- Email: [email protected]
Phase retrieval versus phaseless reconstruction
J. Math. Anal. Appl. 436 (2016) 131–137
Contents lists available at ScienceDirect
Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Phase retrieval versus phaseless reconstruction ✩ Sara Botelho-Andrade ∗ , Peter G. Casazza, Hanh Van Nguyen, Janet C. Tremain Department of Mathematics, University of Missouri, Columbia, MO 65211-4100, United States
a r t i c l e
i n f o
Article history: Received 27 June 2015 Available online 2 December 2015 Submitted by E. Saksman Keywords: Phase retrieval Phaseless reconstruction Complement property Projections Weak phase retrieval Frame theory
a b s t r a c t In 2006, Balan, Casazza, and Edidin [1] introduced the frame theoretic study of phaseless reconstruction. Since then, this has turned into a very active area of research. Over the years, many people have replaced the term phaseless reconstruction with phase retrieval. At a meeting in 2012, Casazza asked: Are these really the same? In this paper, we will show that phase retrieval is equivalent to phaseless reconstruction. We then show, more generally, that phase retrieval by projections is equivalent to phaseless reconstruction by projections in the real case. © 2015 Elsevier Inc. All rights reserved.
1. Introduction Throughout, Rm (resp. Cm ) will denote an m-dimensional real (resp. complex) Hilbert space, and Hm will denote an m-dimensional real or complex Hilbert space. Phase retrieval is an old problem in signal processing and has been studied for over 100 years by electrical engineers. We begin with a few necessary definitions and a brief history the problem before proceeding to the main result. Let x = (a1 , a2 , . . . , am ) and y = (b1 , b2 , . . . , bm ) be vectors in Hm . We say that x, y have the same phase if phase ai = phase bi , for all i = 1, 2, . . . , m. Definition 1.1. Given Φ = {φi }ni=1 , a family of vectors in Hm , let x and y be vectors in Hm such that |x, φi | = |y, φi | , for all i = 1, 2, . . . , n.
(1)
i. We say Φ does phase retrieval if (1) implies there exists a |θ| = 1 so that x and θy have the same phases. ii. We say Φ does phaseless reconstruction if (1) implies there is a |θ| = 1 so that x = θy. ✩
The authors were supported by NSF DMS 1307685; and NSF ATD 1321779; AFOSR: FA9550-11-1-0245.
* Corresponding author. E-mail addresses: [email protected] (S. Botelho-Andrade), [email protected] (P.G. Casazza), [email protected] (H. Van Nguyen), [email protected] (J.C. Tremain). http://dx.doi.org/10.1016/j.jmaa.2015.11.045 0022-247X/© 2015 Elsevier Inc. All rights reserved.
132
S. Botelho-Andrade et al. / J. Math. Anal. Appl. 436 (2016) 131–137
Moreover, in the real case, if θ = 1 we say x and y have the same signs and if θ = −1 we say x and y have opposite signs. Note that from the definition of phase retrieval it is not immediately apparent that x = θy. The analogous definition for phase retrieval by projections follows. Definition 1.2. Given {Pi }ni=1 , a family of projections on Hm , let x and y be vectors in Hm such that Pi x = Pi y, for all i = 1, 2, . . . , n.
(2)
i. We say {Pi }ni=1 does phase retrieval if (2) implies there is a |θ| = 1 so that x and θy have the same phases. ii. We say {Pi }ni=1 does phaseless reconstruction if (2) implies there is a |θ| = 1 so that x = θy. In the setting of frame theory, the concept of phaseless reconstruction was introduced in 2006 by Balan, Casazza, and Edidin [1]. They showed that a generic family of (2m − 1)-vectors in Rm does phaseless reconstruction, however no set of (2m − 2)-vectors can. By generic we are referring to an open dense set in the set of (2m − 1)-element frames in Hm . In the complex case, they showed that a generic set of (4m − 2)-vectors does phaseless reconstruction. Heinosaari, Mazzarella and Wolf improved this result in [7], where they showed that n-vectors doing phaseless reconstruction in Cm requires n ≥ 4m −4 −2α, where α is the number of 1’s in the binary expansion of (m − 1). This bound was further improved when Bodmann [3] showed that phaseless reconstruction in Cm can be done with (4m −4)-vectors. Later, Conca, Edidin, Hering, and Vinzant [5] showed that a generic frame with (4m − 4)-vectors does phaseless reconstruction in Cm . They also showed that if m = 2k + 1 then no n-vectors with n < 4m − 4 can do phaseless reconstruction. It was conjectured that no fewer than (4m −4)-vectors can do phaseless reconstruction. Recently, Vinzant [8] showed that this conjecture does not hold by producing 11 vectors in C4 which do phaseless reconstruction. The pursuit of improving this bound was finally resolved when Edidin [6] showed that a generic family of 2m − 1 subspaces will do phaseless in Rm and if m = 2k + 1, then this bound is sharp. The analogous question for subspaces has still not been answered, however it was shown that there are six 2-dimensional subspaces of R4 which do phase retrieval (Xu [9]). Over the years, we have replaced the phrase “phaseless reconstruction” with “phase retrieval.” Casazza, at a meeting in 2012, raised the question: “Are these really the same?” In this paper we will answer in the affirmative, by showing that in both the real and complex cases phase retrieval implies phaseless reconstruction (and similarly for phase retrieval by projections in Rm ). This problem arose because of the way the engineering version of phase retrieval was translated into the language of frame theory. The engineers are working with the modulus of the Fourier transform and want to recover the phases to be able to invert the Fourier transform and recover the signal. In other words, the engineers only need is to recover the phase. But in the frame theory version of this, for x = (a1 , a2 , . . . , am ), we are really trying to recover two things: (1) Recover the phases of the ai . (2) Recover |ai | (which in the engineering case, is already known). 2. Phase retrieval versus phaseless reconstruction: real case We begin by defining the complement property from [1]. Definition 2.1. A family of vectors {φi }ni=1 in Hm has the complement property if for every I ⊂ [n], either {φi }i∈I spans Hm or {φi }i∈I c spans Hm .
S. Botelho-Andrade et al. / J. Math. Anal. Appl. 436 (2016) 131–137
133
We will prove that phase retrieval implies the complement property for both the real and complex cases and even in a more general setting. First, we need to make a couple observations: Observation 2.2. If {φi }ni=1 does phase retrieval in Hm , then span {φi }ni=1 = Hm . For otherwise, there would exist 0 = x ∈ Hm so that x, φi = 0, φi = 0, for all i = 1, 2, . . . , n, while x and 0 do not have the same phase. Observation 2.3. If x = (a1 , a2 , . . . , am ) and y = (b1 , b2 , . . . , bm ) have the same phases, then ai = 0 if and only if bi = 0. I.e. zero has no phase. The following theorem (from [1]) was proved for the real case and it was stated that the same proof holds in the complex case. In [2] it is mentioned that this argument does not hold in the complex case. We will now show that the proof in [1] holds in the complex case. Theorem 2.4. (See [1].) Let Φ = {φi }ni=1 be vectors in Hm . If Φ = {φi }ni=1 does phaseless reconstruction, then it has complement property. Proof. Assume Φ does not have the complement property but does phaseless reconstruction. Choose I ⊂ [n] so that neither of the sets of vectors {φi }i∈I or {φi }i∈I c spans Hm . Choose x = 1 = y so that x ⊥ φi for i ∈ I and y ⊥ φi for i ∈ I c . Then |x + y, φi | = |x − y, φi |, for all i = 1, 2, . . . , n. Since Φ does phaseless reconstruction, there is a |θ| = 1 so that x + y = θ(x − y), and hence (1 − θ)x = −(1 + θ)y. If θ = 1, then y = 0 and if θ = −1 then x = 0, contradicting the fact that x, y are unit norm. Otherwise, x=
−(1 + θ) y = dy, for d = 0. 1−θ
Now, x, φi = dy, φi = 0, for all i = 1, 2, . . . , n, and hence Φ does not span Hm contradicting Observation 2.2. 2 Remark 2.5. Moreover, in the real case phaseless reconstruction is equivalent to the complement property (as shown in Corollary 2.7). Theorem 2.6. If {Pi }ni=1 are projections onto the subspaces {Wi }ni=1 of Hm which do phase retrieval, then n, Di i for every orthonormal basis {φi,j }D j=1 of Wi , the set {φi,j }i=1,j=1 has complement property. i Proof. Suppose {Wi }ni=1 does phase retrieval for Hm , but there exist an orthonormal basis {φi,j }D j=1 of each n, Di Wi such that the set {φi,j }i=1,j=1 fails the complement property. In other words, there exists I ⊂ {(i, j) : 1 ≤ i ≤ n and 1 ≤ j ≤ Di } so that {φi,j }(i,j)∈I and {φi,j }(i,j)∈I c do not span Hm . Choose vectors x, y ∈ Hm
S. Botelho-Andrade et al. / J. Math. Anal. Appl. 436 (2016) 131–137
134
with x = 1 = y, where x = (a1 , a2 , . . . , am ) and y = (b1 , b2 , . . . , bm ) such that x ⊥ φi,j for all (i, j) ∈ I and y ⊥ φi,j for all (i, j) ∈ I c . Note this choice of vectors forces that for each (i, j) either x, φi,j = 0 or y, φi,j = 0. Fix c = 0. Then for each 1 ≤ i ≤ n, |x + cy, φi,j | = |x − cy, φi,j |, for all i, j. Hence, Pi (x + cy)2 =
Di
2
|x + cy, φi,j | =
j=1
Di
2
|x − cy, φi,j | = Pi (x − cy)2 .
j=1
By the assumption that {Pi }ni=1 does phase retrieval, this implies there is a |θ| = 1 so that x + cy and θ(x − cy) have the same phases. Assume there exists some 1 ≤ i0 ≤ m so that ai0 = 0 = bi0 and let −a c = bi i0 . Then 0
(x + cy)i0 = ai0 + cbi0 = ai0 +
−ai0 bi = 0, bi0 0
while (x − cy)i0 = ai0 −
−ai0 = 2ai0 = 0. bi0
But this contradicts Observation 2.3. It follows that for every 1 ≤ i ≤ m, either ai = 0 or bi = 0. Let {ei }m i=1 be an orthonormal basis for Hm and let I = {1 ≤ i ≤ m : bi = 0}. Then x+y =
i∈I
ai ei +
bi ei , and x − y =
i∈I c
i∈I
ai ei +
(−bi )ei .
i∈I c
By the above argument, there is a |θ| = 1 so that x + y and θ(x − y) have this same phase. But this is clearly impossible. This final contradiction completes the proof. 2 We have a number of consequences of Theorem 2.6. Letting the subspaces Wi be one dimensional, this becomes a theorem about vectors. Corollary 2.7. If Φ = {φi }ni=1 does phase retrieval in Hm , then Φ has the complement property. Hence, in the real case, phase retrieval and phaseless reconstruction for vectors are equivalent properties. We need the following result from [4]: Theorem 2.8. (See [4].) Let {Pi }ni=1 be projections onto the subspaces {Wi }ni=1 of Hm . The following are equivalent: (1) {Pi }ni=1 does phaseless reconstruction. n,Di i (2) For every orthonormal basis {φi,j }D j=1 of Wi , the set {φi,j }i=1,j=1 does phaseless reconstruction. Combining Theorems 2.6 and 2.8, the result follows. Corollary 2.9. In Rm , a family of projections {Pi }ni=1 does phase retrieval if and only if it does phaseless reconstruction.
S. Botelho-Andrade et al. / J. Math. Anal. Appl. 436 (2016) 131–137
135
3. Phase retrieval versus phaseless reconstruction: complex case In the complex case, the complement property is not equivalent to phaseless reconstruction. However, our main result will establish that (in the complex case) phase retrieval implies phaseless reconstruction. The following lemmas are needed for this result. For this section we adopt the notation a, bR to denote Rea, b. M and any u ∈ CM then φn φ∗n u, iuR = 0. Lemma 3.1. Given {φn }N n=1 ⊆ C
Proof. The following calculation gives the result almost immediately: φn φ∗n u, iuR = u, φn φn , iuR = Re(−iu, φn φn , u) = − Re(i|u, φn |2 ) = 0.
2
M Lemma 3.2. Given {φn }N and any u, v ∈ CM then for each φn , n=1 ⊆ C
|u + v, φn |2 − |u − v, φn |2 = 4φn φ∗n u, vR . Proof. Consider the following |u + v, φn |2 = |u, φn |2 + 2 Re(u, φn v, φn ) + |v, φn |2
(3)
|u − v, φn |2 = |u, φn |2 − 2 Re(u, φn v, φn ) + |v, φn |2 .
(4)
and
Then subtracting (4) from (3) we obtain |u + v, φn |2 − |u − v, φn |2 = 4 Re(u, φn v, φn ) = 4φn φ∗n u, vR .
2
∗ Corollary 3.3. If {φn }N n=1 does phaseless reconstruction and φn φn u, vR = 0 for each n then u +v = ω(u −v) 2 Im(ω) for |ω| = 1 and thus v = |1+ω|2 u.
Proof. If u + v = ωu − ωv then v =
ω−1 ω+1 u
= − (1−ω)(1+ω) u= |1+ω|2
Lemma 3.4. Given any u, let v = αiu for α ∈ R and let ω = 1+αi 1−αi (u − αiu) = ω(u − v).
2 Im(ω) |1+ω|2 u. 1+αi 1−αi
2
then |ω| = 1 and u + v = u(1 + αi) =
Lemma 3.5. If x − y = 0 then φφ∗ (x − y), x + yR = 0. Proof. Consider the following calculation, φφ∗ (x − y), x + yR = Re((x + y)∗ φφ∗ (x − y)) = Re(|φ∗ x|2 − x∗ φφ∗ y + y ∗ φφ∗ x − |φy|2 ) = Re(−x∗ φφ∗ y + x∗ φφ∗ y) = 0. The following is a direct calculation.
2
136
S. Botelho-Andrade et al. / J. Math. Anal. Appl. 436 (2016) 131–137
Lemma 3.6. Let a, b ∈ C such that |a| + |b| > 0. If arg(a + b) = arg(eiθ (a − b)), then tan θ =
2 Im(¯ ab) |a|2 − |b|2
for |a| = |b| and θ = π/2 otherwise. Showing equivalence of phase retrieval and phaseless reconstruction for vectors in Cm requires a theorem in [2], which we state below and then follow with the main result of this paper. M Theorem 3.7. (See [2].) Consider Φ = {φn }N and the mapping A : CM /T → RN defined by n=1 ⊆ C 2 ∗ N (A(x))(n) := |x, φn | . Viewing {φn φn u}n=1 as vectors in R2M , denote S(u) := spanR {φn φ∗n u}N n=1 . Then the following are equivalent:
(a) A is injective. (b) dim S(u) ≥ 2M − 1 for every u ∈ CM \ {0}. (c) S(u) = spanR {iu}⊥ for every u ∈ CM \ {0}. Theorem 3.8. Phase retrieval implies phaseless reconstruction for vectors in the complex case. M Proof. Suppose Φ = {φn }N does phase retrieval. Let u, v be non-zero vectors in CM such that n=1 ⊆ C ∗ φn φn u, vR = 0 for all n. Note that Lemma 3.2 ensures that |u + v, φn |2 = |u − v, φn |2 for each n. In order to apply Theorem 3.7 and complete the proof, we must show v = λiu for some λ ∈ R. For simplicity, denote u = (u1 , u2 , . . . , uN ) and v = (v1 , v2 , . . . , vN ). Consider the following cases:
Case 1: uj vj = 0 for all 1 ≤ j ≤ N . Without loss of generality, suppose u = (eiα1 , 0, . . .) and v = (0, eiβ2 , . . .) for some α1 , β1 ∈ R. Since Φ does phase retrieval, we have that u + v has the same phase as eiγ (u − v), with some real constant γ. In particular arg(u1 + v1 ) = arg(eiγ (u1 − v1 )), i.e. arg(eiα1 ) = arg(eiγ eiα1 ). Similarly arg(u2 + v2 ) = arg(eiγ (u2 − v2 )), i.e. arg(eiβ2 ) = arg(−eiγ eiβ2 ). However the first condition implies γ = 0 and the second gives γ = π, a contradiction. Case 1: uj vj = 0 for some 1 ≤ j ≤ N . Without loss of generality, we can assume u1 v1 = 0 and by multiplying by the appropriate constants we may also assume |u1 | = |v1 | = r1 > 0. Then by Lemma 3.6, for each 1 ≤ j ≤ N we have that tan(γ) =
2 Im(uj vj ) . |uj |2 − |vj |2
By assumption |u1 | = |v1 |, therefore γ = π/2 and hence |uj | = |vj | for all 1 ≤ j ≤ N . So we have shown that u = (r1 eiα1 , r2 eiα2 , . . . , rN eiαN )
S. Botelho-Andrade et al. / J. Math. Anal. Appl. 436 (2016) 131–137
137
and v = (r1 eiβ1 , r2 eiβ2 , . . . , rN eiβN ). Now we claim that sin(βj − αj ) = c for all j. Note that arg(2uj + vj ) = arg(eiθ (2uj − vj )) for all j and fixed θ, then by Lemma 3.6 we see that c = tan θ =
4 Im(uj vj ) 4 = sin(βj − αj ), 3rj2 3
∀1 ≤ j ≤ N.
√ For each j, set aj = cos(βj − αj ) = ± 1 − c2 . We can rewrite v = r1 eiα1 ei(β1 −α1 ) , r2 eiα2 ei(β2 −α2 ) , . . . , rM eiαM ei(βM −αM ) and each ei(βj −αj ) = cos(βj − αj ) + i sin(βj − αj ) = aj + ic. Hence v = w + ciu, where w = (a1 r1 eiα1 , a2 r2 eiα2 , . . . , aN rN eiαN ). We must show w = 0. Indeed, for every n we have 0 = φn φ∗n u, w + ciuR = φn φ∗n u, wR + φn φ∗n u, ciuR . By Lemma 3.1 we see that φn φ∗n u, wR = 0 for all n. Note that w = 0 if and only if aj = 0 for all j. This is clear since that if a1 = 0 then, by Lemma 3.2, we have |a1 u + w, φn | = |a1 u − w, φn | for all 1 ≤ n ≤ N , which implies that a1 u + w and a1 u − w have the same phases, contradicting the fact that the first component of a1 u + w is non-zero but the first component of a1 u − w is 0. 2 We end with the following open question: Problem 3.9. Show that phase retrieval and phaseless reconstruction for projections are equivalent in the complex case. Acknowledgment We are indebted to the referee for making substantial improvements to this paper. References [1] R. Balan, P.G. Casazza, D. Edidin, On signal reconstruction without phase, Appl. Comput. Harmon. Anal. 20 (3) (2006) 345–356. [2] A.S. Bandeira, J. Cahill, D. Mixon, A.A. Nelson, Saving phase: injectivity and stability for phase retrieval, Appl. Comput. Harmon. Anal. 37 (1) (2014) 106–125. [3] B. Bodmann, N. Hammen, Stable phase retrieval with low redundancy frames, preprint, arXiv:1302.5487. [4] J. Cahill, P. Casazza, K. Peterson, L. Woodland, Phase retrieval by projections, available online, arXiv:1305.6226. [5] A. Conca, D. Edidin, M. Hering, C. Vinzant, An algebraic characterization of injectivity of phase retrieval, Appl. Comput. Harmon. Anal. 38 (2) (2015) 346–356. [6] E. Edidin, Projections and phase retrieval, preprint. [7] T. Heinosaari, L. Mazzarella, M.M. Wolf, Quantum tomography under prior information, Comm. Math. Phys. 318 (2) (2013) 355–374. [8] C. Vinzant, A small frame and a certificate of its injectivity, preprint, arXiv:1502.0465v1. [9] Z. Xu, The minimal measurement number for low-rank matrices recovery, preprint, available on arXiv:1505.07204v1.
Contents lists available at ScienceDirect
Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Phase retrieval versus phaseless reconstruction ✩ Sara Botelho-Andrade ∗ , Peter G. Casazza, Hanh Van Nguyen, Janet C. Tremain Department of Mathematics, University of Missouri, Columbia, MO 65211-4100, United States
a r t i c l e
i n f o
Article history: Received 27 June 2015 Available online 2 December 2015 Submitted by E. Saksman Keywords: Phase retrieval Phaseless reconstruction Complement property Projections Weak phase retrieval Frame theory
a b s t r a c t In 2006, Balan, Casazza, and Edidin [1] introduced the frame theoretic study of phaseless reconstruction. Since then, this has turned into a very active area of research. Over the years, many people have replaced the term phaseless reconstruction with phase retrieval. At a meeting in 2012, Casazza asked: Are these really the same? In this paper, we will show that phase retrieval is equivalent to phaseless reconstruction. We then show, more generally, that phase retrieval by projections is equivalent to phaseless reconstruction by projections in the real case. © 2015 Elsevier Inc. All rights reserved.
1. Introduction Throughout, Rm (resp. Cm ) will denote an m-dimensional real (resp. complex) Hilbert space, and Hm will denote an m-dimensional real or complex Hilbert space. Phase retrieval is an old problem in signal processing and has been studied for over 100 years by electrical engineers. We begin with a few necessary definitions and a brief history the problem before proceeding to the main result. Let x = (a1 , a2 , . . . , am ) and y = (b1 , b2 , . . . , bm ) be vectors in Hm . We say that x, y have the same phase if phase ai = phase bi , for all i = 1, 2, . . . , m. Definition 1.1. Given Φ = {φi }ni=1 , a family of vectors in Hm , let x and y be vectors in Hm such that |x, φi | = |y, φi | , for all i = 1, 2, . . . , n.
(1)
i. We say Φ does phase retrieval if (1) implies there exists a |θ| = 1 so that x and θy have the same phases. ii. We say Φ does phaseless reconstruction if (1) implies there is a |θ| = 1 so that x = θy. ✩
The authors were supported by NSF DMS 1307685; and NSF ATD 1321779; AFOSR: FA9550-11-1-0245.
* Corresponding author. E-mail addresses: [email protected] (S. Botelho-Andrade), [email protected] (P.G. Casazza), [email protected] (H. Van Nguyen), [email protected] (J.C. Tremain). http://dx.doi.org/10.1016/j.jmaa.2015.11.045 0022-247X/© 2015 Elsevier Inc. All rights reserved.
132
S. Botelho-Andrade et al. / J. Math. Anal. Appl. 436 (2016) 131–137
Moreover, in the real case, if θ = 1 we say x and y have the same signs and if θ = −1 we say x and y have opposite signs. Note that from the definition of phase retrieval it is not immediately apparent that x = θy. The analogous definition for phase retrieval by projections follows. Definition 1.2. Given {Pi }ni=1 , a family of projections on Hm , let x and y be vectors in Hm such that Pi x = Pi y, for all i = 1, 2, . . . , n.
(2)
i. We say {Pi }ni=1 does phase retrieval if (2) implies there is a |θ| = 1 so that x and θy have the same phases. ii. We say {Pi }ni=1 does phaseless reconstruction if (2) implies there is a |θ| = 1 so that x = θy. In the setting of frame theory, the concept of phaseless reconstruction was introduced in 2006 by Balan, Casazza, and Edidin [1]. They showed that a generic family of (2m − 1)-vectors in Rm does phaseless reconstruction, however no set of (2m − 2)-vectors can. By generic we are referring to an open dense set in the set of (2m − 1)-element frames in Hm . In the complex case, they showed that a generic set of (4m − 2)-vectors does phaseless reconstruction. Heinosaari, Mazzarella and Wolf improved this result in [7], where they showed that n-vectors doing phaseless reconstruction in Cm requires n ≥ 4m −4 −2α, where α is the number of 1’s in the binary expansion of (m − 1). This bound was further improved when Bodmann [3] showed that phaseless reconstruction in Cm can be done with (4m −4)-vectors. Later, Conca, Edidin, Hering, and Vinzant [5] showed that a generic frame with (4m − 4)-vectors does phaseless reconstruction in Cm . They also showed that if m = 2k + 1 then no n-vectors with n < 4m − 4 can do phaseless reconstruction. It was conjectured that no fewer than (4m −4)-vectors can do phaseless reconstruction. Recently, Vinzant [8] showed that this conjecture does not hold by producing 11 vectors in C4 which do phaseless reconstruction. The pursuit of improving this bound was finally resolved when Edidin [6] showed that a generic family of 2m − 1 subspaces will do phaseless in Rm and if m = 2k + 1, then this bound is sharp. The analogous question for subspaces has still not been answered, however it was shown that there are six 2-dimensional subspaces of R4 which do phase retrieval (Xu [9]). Over the years, we have replaced the phrase “phaseless reconstruction” with “phase retrieval.” Casazza, at a meeting in 2012, raised the question: “Are these really the same?” In this paper we will answer in the affirmative, by showing that in both the real and complex cases phase retrieval implies phaseless reconstruction (and similarly for phase retrieval by projections in Rm ). This problem arose because of the way the engineering version of phase retrieval was translated into the language of frame theory. The engineers are working with the modulus of the Fourier transform and want to recover the phases to be able to invert the Fourier transform and recover the signal. In other words, the engineers only need is to recover the phase. But in the frame theory version of this, for x = (a1 , a2 , . . . , am ), we are really trying to recover two things: (1) Recover the phases of the ai . (2) Recover |ai | (which in the engineering case, is already known). 2. Phase retrieval versus phaseless reconstruction: real case We begin by defining the complement property from [1]. Definition 2.1. A family of vectors {φi }ni=1 in Hm has the complement property if for every I ⊂ [n], either {φi }i∈I spans Hm or {φi }i∈I c spans Hm .
S. Botelho-Andrade et al. / J. Math. Anal. Appl. 436 (2016) 131–137
133
We will prove that phase retrieval implies the complement property for both the real and complex cases and even in a more general setting. First, we need to make a couple observations: Observation 2.2. If {φi }ni=1 does phase retrieval in Hm , then span {φi }ni=1 = Hm . For otherwise, there would exist 0 = x ∈ Hm so that x, φi = 0, φi = 0, for all i = 1, 2, . . . , n, while x and 0 do not have the same phase. Observation 2.3. If x = (a1 , a2 , . . . , am ) and y = (b1 , b2 , . . . , bm ) have the same phases, then ai = 0 if and only if bi = 0. I.e. zero has no phase. The following theorem (from [1]) was proved for the real case and it was stated that the same proof holds in the complex case. In [2] it is mentioned that this argument does not hold in the complex case. We will now show that the proof in [1] holds in the complex case. Theorem 2.4. (See [1].) Let Φ = {φi }ni=1 be vectors in Hm . If Φ = {φi }ni=1 does phaseless reconstruction, then it has complement property. Proof. Assume Φ does not have the complement property but does phaseless reconstruction. Choose I ⊂ [n] so that neither of the sets of vectors {φi }i∈I or {φi }i∈I c spans Hm . Choose x = 1 = y so that x ⊥ φi for i ∈ I and y ⊥ φi for i ∈ I c . Then |x + y, φi | = |x − y, φi |, for all i = 1, 2, . . . , n. Since Φ does phaseless reconstruction, there is a |θ| = 1 so that x + y = θ(x − y), and hence (1 − θ)x = −(1 + θ)y. If θ = 1, then y = 0 and if θ = −1 then x = 0, contradicting the fact that x, y are unit norm. Otherwise, x=
−(1 + θ) y = dy, for d = 0. 1−θ
Now, x, φi = dy, φi = 0, for all i = 1, 2, . . . , n, and hence Φ does not span Hm contradicting Observation 2.2. 2 Remark 2.5. Moreover, in the real case phaseless reconstruction is equivalent to the complement property (as shown in Corollary 2.7). Theorem 2.6. If {Pi }ni=1 are projections onto the subspaces {Wi }ni=1 of Hm which do phase retrieval, then n, Di i for every orthonormal basis {φi,j }D j=1 of Wi , the set {φi,j }i=1,j=1 has complement property. i Proof. Suppose {Wi }ni=1 does phase retrieval for Hm , but there exist an orthonormal basis {φi,j }D j=1 of each n, Di Wi such that the set {φi,j }i=1,j=1 fails the complement property. In other words, there exists I ⊂ {(i, j) : 1 ≤ i ≤ n and 1 ≤ j ≤ Di } so that {φi,j }(i,j)∈I and {φi,j }(i,j)∈I c do not span Hm . Choose vectors x, y ∈ Hm
S. Botelho-Andrade et al. / J. Math. Anal. Appl. 436 (2016) 131–137
134
with x = 1 = y, where x = (a1 , a2 , . . . , am ) and y = (b1 , b2 , . . . , bm ) such that x ⊥ φi,j for all (i, j) ∈ I and y ⊥ φi,j for all (i, j) ∈ I c . Note this choice of vectors forces that for each (i, j) either x, φi,j = 0 or y, φi,j = 0. Fix c = 0. Then for each 1 ≤ i ≤ n, |x + cy, φi,j | = |x − cy, φi,j |, for all i, j. Hence, Pi (x + cy)2 =
Di
2
|x + cy, φi,j | =
j=1
Di
2
|x − cy, φi,j | = Pi (x − cy)2 .
j=1
By the assumption that {Pi }ni=1 does phase retrieval, this implies there is a |θ| = 1 so that x + cy and θ(x − cy) have the same phases. Assume there exists some 1 ≤ i0 ≤ m so that ai0 = 0 = bi0 and let −a c = bi i0 . Then 0
(x + cy)i0 = ai0 + cbi0 = ai0 +
−ai0 bi = 0, bi0 0
while (x − cy)i0 = ai0 −
−ai0 = 2ai0 = 0. bi0
But this contradicts Observation 2.3. It follows that for every 1 ≤ i ≤ m, either ai = 0 or bi = 0. Let {ei }m i=1 be an orthonormal basis for Hm and let I = {1 ≤ i ≤ m : bi = 0}. Then x+y =
i∈I
ai ei +
bi ei , and x − y =
i∈I c
i∈I
ai ei +
(−bi )ei .
i∈I c
By the above argument, there is a |θ| = 1 so that x + y and θ(x − y) have this same phase. But this is clearly impossible. This final contradiction completes the proof. 2 We have a number of consequences of Theorem 2.6. Letting the subspaces Wi be one dimensional, this becomes a theorem about vectors. Corollary 2.7. If Φ = {φi }ni=1 does phase retrieval in Hm , then Φ has the complement property. Hence, in the real case, phase retrieval and phaseless reconstruction for vectors are equivalent properties. We need the following result from [4]: Theorem 2.8. (See [4].) Let {Pi }ni=1 be projections onto the subspaces {Wi }ni=1 of Hm . The following are equivalent: (1) {Pi }ni=1 does phaseless reconstruction. n,Di i (2) For every orthonormal basis {φi,j }D j=1 of Wi , the set {φi,j }i=1,j=1 does phaseless reconstruction. Combining Theorems 2.6 and 2.8, the result follows. Corollary 2.9. In Rm , a family of projections {Pi }ni=1 does phase retrieval if and only if it does phaseless reconstruction.
S. Botelho-Andrade et al. / J. Math. Anal. Appl. 436 (2016) 131–137
135
3. Phase retrieval versus phaseless reconstruction: complex case In the complex case, the complement property is not equivalent to phaseless reconstruction. However, our main result will establish that (in the complex case) phase retrieval implies phaseless reconstruction. The following lemmas are needed for this result. For this section we adopt the notation a, bR to denote Rea, b. M and any u ∈ CM then φn φ∗n u, iuR = 0. Lemma 3.1. Given {φn }N n=1 ⊆ C
Proof. The following calculation gives the result almost immediately: φn φ∗n u, iuR = u, φn φn , iuR = Re(−iu, φn φn , u) = − Re(i|u, φn |2 ) = 0.
2
M Lemma 3.2. Given {φn }N and any u, v ∈ CM then for each φn , n=1 ⊆ C
|u + v, φn |2 − |u − v, φn |2 = 4φn φ∗n u, vR . Proof. Consider the following |u + v, φn |2 = |u, φn |2 + 2 Re(u, φn v, φn ) + |v, φn |2
(3)
|u − v, φn |2 = |u, φn |2 − 2 Re(u, φn v, φn ) + |v, φn |2 .
(4)
and
Then subtracting (4) from (3) we obtain |u + v, φn |2 − |u − v, φn |2 = 4 Re(u, φn v, φn ) = 4φn φ∗n u, vR .
2
∗ Corollary 3.3. If {φn }N n=1 does phaseless reconstruction and φn φn u, vR = 0 for each n then u +v = ω(u −v) 2 Im(ω) for |ω| = 1 and thus v = |1+ω|2 u.
Proof. If u + v = ωu − ωv then v =
ω−1 ω+1 u
= − (1−ω)(1+ω) u= |1+ω|2
Lemma 3.4. Given any u, let v = αiu for α ∈ R and let ω = 1+αi 1−αi (u − αiu) = ω(u − v).
2 Im(ω) |1+ω|2 u. 1+αi 1−αi
2
then |ω| = 1 and u + v = u(1 + αi) =
Lemma 3.5. If x − y = 0 then φφ∗ (x − y), x + yR = 0. Proof. Consider the following calculation, φφ∗ (x − y), x + yR = Re((x + y)∗ φφ∗ (x − y)) = Re(|φ∗ x|2 − x∗ φφ∗ y + y ∗ φφ∗ x − |φy|2 ) = Re(−x∗ φφ∗ y + x∗ φφ∗ y) = 0. The following is a direct calculation.
2
136
S. Botelho-Andrade et al. / J. Math. Anal. Appl. 436 (2016) 131–137
Lemma 3.6. Let a, b ∈ C such that |a| + |b| > 0. If arg(a + b) = arg(eiθ (a − b)), then tan θ =
2 Im(¯ ab) |a|2 − |b|2
for |a| = |b| and θ = π/2 otherwise. Showing equivalence of phase retrieval and phaseless reconstruction for vectors in Cm requires a theorem in [2], which we state below and then follow with the main result of this paper. M Theorem 3.7. (See [2].) Consider Φ = {φn }N and the mapping A : CM /T → RN defined by n=1 ⊆ C 2 ∗ N (A(x))(n) := |x, φn | . Viewing {φn φn u}n=1 as vectors in R2M , denote S(u) := spanR {φn φ∗n u}N n=1 . Then the following are equivalent:
(a) A is injective. (b) dim S(u) ≥ 2M − 1 for every u ∈ CM \ {0}. (c) S(u) = spanR {iu}⊥ for every u ∈ CM \ {0}. Theorem 3.8. Phase retrieval implies phaseless reconstruction for vectors in the complex case. M Proof. Suppose Φ = {φn }N does phase retrieval. Let u, v be non-zero vectors in CM such that n=1 ⊆ C ∗ φn φn u, vR = 0 for all n. Note that Lemma 3.2 ensures that |u + v, φn |2 = |u − v, φn |2 for each n. In order to apply Theorem 3.7 and complete the proof, we must show v = λiu for some λ ∈ R. For simplicity, denote u = (u1 , u2 , . . . , uN ) and v = (v1 , v2 , . . . , vN ). Consider the following cases:
Case 1: uj vj = 0 for all 1 ≤ j ≤ N . Without loss of generality, suppose u = (eiα1 , 0, . . .) and v = (0, eiβ2 , . . .) for some α1 , β1 ∈ R. Since Φ does phase retrieval, we have that u + v has the same phase as eiγ (u − v), with some real constant γ. In particular arg(u1 + v1 ) = arg(eiγ (u1 − v1 )), i.e. arg(eiα1 ) = arg(eiγ eiα1 ). Similarly arg(u2 + v2 ) = arg(eiγ (u2 − v2 )), i.e. arg(eiβ2 ) = arg(−eiγ eiβ2 ). However the first condition implies γ = 0 and the second gives γ = π, a contradiction. Case 1: uj vj = 0 for some 1 ≤ j ≤ N . Without loss of generality, we can assume u1 v1 = 0 and by multiplying by the appropriate constants we may also assume |u1 | = |v1 | = r1 > 0. Then by Lemma 3.6, for each 1 ≤ j ≤ N we have that tan(γ) =
2 Im(uj vj ) . |uj |2 − |vj |2
By assumption |u1 | = |v1 |, therefore γ = π/2 and hence |uj | = |vj | for all 1 ≤ j ≤ N . So we have shown that u = (r1 eiα1 , r2 eiα2 , . . . , rN eiαN )
S. Botelho-Andrade et al. / J. Math. Anal. Appl. 436 (2016) 131–137
137
and v = (r1 eiβ1 , r2 eiβ2 , . . . , rN eiβN ). Now we claim that sin(βj − αj ) = c for all j. Note that arg(2uj + vj ) = arg(eiθ (2uj − vj )) for all j and fixed θ, then by Lemma 3.6 we see that c = tan θ =
4 Im(uj vj ) 4 = sin(βj − αj ), 3rj2 3
∀1 ≤ j ≤ N.
√ For each j, set aj = cos(βj − αj ) = ± 1 − c2 . We can rewrite v = r1 eiα1 ei(β1 −α1 ) , r2 eiα2 ei(β2 −α2 ) , . . . , rM eiαM ei(βM −αM ) and each ei(βj −αj ) = cos(βj − αj ) + i sin(βj − αj ) = aj + ic. Hence v = w + ciu, where w = (a1 r1 eiα1 , a2 r2 eiα2 , . . . , aN rN eiαN ). We must show w = 0. Indeed, for every n we have 0 = φn φ∗n u, w + ciuR = φn φ∗n u, wR + φn φ∗n u, ciuR . By Lemma 3.1 we see that φn φ∗n u, wR = 0 for all n. Note that w = 0 if and only if aj = 0 for all j. This is clear since that if a1 = 0 then, by Lemma 3.2, we have |a1 u + w, φn | = |a1 u − w, φn | for all 1 ≤ n ≤ N , which implies that a1 u + w and a1 u − w have the same phases, contradicting the fact that the first component of a1 u + w is non-zero but the first component of a1 u − w is 0. 2 We end with the following open question: Problem 3.9. Show that phase retrieval and phaseless reconstruction for projections are equivalent in the complex case. Acknowledgment We are indebted to the referee for making substantial improvements to this paper. References [1] R. Balan, P.G. Casazza, D. Edidin, On signal reconstruction without phase, Appl. Comput. Harmon. Anal. 20 (3) (2006) 345–356. [2] A.S. Bandeira, J. Cahill, D. Mixon, A.A. Nelson, Saving phase: injectivity and stability for phase retrieval, Appl. Comput. Harmon. Anal. 37 (1) (2014) 106–125. [3] B. Bodmann, N. Hammen, Stable phase retrieval with low redundancy frames, preprint, arXiv:1302.5487. [4] J. Cahill, P. Casazza, K. Peterson, L. Woodland, Phase retrieval by projections, available online, arXiv:1305.6226. [5] A. Conca, D. Edidin, M. Hering, C. Vinzant, An algebraic characterization of injectivity of phase retrieval, Appl. Comput. Harmon. Anal. 38 (2) (2015) 346–356. [6] E. Edidin, Projections and phase retrieval, preprint. [7] T. Heinosaari, L. Mazzarella, M.M. Wolf, Quantum tomography under prior information, Comm. Math. Phys. 318 (2) (2013) 355–374. [8] C. Vinzant, A small frame and a certificate of its injectivity, preprint, arXiv:1502.0465v1. [9] Z. Xu, The minimal measurement number for low-rank matrices recovery, preprint, available on arXiv:1505.07204v1.