if f�igi�Icis still a frame� for I any index set of k erasures� I � f � �� Mg 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 de�nition of a frame that if we apply an orthogonal

projection P on H to a frame we will get a frame for PH with the same frame

bounds�

Proposition �� Let f�igMi�� be a frame in HN robust to k erasures and let P

be an orthogonal projection on HN � Then fP�igMi�� is a frame for PHN robust to

k erasures�

Proof� Let I be the index set of erasures� I � f � �� Mg with jI j � k� Now�

f�igi�Ic spans HN and so fP�igi�Ic is a frame for PHN with the same frame

bounds�

The main ingredient for classifying frames robust to one erasure is contained in

the next proposition�

Proposition �� Let f�igMi�� be a set of vectors in HN � The following are

equivalent�

�f�igMi�� is a frame robust to one erasure�

��There are nonzero scalars ai �� for � i �M so that

MXi��

ai�i � ��

Proof� � �� Choose I � f � �� Mg maximal for which there are nonzeroai�s� i � I and

g �Xi�I

ai�i � ��


We claim that jI j � M � We proceed by contradiction� If I �� f � �� Mg� choosem � Ic� Since f�igMi�� is robust to one erasure� there are scalars fbig� not all zero�so that if �m is erased� it can be recovered from the rest as

�m �Xi��m

bi�i�

or

h � �m �Xi�I

bi�i �X

m��i�Icbi�i � ��

We have two cases�

Case I� Assume that bi � � for all i � I �

Then� h � �m �Pm��i�Ic bi�i � �� In this case� g � h � � and has nonzero

coe�cients on every �i� i � I � plus a nonzero coe�cient on �m contradicting the

maximality of I �

Case II� At least one bi �� for some i � I �

Since ai �� for all i � I � we can choose an � � so that

� �� ai�bi� for all i � I�

Now� g � �h � � and has nonzero coordinates on �i� for all i � I � as well as � for a

coordinate on �m� again contradicting the maximality of I �

�� Assume ai �� for all � i �M andXi�I

ai�i � ��

Then for each m � I we have�

�m � �X

m��i�I

aiam

�i�

that is� any vector lost can be recovered using the rest and so f�igMi�� is robust to theerasure of �m� for an arbitrary m � 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 the frame property� Thus� we cannot �nd a nonzero coe�cient aMsuch that

PMi�� ai�i � ��

For example� we know that the MB frame f�ig�i�� is robust to one erasure �The�orem �� from �� Hence� we can �nd a� � a� � a� � such that

P�i�� ai�i � ��

The next results will characterize frames robust to one erasure� Since we men�

tioned the idea of stepping down� we will start from an orthonormal basis feigMi��for HM and project it using a projection operator P �or I �P �� We thus look into

a few facts connecting P and �I � P ��

� If P is an orthogonal projection� then �I � P � is an orthogonal projection as

well�


�� The subspaces P and �I � P � project onto are orthogonal� For any f � H �

h�I � P �f� Pfi � hP ��I � P �f� fi � h�P � P �f� fi � ��

�� f�I � P �eigMi�� is uniform if and only if fPeigMi�� is uniform� To see this wecompute� for all i� j

h�I � P �ei� �I � P �eii � h�I � P �ej � �I � P �ejih�I � P ��I � P �ei� eii � h�I � P ��I � P �ej � eji

h�I � P �ei� eii � h�I � P �ej � eji � hPei� eii � � hPej � ejihP �Pei� eii � hP �Pej � ejihPei� P eii � hPej � P eji�

Here we used the fact that P is an orthogonal projection and thus P � P � � P �P �� Both fPeigMi�� and f�I � P �eigMi�� are NTFs�

By Theorem �� all NTFs are of the form fPeigMi�� where feigMi�� is an or�thonormal 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 PH determines when the frame is a UNTF� From our earlier discussion� if

P is an orthogonal projection on H � we may work either with fPeig or f�I �P �eigto classify UNTFs� We will freely switch between these classes because each of

them has certain advantages for speci�c results� The class fPeig works well forclassi�cations when the dimension of PH is �large� relative to the dimension of H �

On the other hand� f�I�P �eig is easier to work with when the dimension of PH is�small� relative to the dimension of H � Also� as we will see� important information

about fPeig 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 classi�cation for frames robust to one erasure�

Corollary �� Let feigMi�� be an orthonormal basis for HM and let P be an

orthogonal projection on HM � The following are equivalent�

�f�I � P �eigMi�� is a frame robust to one erasure�

��There are nonzero scalars ai �� for � i �M so that

MXi��

ai�I � P �ei � ��

��There is an f � span��i�MPei so that

hf� eii �� for all � i �M�


Proof� The equivalence of � � and �� comes from Proposition �� Let us check

the equivalence of �� and ��

�� Given �� choose f to be

f �

MXi��

aiei�

By our assumption�PM

i�� aiei �PM

i�� aiPei and thus

f �

MXi��

aiPei � PHM �

Moreover� hf� eii � ai �� for all � i �M �

�� Choose f � PHM as in �� and call ai � hf� eii �� for all � i �M �

Now

MXi��

ai�I � P �ei �

MXi��

hf� eii�I � P �ei �

MXi��

hf� eiiei �MXi��

hPf� eiiei � ��

since f � PHM � and thus Pf � f �

Corollary �� classi�es frames which are robust to one erasure� We now extend this

to a classi�cation of UNTFs robust to one erasure�

Corollary �� Let feigMi�� be an orthonormal basis for HM and let P be an

orthogonal projection on HM � The following are equivalent�

�f�I � P �eigMi�� is a frame robust to one erasure�

��There is a g �PM

i�� aiei with nonzero scalars ai �� for all � i � M such

that Pf � hf� gig� for all f � HM � and kgk � �Moreover� if P is rank� � f�I � P �eigMi�� is a uniform frame if and only if jaij � �pM � for all � i �M �

Proof� � � �� is immediate from Corollary ��

For the moreover part� we know that f�I � P �eigMi�� is uniform if and only if

fPeigMi�� is uniform� Also� since P is rank� � for all � i � M and using the

expression expression Pf � hf� gig in Part �� with f � ei we have�

Pei � aig�

Hence� kPeik � kPejk is true if and only if jaij � jaj j� for all � i� j � M �

That is� fPeigMi�� is uniform �and hence f�I � P �eigMi�� is uniform� if and onlyif jaij � jaj j� for all � i� j �M � Finally� kPeik � jaijkgk � jaij � �

pM �

�� 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 classi�cation of UNTFs with k erasures in


this section is useful if we haveM vectors in an N �dimensional space andM �N is

�small�� In the next section we will give another classi�cation of this family which

works best when M �N is �large��

In what follows� I � f � �� Mg will denote the erasure index set�Proposition �� Let feigMi�� be an orthonormal basis for HM � Let P be an or�

thogonal projection of HM onto an L�dimensional subspace H L � Fix I � f � �� Mgwith jI j � k � L and let K � spani�Icei� If f�jgLj�� is any orthonormal basis for

H L and we have the following L�M matrix

A � �h�j � eii� L Mj��i��

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 � K j�I � P �f � �g � �ker �I � P �� K �

Now� apply row reduction �relative to feigMi�� to f�igLi�� using the elements of thecolumns in I � This gives a linearly independent set fgjgsj�� fgjg L

j�s�� spanning

K with

s � row rank of �h�j � eii� Lj��i�I �gj � K for s� � j � L and

�span��j�sgj� � K � ��

Hence�

ker �I � P � � K � spans��j�Lgj �Therefore�

dim �ker �I � P �� K � dim �ker �I � P �jK � L� �s� � � � L� s�

We are looking for the NTFs which are robust to k erasures� Proposition ��

actually yields a stronger result� Namely� it gives necessary and su�cient conditions

for fPeig to be robust to one particular choice of k erasures� We state this strongerresult in two di�erent forms in the next two propositions�

Proposition �� With the notation of Proposition �� the following are equiv�

alent�

�f�I � P �eigMi�� is robust to the erasure of the elements f�I � P �eigi�I ��We have rank �I � P �jK �M � L�

��We have dim �ker �I � P �jK � � L� k�


Proof� � � �� f�I � P �eigMi�� is robust to the erasure of the elementsf�I � P �eigi�I if and only if �I � P �jK is full rank� which must be rank �I � P � �

M � L�

�� By Proposition ��

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 HM onto spani�Iei� Hence�

dim �ker �I � P ��I � Pi�� L�

But� ei � ker �I � Pi�� for all i � I � Hence�

dim �ker �I � P ��I � Pi� � dim �ker �I � P � � k � L�

Therefore�

dim �ker �I � P � � L� k�

Proposition �� gives precise information when a frame is robust to the erasure of

one particular choice of k elements in terms of the coe�cients of f�ig relative tothe unit vector basis of H L � The following corollary is a statement for a �xed choice

of erasures as well�

Corollary �� With the notation of Proposition �� the following are equiv�

alent�

�f�I � P �eigMi�� is robust to the erasure of the elements f�I � P �eigi�I ��The row rank of �h�j � eii�Lj��i�I � k�

Applying the above result to every choice of erasures results in Theorem �� and

a classi�cation 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 �� With the notation of Proposition �� the following are equivalent�

�f�I � P �eigMi�� is robust to k erasures�

��For every I � f � �� Mg with jI j � k� the row rank of A � �h�j � eii�Lj��i�Iequals 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 classi�es when f�I � P �eig is a UNTF without any referenceto erasures�

Theorem �� With the notation of Proposition �� the following are equivalent�


�fPeigMi�� is uniform �and hence f�I � P �eigMi�� is uniform��

��For every � i �M we have�

�A�A�ii �LXj��

jh�j � eiij� � L

M�

That is� the columns of A � f�jgLj�� 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 jth columns of f�jgLj��

Proof� � � �� We know that f�I�P �eigMi�� is uniform if and only if fPeigMi��which� in turn� is uniform if and only if the diagonal elements of the matrix for P

relative to feigMi�� have constant modulus� We have�

Pei �

LXn��

hei� �ni�n �

MXm��

�LX

n��

hei� �nih�n� emi�em�

The diagonal element of Pei is�

LXn��

hei� �nih�n� eii �LX

n��

jh�n� eiij��

So fPeigMi�� is a uniform tight frame if and only if for all � i� j �M we have

LXn��

jh�n� eiij� �LX

n��

jh�n� ejij��

For the moreover part� we compute for � i �� j �M �

h�I � P �ei� �I � P �eji � hei� eji � hPei� eji � hei� P eji� hPei� P eji

� hPei� P eji � hei� P eji

hei�LX

n��

hej � �ni�ni � hei�MXm��

�LX

n��

hej � �nih�n� emi�emi �

LXn��

hej � �nih�n� eii�

The right�hand side of the above equality is precisely the inner product of the ith

and jth columns of f�jgLj��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 �� In the real case� if M � �� there do not exist rank� orthogonal

projections P�� P� with P�P� � � so that f�I � P��eigMi�� is uniform and f�I �P��I � P��eigMi�� is uniform and robust to two erasures�


Proof� Assume such P�� P� exist� Since f�I � P��eigMi�� is uniform� by The�orem �� there is a vector ��

PMi�� aiei � HM with P�f � hf� ��i�� for all

f � HM and jaij � jaj j� for all � i� j � M � Now choose �� PM

i�� biei � HM

so the P�f � hf� ��i�� for all f � HM � Since P�� P� are rank� � P�P� � � implies

P�P� � �� That is� h�� i � �� Let P � P��P�� Then �I �P��I �P�� I �P �

Since f�I � P �gMi�� is uniform� by Theorem �� we have

a�i � b�i � a�j � b�j � for all a � i� j �M�

Hence� b�i � b�j � for all � i� j �M � Since k��k � k��k � � it follows that

a�i � b�i �

M� for all � i� j �M�

Since f�I � P �eigMi�� is robust to two erasures� by Theorem ��

aibj � ajbi �� for all � i� j �M� ��

However� for all i� j we have ai � aj � bi� Hence� there is some i� j witheither ai � aj and bi � bj or ai � �aj and bi � �bj � In either case �� fails�Corollary �� is heavily dependent on having a real Hilbert space� In the complex

case� we will show in Corollary �� that the GHFs have the property that we can step

down to the frame by successively applying rank� projections to the orthonormal

basis� Corollary �� 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 classi�cation of when we can step down from k� erasures to k� erasures

and so on� One reason is that this has implications for entangled states in quantum

detection theory � ��

This raises the question of whether we can step down one dimension at a time

as in Corollary �� if we only ask for each level to be uniform� The answer is no as

the next example shows�

Example �� Let feig�i�� be an orthonormal basis for H � and let

�� p�e� �

�e� �

�e��

and

�� p�e� �

�e� �

� �e��

Then h�� i � �� and if P is the orthogonal projection of H � onto the span of

f�� g� then by Theorem �� f�I�P �eig�i�� is a uniform normalized tight frame�However� there does not exist a rank� orthogonal projection P� of H � into range

of P so that f�I � P��eig�i�� is a uniform frame�


Proof� By Theorem �� in order for such a P� to exist� there must exist a vector

in PH � of the formP�

i�� biei with jbij � jbj j� for all � i� j � �� For any a�� a��

a�� a�� a�p�e� �

a�p�e� �

a� � a��

e� �a� � a��

e��

If �ja� � a�j�� ja� � a�j�� then one of ja�j� ja�j equals � so ja�j �� ja�j�The next question is whether we can step down one dimension at a time as in

Corollary �� if all we want is for our frame to be robust to j erasures at the jth

level but are willing to give up uniformity at each level� Surprisingly� the answer

here is yes�

Proposition �� Let P be an orthogonal projection of HM onto an L�dimensional

subspace H L � Let feigMi�� be an orthonormal basis for HM � If f�I � P �eigMi�� is

robust to k erasures �k � L�� then there are mutually orthogonal rank� projections

fPigLi�� taking HM into H L so that f�I �Pj��I � Pj�� I �P��eigMi�� is robust

to j erasures� for all � j � k� and robust to k erasures for all k � j � L�

Proof� See Appendix A��

�� UTFs Robust to More than One Erasure� An Alternative

Approach

In the previous sections we used the orthogonal projection P on HM to classify

when f�I � P �eigMi�� is robust to k erasures and when it is uniform� However� ifthe dimension of PHM is large �that is� close to the dimension of HM � these results

become di�cult to implement since they require knowledge about the orthonormal

bases for the large�dimensional space PHM � In this case� f�I � P �eigMi�� is a largenumber of vectors in a small�dimensional space �I � P �HM � So it is easier in this

case to work directly with �I � P �H and f�I � P �eig� Or� equivalently� with PH

and fPeig� where dim PH is small�

Proposition �� Let f�igNi�� be an orthonormal sequence in HM � let feigMi�� bean orthonormal basis for HM and let P be the orthogonal projection of HM onto the

span of f�igNi�� Then the normalized tight frame fPeigMi�� is unitarily equivalent

to fgigMi�� where for � j �M we have

gj �

NXi��

h�i� ejiei�

That is� fPeigMi�� is the frame obtained by turning the columns of f�igNi�� into row

vectors in HN �

Proof� For any � i �M we have�

Pei �

LXn��

hei� �ni�n�


Since f�ngLn�� is an orthonormal sequence� the operator T�n � en� for � n � L

is a unitary operator which takes Pei to fgjgMj�� where for � j �M and we have

gj �

NXi��

h�i� ejiei�

Corollary �� Given the conditions in Proposition �� we have�

�fgjgMj�� is a uniform frame for HN if and only if

NXi��

jh�I � ejij� � N

M� for all � j �M�

��The following are equivalent�

�a� fgjgMj�� is robust to the erasure of fgjgj�I for some I � f � �� Mg��b� We have that

�h�i� eji�Ni��j�Ichas row rank N �

Hence� fgjgMj�� is robust to any choice of k erasures� k �M �N � if and only if

every set of M � k columns of the matrix

A � �h�i� eji� N Mi��j��

has row rank N �

Finally� the angle between gi and gj is given by the inner product of the ith and

jth 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 �eigMi�� a GHF� One important propertyof GHFs is contained in the following result which says they have the step�down

property of Section ��

Corollary �� Let P be an orthogonal projection of HM onto an L�dimensional

subspace H L and let feigM��i�� be an orthonormal basis for HM � If f�I � P �eigM��

i��

is a GHF then there is an orthogonal sequence of rank� projections fPigLi�� of HMinto H L so that for all � j � L and for all permutations � of f � �� Lg we

have that

fj��Ym��

�I � P��m��eigM��i��

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 nota�

tional changes but obscures the basic ideas of the proof� By Proposition �� there


is a unique way to get HTFs� Namely� let feigM��i�� be the natural orthonormal basis

for HM � Let fwigM��i�� be distinctMth roots of unity and consider the orthonormal

basis f�igM��i�� for HM given by�

�i �

M��Xj��

wji ej �

Without loss of generality� we may as well assume that H L � span f�igLi��Now� turning the row vectors of f�igM��

i�L into column vectors gives a HTF for

HM�L � Moreover� again by Proposition �� this HTF is unitarily equivalent tof�I �P �eigM��

i�� where I �P is the orthogonal projection onto �H L �� Now� let Pi

be the orthogonal rank� projection of HM onto span �i for � � i � L� � Fix apermutation � of f � �� Lg and let Ij � f�� j � �g� Then� againby Proposition ��

fj��Ym��

�I � P��m��eigM��i��

is unitarily equivalent to the HTF obtained by turning the row vectors of f�igi��Ijinto column vectors�

It is possible that the property in Corollary �� characterizes GHFs� but we do

not have a proof for this� We can show that the use of permutations is necessary

in Proposition �� That is� there are UNTFs which have the step�down property

for erasures and for uniformness for one �xed ordering of the rank� projections Piwhile failing to be equivalent to a GHF� This is the point of the next example�

Example �� In C � � let

��

�Xi��

ei� ��

�Xi��

wiei� �� e� � e� � e� � e��

where w� � � w� � � � w� � i and w� � �i� Let fPig�i�� be the rank� orthogonalprojections of C � onto the span of �i� Then it follows easily from our results that�

� f�I � P��eig�i�� is a UNTF which is robust to one erasure�� f�I � P��I � P��eig�i�� is a UNTF which is robust to � erasures�� f�I�P��I�P��I�P��eig�i�� is a UNTF which is robust to � erasures� but is

not a harmonic frame since f�I �P��I �P��eig�i�� is a UNTF which is not robust

to � erasures�

Example �� If �i � � � wi� in C� for � i � M � where jwij � and the

fwigMi�� are distinct� then f�igMi�� is a UNTF for C � if and only ifPM

i�� wi � ��

In this case� if Pi is an orthogonal rank� projection onto the span �i� then f�I �Pj�eigMi�� is a UNTF which is robust to one erasure while f�I � Pj��I � Pk�eigMi��is a UNTF which is robust to � erasures for all � j �� k �M �


Proof� This family is certainly uniform� The assumptionPM

i�� wi � � guaran�

tees that the vectors

� � � � � � � � and �w�� w�� wM �

are orthogonal in HM �and conversely�� Hence� f�igMi�� is a UNTF for C � by Propo�sition ��

Example �� In general� the frames constructed in Example �� are not equivalent

to harmonic frames even after a permutation�

Proof� The reason is that f�jgM��j�� is a GHF implies that

h�j � �j��i � h�j�� j��i� for all � � j �M � ��

�see Proposition �� This would imply in Example �� that

h�j � �j��i � � wjwj�� wj��wj�� h�j�� j��i�

Hence� wjwj�� wj��wj�� Now� let w� � � w� � � �� p��i and w� �

� �� p��i� Then jwj j � andP�

j�� wj � �� Hence� by Example ��

f� � wj�g�j�� is a UNTF for C � � However� for any permutation � of f � �� g� wedo not have w��j�w��j�� w��j��w��j�� for all � � j � � �with w� � w��

It is not hard to see that in the case of R� � the condition in Corollary �� is su�cient�

APPENDIX� PROOFS

A�� PROOF OF PROPOSITION ��

First note that if fcigNi�� are distinct Mth roots of c with jcj � � then

M��Xk��

cki � ��M��Xk��

jcki j� � M�M��Xk��

i��l

�cicl�k � �� A� �

and thus

M��Xk��

�cicl�k � M�i�l� �A��

We now check the necessity of our condition for a frame to be a general harmonic

frame� If f�kgMk�� is a general harmonic frame for HN then by the de�nition�

�k � �ck�b�� ck�b�� ckNbN� �

NXi��

cki biei�


where feigNi�� is the natural unit vector orthonormal basis of HN � jbij � �pM �

and fcigNi�� are distinct Mth roots of c with jcj � � Now let

��

nXi��

biei�

De�ne a unitary operator U on HN by Uei � ciei� Then� for all � � k �M � wehave Ukei � cki ei� and thus� U

k�� k for all � � k � M � � So our frame hasthe form given in the proposition�

The su�ciency of the condition is checked similarly� If we assume that f�kg isof the form described in the proposition� then�

k�kk � kUk��k � k��k �r

N

M�

since U is a unitary operator� So f�kg is a uniform frame� To see that f�kg is anormalized tight frame� we let f �

PNi�� aiei � HN and compute�

M��Xk��

jhf� Uk��ij� �M��Xk��

jNXi��

aicki bij� �

�

M��Xk��

NXi��

jaicki bij� �Xm��

amb�

M��Xk��

�cmc��k

� MNXi��

jaij�jbij� � � � MNXi��

jaij�j pMj� � kfk��

where in the next to last equality we used the fact that the fcig is a family ofdistinct Mth roots of c �see �A�� Hence� f�kg is a UNTF� Thus� using � �� andwriting f � ej � we see that

kejk� � �

MXk��

jhej � Uk��ij� �MXk��

jhej �NXi��

cki bieiij� �MXk��

jckj bj j� � M jbj j��

So jbj j � �pM � for all j� � � j � N � � This further means that

�� b�� b�� bN�

with jbj j � �pM and since �k � Uk��

�k � �ck�b�� ck�b�� c

kNbN ��

A�� PROOF OF THEOREM ��

We prove the theorem in steps�

Step I� We �rst prove that there is a constant jcj � so that UM � cI � This will

show that the operators fU igM��i�� in fact� form a � Since the frame is an NTF� for


every f � HN we have�

f �

M��Xk��

hf� Uk��iUk��

Hence�

Uf �

M��Xk��

hUf� Uk��iUk��

�

M��Xk��

hf� Uk��iUk�� since U� � U��

� U

�M��Xk��


�� U

�M��Xk��


�� A��

Since U is one�to�one�

f �M��Xk��

hf� Uk��iUk�� M��Xk��

hf� Uk��iUk�� A��

From �A�� and �A�� it follows that�

hf� UM��iUM�� hf� U��iU��

Applying U we have

hf� UM��iUM�� hf� U��iU�� hf� U��i�� A��

Hence� there is a c � C so that UM�� c�� Replacing f by U��f in �A�� gives�

hU��f� UM��iUM�� hf� UM��iUM��

� hf� c��ic�� jcj�hf� ��i�� hU��f� U��i�� hf� ��i��

So jcj� � � Also� for all � � k �M � �

UMUk�� UkUM�� Ukc�� cUk��

Since fUk��g spans HN � that is� for any f � HN f �P

khf� Uk��iUk�� it follows

that UM � cI � This completes Step I�

Step II� We want to prove�

� U is diagonalizable with respect to an orthonormal basis feigNi�� with diag�onal elements fcigNi�� with ci an Mth root of c�

�� PN

i�� biei and jbij �q

�M � for all � i � N �

These two steps give us frame elements as in De�nition ��


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 feigNi�� with diagonalelements fcigNi�� Writing �� as in �� we have that

Uk�� NXi��

cki biei�

Since UM � cI � it follows that each ci is an Mth root of c� Also� since the frame is

an NTF� for all � j � N we have

�

MXk��

jhej � Uk��ij� �MXk��

jhej �NXi��

cki bieiij� �NXk��

jckj bj j� � M jbj j��

So jbj j � �pM � This completes Step II�

Step III� We �nally prove that the cj are distinct Mth roots of c� This will

complete the proof�

Fix � m �� M � Since the columns of a normalized tight frame are orthogonal

vectors� that is� F �F � I � we have for columns m� ��

� �

M��Xk��

bmckmb�c

k� � bmb�

M��Xk��

�cmc��k�

Hence�

M��Xk��

�cmc��k � ��

Thus� cm and c� are distinct Mth roots of c� This completes the proof of Step III

and the theorem�

A�� PROOF OF PROPOSITION ��

For all � j � L� let

gj �MXi��

hgj � eiiei�

be an orthonormal basis for H L � We will construct the projections fPjgLj�� by�nite induction� We start with P�� Since f�I � P �eigMi�� is robust to erasure� byTheorem �� for every � i � M there exists a � j � L so that hgj � eii �� Choose scalars a�� a� � � � � aL so that for any � j � L� if hgj � eii �� for some

� i �M � then

jajhgj � eiij � �

j��Xk��

jakhgk� eiij�

It follows easily that

��

PLj�� ajgj

kPLj�� ajgjk

�


then h�� eii �� for all � i �M � Now� let P� be the orthogonal projection onto

the span f��g� then f�I � P��eigMi�� is robust to erasure by Corollary �� Now assume we have found orthonormal vectors f�jgnj�� n � k � so that the

rank� projections Pj onto span f�jg satisfy the proposition and span f�jgnj�� span fgjgnj�� We will construct �n�� and Pn�� Let fgjgLj�� be an orthonormalbasis for the orthogonal complement of span f�jgnj�� in H L � Fix � i� � i� �

� � � � in�� M � Since

f�I � Pn��I � Pn�� I � P��eigMi��

is robust to n erasures� if we row reduce f�jgnj�� using the columns fi�� i�� ing�and switching rows if necessary� we obtain vectors fhjgnj�� with span fhjgnj�� span f�jgnj�� and for every � j � n we have�

hhj � eili � �j��

Now we verify the following claim�

Claim� For all but �nitely many � � x � � if

hn��

LXj�n��

xjgj �

then the row rank of

�hhj � ei�i�n��j��

is n� �

Proof of Claim� Fix � � x � � We row reduce this matrix by taking� for all

� � �M �

�hLX

j�n��

xjgj � ei�ih� � hn��

We then arrive at a matrix with s in the jth row and the ijth column for � j � n�

zeroes otherwise� zeroes in the �n � �st row for columns ij for all � j � n� and

in the �n� � n� �st position we have�

hhn�� ein��i �nX��

hLX

j�n��

xjgj � ei�ihh�� ei�i�

If this number is nonzero� then our matrix is of rank n� � Now we check what it

takes for this to be nonzero�


hLX

j�n��

xjgj � ei�ihh�� ei�i

�

LXj�n��

xjhgj � ein��i �LX

j�n��

xjnXj��

hgj � ei�ihh�� ei�i


�

LXj�n��

xj

hgj � ein��i �

nX��

hgj � ei�ihh�� ei�i��

By the hypotheses on the proposition�

�hgj � ei�i� L n��j��

has rank n� � It follows that there is a n� � j � L so that

hgj � ein��i �nX��

hgj � ei�ihh�� ei�i ��

Hence� the number of x�s with


hLX

j�n��

xjgj � ei�ihh�� ei�i � ��

is �nite� This establishes the claim�

Applying the claim to all choices of � i� � i� � � � � in�� M � we see that for

all but �nitely many � � x � � for all � i� � i� � � � � � in�� L� the rank of

�hhj � ei�i� M n��j��

is n� � Let

�n�� hn��khn��k �

and let Pn�� be the orthogonal projection of HM onto span f�n��g� Now�

f�I � Pn��I � Pn� � � � �I � P��eiiMi��

is robust to �n � � erasures by Theorem �� This completes the proof of the

proposition�

ACKNOWLEDGMENT

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

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equal-norm tight frames with erasures

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