Norm retrieval and phase retrieval by projections

0.2 MB·

by  Peter G. Casazza

NORM RETRIEVAL AND PHASE RETRIEVAL BY
PROJECTIONS
7
1 PETER G. CASAZZA, DORSA GHOREISHI, SHANI JOSE, JANET C. TREMAIN
0
2
Abstract. We make a detailed study of norm retrieval. We give several
n
a classificationtheoremsfor normretrievalandgivealargenumber ofexam-
J ples to gowith the theory. One consequenceis anew resultaboutParseval

7 frames: If a Parseval frame is divided into two subsets with spans W1,W2
2 and W1 W2 = 0 , then W1 W2.
∩ { } ⊥

]
A
F
. 1. Introduction
h
t
a Signal reconstruction is an important problem in engineering and has a
m wide variety of applications. Recovering signals when there is partial loss
[ of information is a significant challenge. Partial loss of phase information

1 occurs in application areas such as speech recognition [4, 17, 18], and optics
v applications such as X-ray crystallography [3, 13, 14], and there is a need to
4
do phase retrieval efficiently. The concept of phase retrieval for Hilbert space
1
0 frames was introduced in 2006 by Balan, Casazza, and Edidin [2], and since
8
thenithasbecomeanactiveareaofresearch insignal processing andharmonic
0
. analysis.
1
0 Phase retrieval has been defined for vectors as well as for projections and in
7 general deals with recovering the phase of a signal given its intensity measure-
1
ments from a redundant linear system. Phase retrieval by projections, where
:
v
the signal is projected onto some higher dimensional subspaces and has to be
i
X recovered from the norms of the projections of the vectors onto the subspaces,
r appears in real life problems such as crystal twinning [12]. We refer the reader
a
to [8] for a detailed study of phase retrieval by projections.
Another related problem is that of phaseless reconstruction, where the un-
known signal is reconstructed from the intensity measurements. Recently, the
two terms phase retrieval and phaseless reconstruction were used interchange-
ably. However, it is not clear from their respective definitions how these two
are equivalent. Recently, in [5] the authors proved the equivalence of phase re-
trieval and phaseless reconstruction in real as well as in complex case. Due to
The first, second and fourth authors were supported by NSF DMS 1609760; NSF ATD
1321779;andARO W911NF-16-1-0008. Partofthis researchwascarriedoutwhile the first
and fourth authors were visiting the Hong Kong University of Science and Technologyon a
grantfrom(ICERM)InstituteforcomputationalandexperimentalresearchinMathematics.
1
2 CASAZZA,GHOREISHI,JOSE, TREMAIN
this equivalence, in this paper, we restrict ourselves to proving results regard-
ing phase retrieval. Further, a weaker notion of phase retrieval and phaseless
reconstruction was introduced in [6].
In this work, we consider the notion of norm retrieval which was recently
introduced by Bahmanpour et.al. in [1], and is the problem of retrieving
the norm of a vector given the absolute value of its intensity measurements.
Norm retrieval arises naturally from phase retrieval when one utilizes both
a collection of subspaces and their orthogonal complements. Here we study
norm retrieval and certain classifications of it. We use projections to do norm
retrieval and to extend certain results from [16] for frames. We provide a
complete classification of subspaces of RN which do norm retrieval. Various
examples for phase and norm retrieval by projections are given. Further, a
classification of norm retrieval using Naimark’s theorem is also obtained.
We organize the rest of the paper as follows. In Section 2, we include
basic definitions and results of phase retrieval. Section 3 introduces the norm
retrieval and properties. Section 4 provides the relationship between phase
and norm retrieval and related results. Detailed classifications of vectors and
subspaces which do norm retrieval are provided in Section 5.
2. Preliminaries
We denote by HN a N dimensional real or complex Hilbert space, and we
write RN or CN when it is necessary to differentiate between thetwo explicitly.
Below, we give the definition of a frame in HN.
Definition 2.1. A family of vectors Φ = φ M in HN is a frame if there
are constants 0 < A B < so that for a{ll ix}i=1HN,
≤ ∞ ∈
n
A x 2 x,φ 2 B x 2. (1)
i
k k ≤ |h i| ≤ k k
i=1
X
The following definitions and terms are useful in the sequel.
The constants A and B are called the lower and upper frame
•
bounds of the frame, respectively.
If A = B, the frame is called an A-tight frame (or a tight frame). In
•
particular, if A = B = 1, the frame is called a Parseval frame.
Φ is an equal norm frame if φ = φ for all i,j and is called a
i j
• k k k k
unit norm frame if φ = 1 for all i = 1,2, n.
i
k k ···
If, only the right hand side inequality holds in (1), the frame is called
•
a B-Bessel family with Bessel bound B.
Note that in a finite dimensional setting, a frame is a spanning set of vectors
intheHilbertspace. Werefer to[10]foranintroductiontoHilbertspaceframe
theory and applications.
NORM RETRIEVAL AND PHASE RETRIEVAL BY PROJECTIONS 3
Let Φ = φ N be a frame in HN. The analysis operator associated with
Φ is defined{ais}it=h1e operator T : HN ℓM to be
→ 2
M
Tx = x,φ e = x,φ M , for all x HN.
h ii i {h ii}i=1 ∈
i=1
X
Here, e M is understood to be the natural orthonormal basis for ℓM. The
{ i}i=1 2
adjoint T∗ of the analysis operator T is called the synthesis operator of the
frame Φ. It can be shown that T∗(e ) = φ .
i i
The frame operator for the frame Φ is defined as S : T∗T : HN HN.
→
That is,
M
Sx = T∗T(x) = x,φ φ for all x HN.
i i
h i ∈
i=1
X
Note that the frame operator S is a positive, self-adjoint and invertible
operator satisfying the operator inequality AI S BI, where A and B are
the frame bounds and I denotes the identity on≤HN≤. Frame operators play an
important role since they are used to reconstruct the vectors in the space. To
be precise, any x HN can be written as
∈
M M
x = SS−1x = S−1Sx = S−1x,φ φ = x,S−1φ φ . (2)
i i i i
i=1 i=1
X(cid:10) (cid:11) X(cid:10) (cid:11)
The frame operator of a Parseval frame is the identity operator. Thus, if
φ M is a Parseval frame, it follows from equation (2) that
{ i}i=1
M
x = x,φ φ , x HN.
i i
h i ∈
i=1
X
We concentrate on norm retrieval and its classifications in this paper. We
now see the basic definitions of phase retrieval formally, starting with phase
retrieval by projections. Throughout the paper, the term projection is used
to describe orthogonal projection (orthogonal idempotent operator) onto sub-
spaces.
Definition 2.2. Let W M be a collection of subspaces in HN and let P M
{ i}i=1 { i}i=1
be the projections onto each of these subspaces. We say that W M (or
P M ) yields phase retrieval if for all x,y HN satisfying P{x i}=i=1P y
{ i}i=1 ∈ k i k k i k
for all i = 1,2, ,M then x = cy for some scalar c such that c = 1
··· | |
Phase retrieval by vectors is a particular case of the above.
Definition 2.3. Let Φ = φ M HN be such that for x,y HN
{ i}i=1 ∈ ∈
x,φ = y,φ , for all i = 1,2, ,M.
i i
|h i| |h i| ···
4 CASAZZA,GHOREISHI,JOSE, TREMAIN
Φ yields phase retrieval with respect to an orthonormal basis e N if there
{ i}i=1
is a θ = 1 such that x = cy , for all i = 1,2, ,N, where x = x,e .
i i i i
| | ··· h i
Orthonormal bases fail to do phase retrieval, since in any given orthonormal
basis, the corresponding coefficients of a vector are unique. One of the funda-
mental properties to identify the minimum number of vectors required to do
phase retrieval is the complement property.
Definition 2.4 ([2]). A frame Φ = φ M in HN satisfies the complement
{ i}i=1
property if for all subsets I 1,2, ,M , either φ or φ spans
i i∈I i i∈Ic
the whole space HN. ⊂ { ··· } { } { }
It is proved in [2] that phase retrieval is equivalent to the complement prop-
erty in RN. Further, it is proven that a generic family of (2N 1)-vectors
in RN does phase retrieval, however no set of (2N 2)-vectors−can. Here,
−
generic refers to an open dense set in the set of (2N 1)-element frames in
HN. Full spark is another important notion of vecto−rs in frame theory. A
formal definition is given below:
Definition 2.5. Given a family of vectors Φ = φ M in HN, the spark of
{ i}i=1
Φ is defined as the cardinality of the smallest linearly dependent subset of Φ.
When spark(Φ) = N + 1, every subset of size N is linearly independent, and
in that case, Φ is said to be full spark.
Note from the definitions that full spark frames with M 2N 1 have the
≥ −
complement property and hence do phase retrieval. Moreover, if M = 2N 1
−
then the complement property clearly implies full spark.
Next result, known as Naimark’s theorem, characterizes Parseval frames in
a finite dimensional Hilbert space. This theorem facilitates a way to construct
Parseval frames, and crucially it is the only way to obtain Parseval frames.
Later, we use this to obtain a classification of frames which do norm retrieval.
The notation [M] = 1,2, ,M is used throughout the paper.
{ ··· }
Theorem 2.6 (Naimark’s Theorem). [9] A frame φ M is a Parseval frame
for RN if and only if RN ℓM with orthonorma{l bia}si=is1 e M so that the
orthogonal projection P ont⊂o R2N satisfies: Pe = φ for eve{ryi}ii=1 [M].
i i
∈
3. Beginnings of Norm Retrieval
In this section, we provide the definition of norm retrieval along with certain
related results, and pertinent examples.
Definition 3.1. Let W M be a collection of subspaces in HN and let P M
{ i}i=1 { i}i=1
bethe orthogonalprojectionsonto each of these subspaces. We saythat W M
(or P M ) yields norm retrieval if for all x,y HN satisfying {P xi}i==1
{ i}i=1 ∈ k i k
P y for all i = 1,2, ,M then x = y .
i
k k ··· k k k k
NORM RETRIEVAL AND PHASE RETRIEVAL BY PROJECTIONS 5
In particular, a set of vectors φ M in HN does norm retrieval, if for
x,y HN satisfying x,φ = y,{φ i}if=o1r all i = 1,2, ,M then x = y .
i i
∈ |h i| |h i| ··· k k k k
Remark 3.2. It is immediate that a family of vectors doing phase retrieval
does norm retrieval.
An obvious choice of vectors which do norm retrieval are orthonormal bases.
For, let e N be an orthonormal basis in HN. Now, for x HN, x,φ =
{ i}i=1 ∈ |h ii|
x,e = x . Thus
i i
|h i| | |
N N
x,φ 2 = x 2 = x 2.
i i
|h i| | | k k
i=1 i=1
X X
The following theorem provides a sufficient condition under which the sub-
spaces spanned by the canonical basis vectors do norm retrieval.
Theorem 3.3. Let e N be an orthonormal basis in HN. Let W k be
{ i}i=1 { j}j=1
subspaces of HN where each W = span e , I [N]. If there exists m
j { i}i∈Ij j ⊆
such that for all j, j : e W = m, then W k does norm retrieval.
|{ i ∈ j}| { j}j=1
Proof. Let P be orthogonal projections onto W , for all j. Now, by assump-
j j
tion, we have
k k N
P x 2 = x,e 2 = m x,e 2 = m x 2.
j i j
k k |h i| |h i| k k
Xj=1 Xj=1 Xi∈Ij Xj=1
(cid:3)
It is easy to see that tight frames do norm retrieval.
Theorem 3.4. Tight frames do norm retrieval.
Proof. et φ M in HN be an A-tight frame. Now, if
{ i}i=1
x,φ = y,φ , for all i = 1,2, ,M,
i i
|h i| h i| ···
then
M M
A x 2 = x,φ 2 = y,φ 2 = A y 2.
i i
k k |h i| |h i| k k
i=1 i=1
X X (cid:3)
Observe that if φ M HN does norm retrieval so does φ M ψ K
{ i}i=1 ∈ { i}i=1∪{ j}j=1
for any ψ HN. This is generalized in the following proposition.
j
∈
Proposition 3.5. If P M does norm retrieval, then so does P M
{ i}i=1 { i}i=1 ∪
Q K for any projections Q . In particular, if a frame Φ = φ M con-
{ i}i=1 i { i}i=1
tains an orthonormal basis, then it does norm retrieval. Moreover, in this
case, φ⊥ M does norm retrieval.
{ i }i=1
6 CASAZZA,GHOREISHI,JOSE, TREMAIN
Proof. Let e N beanorthonormalbasisforHN andletP betheprojections
{ i}i=1 i
onto φ ⊥, for each i. Given x HN, we have
i
∈
N N N
P x 2 = x,e 2 = (N 1) x,e 2 = (N 1) x 2.
i j j
k k |h i| − |h i| − k k
i=1 i=1 j6=i i=1
X XX X
(cid:3)
The above proposition does not hold if the number of hyperplanes is strictly
less than N. This is proved in the next theorem.
Theorem 3.6. If ϕ N is an orthonormal basis for RN then W where
{ i}i=1 { i}i∈I
W = ϕ⊥ cannot do norm retrieval for I [N 1].
i i ⊆ −
Proof. Without loss of generality consider the collection W N−1 (for N > 2).
{ i}i=1
Now, let x = N ϕ and y = N−1 N−1ϕ so that P x 2 = ϕ 2 =
i=1 i N−2 i=1 i k j k i6=j| i|
N 1 and P y 2 = N−1 Nq−1 ϕ 2. Thus, P x 2 = P y 2. However
− kPj k N−2 i=1 | i|P k j k k j kP
i6=j
x 2 = N and y 2 = (N−1)2Pwhich proves the theorem.
k k k k N−2
(cid:3)
Now, we strengthen the above result by not requiring the vectors to be
orthogonal. To prove this, we need the following lemma.
Lemma 3.7. If φ N are independent vectors in RN, then there is a vector
φ RN satisfying{: i}i=1
∈
φ,φ = c = 0, for all i = 1,2, ,N.
i
|h i| 6 ···
Proof. We do this by induction on N with the case N = 2 obvious. So assume
this holds for N 1. Given φ N , we can find a φ span φ N−1 with
− { i}i=1 ∈ { i}i=1
φ = 1 and satisfying
k k
φ,φ = c = 0, for all i = 1,2, ,N 1.
i
|h i| 6 ··· −
Choose ψ span φ N−1 and note that linear independence of the φ
⊥ { i}i=1 i
implies
ψ,φ = 0.
N
|h i| 6
Consider φ+λψ. For i = 1,2, ,N 1,
··· −
φ+λψ,φ = φ,φ +λ ψ,φ
i i i
|h i| |h i h i|
= ψ,φ
i
|h i|
= c
Also,
φ+λψ,φ = φ,φ +λ ψ,φ .
N N N
h i h i h i
NORM RETRIEVAL AND PHASE RETRIEVAL BY PROJECTIONS 7
As λ varies from to + , the right hand side varies from to +
−∞ ∞ −∞ ∞
and for some λ, we have
φ,φ +λ ψ,φ = c.
N N
|h i h i|
(cid:3)
Proposition 3.8. If φ , ,φ RN are independent and unit norm
1 N−1
{ ··· } ∈
and W = φ⊥ , for all i = 1,2, ,N 1, then W N−1 cannot do norm
{ i} { i } ··· − { i}i=1
retrieval.
Proof. Let P be the projection onto W and choose
i i
x N−1W , with x = 1.
∈ ∩i=1 i k k
By the assumption, there is a vector φ span φ N−1, with φ = 1,
∈ { i}i=1 k k
and φ,φ = c = 0, for all i = 1, ,N 1. Let y = λx + µφ, where
i
|h i| 6 ··· −
λ2 + (1 c2)µ2 = 1. Note that x φ for all i implies that φ x, and so
i
− ⊥ ⊥
y 2 = λ2 +µ2 = 1.
k k 6
Now, for all i = 1,2, ,N 1,
··· −
P y 2 = y 2 y,a 2
i i
k k k k −|h i|
= λ2 +µ2 µ2c2
−
= λ2 +(1 c2)µ2
−
= 1
= x 2
k k
= P x 2.
i
k k
But x 2 = 1 while y 2 = 1, and so norm retrieval fails. (cid:3)
k k k k 6
However, in the following theorem, we show that three proper subspaces of
codimension one can do norm retrieval in RN.
Theorem 3.9. In RN three proper subspaces of codimension one can do norm
retrieval.
Proof. Let e N be an orthonormal basis for RN. Let
{ i}i=1
φ = e φ = e φ = (e e )/√2
1 1 2 2 3 1 2
−
We claim φ⊥ 3 does norm retrieval. Let P be the orthogonal projection
{ i }i=1 i
onto φ⊥. Let x = (a , ,a ). We then have that
i 1 ··· N
N N
P x 2 = a2 + a2, P x 2 = a2 + a2
|| 1 || 2 k || 2 || 1 k
k=3 k=3
X X
a +a 2 N a2 +2a a +a2 N
P x 2 = 1 2 + a2 = 1 1 2 2 + a2
|| 3 || √2 k 2 k
(cid:18) (cid:19) k=3 k=3
X X
8 CASAZZA,GHOREISHI,JOSE, TREMAIN
Case 1: If a = 0 or a = 0, we know that x 2 = P x 2 or x 2 = P x 2
1 2 1 2
|| || || || || || || ||
respectively.
Case 2: Assume both a = 0 and a = 0. We then know both of the
1 2
6 6
equalities below:
(a +a )2 1 N
1 2 P x 2 + P x 2 = c a2
− 2 · a2|| 1 || || 3 || k
2 k=3
X
(a +a )2 1 N
1 2 P x 2 + P x 2 = d a2
− 2 · a2|| 2 || || 3 || k
1 k=3
X
where
(a +a )2 (a +a )2
1 2 1 2
c = +1 and d = +1
− 2a2 − 2a2
2 1
If either c or d is nonzero, then the proof is complete as in that case, we can
express x 2 as a linear combination of P x 2, P x 2, and P x 2.
1 2 3
|| || || || || || || ||
Now, suppose that c = d = 0. If c = 0, then (a +a )2 = 2a2 and if d = 0,
1 2 2
then (a +a )2 = 2a2. This implies that
1 2 1
2(a +a )2 = 2a2 +2a2
1 2 1 2
which holds only if either a or a or both is zero which contradicts our as-
1 2
(cid:3)
sumption.
It follows that in R3, two 2-dimensional subspaces cannot do norm retrieval
but three 2-dimensional subspaces can do norm retrieval.
Proposition 3.10. For every K N, there exist subspaces W K+1 of HN
≤ { i}i=1
which do norm retrieval and W⊥ K+1 span a K dimensional space.
{ i }i=1
Proof. ChooseanorthonormalbasisofHN,say e N . LetW = span e N−K
{ i}i=1 1 { i}i=1
N
and W = span W ,e for all 2 i K +1. If x = a e , then
i { 1 N−K+i−1} ≤ ≤ j=1 j j
P x 2 = N−K a 2 and P x 2 = N−K a 2 + a 2 for 2 i
k 1 k j=1 | j| k i k j=1 | j| | N−K+i−P1| ≤ ≤
K+1. Therefore x 2 = K+1 P x 2 (K 1) P x 2. Since W⊥ W⊥ for
P k k i=2 k i kP− − k 1 k i ⊆ 1
all i, it’sclear that W⊥ K+1 isspanned by e N , which has dimension
K . { i }iP=1 { i}i=M−K+1 (cid:3)
The following proposition shows a relationship between subspaces doing
norm retrieval and the sum of the dimensions of the subspaces. The impor-
tance of this proposition is that we are looking for conditions on subspaces to
do norm retrieval. To do so, the dimension of the subspaces is one of the tools
we have.
Proposition 3.11. If W M in RN does norm retrieval then M dimW
{ i}i=1 i=1 i ≥
N. Moreover, if M k = LN then there exist W M doing norm retrieval
i=1 i { i}i=1 P
where dimW = k for each 1 i M.
i i
P ≤ ≤
NORM RETRIEVAL AND PHASE RETRIEVAL BY PROJECTIONS 9
M
Proof. If dimW < N then we may pick non-zero x W for each i so
i=1 i ⊥ i
that P x = 0 for all i and therefore W M fails norm retrieval.
k i Pk { i}i=1
For the moreover part, let g N be an orthonormal basis. We represent
{ i}i=1
this basis L-times as a multiset:
φ LN =: g , ,g ,g , ,g , ,g , ,g ,
{ i}i=1 { 1 ··· N 1 ··· N ··· 1 ··· N}
and index it as: e LN. We may pick a partition of [LN] in the following
{ i}i=1
manner:
I = 1,2, ,k , I = k +1, ,k +k , I = k +k +1, ,k +k +k , .
1 1 2 1 1 2 3 1 2 1 2 3
{ ··· } { ··· } { ··· } ···
N
Now define W = span e with projection P . Then if x = a e then
i { j}j∈Ii i j=1 j j
M M N P
P x 2 = a 2 = L a 2 = L x 2.
i j j
k k | | | | k k
Xi=1 Xi=1 Xj∈Ii Xj=1
(cid:3)
Hence the result.
M
As we have seen, the above proposition may fail if k = LN.
i=1 i 6
P
4. Phase retrieval and Norm Retrieval
Inthissection, weprovideresultsrelatingphaseretrievalandnormretrieval.
The following theorem of Edidin [11] is significant in phase retrieval as it gives
a necessary and sufficient condition for subspaces to do phase retrieval.
Theorem 4.1 ( [11]). A family of projections P M in RN does phase re-
trieval if and only if for every 0 = x RN, t{hei}vie=c1tors P x M span the
6 ∈ { i }i=1
space.
Corollary 4.2. Let W M be a collectionof subspacesof RN with P denoting
{ i}i=1 i
the projection onto W for each 1 i M. If W M does phase retrieval in
RN then for every I i [M] with ≤I ≤N 2,{thei}ci=ol1lection W⊥ spans
RN. ⊂ | | ≤ − { i }i∈Ic
Proof. If not, pick non-zero x W⊥ for all i Ic. This implies x W
⊥ i ∈ ∈ ∩i∈Ic i
and therefore P (x) N contains at most N 1 distinct vectors and can not
span RN. This{cointr}aid=i1cts the theorem 4.1. − (cid:3)
Corollary 4.3. If W M in HN does phase retrieval, then W⊥ M spans
{ i}i=1 { i }i=1
the space.
Proof. If (W⊥) does not span, then there exists 0 = x W . So P x = x
i 6 ∈ ∩ i i
for all i = 1,2, ,M, and so P (x) does not span. Thus, by Theorem 4.1,
i
(W ) does not d·o··phase retriev{al. } (cid:3)
i
10 CASAZZA,GHOREISHI,JOSE, TREMAIN
The following example shows that it is possible for subspaces to do norm
retrieval even if W⊥ do not span the space which we see as one of main
{ i }
differences between phase retrieval and norm retrieval.
Example 4.4. Let e 3 be a orthonormal basis for R3, then let
{ i}i=1
W = span e ,e W ⊥ = span e
1 1 2 1 3
{ } { }
W = span e ,e W ⊥ = span e
2 2 3 2 1
{ } { }
W = span e W ⊥ = span e ,e
3 2 3 1 3
{ } { }
Then, W 3 does norm retrieval since x 2 = P x 2+ P x 2 P x 2.
But W {⊥, ii}i==11,2,3 do not span R3. k k k 1 k k 2 k −k 3 k
i
No{te th}at if W = HN, then W itself doesnorm retrieval whileW⊥ = 0 .
1 { 1} 1 { }
Any collection of subspaces which does phase retrieval yields norm retrieval,
which follows from the definitions. However, the converse need not hold true
always. For instance, any orthonormal basis does norm retrieval in RN. But it
has too few vectors to do phase retrieval as it requires at least 2N 1 vectors
to do phase retrieval in RN. −
Given subspaces W M of HN which yield phase retrieval, it is not neces-
{ i}i=1
sarily true that W⊥ M do phase retrieval. The following result proves that
{ i }i=1
norm retrieval is the condition needed to pass phase retrieval to orthogonal
complements. Though the result is already proved in [1], we include it here
for completeness.
Lemma 4.5. Suppose subspaces W M , with respective projections P M ,
{ i}i=1 { i}i=1
doesphaseretrieval. Then W⊥ M doesphaseretrievalif and onlyif W⊥ M
{ i }i=1 { i }i=1
does norm retrieval.
Proof. Assume that (I P )x = (I P )y for all i = 1,2, ,M and P M
k − i k − i ··· { i}i=1
does norm retrieval. I.e. x = y . Then
k k k k
(I P )(x) 2 = x 2 P x 2 = y 2 P y 2 = (I P )(y) 2.
i i i i
k − k k k −k k k k −k k k − k
Since x = y , we have
k k k k
P x = P y for all i = 1,2, ,M.
i i
k k k k ···
Since P M does phase retrieval, it follows that x = cy for some c = 1.
{ i}i=1 | | (cid:3)
The other direction of the theorem is clear.
Next is an example of a family of subspaces W M which does phase
{ i}i=1
retrieval but complements fail phase retrieval and hence fail norm retrieval [8].