This paper explores balanced frames, especially balanced unit norm tight frames, highlighting their advantages in signal reconstruction, error detection, and robustness against noise in finite-dimensional Hilbert spaces.
Contribution
It introduces the concept of balanced frames, analyzes their properties, and provides methods for constructing and characterizing balanced unit norm tight frames.
The paper provides methods to find and construct balanced frames.
Abstract
So far there has not been paid attention to frames that are balanced, i.e. those frames which sum is zero. In this paper we consider balanced frames, and in particular balanced unit norm tight frames, in finite dimensional Hilbert spaces. Here we discover various advantages of balanced unit norm tight frames in signal processing. They give an exact reconstruction in the presence of systematic errors in the transmitted coefficients, and are optimal when these coefficients are corrupted with noises that can have non-zero mean. Moreover, using balanced frames we can know that the transmitted coefficients were perturbed, and we also have an indication of the source of the error. We analyze several properties of these types of frames. We define an equivalence relation in the set of the dual frames of a balanced frame, and use it to show that we can obtain all the duals from the balanced…
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Balanced frames: a useful tool in signal processing with good properties
Sigrid B. Heineken1
,
Patricia M. Morillas2,∗
and
Pablo Tarazaga2,3
*1**Departamento de Matemática, FCEyN, Universidad de Buenos Aires, Pabellón I, Ciudad Universitaria, IMAS, UBA-CONICET, C1428EGA C.A.B.A., Argentina
2 Instituto de Matemática Aplicada San Luis, UNSL-CONICET, Ejército de los Andes 950, 5700 San Luis, Argentina
3 Departamento de Matemática, FCFMyN, UNSL, Ejército de los Andes 950, 5700 San Luis, Argentina*
Abstract.
So far there has not been paid attention to frames that are
balanced, i.e. those frames which sum is zero. In this paper we
consider balanced frames, and in particular balanced unit norm tight
frames, in finite dimensional Hilbert spaces.
Here we discover various advantages of balanced unit norm tight
frames in signal processing. They give an exact reconstruction in
the presence of systematic errors in the transmitted coefficients,
and are optimal when these coefficients are corrupted with noises
that can have non-zero mean. Moreover, using balanced frames we can
know that the transmitted coefficients were perturbed, and we also
have an indication of the source of the error.
We analyze several properties of these types of frames. We define an
equivalence relation in the set of the dual frames of a balanced
frame, and use it to show that we can obtain all the duals from the
balanced ones. We study the problem of finding the nearest balanced
frame to a given frame, characterizing completely its existence and
giving its expression. We introduce and study a concept of
complement for balanced frames. Finally, we present many examples
and methods for constructing balanced unit norm tight frames.
A spanning set of vectors in a finite dimensional Hilbert space is
called a frame. The redundancy of these spanning sets is
the crucial property in their vast types of applications in many
different areas of pure and applied mathematics and sciences, such
as efficient representation of vectors and operators, signal
processing, coding theory, communication theory, sampling theory,
quantum information, and computing among others (see e.g.
[11, 12, 24, 29, 34]).
In this paper we study balanced frames, i.e. those frames
which sum is zero, and several particular cases of them, especially
balanced unit norm tight frames (see e.g.
[5] for the concept of unit norm
tight frame).
We show that although non balanced unit norm tight frames are
optimal in many situations that appear in applications (see e.g.
[5, 10, 11, 16, 34] and the references therein), balanced unit norm tight frames
are even optimal in cases where the non balanced are not the best
ones.
In applications, a signal f is usually represented by a sequence
of numbers which are measurements of f. In frame theory, these
measurements are expressed as inner products of f with the
elements of a frame, and will be called frame coefficients.
As we will explain in this work, the reconstructions using balanced
frames are robust against systematic errors in the frame
coefficients. Systematic errors can come from a wrong calibration of
instruments, inexact methods of observation, or interference of the
environment in the measurement, transmission or reception processes.
A systematic error can be produced, for example, by the incorrect
zeroing of an instrument. Another example are measurements by radar
that can be systematically overestimated if we do not take into
account the slowing down of the waves in the air. Systematic errors
are not random, and cannot be reduced by taking the average of many
readings. Considering this, it is important to highlight that
balanced frames are immune to these type of errors. This means that
in the presence of a systematic error in the frame coefficients,
balanced frames still give the exact reconstruction.
In signal processing, the frame coefficients can be perturbed with
additive noises. It has been shown
[16] that if the mean of these
noises is zero, the reconstruction of the signal with unit norm
tight frames is optimal. We prove that if we use balanced unit norm
tight frames, these noises can have a nonzero mean but the
reconstruction is still optimal. Thus we can deal with noises of
different sources. If the mean is non-zero we are under the presence
of non-white noises. Nonzero mean noises appear naturally
in certain applications. Digital watermarking is an application for
which the zero mean assumption for the noises is not realistic
[23]. It is a useful tool for multimedia
copyright protection, access control, annotation and authentication
[27, 14, 25]. In certain cases such as median filtering, a standard
signal processing method for denoising, the noises in the
watermarking channel are additive with a non-zero mean.
Given a frame, each element of the Hilbert space can be expressed as
linear combinations of the elements of the frame using the so called
dual frames. As we will see, another advantage of balanced
frames is that they are resilient against a perturbation of the dual
frame by a constant vector, i.e. if we sum to each element of the
dual frame a fixed vector, we still obtain a dual frame. We use this
fact to define an equivalence relation in the set of dual frames of
a given balanced frame and prove that all the dual frames can be
obtained from the balanced ones.
We show that balanced frames are robust against one erasure, that
is, they remain to be a frame if we delete any of its elements. The
dual frames of these subfamilies are easy to obtain from the dual
frames of the original family.
If we use a balanced frame the sum of the frame coefficients is
always zero. So, if the transmitted numbers do not have zero sum we
know that they were perturbed. Moreover, as we will explain, if we
use balanced frames we can have an indication of when we are in the
presence of a systematic error, of random additive noises or of
other sources of perturbation as e.g. erasures.
In [8, 34] it is proved that
real balanced unit norm tight frames are spherical
2-designs, a mathematical object applied in different areas. We
want to point out that in contrast to what usually occurs in the
context of spherical 2-designs, we are not necessarily interested
in working with the minimum possible number of elements since, as
observed before, from the point of view of frame theory redundancy
is convenient for the applications. It can be seen in
[6, 7] that
balanced unit norm tight frames have advantages for sigma-delta
quantization. In
[13]
tight frames are characterized using balanced sequences via diagram
vectors. Balanced frames are mentioned in [34] in
the definition of simple lift. But this concept has not so far been
developed neither their multiple advantages noticed.
1.1. Contents.
In Section 2, we briefly
review frames.
In Section 3 we analyze the various advantages of balanced frames
and balanced unit norm tight frames for applications which were
mentioned before.
In Section 4 we show that balanced frames and in particular balanced
equal norm frames and balanced unit norm tight frames behave well,
in the sense that they are invariant under various transformations.
We find several characterizations of them and analyze properties of
their dual frames.
In Section 5 we study the closest balanced frame to a given frame in
the ℓ1 and ℓ2 norms. We give necessary and
sufficient conditions for the closest balanced frame to exist, and
for the case it exists we give its expression.
In Section 6 we introduce a concept of complement that is more
suitable for balanced frames than the definition used so far for
frames in general. Properties of this new notion are given.
In Section 7 we give many examples of balanced unit norm tight
frames such as those corresponding to roots of unity in
R2, certain types of harmonic frames, frames obtained
from Hadamard matrices, partition frames and some that are spherical
t-designs.
Finally, in Section 8, we present several explicit and painless
methods for constructing balanced unit norm tight frames.
2. Preliminaries
In this section we recall some concepts of frame theory
[11, 12, 24, 29, 34]. We refer to the mentioned
works for more details. We begin introducing some notation.
2.1. Notation
Let d,K∈N. Let Hd be a Hilbert space
of dimension d over a field F where
F=R or F=C. We write
⟨.,.⟩ and ∥.∥ for the inner product and the norm
in Hd, respectively. Let
L(Hd,HK) be the space of linear
transformations from Hd to HK (we write
L(Hd) for
L(Hd,Hd)). Let
Gl(Hd) (U(Hd)) be
set of invertible (unitary) elements in
L(Hd). If T∈L(Hd,HK), then im(T),
ker(T) and T∗ denote the range, the kernel and the
adjoint of T, respectively. If T∈L(Hd)
and (fk)k=1K is a sequence in Hd, we will
write T(fk)k=1K for (Tfk)k=1K. The elements of
FK will be considered as column vectors. We write e
for the real vector which entries are all equal to 1.
2.2. Frames
To a sequence F=(fk)k=1K in Hd we
associate the synthesis operator
TF:FK→Hd,TFc=∑k=1Kckfk,
the analysis operator
TF∗:Hd→FK, TF∗f=(⟨f,fk⟩)k=1K,
the frame operator
SF=TFTF∗,
and the Gram operator
GF=TF∗TF.
Definition 2.1**.**
Let F=(fk)k=1K be a sequence in
Hd. F is a frame for
Hd if spanF=Hd.
Proposition 2.2**.**
Let F=(fk)k=1K be a sequence in
Hd. The following assertions are equivalent:
(1)
F* is a frame for Hd.*
2. (2)
TF* is onto.*
3. (3)
TF∗* is one to one.*
4. (4)
SF* is invertible.*
5. (5)
rank(GF)=d.
6. (6)
There exist α,β>0 such that
[TABLE]
We call α and β in (2.1) the frame
bounds. The optimal lower frame bound is
λmin(SF)=∥SF−1∥−1 and
the optimal upper frame bound is
λmax(SF)=∥SF∥=∥TF∥2
where λmin(SF) and
λmax(SF) are the smallest and largest
eigenvalues of SF, respectively.
Definition 2.3**.**
Let F=(fk)k=1K be a sequence in
Hd. We say that:
(1)
F is balanced (B) if ∑k=1Kfk=0.
2. (2)
F is real if GF is a real matrix.
3. (3)
F is equal-norm (EN) if ∣∣fk∣∣=∣∣fk′∣∣ for k,k′=1,…,K. F is unit norm (UN) if ∣∣fk∣∣=1 for k=1,…,K.
4. (4)
F is isogonal if F is EN and there exists an a∈R such that ⟨fk,fl⟩=a for k,l∈{1,…,K}, k=l.
Isogonal vectors appear in [28] in relation with
the structure of soap films and bubbles. They are a particular case
of equiangular frames [18, 33], i.e., EN frames for which there exists
a∈R such that ∣⟨fk,fl⟩∣=a for
k,l∈{1,…,K}, k=l. A unit norm frame is a
(spherical) m-distance frame if the inner products between
distinct vectors take m real values [34]. A unit
norm isogonal frame is a 1-distance frame. For the case m=2 see
e.g. [4].
Definition 2.4**.**
Let F=(fk)k=1K be a frame for Hd.
(1)
F is an
α-tight frame (α-TF) if
SF=αI. F is a Parseval
frame (PF) if SF=I.
2. (2)
F is maximally robust to erasures if every subset
of F with d elements is a basis for Hd.
3. (3)
F is a simplex frame if GF=I−K1eet.
Maximally robust to erasures frames appeared first in
[30]. They are also known as
generic frames [9] and full spark
frames [1].
If Hd=Rd and F is a simplex
frame, then F corresponds to the d+1 vertices of the
regular d-simplex in Rd. We note that a 1-simplex
is a line segment, a 2-simplex is a triangle, a 3-simplex is a
tetrahedron and a 4-simplex is a pentachoron or pentatope.
The following proposition collects some properties of frames.
Proposition 2.5**.**
Let F=(fk)k=1K be a frame for Hd.
(1)
If F is an α-UNTF then
α=dK.
2. (2)
F* is a UNPF if and only if F is an orthonormal
basis.*
3. (3)
F* is a PF with K=d if and only if F is an orthonormal
basis.*
4. (4)
F* is an α-TF if and only if
α1F is a Parseval frame.*
5. (5)
F* is Parseval if and only if GF is an
orthogonal projection.*
6. (6)
SF−1/2F* is a PF for
Hd.*
7. (7)
If F is a simplex frame then K=d+1 and F is an isogonal
PF.
8. (8)
Let W be a subspace of
Hd and πW be the orthogonal
projection onto W. If F is an α-TF
for Hd then πWF is an
α-TF for W.
Definition 2.6**.**
Two frames F and G are complements
of each other if the sum of the Gramians of
SF−1/2F and
SG−1/2G is the identity I.
The complement of a frame of K vectors for a space of dimension
d is a frame of K vectors for a space of dimension K−d.
Definition 2.7**.**
Let F=(fk)k=1K and
G=(gk)k=1K be frames for Hd. Then
G is a dual frame of F if the
following reconstruction formula holds
f=∑k=1K⟨f,fk⟩gk, for all
f∈Hd,
or equivalently,
TGTF∗=I.
Let F=(fk)k=1K be a frame for Hd.
Then (SF−1fk)k=1K is the
canonical dual frame of F.
If F is an α-tight frame for Hd we
have the following reconstruction formula
f=α1∑k=1K⟨f,fk⟩fk, for all f∈Hd.
Proposition 2.8**.**
Let F=(fk)k=1K be a frame for Hd
and G=(gk)k=1K be a sequence in
Hd. The following assertions are equivalent:
(1)
G* is a dual frame of F.*
2. (2)
TG=SF−1TF+R* with
R∈L(FK,Hd) and RTF∗=0.*
3. (3)
TG=SF−1TF+W(I−TF∗SF−1TF)* with W∈L(FK,Hd).*
Note that a sequence (fk)k=1K in Hd is a
BUNTF if and only if (Kdfk)k=1K is a
BENPF. In view of this, we will work either with BUNTFs or BENPFs
according to convenience.
3. Applications of balanced frames
In this section we describe various advantages of balanced frames
and BUNTFs for applications. We will see that we can gain already
very good properties assuming only balancedness, which is a
condition that can be easily obtained. Note e.g. that if
(fk)k=1K is a frame then (f1,…,fK,−∑k=1Kfk) is a balanced frame.
As we mentioned in the introduction, the measurements of a signal
f, that in frame theory are expressed as inner products of f
with the elements of a frame, are used to represent it and will be
called frame coefficients. In obtaining these measurements,
or in the transmission or reception of them, different errors or
erasures can occur.
3.1. Robustness of the reconstructions under systematic errors
Let F=(fk)k=1K be a frame for Hd.
Any f∈Hd can be represented as a linear
combination f=∑k=1Kckfk. If F is
balanced we can change the numbers ck by summing to each of
them a constant c and the reconstruction will still be the desired
one, i.e., f=∑k=1K(ck+c)fk. This situation occurs
in the presence of systematic errors. The previous
considerations show that the reconstruction using balanced frames is
not affected by these type of errors. This is a very important fact,
because repeating the readings numerous times and taking the average
of them will not decrease systematic errors. Note that c can vary
with f, as it happens with the ck, and it can also be random.
3.2. Reconstruction error bounds
Let F=(fk)k=1K be a BF for Hd.
Assume that (⟨f,fk⟩)k=1K is perturbed by the
additive noises(ak)k=1K, i.e. we have the
sequence (⟨f,fk⟩+ak)k=1K. Then the
reconstruction f^ is
Since the frame is balanced we can give error bounds assuming that
the noises are near a constant that is not necessarily equal to
zero:
Proposition 3.1**.**
Let f∈Hd and (fk)k=1K be a
balanced frame for Hd. The following statements
about the norm of the reconstruction error hold:
(1)
Suppose that there exists μ such that
∣ak−μ∣<ϵ for each k=1,…,K. Let
λmax=λ1≥λ2≥…≥λd=λmin>0 be the eigenvalues of SF.
Then ∣∣f−f^∣∣<K/λminϵ.
Furthermore, if (fk)k=1K is a BUNTF for Hd
we can assert that ∣∣f−f^∣∣<dϵ and
∣∣f−f^∣∣∞<dϵ.
2. (2)
Assume that there exists μ such that
(∑k=1K∣ak−μ∣2)1/2<ϵ. If
(fk)k=1K is a BUNTF for Hd then
∣∣f−f^∣∣<d/Kϵ.
Proof.
For the first inequality we can argue similarly as in the proof of
[17, Proposition 2.1]. Consider the
canonical dual of F,
G=SF−1F. Then
TG=SF−1TF. The
reconstruction error is
For (2) observe that ∣∣f−f^∣∣=∣∣∑k=1K(ak−μ)Kdfk∣∣≤KddK(∑k=1K∣ak−μ∣2)1/2<Kdϵ.
∎
3.3. Presence of random additive noises without the zero mean assumption
We can analyze the behavior of the reconstruction error using a
statistical model for noise. Let F=(fk)k=1K be
a BUNF for Rd with frame bounds α,β.
Assume now that (⟨f,fk⟩)k=1K is perturbed
with additive noises (ηk)k=1K, and that each noise
ηk is a random variable with mean E[ηk]=μ and
variance E[(ηk−μ)2]=σ2. Suppose also that the
noises ηk and ηl are uncorrelated for k=l,
i.e. cov(ηk,ηl)=E[(ηk−μ)(ηl−μ)]=δk,lσ2
for each k,l. As before, the receiver will reconstruct the signal
as
The advantage of considering the balanced case in what follows is
that the mean of the noises is not required to be zero, an
assumption needed for the non balanced case which has been
considered so far in the literature.
The mean square error is
MSE:=d1E[∣∣∑k=1KηkSF−1fk∣∣2].
Since F is balanced, we can write
MSE=d1E[∣∣∑k=1K(ηk−μ)SF−1fk∣∣2].
The assumptions on the noises lead to
MSE=d1σ2∑k=1K∣∣SF−1fk∣∣2.
So,
dβ2Kσ2≤MSE≤dα2Kσ2.
If d and K are fixed it can be proved, as in
[16, Theorem 3.1] but now without
assuming μ=0, that the MSE is minimal if and only if the frame
is tight and that in this case MSE=Kdσ2.
Sometimes the reconstruction is done using the orthogonal projection
of (⟨f,fk⟩+ηk)k=1K onto R(TF∗) given by p=TF∗f+TF∗SF−1TF(ηk)k=1K. Since the frame is
balanced,
p=TF∗f+TF∗SF−1TF(ηk−μe)k=1K=TF∗f+p.
So, as in [12, Section 8.5] but again
without assuming μ=0, it can be proved that,
βσ2≤E[∣p(k)∣2]≤ασ2
where the equality holds if (fk)k=1K is a tight
frame. In this case,
E[∣p(k)∣2]=Kdσ2.
Note that when considering BUNTFs, if the number of elements of the
frame increases (higher redundancy) both the MSE and the mean of
∣p(k)∣2 decrease. This shows the advantage of
using redundant BUNTFs.
3.4. Resilience of the dual frames against fixed perturbations
Let (fk)k=1K,(gk)k=1K be sequences in
Hd where (fk)k=1K is balanced. Then
∑k=1K⟨f,gk⟩fk=∑k=1K⟨f,(gk+g)⟩fk for each g∈Hd. As a consequence of this we obtain:
Proposition 3.2**.**
If F is a balanced frame for Hd and
G=(gk)k=1K is a dual frame of F,
then (gk+g)k=1K is also a dual frame of F for
each g∈Hd.
Proposition 3.2 says that for a balanced frame
F the reconstruction is not altered if we use a dual
which is perturbed by a fixed vector, and can also be used to define
an equivalence relation in the set of dual frames of F.
Definition 3.3**.**
Let F be a balanced frame for Hd. We say
that two dual frames G=(gk)k=1K and
G=(gk)k=1K of
F are equivalent if there exists g∈Hd such that gk=gk+g for each k=1,…,K.
It is clear that if there exists a balanced frame in an equivalence
class, it is the unique balanced one in this class. Let
[G]={(gk+g)k=1K:g∈Hd} be the
equivalence class of the dual frame G of F.
If G is not balanced, the dual frame
(gk−K1TGe)k=1K is equivalent to
G and is balanced. This shows that each of the
equivalence classes contains a unique dual frame which is balanced
and can be considered as the representative of the class. Thus, in
order to obtain all the dual frames of F, we only need
to compute those that are balanced, the others will be in their
equivalence classes.
3.5. Presence of erasures
When some of the frame coefficients are no longer accessible after
the transmission, we say that an erasure occurs.
Part (1) of the following proposition says that if one of the frame
coefficients is deleted (or is set equal to zero) we can still
recover f exactly. It also says that a balanced frame
(fk)k=1K remains to be a frame if we delete one of its
elements. Joining both parts of the proposition we have a
characterization of balanced frames.
Proposition 3.4**.**
Let (fk)k=1K be a frame for Hd and
(fk)k=1K be one of its duals. The following
assertions holds:
(1)
If (fk)k=1K is balanced, then
for each l∈{1,…,K}, (fk)k=1,k=lK and
(fk−fl)k=1,k=lK are dual
frames.
2. (2)
If there exists l∈{1,…,K} such that fl=0, and (fk)k=1,k=lK
and (fk−fl)k=1,k=lK are
dual frames, then (fk)k=1K is balanced.
Proof.
(1) Suppose that (fk)k=1K is balanced. Let l∈{1,…,K}. Then for each f∈Hd,
If (fk)k=1K is a balanced frame, then ∑k=1K⟨f,fk⟩=0. So if the transmitted numbers
(ck)k=1K satisfy ∑k=1Kck=0, we know
that (ck)k=1K comes from a perturbation of the frame
coefficients (⟨f,fk⟩)k=1K. In this way we
can easily detect the presence of a problem.
Furthermore, using balanced frames we can have a hint about the
source of the error. If we are in the presence of a systematic error
i.e. ck=⟨f,fk⟩+c for some constant c, then
∑k=1Kck=Kc independently of the signal f. In this
case, although we can know the error, it is not necessary to correct
it because the reconstruction with a balanced dual frame will be the
desired one. If the perturbation is due to random additive noises
(ηk)k=1K with ∣ηk−η∣≤σ for each
k, then ck=⟨f,fk⟩+ηk and
∑k=1Kck=∑k=1Kηk fluctuates without
any apparent pattern between two fixed values, also independently of
the signal. If instead the sum of the transmitted numbers is non
zero and varies with the signal, we can suspect that the error
arises from other sources. For example, assume that erasures occur,
i.e. we only receive (⟨f,fk⟩)k∈I where I
is a proper subset of {1,…,K}. In this case ∑k∈I⟨f,fk⟩ generally varies with the signal f.
3.7. BUNTFs for Rd and real spherical 2-designs
Real spherical t-designs appear in relation with cubature
formulas on the sphere in Rd [3]. They are sets of points on the unit sphere
Sd−1 of Rd such that the integral on
Sd−1 of any homogeneous polynomial of total degree
less than or equal to t in d variables is equal to the mean of
the values of the polynomial over these points. In other words, they
approximate the unit sphere in the sense that computing the average
of these polynomials only over these sets of points is identical to
taking the average over the entire unit sphere. The following result
can be found in different forms e.g. in [8, 32] (see also [34]).
Proposition 3.5**.**
A sequence (fk)k=1K of unit vectors in Rd is a
spherical 2-design if and only if it is a BUNTF for Rd.
Spherical t-designs are used in approximation theory, in numerical
interpolation, integration, and regularized least squares
approximation. They have connections with many areas of mathematics
such as analysis and statistics (in particular with orthogonal
polynomials and moment problems), algebraic combinatorics
(association schemes, design theory, coding theory), group theory
(spherical designs which are orbits of a finite group in the real
orthogonal group O(n)), number theory (designs that are shells of
Euclidean lattices are related with modular forms and the Lehmer’s
conjecture about the zeros of the Ramanujan r function), geometry
(sphere packing problems) and optimization (Delsarte’s linear
programming method).
3.8. Balanced sequences and tight frames
There is a strong connection between balanced sequences and tight
frames via diagram vectors. Diagram vectors can be used for
determining whether a frame for R2 is tight or not
[19]. This notion has been
extended to Fd in
[13]
(we refer the reader to this paper for the definition of diagram
vectors). By [13, Proposition
3.6] and
Proposition 4.4 below, we have:
Proposition 3.6**.**
Let c1,…,cK be nonnegative numbers, which are not all
zero. Let (fk)k=1K in Fd and let
(fk)k=1K in Fd the
corresponding diagram vectors. Then (ckfk)k=1K is a
tight frame for Fd if and only if
(ck2fk)k=1K is balanced.
This result shows that balanced sequences are useful in the study of
tight frames.
4. Properties
In this section we study properties of BFs and in particular of BPFs
and BUNTFs. We consider their behavior under transformations, give
several characterizations and analyze duality.
4.1. Invariance under certain transformations
Given a frame, it is important which properties are preserved under
certain transformations. The following are analogous to those
presented in [30], here we analyze them
regarding balancedness.
Proposition 4.1**.**
(1)
Let A∈Gl(Hd) and B∈Gl(FK). If Be=e, then F is a
balanced frame if and only if AFB is a balanced frame.
2. (2)
Let a=0, U∈U(Hd) and V∈U(FK) such that Ve=e. Then aUFV is a BTF if and only if F is a BTF.
3. (3)
Let a=0, U∈U(Hd). Then aUF (UF) is a BENF (BUNF) if and only
if F is a BENF (BUNF).
4. (4)
Let A∈G(Hd). Then F is a maximally robust to erasures BF if
and only if AF is a maximally robust to erasures BF.
5. (5)
F* is a BUNTF if and only if F
is a BUNTF.*
6. (6)
Let W be a subspace of
Hd and πW be the orthogonal
projection onto W. If F is an α-BTF
for Hd then πWF is an
α-BTF for W.
7. (7)
Let A∈L(Hd,Hn) be an
isometry, i.e., A∗A=I, then F is an α-BUNTF
for Hd if and only if AF is an
α-BUNTF for Hn.
Proof.
In each case, the proof follows straightforward from the
definitions. To illustrate we show (1):
If A∈L(Hd) is injective and B∈L(FK) is such that Be=e, then
TFe=0 if and only if ATFBe=0.
Moreover, if A and B are invertible, TF is onto
if and only if ATFB is onto.
∎
In view of Proposition 4.1, we define an
equivalence relation:
Definition 4.2**.**
Two frames F and G for Hd are
unitary equivalent if and only if there exists a unitary
operator U∈L(H) such that
G=UF.
In the previous equivalence relation, the permutation or numbering
of the elements of F or G will not be
considered. Two PFs are unitary equivalent if and only if they have
the same Gram matrix [34, Corollary 2.1.].
As a consequence of Proposition 4.1 and
Proposition 2.5:
Corollary 4.3**.**
Let F be a frame for Hd. The following
assertions are equivalent:
(1)
F* is a BF for Hd.*
2. (2)
SF−1F* is a BF for
Hd.*
3. (3)
SF−1/2F* is a BPF for
Hd.*
4.2. Some characterizations
The following proposition gives several equivalences for a sequence
to be balanced.
Proposition 4.4**.**
Let F=(fk)k=1K be a sequence in
Hd. The following assertions are equivalent:
(1)
F* is balanced.*
2. (2)
TFe=0.
3. (3)
GFe=0.
4. (4)
∑k=1K⟨fl,fk⟩=0* for each l∈{1,…,K}.*
5. (5)
∑k,l=1K⟨fk,fl⟩=0.
6. (6)
∑k=1K⟨f,fk⟩=0* for each
f∈Hd.*
7. (7)
∑k=1K∣∣f−fk∣∣2=∑k=1K∣∣fk∣∣2+K∣∣f∣∣2* for each
f∈Hd.*
8. (8)
∑k=1,k=lK∣∣fl−fk∣∣2=∑k=1K∣∣fk∣∣2+K∣∣fl∣∣2*
for each l∈{1,…,K}.*
Proof.
Taking into account the definition of balanced sequences and that
ker(TF)=ker(GF), (1)⇔(2)⇔(3) follows immediately.
Considering the entries of the matrix GF, it is
immediate that (3)⇔(4).
Observe that ∑k,l=1K⟨fk,fl⟩=∣∣TFe∣∣2. Thus we have (2)⇔(5).
For (1)⇔(6) we note that ∑k=1Kfk=0
if and only if ⟨f,∑k=1Kfk⟩=0 for each f∈Hd.
(6)⇔(7) follows from
∣∣f−fk∣∣2=∣∣fk∣∣2−2Re(⟨f,fk⟩)+∣∣f∣∣2 and ∣∣if−fk∣∣2=∣∣fk∣∣2+2Im(⟨f,fk⟩)+∣∣f∣∣2. Similarly it can be proved (4)⇔(8).
∎
From Proposition 4.4 we can obtain the
next well-known basic result about simplex frames:
Corollary 4.5**.**
If F is a simplex frame then F is balanced.
There exists a bijective correspondence between the BUNTFs for
Hd and the BUNTFs for its dual space. This is a
consequence of the following result which follows from the Riesz
representation theorem and Proposition 4.4:
Corollary 4.6**.**
F=(fk)k=1K* is a BUNTF for Hd if
and only if (⟨.,fk⟩)k=1K is a BUNTF for
the dual space Hd∗.*
Proposition 3.4 says that a BF is
one-robust to erasures. This suggests the following version of
[10, Corollary 5.1]:
Proposition 4.7**.**
Let (ek)k=1K be an orthonormal basis for HK
and πW be the an orthogonal projection onto a subspace
W of HK. The following are equivalent:
(1)
(πWek)k=1K* is a balanced Parseval frame for W.*
2. (2)
∑k=1Kek∈W⊥.
3. (3)
There exists f∈W⊥ such that ⟨f,ek⟩=1 for each k=1,…,K.
Proof.
By Proposition 2.5,
(πWek)k=1K is a Parseval frame for
W.
(1)⇒(2) Since (πWek)k=1K is
balanced, πW∑k=1Kek=0. This shows that
∑k=1Kek∈W⊥.
(2)⇒(3) Take f=∑k=1Kek.
(3)⇒(1) Let f∈W⊥ such that
⟨f,ek⟩=1 for each k=1,…,K. Then
∑k=1KπWek=πW∑k=1K⟨f,ek⟩ek=πWf=0.
∎
We have the following versions of the Naimark characterization for
BPFs:
Theorem 4.8**.**
A sequence (fk)k=1K in Hd is a BPF for
Hd if and only if there is a larger Hilbert space
HK⊇Hd and an orthonormal basis
(gk)k=1K for HK satisfying
∑k=1Kgk∈Hd⊥ so that
fk=πHdgk for each k=1,…,K.
Proof.
Any Parseval frame F=(fk)k=1K for
Hd is unitary equivalent to
(GFek)k=1K where (ek)k=1K is the
standard basis of FK (see [34, Theorem 2.2.]). In this case, GF is the orthogonal
projection onto im(GF) that has dimension
d. If F is also balanced then, by Proposition 4.4, e=∑k=1Kek∈im(GF)⊥. Let U∈L(im(GF),Hd) unitary such that
fk=UGFek for each k=1,…,K. Let
HK such that HK⊇Hd
and HK=Hd⊕Hd⊥. Let
U∈L(FK,HK) unitary such
that U∣im(GF)=U. Let
gk=Uek for each k=1,…,K. Then
πHd=UGFU∗,
(gk)k=1K is an orthonormal basis for HK,
∑k=1Kgk∈Hd⊥ and
fk=πHdgk for each k=1,…,K.
∎
Remark 4.9*.*
The previous proof is constructive. The theorem follows also from
Proposition 4.7 and
Naimark’s theorem [11].
Theorem 4.10**.**
A sequence (fk)k=1K in Hd is a BPF for
Hd if and only if there is a larger Hilbert space
HK−1⊇Hd and a simplex frame
(gk)k=1K for HK−1 so that
fk=πHdgk for each k=1,…,K.
Proof.
Let F=(fk)k=1K be a balanced Parseval frame for
Hd. Similar to the proof of Theorem 4.8, F is unitary equivalent to
(GF(ek−K1e))k=1K where
(ek)k=1K is the standard basis of FK. The
sequence (ek−K1e)k=1K is a simplex frame for
span{e}⊥ that has dimension K−1. The rest
follows as in proof of Theorem 4.8.
∎
Let (fk)k=1K be a sequence in Hd and I⊆{1,…,K} such that fk⊥fl=0 for k∈I, l∈Ic. Let W:=span(fk)k∈I. Then (fk)k=1K is BUNTF for Hd if and
only if (fk)k∈I is a BUNTF for W and
(fk)k∈Ic is a BUNTF for W⊥.
The frame graph (or correlation network) of a sequence
(fk)k=1K in Hd is the graph with vertices
(fk)k=1K and an edge between fk and fk′, k=k′, if and only if ⟨fk,fk′⟩=0
[34]. Each frame can be uniquely decomposed into a
union of frames for orthogonal subspaces, each corresponding to the
vertices of a connected component of the frame graph.
Proposition 4.11 gives a
characterization of BUNTFs in terms of the cycles in its frame
graph:
Theorem 4.12**.**
A sequence (fk)k=1K in Hd is a BUNTF if and
only if the vertices of each of the connected components in its
frame graph is a BUNTF for its span.
4.3. Duals of a balanced frame
As was explained in section 3.4, in order to obtain the duals of a
balanced frame it is sufficient to consider the balanced ones.
Proposition 2.8 leads to different
characterizations of balanced dual frames of a given BF:
Proposition 4.13**.**
Let F=(fk)k=1K be a BF for Hd and
G=(gk)k=1K be a sequence in Hd.
Then the following are equivalent:
(1)
G* is a balanced dual frame of F.*
2. (2)
TG=SF−1TF+R* where R∈L(FK,Hd), RTF∗=0 and Re=0.*
3. (3)
(gk)k=1K=(SF−1fk+rk)k=1K* with (rk)k=1K in Hd such that
∑k=1K⟨f,fk⟩rk=0 for each f∈Hd and ∑k=1Krk=0.*
4. (4)
TG=SF−1TF+R, where
R∈L(FK,Hd) and
span{e}⊕im(TF∗)⊂ker(R).
5. (5)
TG=SF−1TF+R, where
R∈L(FK,Hd) and
im(R∗)⊕im(TF∗)⊂span{e}⊥.
6. (6)
TG=SF−1TF+W(I−TF∗SF−1TF), where W∈L(FK,Hd) and We=0.
7. (7)
(gk)k=1K=(SF−1fk+hk+∑l=1K⟨SF−1fk,fl⟩hl)k=1K*
with (hk)k=1K in Hd such that
∑k=1Khk=0.*
Corollary 4.14**.**
Let F=(fk)k=1K be a BF for Hd.
Let G be a balanced dual frame of F with
TG=SF−1TF+R, where R∈L(FK,Hd). Then rank(R)≤K−d−1.
As a consequence of Proposition 4.13 we also have the following uniqueness result:
Corollary 4.15**.**
Let F=(fk)k=1K be a BF for Hd.
Then K=d+1 if and only if (SF−1fk)k=1K
is the unique balanced dual frame of (fk)k=1K.
The above corollary suggests that in the family of BFs, those BFs
with K=d+1 can be seen as the analogous to the bases in the family
of frames.
The existence of distinct types of dual frames of a given Parseval
frame is studied in [10]. In particular,
it is shown that a Parseval frame is itself its unique Parseval dual
frame. They also consider for a given Parseval frame its tight dual
frames. Here we are interested in balanced tight dual frames of a
given balanced Parseval frame.
Theorem 4.16**.**
Let F=(fk)k=1K be a BPF for Hd.
If K≤2d the unique balanced tight dual frame of F
is F. If K>2d there exist infinite non unitary
equivalent balanced tight dual frames of F.
Proof.
Let G be a balanced α-tight dual frame of
F. Assume TG=TF+R with R∈L(FK,Hd) such that
RTF∗=0 and Re=0. Then
αIHd=IHd+RR∗.
Thus, α=1 (indeed α>1) if and only if
rank(R)=d, and α=1 if and only if R=0.
If rank(R)=d, by Corollary 4.14, K≥2d+1. So, if K≤2d the unique balanced tight dual frame of
F is F.
Let {e1,…,ed} be any orthonormal basis of
Hd. If K>2d, we can consider any set of equal norm
orthogonal vectors {s1,…,sd}⊂(span{e}⊕im(TF∗))⊥. Let
R=(rk)k=1K with
rk=∑i=1dsi(k)ei for k=1,…,K. Then
where ρ=∣∣si∣∣2 for i=1,…,d. Thus
(fk+rk)k=1K is a balanced (ρ+1)-tight dual frame
of F with Gram matrix GF+ρI=GF. This shows that if K>2d there exist infinite non unitary equivalent balanced tight dual frames of
F.
∎
5. The closest balanced frame to a given frame
A natural question that arises is: Given a frame, is there a
balanced frame that is closest to it in some norm and how do we find
it? The first step to answer this question is the following theorem,
that describes the ℓ1-norm closest balanced sequence to a
given sequence of elements in Hd. We consider the
ℓ1-norm of a sequence in Hd given by
∣∣(fk)k=1K∣∣1:=∑k=1K∣∣fk∣∣.
Theorem 5.1**.**
Let (fk)k=1K be a sequence in Hd. Then
∣∣∑k=1Kfk∣∣=inf{∑k=1K∣∣fk−gk∣∣:(gk)k=1K
is a balanced sequence in Hd},
and the infimum is attained for the sequences of the form
(fk−pk∑l=1Kfl)k=1K, where 0<pk<1
for each k=1,…,K and ∑k=1Kpk=1.
Proof.
Let (gk)k=1K be a balanced sequence in Hd,
and 0<pk<1 for each k=1,…,K with
∑k=1Kpk=1. We have,
Now suppose that (gk)k=1K is a balanced sequence in
Hd and
∑k=1K∣∣fk−gk∣∣=∣∣∑k=1Kfk∣∣. Then
∑k=1K∣∣fk−gk∣∣=∣∣∑k=1K(fk−gk)∣∣, and
this happens if and only if there exist positive real numbers
c1,…,cK−1 such that
fk+1−gk+1=ck(f1−g1) for each k=1,…,K−1.
Setting p1=1+c1+…+cK−11 and
pk+1=1+c1+…+cK−1ck for each k=1,…,K−1 we have ∑k=1Kpk=1, 0<pk<1 and
gk=fk−pk∑l=1Kfl for each k=1,…,K.
∎
Now we analyze the problem for the ℓ2-norm. Given a sequence
(fk)k=1K in Hd, the next theorem asserts
that (fk−K1∑l=1Kfl)k=1K is the
balanced sequence in Hd closest to (fk)k=1K
in the ℓ2-norm, where
∣∣(fk)k=1K∣∣2:=(∑k=1K∣∣fk∣∣2)1/2.
In its proof we use the following equality:
[TABLE]
Theorem 5.2**.**
Let (fk)k=1K be a sequence in Hd. Then
K1∣∣∑l=1Kfl∣∣2=inf{∑k=1K∣∣fk−gk∣∣2:(gk)k=1K
is a balanced sequence in Hd},
and the infimum is attained for
gk=fk−K1∑l=1Kfl for each k=1,…,K.
Proof.
Let (gk)k=1K be a balanced sequence in Hd.
Using (5.1),
[TABLE]
Now suppose that (gk)k=1K is a balanced
sequence in Hd and
∑k=1K∣∣fk−gk∣∣2=K1∣∣∑k=1Kfk∣∣2.
Then
∑k=1K∣∣fk−gk∣∣2=K1∣∣∑k=1K(fk−gk)∣∣2.
So, by (5.1),
f1−g1=…=fK−gK. Therefore,
gk=fk−K1∑l=1Kfl for each k=1,…,K.
∎
Note that if F=(fk)k=1K and
G=(gk)k=1K, then
∑k=1K∣∣fk−gk∣∣2=∥TF−TG∥F2,
where ∥.∥F denotes the Frobenius norm. In order to apply the
above theorems to frames we have the following result.
Lemma 5.3**.**
Let (p1,…,pK)t∈RK, where 0<pk<1 for each k=1,…,K and
∑k=1Kpk=1. If F=(fk)k=1K is a
frame for Hd, then
(fk−pk∑l=1Kfl)k=1K is a BF for
Hd if and only if (p1,…,pK)t∈/im(TF∗).
Proof.
The synthesis operator of
(fk−pk∑l=1Kfl)k=1K is
TF(I−e(p1,…,pK)). If F is a
frame for Hd, by the Sylvester inequality
[22], d−1≤rank(TF(I−e(p1,…,pK)))≤d,
and by the Wedderburn’s rank-one reduction formula
[22],
rank(TF−TFe(p1,…,pK))=d−1 if and only if (p1,…,pK)t∈im(TF∗).
∎
Remark 5.4*.*
Let F=(fk)k=1K be a frame for Hd.
Let (p1,…,pK)t∈RK, where 0<pk<1 for each k=1,…,K and ∑k=1Kpk=1. If
there exists f∈Hd such that (p1,…,pK)t=TF∗f, then
(fk−pk∑l=1Kfl)k=1K is a BF for
span{f}⊥. Conversely, if there exists f∈Hd, f∈/ker(TF∗), such
that (fk−pk∑l=1Kfl)k=1K is a BF for
span{f}⊥, then (p1,…,pK)t=γTF∗f for some γ∈F, γ=0.
We can now give the answer to the question we posed at the beginning
of this section. The following theorem gives necessary and
sufficient conditions for the closest balanced frame to a given
frame to exist, and in this case gives its expression.
Theorem 5.5**.**
Let F=(fk)k=1K be a frame for Hd.
Then
(1)
There exist ℓ1-norm closest to
F balanced frames for Hd if and only if
F is not a basis, and in this case they are the frames
(fk−pk∑l=1Kfl)k=1K where 0<pk<1,
∑k=1Kpk=1 and (p1,…,pK)t∈/im(TF∗).
2. (2)
There exist an
ℓ2-norm closest to F balanced frame for
Hd if and only if e∈/im(TF∗), and in this case it is the frame
(fk−K1∑l=1Kfl)k=1K.
Proof.
If F is a basis, clearly there does not exist an
ℓ1-norm (ℓ2-norm) closest to F balanced
sequence which is a frame for Hd, since there does not
exist balanced frames for Hd with K=d elements. Thus
we suppose that F is not a basis.
(1) Since the set {(p1,…,pK)t∈RK:0<pk<1 and ∑k=1Kpk=1}∩im(TF∗)c has an infinite number of
points, the conclusion follows from Theorem 5.1 and
Lemma 5.3.
(2) By Theorem 5.2 and Lemma 5.3, if e∈/im(TF∗)
then (fk−K1∑l=1Kfl)k=1K is the
ℓ2-norm closest to F balanced frame for
Hd.
Now suppose that e∈im(TF∗). Let
G=(gk)k=1K be any BF for Hd. We
are going to prove that we can always find another BF for
Hd closer to F than G in the
ℓ2-norm.
Suppose without loss of generality that
F2=(fk)k=2K still generates
Hd, that is, F2 is a frame for
Hd. So, TF2∗ is injective.
For ϵ=0 let
Fϵ=(fϵ,k)k=1K where
fϵ,1=ϵf1, fϵ,2=f2, ….,
fϵ,K=fK.
Since e∈im(TF∗), none of the elements
of F is the null vector. Also, by Lemma 5.3,
F=(fk−K1∑l=1Kfl)k=1K
is not a frame for Hd. So
G=F and, by Theorem 5.2,
K1∣∣∑k=1Kfk∣∣2<∑k=1K∣∣fk−gk∣∣2.
Take ϵ such that 0<∣1−ϵ∣<∣∣f1∣∣11−K1∑k=1K∣∣fk−gk∣∣2−K1∣∣∑k=1Kfk∣∣2.
Let f∈Hd such that TF∗f=e. Then
TF2∗f=(1,…,1)t. If there would exist
g∈Hd such that
TFϵ∗g=e, then
TF2∗g=(1,…,1)t. Since
TF2∗ is injective g=f. So, ⟨f,f1⟩=⟨f,ϵf1⟩=1. Thus ϵ=1 which is absurd. This shows that e∈/im(TFϵ∗). Hence, by Lemma 5.3,
Fϵ=(fϵ,k−K1∑l=1Kfϵ,l)k=1K
is a BF for Hd.
We have,
[TABLE]
Hence Fϵ is a BF for
Hd closer to F in the ℓ2-norm than
G.
∎
6. A new concept of complement for balanced frames
Let F be a BENPF for Hd and G
be any of its complements. Then G is an ENPF for
Hd. Since GGe=e, by Proposition 2.8G is not balanced. Morevover,
since e∈im(TG∗), Theorem 5.5 tells us that although G has
closest balanced frames in the ℓ1-norm, it has not a closest
balanced frame in the ℓ2-norm.
In order to have complementary frames in the same class, we define
an alternative concept of complements for BPFs. To this end, we
first state the following proposition whose proof is
straightforward.
Proposition 6.1**.**
Let F=(fk)k=1K be a BPF for Hd.
Then I−GF−K1eet is the orthogonal
projection onto (im(GF)⊕span{e})⊥.
Note that rank(I−GF−K1eet)=K−d−1
and (I−GF−K1eet)e=0. Based on
Proposition 6.1 and Theorem 4.10 we introduce the following definition:
Definition 6.2**.**
Two PFs F and G are B-complements
of each other if the sum of their Gramians is I−K1eet.
In view of Proposition 6.1, the B-complement of
a BPF of K vectors for a space of dimension d is a BPF of K
vectors for a space of dimension K−d−1. For future references we
state the following lemma that follows immediately from the
definitions of simplex frame and of B-complement:
Lemma 6.3**.**
F* is a simplex frame with K elements if and only if
its B-complement is the frame for the zero vector space given by the
zero vector repeated K times.*
Note that the sum of the Gram matrices of two complementary PFs is
I, which is the Gram matrix corresponding to an orthonormal basis.
By Proposition 2.5, an orthonormal basis
can be seen as a “limit case” of a PF: it is a UNPF or a PF with
K=d. In the case of two B-complementary BPFs, the sum of their
Gram matrices is I−K1eet, which is the Gram matrix
corresponding to simplex frames. We can think that in the family of
BFs, simplex frames are the analogous to othonormal basis in the
family of frames. This follows from Theorem 6.4 below, which shows that a simplex frame can be seen
as a “limit case” of BPF: it is a BENPF which elements have norm
equal to d+1d, or a BPF with K=d+1.
Theorem 6.4**.**
Let F=(fk)k=1K be a sequence in
Hd. The following assertions are equivalent:
(1)
F* is a simplex frame for Hd.*
2. (2)
F* is a BPF for Hd and K=d+1.*
3. (3)
F* is a BPF for Hd and ∣∣fk∣∣2=d+1d for each k=1,…,K.*
4. (4)
F* is an isogonal PF for Hd with K>d and ∣∣fk∣∣2=⟨fk,fl⟩ for each k=1,…,K, k=l.*
5. (5)
F* is a BPF with
ker(TF)=span{e}.*
Proof.
If F is a simplex frame for Hd then, by
Proposition 2.5 and Corollary 4.5, F is an isogonal BPF for
Hd and K=d+1. We also have
diag(GF)=1−K1=d+1d and
ker(TF)=ker(GF)=span{e}.
So (1) implies the rest of the assertions.
(2)⇒(1). Suppose that F is a BPF with
K=d+1. Let G be a B-complement of F. Then
rank(GG)=K−d−1=0. So, by Lemma 6.3, F is a simplex frame.
(3)⇒(2). Suppose that F is BPF for
Hd and ∣∣fk∣∣2=d+1d for each k=1,…,K. Then (dd+1fk)k=1K is a
dd+1-BUNTF. From Proposition 2.5, dd+1=dK. So, K=d+1.
(4)⇒(1). By hypotheses,
GF2=GF and there exists a,c∈R, a=c, such that GF=(c−a)I+aeet.
GF is a circulant matrix, so its eigenvalues are
c+a(K−1) and c−a with multiplicity K−1 [22]. Since rank(GF)=d, K>d and c−a=0, we have a=−K−1c and K−1=d. Thus
GF=K−1c(KI−eet). Since
GF2=GF, c=KK−1. Therefore,
GF=I−K1eet and F is a simplex
frame.
(5)⇒(1). By the hypotheses, GF is an
orthogonal projection matrix and
im(GF)=span{e}⊥, so
GF=I−K1eet and F is a simplex
frame.
∎
Some of the points of the previous theorem can be seen as variations
of statements that appear in [34]. By
Corollary 4.15 and Theorem 6.4, the canonical dual, which in this
case it is itself, is the unique balanced dual of a simplex frame.
Moreover, by Theorem 6.4 and
Corollary 4.3:
Corollary 6.5**.**
F* is a BF for Hd with K=d+1 if and only
if SF−1/2F is a simplex frame for
Hd.*
In what follows we consider properties of B-complementary BPFs that
are analogous to properties of complementary PFs that can be found
in [34].
Let F be a BPF for Hd. The B-complements
of F are unitary equivalent. Let G be a
B-complement of F. Then F is equal-norm (or
isogonal or real) if and only if G is. F and
G can not be unitarily equivalent.
We note that if F is a BPF for Fd with K
elements and (K1e,v1,…,vK−d−1) is an
orthonormal basis for ker(TF), then the
columns of the matrix which rows are vk∗, k=1,…,K−d−1, constitutes a B-complement BPF of F.
We now introduce B-complementary BFs:
Definition 6.6**.**
Two BFs F and G are B-complements
if the PFs SF−1/2F and
SG−1/2G are B-complements.
Let F=(fk)k=1K and
G=(gk)k=1K be BFs for Hd1 and
Hd2, respectively. Then the following are
equivalent:
(1)
F* and G are B-complements.*
2. (2)
im(GF)⊕im(GG)=span{e}⊥.
3. (3)
dim(Hd1)+dim(Hd1)=K−1* and TGTF∗=0.*
4. (4)
The inner sum F⊕G=(fk,gk)k=1K is a BF for Hd1⊕Hd2 with K=d1+d2+1 and TGTF∗=0.
5. (5)
TG=TG(I−K1eet−TF∗SF−1TF).
Remark 6.8*.*
In case that F=(fk)k=1K and
G=(gk)k=1K are BPFs, (4) of the previous
proposition becomes:
F⊕G=(fk,gk)k=1K is a simplex
frame for Hd1⊕Hd2.
This concept can be applied to construct BUNTFs of K vectors for a
space of dimension K−d−1 from BUNTFs of K vectors for a space of
dimension d.
7. Examples of balanced unit norm tight frames
The aim of this section is to present various examples of BUNTFs and
some of their properties. Sometimes we identify a frame
F for Fd of K elements with the matrix
that represents TF in the standard bases of
Fd and FK.
7.1. The case F=R and d=2
As a consequence of Proposition 3.5 and
[20, Lemma 1] we obtain:
Theorem 7.1**.**
The following are equivalent:
(1)
((cosθk,sinθk)t)k=1K* is a
BUNTF for R2.*
2. (2)
∑k=1Keiθk=∑k=1Ke2iθk=0.
3. (3)
∑k=1Keiθk=∑1≤k1<k2≤Ke2iθk1e2iθk2=0.
By Theorem 7.1, the set of vectors coming from the
Kth roots of unity are BUNTFs for R2:
Corollary 7.2**.**
If K≥3 and (eiθk)k=1K are the Kth roots
of unity, ((cosθk,sinθk)t)k=1K is a
BUNTF for R2.
In [20, Theorem A] the types of spherical t-designs
in R2 are described. From this result and
Proposition 3.5 we have:
Theorem 7.3**.**
For K=3,4,5, there is one equivalence class of BUNTF for
R2 with K elements. For K≥6, there are
infinite equivalence classes of BUNTF for R2 with K
elements.
We have that for K=3,4,5 the class corresponding to the frame
coming from the Kth roots of unity is the unique equivalence class
of BUNTFs for R2 with K elements. We can see that
for K≥6 there are infinite equivalence classes as follows.
Note first that, by Corollary 7.2, we
always have the class corresponding to the Kth roots of unity. Now
write K=3n+s where s∈N0,0≤s<3. Then, if s=0
there are the classes corresponding to the union of n rotations of
the third roots of unity, and there are infinitely many of such
classes. If s=1, then K=3n+1=3(n−1)+4. So we have the classes
corresponding to the union of the 4th roots of unity and n−1
rotations of the third roots of unity. If s=2, then K=3(n−2)+8
and similar arguments can be used. Writing e.g. K=4m+r where r∈N0,0≤r<4 and m∈N, or using other
decompositions of K, we can see that there exist more equivalence
classes of BUNTFs.
In what follows we consider several examples of tight frames, some
of them well-known, indicating in which cases they turn out to be
balanced.
7.2. Balanced harmonic frames
Let F be the unitary matrix of order K which entries are
F(k,l)=K1eK2πi(k−1)(l−1), called
Fourier matrix. The ENPFs T consisting of a d×K
submatrix of F are a particular case of the so called
harmonic frames. To obtain real ENPFs we must select real
rows and complex conjugate pairs of rows from the Fourier matrix
F. If T does not contain the first row of F, then T is also
balanced. More general, unlifted harmonic frames are BENPFs
and B-complements of unlifted harmonic frames are unlifted harmonic
frames. See [34, Chapter 11] for a detailed
treatment of harmonic frames.
7.3. BENTFs from Hadamard matrices
A Hadamard matrixH has orthogonal rows and entries ±1
[21]. The smallest examples of Hadamard matrices
are:
A way for contructing Hadamard matrices is the following:
if H is a Hadamard matrix, \left(\begin{array}[]{cc}H&H\\
H&-H\\
\end{array}\right) is a Hadamard matrix. Hadamard matrices obtained in this manner
are known as Sylvester-Hadamard matrices. If H has order
K and we select a submatrix T of order d×K from H, we
can get a BENTF.
7.4. Crosses and eutactic stars
The set (±u1,…,±uK), where (u1,…,uK) is an orthonormal basis for Hd, is a
BUNTF for Hd. By Proposition 4.1
the set (±πWu1,…,±πWuK), where W is a subspace of
Hd, is a BTF for W. If
Hd=Rd, (±u1,…,±uK) is
known as a cross and (±πWu1,…,±πWuK) is known as an eutactic star
(see [15]).
7.5. Partition frames
Let η=(η1,…,ηn)∈Zn be a
partition of K, i.e., K=η1+…+ηn and 1≤η1≤…≤ηn. The η-partition
frame for Rd with d=K−n, is the complement of the
PF
[η1e1…η1e1…ηnen…ηnen],
of K vectors for Rn. Here each
ηjej is repeated ηj times. An
η-partition frame for Rd has Gram matrix
where Bj is the ηj×ηj orthogonal
projection matrix with ηjηj−1 as diagonal
elements and ηj−1 as non diagonal elements. See
[34] for more details of partition frames.
An η-partition frame is balanced and Parseval. If n∣K and
η1=…=ηn we obtain an equal norm frame.
A B-complement G of an (η1,…,ηn)-partition frame F of K elements for
Rd has Gram matrix GG=(Ci,j) where
Ci,j is an ηi×ηj matrix such that the
entries of Ci,j are ηiKK−ηi if i=j and
−K1 if i=j. For example, if n=1, F
is a simplex frame and G is the zero vector repeated K
times. If n=2, G is the BPF of K elements for
R1 consisting of −η1Kη2
and η2Kη1 (or
η1Kη2 and
−η2Kη1) repeated η1 and
η2 times, respectively.
7.6. BUNTFs from spherical designs
We recall that any spherical (t+1)-design is a spherical
t-design and a real spherical 2-design is a BUNTF. In
[3], several examples of spherical
t-design are presented. They include regular K-gons on
S1⊂R2, platonic solids in
R3, regular potytopes and roots systems in
Rd, and the set of minimal vectors of the Leech
lattice in R24.
8. Construction methods for balanced unit norm tight frames
In this section we present explicit and painless constructions of an
infinite variety of BUNTFs.
We begin by showing under which conditions some well-known methods
for constructing frames lead to the obtention of BUNTFs. For
properties of these methods see [34, Chapter 5].
We have the inner product in the orthogonal direct sum
Hd1⊕Hd2 given by ⟨(f1,g1),(f2,g2)⟩:=⟨f1,f2⟩+⟨g1,g2⟩ for each (f1,g1),(f2,g2)∈Hd1⊕Hd2. The
following results give ways to obtain a BUNTF combining two or more
BUNTFs.
First we obtain BUNTFs as a disjoint union of BUNTFs:
Proposition 8.1**.**
Let F=(fk)k=1K be a sequence in
Hd1 and G=(gl)l=1L be a
sequence in Hd2. Then the disjoint union
F∪˙G:=((fk,0)k=1K,(0,gl)l=1L) is a BUNTF for Hd1⊕Hd2 if and only F is a BUNTF for
Hd1, G is a BUNTF for
Hd2 and d1K=d2L.
Proof.
Noting that TF∪˙G=TF⊕TG and
SF∪˙G=SF⊕SG, F∪˙G is a BUNTF
for Hd1⊕Hd2 if and only
F is a BUNTF for Hd1, G is
a BUNTF for Hd2, and
d1+d2K+L=d1K=d2L. This last
condition is equivalent to d1K=d2L.
∎
Note that in view of Theorem 4.12,
each BUNTF is the disjoint union of BUNTFs for orthogonal subspaces,
given by the vertices of each connected component of the frame
graph. This decomposition is unique.
Now we construct BUNTFs as the inner direct sum of BUNTFs:
Proposition 8.2**.**
Let α,β∈F∖{0}. Let
F=(fk)k=1K be a sequence in
Hd1 and G=(gk)k=1K be a
sequence in Hd2 be UN. Then the inner direct sum
αF⊕βG:=((αfk,βgk))k=1K is a BUNTF for Hd1⊕Hd2 if and only if F is a BUNTF for
Hd1, G is a BUNTF for
Hd2, TFTG∗=0,
∣α∣2=d1+d2d1 and
∣β∣2=d1+d2d2.
Proof.
We have TαF⊕βG(c)=(αTF(c),βTG(c)) for all c∈FK
and
for all f∈Hd1,g∈Hd2. Therefore
αF⊕βG is a BUNTF if and only if
F is a BUNTF for Hd1, G is
a BUNTF for Hd2,
TFTG∗=0 and
∣α∣2d1K=∣β∣2d2K=d1+d2K.
∎
See [34, Lemma 5.1] for equivalent conditions to
TFTG∗=0. In particular, this
condition implies that K≥d1+d2.
Another way to construct BUNTFs is to take the sum of
BUNTFs in the following sense:
Proposition 8.3**.**
Let α,β∈F∖{0}. Let
F=(fk)k=1K be a UN sequence in
Hd1 and G=(gl)l=1L be a UN
sequence in Hd2. Then the sum αF+βG:=((αfk,βgl))k,l=1K,L is a BUNTF for Hd1⊕Hd2 if and only F is a BUNTF for
Hd1, G is a BUNTF for
Hd2, ∣α∣2=d1+d2d1 and
∣β∣2=d1+d2d2.
Proof.
For each l=1,…,L, set
Hl:=(hl,k)k=1K where hl,k=gl for
each k=1,…,K. Let E:FL→FK given by E(c)=(∑l=1Lcl,…,∑l=1Lcl). The synthesis operator is given by
where f∈Hd1 and g∈Hd2, respectively. It results that
αF+βG is a BUNTF for
Hd1⊕Hd2 if and only if
F is a BUNTF for Hd1, G is
a BUNTF for Hd2,
L∣α∣2d1K=K∣β∣2d2L=d1+d2KL,
or equivalently, ∣α∣2=d1+d2d1 and
∣β∣2=d1+d2d2.
∎
In the tensor product Hd1⊗Hd2 we have the inner product given by ⟨f1⊗g1,f2⊗g2⟩:=⟨f1,f2⟩⟨g1,g2⟩ for each f1⊗g1,f2⊗g2∈Hd1⊗Hd2. Here we build BUNTFs as a tensor
product of BUNTFs.
Proposition 8.4**.**
Let F=(fk)k=1K be a sequence in
Hd1 and G=(gl)l=1L be a
sequence in Hd2. Then the tensor product
F⊗G:=(fj⊗gk)k,l=1K,L is a BUNTF for Hd1⊗Hd2 if and only F is a TF for
Hd1, G is a TF for
Hd2, F or G is balanced,
and ∣∣fj∣∣∣∣gk∣∣=1 for all k=1,…,K, l=1,…,L.
Proof.
We have TF⊗G=TF⊗TG and SF⊗G=SF⊗SG. By
[34, Corollary 5.1], F⊗G is a UNTF for Hd1⊗Hd2 if and only F is a TF for
Hd1, G is a TF for Hd2
and ∣∣fj∣∣∣∣gk∣∣=1 for all k=1,…,K, l=1,…,L.
Let (em)m=1d1 be an orthonormal basis for
Hd1 and (en)n=1d2 be an orthonormal
basis for Hd2. Since ⟨TF⊗TG(e),em⊗en⟩=⟨TF(e),em⟩⟨TG(e),en⟩ for each m=1,…,d1 and n=1,…,d2,
F⊗G is balanced if and only if
F or G is balanced.
∎
8.1. Other constructions
For sequences F=(fk)k=1K,G=(gl)l=1L in Hd we consider the
unionF∪G:=((fk)k=1K,(gl)l=1L). In this
subsection we introduce other techniques for constructing BUNTFs
that combine unions and direct sums. Among them, there are methods
that can be applied to obtain the five platonic solids in
R3.
The next theorem generalizes the method in [31] for obtaining the vertices of the tetrahedron and of the
dodecahedron in R3 starting from the third roots of
the unity and from the fifth roots of the unity in a plane,
respectively.
Theorem 8.5**.**
Let α,β∈F∖{0}. Assume that
F=(fk)k=1K is a BUNTF for Hd1,
G=(gk)k=1K where gk=h with h∈Hd2 and ∣∣h∣∣=1 for each k=1,…,K. Let
y∈Hd2 with ∣∣y∣∣=1. Then (αF⊕βG)∪(0,y) is a BUNTF for
Hd1⊕Hd2 if and only if
d2=1, y=−h, K=d1+1, ∣α∣2=1−K21
and ∣β∣2=K21.
Proof.
The sequence (αF⊕βG)∪(0,y) is balanced if and only if y=−Kβh. Since
∣∣h∣∣=∣∣y∣∣=1, β=K1. Consequently, y=−h.
Taking into account that F is balanced,
TFTG∗g=∑k=1K⟨g,gk⟩fk=⟨g,h⟩∑k=1Kfk=0
for each g∈Hd2. So, (αF⊕βG)∪(0,y) is a TF for Hd1⊕Hd2 if and only if ∣α∣2d1K=d1+d2K+1 and (K∣β∣2+1)⟨g,h⟩h=d1+d2K+1g for each g∈Hd2. The last condition implies that [math] is the only
element orthogonal to h, therefore d2=1. Consequently,
(αF⊕βG)∪(0,y) is a TF for
Hd1⊕Hd2 if and only if
∣α∣2=K(d1+1)d1(K+1) and (K∣β∣2+1)=d1+1K+1, i.e.,
∣β∣2=K(d1+1)K−d1. The two expressions for
∣β∣2 must be the same, i.e.,
K(d1+1)K−d1=K21, and this is equivalent
to K=1 or K=d1+1. The first case cannot happen because
F is balanced. So, K=d1+1.
Since F is UN, ∣∣h∣∣=∣∣y∣∣=1 and
∣α∣2+∣β∣2=1, we have that (αF⊕βG)∪(0,y) is UN.
∎
The proofs of the following results use arguments similar to the
previous ones, so we omit them.
The vertices of the octahedron in R3 form a BUNTF that
can be obtained adding orthogonally two antipodal points to the
BUNTF consisting of the 4th-roots of unity in a plane (see
[31]). The next theorem generalizes this
construction to an arbitrary direct sum of two Hilbert spaces. Let a
BUNTF for Hd1 be immersed in a direct sum
Hd1⊕Hd2, and add to it one
unit-norm vector of Hd1⊕Hd2
and its opposite. We show that the resulting set is a BUNTF for
Hd1⊕Hd2 if and only if
Hd2 is 1-dimensional, and the added vector is
orthogonal to the elements of the given frame in Hd1⊕Hd2.
Theorem 8.6**.**
Let F=(fk)k=1K be a BUNTF for
Hd1, x∈Hd1 and y∈Hd2, y=0. Then ((fk,0))k=1K∪(x,y)∪(−x,−y) is a BUNTF for Hd1⊕Hd2 if and only if x=0, d2=1, ∣∣y∣∣=1 and
K=2d1.
The procedure of the following theorem can be thought as a kind of
symmetric simple lift (see [34, Definition 5.2] for
the notion of lift and simple lift). It also can
be seen as a generalization of the procedure used in
[31] for obtaining the vertices of the
hexahedron and of the dodecahedron in R3 starting from
the BUNTFs in a plane consisting of the fourth roots of the unity
and of the fifth roots of the unity, respectively.
Theorem 8.7**.**
Let α∈F∖{0}. Let
F=(fk)k=1K,G=(gk)k=1K be
BUNTFs for Hd and
Hk=(βkhl)k=1K where βk∈F for each k=1,…,K, hl∈H1,l, ∣∣hl∣∣=1 and dim(H1,l)=1 for all l=1,…,r. Then (αF⊕H1⊕…⊕Hr)∪(αG⊕(−H1)⊕…⊕(−Hr))
is a BUNTF for Hd⊕H1,1⊕…⊕H1,r if and only if r=1,
∑k=1Kβkfk=∑k=1Kβkgk,
∣α∣2=d+1d and ∣βk∣2=d+11 for
each k=1,…,K.
Note that one choice for βk in Theorem 8.7 is βk=d+11 for each k=1,…,K.
The next method to construct UNTFs can be seen as a partial simple
lift.
Proposition 8.8**.**
Assume α,β∈F∖{0} and
Ld1>K. Let F=(fk)k=1K,G=(gl)l=1L be a UNTF and a BUNTF for
Hd1, respectively. Let
H=(hl)l=1L where hl=h∈Hd2 and ∣∣h∣∣=1 for each l=1,…,L. Then
((fk,0))k=1K∪(αG⊕βH) is a UNTF for Hd1⊕Hd2 if and only if d2=1,
∣α∣2=(d1+1)Ld1L−K and
∣β∣2=(d1+1)LK+L.
Now we consider a variation of the previous method for obtaining a
BUNTF. It can be seen as a symmetric partial simple lift.
Theorem 8.9**.**
Assume α,β∈F∖{0} and
K<2d1L. Let F=(fk)k=1K,
G=(gl)l=1L and
G=(gl)l=1L be BUNTFs
for Hd1. Let H=(hl)l=1L where
hl=h∈Hd2 and ∣∣h∣∣=1 for each l=1,…,L. Then ((fk,0))k=1K∪(αG⊕βH)∪(αG⊕(−βH)) is a BUNTF for Hd1⊕Hd2 if and only if
TGe=TGe, d2=1,
∣α∣2=2(d1+1)L2d1L−K and
∣β∣2=2(d1+1)L2L+K.
The following theorem generalizes [2, Theorem 3],
which is about t-designs in R3, for the case t=2.
Theorem 8.10**.**
Assume αm,βm∈F∖{0} such
that ∣αm∣2+∣βm∣2=1 for each m=1,…,M. Let Fm=(fm,k)k=1K be BUNTFs for
Hd1 for each m=1,…,M. Let
G=(gk)k=1K where gk=g∈Hd2 and ∣∣g∣∣=1 for each k=1,…,K. Then
⋃m=1M(αmFm⊕βmG) is a BUNTF for Hd1⊕Hd2 if and only if d2=1, ∑m=1Mβm=0 and ∑m=1M∣βm∣2=d1+1∣M∣.
Remark 8.11*.*
An example of scalars βm as in Theorem 8.10 is
βm=c(d1+1)∣M∣βm where
βm∈F, ∣M∣1≤∣βm∣2≤∣M∣d1+1 for each m=1,…,M, ∑m=1Mβm=0 and c=∑m=1M∣βm∣2. Another option is to consider
any row of TF where F is a BTF for
Fd1+1 with M elements.
Remark 8.12*.*
Observe that we can vary F, G,
H, etc., in all the above constructions obtaining in
this manner an infinite number of non unitary equivalent BUNTFs. We
can also combine these methods generating a great variety of them.
We note that there exist BUNTFs of K points for Rd
with K≥2 unless K≤d or K=d+2 and K is odd.
This is a consequence of Proposition 3.5
and results of [26].
Acknowledgement
This research has been supported by Grants PIP 112-201501-00589-CO
(CONICET), PROICO 03-1618 (UNSL), PICT-2014-1480 and UBACyT
20020130100422BA. We thank the anonymous referee for valuable
comments that helped to improve the paper.
Bibliography34
The reference list from the paper itself. Each links out to its DOI / PubMed record.
1[1] Alexeev B., Cahill J., Mixon D.G.: Full spark frames. J. Fourier Anal. Appl. 18, 1167-1194 (2012).
2[2] Bajnok B.: Construction of designs on the 2 2 2 -sphere. European J. Combin. 12, 377-382 (1991).
3[3] Bannai E., Bannai E.: A survey on spherical designs and algebraic combinatorics on spheres. European J. Combin. 30(6), 1392-1425 (2009).
4[4] Barg A. , Glazyrin A. , Okoudjou K., Yu W.-H.: Finite two-distance tight frames. Linear Algebra Appl. 475, 163-175 (2015).
6[6] Benedetto J., Yilmaz Ö., Powell A.: Sigma-delta quantization and finite frames. In: Proc. IEEE International Conference on Acoustics, Speech, and Signal Processing, vol. 3, ICASSP’04, IEEE, Philadelphia, PA, pp. 937-940 (2004).
7[7] Bodmann B.G., Paulsen V.I.: Frame paths and error bounds for sigma-delta quantization. Appl. Comp. Harm. Anal. 22, 176-197 (2007).
8[8] Bodmann B.G., Paulsen V.I.: Optimal frames for erasures. Linear Algebra Appl. 377, 31-51 (2004).