Characterization of (Weak) Phase Retrieval Dual Frames

Abstract

Recovering a signal up to a unimodular constant from the magnitudes of linear measurements has been popular and well studied in recent years. However, numerous unsolved problems regarding phase retrieval still exist. Given a phase retrieval frame, may the family of phase retrieval dual frames be classified? And is such a family dense in the set of dual frames? Can we present the equivalent conditions for a family of vectors to do weak phase retrieval in complex Hilbert space case? What is the connection between phase, weak phase and norm retrieval? In this context, we aim to deal with these open problems concerning phase retrieval dual frames, weak phase retrieval frames, and specially investigate equivalent conditions for identifying these features. We provide some characterizations of alternate dual frames of a phase retrieval frame which yield phase retrieval in finite dimensional Hilbert spaces. Moreover, for some classes of frames, we show that the family of phase retrieval dual frames is open and dense in the set of dual frames. Then, we study weak phase retrieval problem. Among other things, we obtain some equivalent conditions on a family of vectors to do phase retrieval in terms of weak phase retrieval.

1 Introduction and Preliminaries

Signal reconstruction without using phase is a longstanding conjecture, specially with regard to speech recognition systems which was first introduced from mathematical point of view by Balan et al. [8], and has been a very popular topic in recent years due to so many applications of X-ray, electron microscopy, optics, image processing and much more [5, 14, 20, 21]. To present the problem in a more precise approach, we first briefly state some basic definitions and preliminaries in relevant areas. Then we propound a number of problems to address phase retrieval frames in some new aspects.

Throughout this paper, we suppose that ${\mathcal {H}}$ denotes a separable Hilbert space, ${\mathcal {H}}_{n}$ denote an n-dimensional real or complex Hilbert space and we use ${\mathbb {R}}^{n}$ and ${\mathbb {C}}^{n}$ whenever it is necessary to differentiate between the two. Also, we consider the notations; $I_{m}=\{1, 2,..., m\}$ and $\{\delta _{i}\}_{i\in I_{n}}$ as the standard orthonormal basis of ${\mathcal {H}}_{n}$.

A family of vectors $\Phi :=\{\phi _{i}\}_{i\in I}$ in ${\mathcal {H}}$ is called a frame if there exist the constants $0<A_{\Phi }\le B_{\Phi }<\infty$ such that

$$\begin{aligned} A_{\Phi }\Vert f\Vert ^{2}\le \sum _{i\in I}\vert f,\phi _{i}\rangle \vert ^{2}\le B_{\Phi }\Vert f\Vert ^{2},\qquad (f\in {\mathcal {H}}). \end{aligned}$$

(1.1)

The constants $A_{\Phi }$ and $B_{\Phi }$ are called frame bounds. The sequence $\{\phi _{i}\}_{i\in I}$ is said to be a Bessel sequence whenever the right hand side of (1.1) holds. A frame $\{\phi _{i}\}_{i\in I}$ is called A-tight frame if $A=A_{\Phi }=B_{\Phi }$, and in the case of $A_{\Phi }=B_{\Phi }=1$ it is called a Parseval frame. Given a frame $\Phi =\{\phi _{i}\}_{i\in I}$, its Grammian matrix formed by the inner product of the frame vectors is as $G_{\Phi }=[\langle \phi _{i},\phi _{j}\rangle ]_{i,j}$. The frame operator is defined by $S_{\Phi }f=\sum _{i\in I}\langle f,\phi _{i}\rangle \phi _{i}$. It is a bounded, invertible, and self-adjoint operator [13]. Also, the synthesis operator $T_{\Phi }: l^{2}\rightarrow {\mathcal {H}}$ is defined by $T_{\Phi }\lbrace c_{i}\rbrace = \sum _{i\in I} c_{i}\phi _{i}$. The frame operator can be written as $S_{\Phi }= T_{\phi }T_{\phi }^{*}$ where $T_{\Phi }^{*}: {\mathcal {H}}\rightarrow l^{2}$, the adjoint of T, given by $T_{\Phi }^{*}f= \lbrace \langle f,\phi _{i}\rangle \rbrace _{i\in I}$ is called the analysis operator. The family $\{S_{\Phi }^{-1}f_{i}\}_{i\in I}$ is also a frame for ${\mathcal {H}}$, called the canonical dual frame. In general, a frame $\{\psi _i\}_{i\in I}\subseteq {\mathcal {H}}$ is called an alternate dual or simply a dual for $\{\phi _{i}\}_{i\in I}$ if $f=\sum _{i\in I}\langle f,\psi _{i}\rangle \phi _i,$ for $f\in {\mathcal {H}}$. All frames have at least a dual, the canonical dual, and redundant frames have an infinite number of alternate dual frames. We denote the excess of a frame $\Phi$ by $E(\Phi )$. It is known that every dual frame is of the form $\{S_{\Phi }^{-1}\phi _{i}+u_{i}\}_{i\in I}$, where $\{u_{i}\}_{i\in I}$ is a Bessel sequence that satisfies $\sum _{i\in I}\langle f,u_{i}\rangle \phi _{i}=0,$ for all $f\in {\mathcal {H}}$. Also recall that, two frames $\Phi$ and $\Psi$ are equivalent if there exists an invertible operator U on ${\mathcal {H}}$ so that $\Psi =U\Phi$. See [13, 16] for more detailed information on frame theory and [3, 4, 7, 17,18,19] for the importance of duality principle.

Consider a frame $\{\phi _{i}\}_{i\in I_{m}}$ in a Hilbert space ${\mathcal {H}}_{n}$. A finite set of indices $\sigma \subset I_{m}$ satisfies the minimal redundancy condition (MRC) whenever $\{\phi _{i}\}_{i\in \sigma ^{c}}$ remains to be a frame for ${\mathcal {H}}_{n}$. Furthermore, we say $\Phi$ satisfies MRC for r-erasures if every subset $\sigma \subset I_{m}$, $\vert \sigma \vert =r$ satisfies MRC for $\Phi$. The spark of a matrix is the size of the smallest linearly dependent subset of the columns and the spark of a family $\{\phi _{i}\}_{i\in I_{m}}$ is defined as the spark of its synthesis matrix $\Phi$. Also, the family $\{\phi _{i}\}_{i\in I_{m}}$, $m\ge n$ is said full spark if it has the spark $n+1$. It is shown that if $m\ge n$, then the set of full spark frames is open and dense in the set of all frames [2].

Suppose now the nonlinear mapping

$$\begin{aligned} {\mathbb {M}}_{\Phi }: {\mathcal {H}}\rightarrow l^{2}(I), \qquad {\mathbb {M}}_{\Phi }(f)=\lbrace \vert \langle f,\phi _{i}\rangle \vert ^{2}\rbrace _{i\in I} \end{aligned}$$

(1.2)

obtained by taking the absolute value element wise of the analysis operator. Let us denote by ${\mathbb {H}}={\mathcal {H}}/\sim$ considered by identifying two vectors which are different in a phase factor, i.e., $f \sim g$ whenever there exists a scalar $\theta$ with $\vert \theta \vert =1$ so that $g=\theta f$. Obviously in a real Hilbert space we have ${\mathbb {H}}={\mathcal {H}}/\{1,-1\}$ and in the complex case ${\mathbb {H}}={\mathcal {H}}/{\mathbb {T}}$, where ${\mathbb {T}}$ is the complex unit circle. So, the mapping ${\mathbb {M}}_{\Phi }$ can be extended to ${\mathbb {H}}$ as ${\mathbb {M}}_{\Phi }({\hat{f}})=\lbrace \vert \langle f,\phi _{i}\rangle \vert ^{2}\rbrace _{i\in I}$, $f\in {\hat{f}}=\{g\in {\mathcal {H}}: g \sim f\}$. The injectivity of the nonlinear mapping ${\mathbb {M}}_{\Phi }$ leads to the reconstruction of every signal in ${\mathcal {H}}$ up to a constant phase factor from the modula of its frame coefficients. In [8], the authors investigated the injectivity of ${\mathbb {M}}_{\Phi }$ in finite dimensional real Hilbert spaces and moreover, it was proven that $4n-2$ measurements suffice for the injectivity in n-dimensional complex Hilbert spaces. Indeed, the injectivity of the mapping ${\mathbb {M}}_{\Phi }$ is equivalent to the following definition:

Definition 1.1

A family of vectors $\Phi =\{\phi _{i}\}_{i\in I}$ in ${\mathcal {H}}$ does phase retrieval if whenever $f,g\in {\mathcal {H}}$ satisfy

$$\begin{aligned} \vert \langle f,\phi _{i}\rangle \vert =\vert \langle g,\phi _{i}\rangle \vert , \qquad (i\in I) \end{aligned}$$

(1.3)

then there exists a scalar $\theta$ with $\vert \theta \vert =1$ so that $f= \theta g$.

If for every $f,g\in {\mathcal {H}}$, which satisfy (1.3) we get $\Vert f\Vert =\Vert g\Vert$, then it said $\Phi$ to do norm retrieval. Clearly, if $\Phi$ does phase (norm) retrieval, then so does $\alpha _{i}f_{i}$ for every $0<\alpha _{i}$, $i\in I$. Also, tight frames do norm retrieval. Moreover, phase retrieval implies norm retrieval, but the converse fails. For example every orthonormal basis does norm retrieval, but fails at phase retrieval.

The following result states that for two equivalent frames $\Phi$ and $\psi$, the injectivity of ${\mathbb {M}}_{\Phi }$ and ${\mathbb {M}}^{\Psi }$ are the same.

Theorem 1.2

[8] A family $\Phi =\{\phi _{i}\}_{i\in I}$ in ${\mathcal {H}}$ does phase retrieval if and only if $\{U\phi _{i}\}_{i\in I}$ does phase retrieval for every invertible operator U on ${\mathcal {H}}$.

Applying the above theorem, shows that a frame does phase retrieval if and only if its canonical dual does phase retrieval.

Definition 1.3

[8] A family of vectors $\Phi =\{\phi _{i}\}_{i\in I}$ in ${\mathcal {H}}$ has the complement property if for every $\sigma \subset I$ either ${\overline{span}}\{\phi _{i}\}_{i\in \sigma }={\mathcal {H}}$ or ${\overline{span}}\{\phi _{i}\}_{i\in \sigma ^{c}}={\mathcal {H}}$.

As we stated, a fundamental classification of frames which do phase retrieval was presented for finite dimensional real case in [8] and then for infinite dimensional case in [11] as follows:

Theorem 1.4

A family $\Phi =\{\phi _{i}\}_{i\in I}$ in a real Hilbert space ${\mathcal {H}}$ does phase retrieval if and only if it has the complement property.

As an immediate result of the above theorem, every phase retrieval frame in real Hilbert space ${\mathcal {H}}$ is satisfied in MRC for $(n-1)$-erasures.

Proposition 1.5

[8] If $\Phi =\{\phi _{i}\}_{i\in I_{m}}$ does phase retrieval in ${\mathbb {R}}^{n}$, then $m\ge 2n-1$. If $m\ge 2n-1$ and $\Phi$ is full spark then $\mathcal {\phi }$ does phase retrieval. Moreover, $\{\phi _{i}\}_{i\in I_{2n-1}}$ does phase retrieval if and only if $\Phi$ is full spark.

Two vectors $x=\{x_{i}\}_{i\in I_{n}}$ and $y=\{y_{i}\}_{i\in I_{n}}$ in ${\mathcal {H}}_{n}$ weakly have the same phase if there is a $\vert \alpha \vert =1$ so that $phase (x_{i})=\alpha phase (y_{i})$, for all $i\in I_{n}$ which $x_{i}\ne 0 \ne y_{i}$.

Definition 1.6

A family $\Phi =\{\phi _{i}\}_{i\in I_{m}}$ in ${\mathcal {H}}_{n}$ does weak phase retrieval if for any $x, y \in {\mathcal {H}}_{n}$ with $\vert \langle x,\phi _{i}\rangle \vert =\vert \langle y,\phi _{i}\rangle \vert$ for all $i\in I_{m}$, then x and y weakly have the same phase.

It is shown that if $\Phi =\{\phi _{i}\}_{i\in I_{m}}$ does weak phase retrieval in ${\mathbb {R}}^{n}$, then $m\ge 2n-2$ [10]. Also, clearly phase retrieval implies weak phase retrieval property, although the converse does not hold in general. As a simple example $\{(1,1), (1,-1)\}$ does weak phase retrieval for ${\mathbb {R}}^{2}$, see [1], but clearly does not phase retrieval.

Now, we state the concept of a generic set; A subset $\Omega \subseteq {\mathbb {R}}^{n}$ is called generic whenever there exists a nonzero polynomial $p(x_{1},...,x_{n})$ so that

$$\begin{aligned} \Omega ^{c}\subseteq \{(x_{1},...,x_{n})\in {\mathbb {R}}^{n}: p(x_{1},...,x_{n})=0\}. \end{aligned}$$

It is known that generic sets are open, dense and full measure. Furthermore, a generic set in ${\mathbb {C}}^{n}$ is defined as a generic set in ${\mathbb {R}}^{2n}$.

Applying the linear operator introduced in [6] gives an equivalent condition for injectivity of ${\mathbb {M}}_{\Phi }$, defined by (1.2). More precisely, let $\{\phi _{i}\}_{i\in I_{m}}$ be a family of vectors in ${\mathbb {C}}^{n}$ and $H_{n\times n}$ denotes the space of all $n\times n$ Hermitian matrices. Consider the operator $\Lambda _{\Phi }:H_{n\times n}\rightarrow {\mathbb {R}}^{m}$ by $\Lambda _{\Phi }(A)=\{\langle A, \phi _{i}\otimes \phi _{i}\rangle \}_{i\in I_{m}}$, in which $\phi _{i}\otimes \phi _{i}$ is the rank one projection onto $span\{\phi _{i}\}$. It is easily proven that $\Lambda _{\Phi }(f\otimes f)={\mathbb {M}}_{\Phi }(f)$ and this equality has a key role for characterizing the injectivity of ${\mathbb {M}}_{\Phi }$, [6, 12], See also [15]. So, we get the following equivalent conditions for a frame to do phase retrieval.

Corollary 1.7

Let $\Phi =\{\mathcal {\phi }_{i}\}_{i\in I_{m}}$ be a frame in ${\mathcal {H}}_{n}$, then the followings are equivalent;

• (i) $\Phi$ does phase retrieval.

• (ii) ${\mathbb {M}}_{\Phi }$ is injective.

• (iii) $\Lambda _{\Phi }\vert _{B_{1}}$ is injective, where $B_{1}$ denotes rank one matrices.

• (iv) There exists no rank 2 matrix in the null space of $\Lambda _{\Phi }$.

Proof

$(i) \Leftrightarrow (ii)$ is obvious by the definition. For $(ii) \Leftrightarrow (iv)$ we just note that, due to the completeness of $\Phi$ in ${\mathcal {H}}_{n}$, the rank one matrices $B_{1}=\{f\otimes f: \quad f\in {\mathcal {H}}_{n}\}$ can not be in the null space of $\Lambda _{\Phi }$. Indeed, $\Lambda _{\Phi }(f\otimes f)={\mathbb {M}}_{\Phi }(f)=0$ implies that $f\perp \phi _{i}$, for all $i\in I_{m}$ and so $f=0$. So, as a result of Lemma 5.5 of [6], the injectivity of ${\mathbb {M}}_{\Phi }$ is equivalent to the statement that; there is no rank 2 matrix in the null space of $\Lambda _{\Phi }$. On the other hand, the injectivity of ${\mathbb {M}}_{\Phi }$ along with the fact that the mapping $\gamma : {\mathcal {H}}_{n}\rightarrow B_{1}$, $\gamma (f)=f\otimes f$ is invertible, deduce that the linear operator $\Lambda _{\Phi }\vert _{B_{1}}$ is also injective. Moreover, for a left inverse ${\mathbb {L}}_{\Phi }$ of ${\mathbb {M}}_{\Phi }$, the mapping $\gamma {\mathbb {L}}_{\Phi }$ is a left inverse for $\Lambda _{\Phi }\vert _{B_{1}}$. Conversely, for a left inverse $\Gamma _{\Phi }$ of $\Lambda _{\Phi }\vert _{B_{1}}$, the mapping $\gamma ^{-1}\Gamma _{\Phi }$ is a left inverse for ${\mathbb {M}}_{\Phi }$, this completes the proof of $(ii) \Leftrightarrow (iii)$. $\square$

This paper is organized as follows: Sect. 2 is devoted to characterizing phase retrieval dual frames and full spark dual frames. In this section, for some classes of frames, we show that the family of all ( full spark) phase retrieval dual frames is open and dense in the set of all dual frames. Moreover, we present several examples in this regard. In Sect. 3, we survey weak phase retrieval problem and investigate some equivalent conditions for identifying phase and weak phase retrieval frames. We also, obtain a sufficient conditions on a family of weak phase retrieval frames to constitute a frame and its canonical dual yields weak phase retrieval, as well.

2 Phase Retrieval Dual Frames

In this section, we address the problem that, given a phase retrieval frame $\Phi$, characterize phase retrieval dual frames of $\Phi$ in finite dimensional Hilbert spaces. For some classes of frames we show that phase retrieval dual frames of a given frame are dense in the set of all dual frames. For a frame $\Phi$ we denote the set of all its dual frames of $\Phi$ by $D_{\Phi }$ and the subset of phase retrieval dual frames is denoted by $PD_{\Phi }$. We first investigate the relationship between phase retrieval duals of equivalent frames, which is a very useful tool for the main results of this section.

Lemma 2.1

Suppose that $\Phi =\{\phi _{i}\}_{i\in I_{m}}$ is a frame for ${\mathcal {H}}_{n}$ and ${\mathcal {T}}$ is an invertible operator on ${\mathcal {H}}_{n}$. Then,

• (i) $D_{{\mathcal {T}}\Phi } = ({\mathcal {T}}^{*})^{-1} D_{\Phi }$.

• (ii) $PD_{{\mathcal {T}}\Phi } = ({\mathcal {T}}^{*})^{-1} PD_{\Phi }$.

Proof

It is known that ${\mathcal {T}}\Phi$ is a frame with $S_{{\mathcal {T}}\Phi }= {\mathcal {T}}S_{\Phi }{\mathcal {T}}^{*}$ [13], and so a simple computation assures that $D_{{\mathcal {T}}\Phi } = \{({\mathcal {T}}^{*})^{-1}G: G\in D_{\Phi }\}= ({\mathcal {T}}^{*})^{-1} D_{\Phi }$.

Moreover, (ii) is given as an immediate result of (i) along with Theorem 1.2. $\square$

Remark 2.2

If $\Phi =\{\phi _{i}\}_{i\in I_{m}}$ is a frame for ${\mathcal {H}}_{n}$ so that $E(\Phi )=k$, which $E(\Phi )$ denotes the excess of $\Phi$, then there exist $i_{1},..., i_{k}$ so that $\Phi {\setminus } \{\phi _{i_{j}}\}_{j=1}^{k}$ constitutes a Riesz basis for ${\mathcal {H}}_{n}$. Therefore, applying the above notations and without loss of generality, we may consider $\Phi =\{\phi _{i}\}_{i\in I_{n}}\cup \{\phi _{i}\}_{i=n+1}^{m}$, where $\{\phi _{i}\}_{i\in I_{n}}$ is a Riesz basis for ${\mathcal {H}}_{n}$. In this point of view, $\Phi$ is indeed equivalent to a form as $\{\delta _{i}\}_{i\in I_{n}}\cup \{{\tilde{\phi }}_{i}\}_{i=n+1}^{m}$ with redundant elements $\{{\tilde{\phi }}_{i}\}_{i=n+1}^{m}$.

In the next theorem we identify the set of dual frames of a frame $\{\phi _{i}\}_{i\in I_{2n-1}}$ in n-dimensional real space and show that $PD_{\phi }$ is an open and dense subset in $D_{\phi }$.

Theorem 2.3

Let $\phi =\{\phi _{i}\}_{i\in I_{2n-1}}$ be a phase retrieval frame in ${\mathbb {R}}^{n}$. Then $PD_{\phi }$ is an open and dense subset in $D_{\phi }$.

Proof

First let $\{\phi _{i}\}_{i\in I_{n}}$ be the standard orthonormal basis of ${\mathbb {R}}^{n}$ and $\phi _{j}=\sum _{i\in I_{n}} a_{i}^{j} \phi _{i}$, $j= n+1,...,2n-1$, for some non-zero coefficients $\{a_{i}^{j}\}_{i\in I_{n}}$. Due to the fact that

$$\begin{aligned} D_{\Phi } = \left\{ \{S_{\Phi }^{-1}\phi _{i}+u_{i}\}_{i\in I_{2n-1}}: \sum _{i\in I_{2n-1}}\langle f,\phi _{i}\rangle u_{i}=0, \textit{for all } f\in {\mathbb {R}}^{n}\right\} \end{aligned}$$

by putting $f=\phi _{j}$, $j=1,..., 2n-1$ we observe that $(2n-1) \times n$ matrix $[u_{1} \vert u_{2} \vert ... \vert u_{2n-1}]^{T}$ is in the null space of $G_{\Phi }^{T}$, the transposed Gram matrix of $\Phi$. The fact that, $dim (null G_{\Phi } )=n-1$; assures that just $n-1$ vectors of $\{u_{i}\}_{i\in I_{2n-1}}$ can be independent. More precisely,

$$\begin{aligned} u_{1}+\sum _{i=n+1}^{2n-1}\langle \phi _{1},\phi _{i}\rangle u_{i}=0,\quad ... \quad u_{n}+\sum _{i=n+1}^{2n-1}\langle \phi _{n},\phi _{i}\rangle u_{i}=0. \end{aligned}$$

So, by choosing $\{u_{j}\}_{j=n+1}^{2n-1}$ we get

$$\begin{aligned} u_{i}= - \sum _{j=n+1}^{2n-1}a_{i}^{j} u_{j}, \quad (i\in I_{n}), \end{aligned}$$

(2.1)

in which $a_{i}^{j}=\langle \phi _{i},\phi _{j}\rangle$, for $n+1\le j\le 2n-1$. Hence, every dual frame is uniquely constructed by the following vector

$$\begin{aligned} {\mathcal {U}} = \left[ u_{n+1},..., u_{2n-1}\right] \in {\mathbb {R}}^{n(n-1)}. \end{aligned}$$

(2.2)

Considering $D_{\Phi }$ as a metric space by $d(G, H)=\sum _{i\in I_{2n-1}}\Vert g_{i}-h_{i} \Vert$, we define the mapping $\xi :D_{\Phi }\rightarrow {\mathbb {R}}^{n(n-1)}$ by $\xi (G) = {\mathcal {U}}_{g}$, where ${\mathcal {U}}_{g}\in {\mathbb {R}}^{n(n-1)}$ is the unique sequence associated to $G\in D_{\Phi }$ as in (2.2). Then, clearly $\xi$ is well-defined and injective. Also, take ${\mathcal {A}} \in {\mathbb {R}}^{n(n-1)}$, then put $u_{n+1}=\{{\mathcal {A}}_{1},..., {\mathcal {A}}_{n}\}$,..., $u_{2n-1}=\{{\mathcal {A}}_{n^{2}-2n+1},..., {\mathcal {A}}_{n(n-1)}\}$ and construct $u_{i}$, $i\in I_{n}$ by (2.1). Thus we get $\{S_{\Phi }^{-1}\phi _{i}+u_{i}\}_{i\in I_{2n-1}} \in D_{\Phi }$, i.e., $\xi$ is a bijective map. Moreover, $\xi$ is a Lipschitz function. Let $G=\{S_{\Phi }^{-1}\phi _{i}+u_{i}\}_{i\in I_{2n-1}}$ and $H=\{S_{\Phi }^{-1}\phi _{i}+v_{i}\}_{i\in I_{2n-1}}$ be dual frames of $\Phi$ with the associated ${\mathcal {U}}_{g}$ and ${\mathcal {U}}_{h}$ obtained as in (2.2), respectively. Then,

$$\begin{aligned} \Vert \xi (G)-\xi (H)\Vert = \Vert {\mathcal {U}}_{g}-{\mathcal {U}}_{h}\Vert = \sum _{i=n+1}^{2n-1} \Vert u_{i}-v_{i}\Vert \le \sum _{i\in I_{2n-1}} \Vert u_{i}-v_{i}\Vert =d(G, H). \end{aligned}$$

What is more, $\xi$ is a bi-Lipschitz function, but we will not need this fact. Now, we note that the only cases in which a dual frame $\{g_{i}\}_{i\in I_{2n-1}}$ fails to do phase retrieval is associated to $det [g_{i_{1}}\vert \quad ... \quad \vert g_{i_{n}}] =0$ for some index set $\{i_{j}\}_{j=1}^{n}\subset I_{2n-1}$, by Theorem 1.4. Multiplying these equations yields an $n(n-1)$-variable polynomial in terms of $u_{n+1}$,..., $u_{2n-1}$, denoted by $P_{\mathcal {\phi }}$. Therefore,

$$\begin{aligned} PD_{\Phi }=\xi ^{-1}\{{\mathcal {U}}\in {\mathbb {R}}^{n(n-1)}: P_{\mathcal {\phi }}({\mathcal {U}})\ne 0\}, \end{aligned}$$

(2.3)

that means $PD_{\Phi }$ is open in $D_{\Phi }$. To show $PD_{\Phi }$ is, moreover, dense in $D_{\phi }$, let $\epsilon >0$ and $G=\{g_{i}\}_{i\in I_{2n-1}} = \{S_{\Phi }^{-1}\Phi _{i}+u_{i}\}_{i\in I_{2n-1}}$ be a dual frame in $D_{\Phi }{\setminus } PD_{\Phi }$. Then G is dependent just in vectors $\{u_{i}\}_{i=n+1}^{2n-1}$ such that $P_{\mathcal {\phi }}(\{u_{i}\}_{i=n+1}^{2n-1})=0$. Take a sequence $\{\{v_{i}^{k}\}_{i=n+1}^{2n-1}\}_{k\in {\mathbb {N}}}$ out of the roots of the polynomial $P_{\mathcal {\phi }}$ in ${\mathbb {R}}^{n(n-1)}$ so that $\lim _{k\rightarrow \infty } \{v_{i}^{k}\}_{i=n+1}^{2n-1}=\{u_{i}\}_{i=n+1}^{2n-1}$. Also, put $v_{i}^{k}=-\sum _{j=n+1}^{2n-1} a_{i}^{j} v_{j}^{k}$, for all $1\le i\le n$ and $k\in {\mathbb {N}}$. Then, $\{v_{i}^{k}\}_{i\in I_{2n-1}}$ satisfies (2.1) and so the sequence $\{h_{i}^{k}\}_{k\in {\mathbb {N}}, i\in I_{2n-1}}$ associated to $\{\{v_{i}^{k}\}_{i=n+1}^{2n-1}\}_{k\in {\mathbb {N}}}$ defined by $h_{i}^{k}= S_{\Phi }^{-1}\Phi _{i}+v_{i}^{k}$, $i\in I_{2n-1}$ constitutes a dual frame of $\Phi$, for all $k\in {\mathbb {N}}$. Furthermore, there exists $k_{0}\in {\mathbb {N}}$ so that for $k \ge k_{0}$

$$\begin{aligned} d\left( \{v_{i}^{k}\}_{i=n+1}^{2n-1}\},\{u_{i}\}_{i=n+1}^{2n-1}\}\right) = \sum _{i=n+1}^{2n-1}\Vert v_{i}^{k}-u_{i}\Vert < \epsilon . \end{aligned}$$

Hence, we can write

$$\begin{aligned} d(\{h_{i}^{k}\}_{k\in {\mathbb {N}}, i\in I_{2n-1}},\{g_{i}\}_{i\in I_{2n-1}})= & {} \sum _{i\in I_{2n-1}} \Vert v_{i}^{k}-u_{i}\Vert \\= & {} \sum _{i\in I_{n}} \Vert v_{i}^{k}-u_{i}\Vert +\sum _{i=n+1}^{2n-1} \Vert v_{i}^{k}-u_{i}\Vert \\= & {} \sum _{i\in I_{n}} \Vert -\sum _{j=n+1}^{2n-1}a_{i}^{j}(v_{j}^{k}-u_{j})\Vert +\sum _{i=n+1}^{2n-1} \Vert v_{i}^{k}-u_{i}\Vert \\\le & {} \sum _{i\in I_{n}}\sum _{j=n+1}^{2n-1}\vert a_{i}^{j}\vert \Vert v_{j}^{k}-u_{j}\Vert +\sum _{i=n+1}^{2n-1} \Vert v_{i}^{k}-u_{i}\Vert \\\le & {} (\alpha +1) \sum _{j=n+1}^{2n-1} \Vert v_{j}^{k}-u_{j}\Vert < \epsilon (\alpha +1), \end{aligned}$$

for $k \ge k_{0}$, i.e., $PD_{\Phi }$ is dense in $D_{\Phi }$.

Now, applying Lemma 2.1 and Remark 2.2 gives the result in general case. In fact, every frame $\Psi = \{\psi _{i}\}_{i\in I_{2n-1}}$ is equivalent to a frame in the form of $\Phi = \{\delta _{i}\}_{i\in I_{n}}\cup \{\phi _{i}\}_{i=n+1}^{2n-1}$, as we discussed in Remark 2.2. So, there exists an invertible operator ${\mathcal {T}}$ on ${\mathbb {R}}^{n}$ so that $\Psi ={\mathcal {T}}\Phi$, consequently by Lemma 2.1 we get

$$\begin{aligned} \overline{PD_{\Psi }}= \overline{PD_{{\mathcal {T}}\Phi }} = \overline{({\mathcal {T}}^{*})^{-1} PD_{\Phi }}\supseteq ({\mathcal {T}}^{*})^{-1}\overline{ PD_{\Phi }}=({\mathcal {T}}^{*})^{-1}D_{\Phi }=D_{\Psi }, \end{aligned}$$

i.e., $\overline{PD_{\Psi }}=D_{\Psi }$. And using (2.3)

$$\begin{aligned} PD_{\Psi }= ({\mathcal {T}}^{*})^{-1} PD_{\Phi } =({\mathcal {T}}^{*})^{-1}\xi ^{-1}\{{\mathcal {U}}\in {\mathbb {R}}^{n(n-1)}: P_{\mathcal {\phi }}({\mathcal {U}})\ne 0\}, \end{aligned}$$

i.e., $PD_{\Psi }$ is open in $D_{\Psi }$. This completes the proof. $\square$

An analogous approach to the proof of Theorem 2.3 deduces that for any full spark frame $\Phi$ in ${\mathbb {R}}^{n}$ with $E(\Phi )=1$, the set of all full spark dual frames is embedded into a generic set in ${\mathbb {R}}^{n}$.

Corollary 2.4

Let $\Phi =\{\phi _{i}\}_{i=1}^{n+1}$ be a full spark frame for ${\mathbb {R}}^{n}$. Then the set of all full spark dual frames of $\Phi$ is an open and dense subset of $D_{\Phi }$ and is embedded into a generic set in ${\mathbb {R}}^{n}$.

Proof

Applying Remark 2.2, a full spark frame $\Phi =\{\phi _{i}\}_{i=1}^{n+1}$ of ${\mathbb {R}}^{n}$ is equivalent to ${\tilde{\Phi }}:=\{\delta _{i}\}_{i=1}^{n}\cup \{\sum _{i=1}^{n}\alpha _{i}\delta _{i}\}$ for non-zero scalars $\alpha _{i}$, $i\in I_{n}$, and in this case the frame operator is as follows:

$$\begin{aligned} S_{{\tilde{\Phi }}} = \left[ \begin{array}{ccc} 1+\alpha _{1}^{2} \quad \alpha _{1}\alpha _{2} \quad ... \quad \alpha _{1}\alpha _{n}\\ \\ \alpha _{1}\alpha _{2} \quad 1+\alpha _{2}^{2} \quad ... \quad \alpha _{2}\alpha _{n}\\ .\\ .\\ \alpha _{1}\alpha _{n}\quad \alpha _{2}\alpha _{n} \quad ... \quad 1+\alpha _{n}^{2} \\ \end{array} \right] \end{aligned}$$

Also, every dual frame of ${\tilde{\Phi }}$ is in the form of $\{S_{{\tilde{\Phi }}}^{-1}{\tilde{\phi }}_{i}+u_{i}\}_{i=1}^{n+1}$ so that $T_{u}T^{*}_{{\tilde{\Phi }}}=0$. By taking $u_{n+1}=[x_{1},..., x_{n}]^{T}$, we get $u_{i}=-\alpha _{i}u_{n+1}$ for all $i\in I_{n}$. Hence, $D_{{\tilde{\Phi }}}$ is the set of all $\{g_{i}\}_{i=1}^{n+1}$ so that

$$\begin{aligned} g_{i}^{k} ={\left\{ \begin{array}{ll} \begin{array}{ccc} -\alpha \alpha _{i}\alpha _{k}-\alpha _{i}x_{k}&{} \; {k\ne i}, \\ \alpha (1+\sum _{j\ne i}\alpha _{j}^{2})-\alpha _{i}x_{k}&{} \; {k=i}, \\ \end{array} \end{array}\right. } \end{aligned}$$

where $\alpha =\dfrac{1}{det S_{{\tilde{\Phi }}}}=\dfrac{1}{1+\sum _{i=1}^{n}\alpha _{i}^{2}}$, $g_{i}^{k}$ denotes $k^{th}$ coordinate of $g_{i}$, $i\in I_{n}$, and

$$\begin{aligned} g_{n+1}=\{\alpha \alpha _{i}+x_{i}\}_{i=1}^{n}. \end{aligned}$$

This shows that, every dual frame is associated to n-variable $\{x_{i}\}_{i\in I_{n}}$ and a dual frame is full spark frame except the cases that the sequence $u_{n+1}=\{x_{i}\}_{i\in I_{n}}$ belongs to the roots of the polynomial constructed by $det [g_{i_{1}}\vert \quad ... \quad \vert g_{i_{n}}]=0$, for some index set $\{i_{j}\}_{j=1}^{n}\subset I_{n+1}$. Multiplying these $n+1$ polynomials yield an n-variable polynomial $P_{{\tilde{\Phi }}}(x_{1},...,x_{n})$. So, every full spark dual frame of ${\tilde{\Phi }}$ is obtained by a generic choice of $u_{n+1}$ out of the roots of the polynomial $P_{{\tilde{\Phi }}}$. Hence, by a similar approach to the proof of Theorem 2.3, and considering the bijection map $\xi :D_{{\tilde{\Phi }}}\rightarrow {\mathbb {R}}^{n}$ defined as $\xi (G) = u_{n+1}$, where $G=\{S_{{\tilde{\Phi }}}^{-1}{\tilde{\phi }}_{i}+u_{i}\}_{i=1}^{n+1}$, the complement of all full spark dual frames of $\Phi$ in $D_{\Phi }$ is equivalent to the set of roots of $P_{{\tilde{\Phi }}}$, and so, the set of full spark dual frames of $\Phi$ is embedded into a generic set in ${\mathbb {R}}^{n}$ by $\xi$. In general case, if $\Phi ={\mathcal {T}}{\tilde{\Phi }}$, for an invertible operator ${\mathcal {T}}$ on ${\mathbb {R}}^{n}$ by using Lemma 2.1, the set of full spark dual frames of $\Phi$, $FD_{\Phi }$, is the same $({\mathcal {T}}^{*})^{-1}FD_{{\tilde{\Phi }}}$, and this completes the proof. $\square$

2.1 Examples

Example 2.5

Suppose that $\phi =\{\phi _{i}\}_{i=1}^{3}$ is a full spark frame for ${\mathbb {R}}^{2}$. Since $E(\mathcal {\phi })=1$ and equivalent frames do the same phase retrieval we assume that $\phi$ is equivalent to $\{\delta _{1},\delta _{2},\alpha _{1}\delta _{1}+\alpha _{2}\delta _{2}\}$, in which both $\alpha _{1}$ and $\alpha _{2}$ are non-zero. In this case,

$$\begin{aligned} D_{\phi } = \left\{ \left[ \begin{array}{ccc} \alpha (1+\alpha _{2}^{2})-\alpha _{1}x\\ \\ -\alpha \alpha _{1}\alpha _{2}-\alpha _{1}y \\ \end{array} \right] , \left[ \begin{array}{ccc} -\alpha \alpha _{1}\alpha _{2}-\alpha _{2}x\\ \\ \alpha (1+\alpha _{1}^{2})-\alpha _{2}y \\ \end{array} \right] , \left[ \begin{array}{ccc} \alpha \alpha _{1}+x\\ \\ \alpha \alpha _{2}+y \end{array} \right] ; x,y\in {\mathbb {R}}\right\} \end{aligned}$$

where $\alpha =\dfrac{1}{1+\alpha _{1}^{2}+\alpha _{2}^{2}}$. It is worthy of note that, all dual frames in this case are full spark and so phase retrieval except dual frames obtained by $(x,y)\in {\mathbb {R}}^{2}$ on three distinct lines with slopes of 0, $\infty$ and $-\alpha _{1}/\alpha _{2}$ such as $x=-\alpha \alpha _{1}$, $y=-\alpha \alpha _{2}$ and $\alpha _{1}x+\alpha _{2}y=\alpha$. Considering

$$\begin{aligned} (x,y)\in {\mathbb {R}}^{2}\setminus \{(x,y): (x+\alpha \alpha _{1})(y+\alpha \alpha _{2})(\alpha _{1}x+\alpha _{2}y-\alpha )=0\} \end{aligned}$$

we get a generic choice of $\{u_{i}\}_{i\in I_{4}}$ in ${\mathbb {R}}^{2}$. The fact that all dual frames of $\phi$ are translations of this family by $\{S_{\phi }^{-1}\phi _{i}\}_{i\in I_{4}}$ shows that full spark dual frames are dense in $D_{\Phi }$ and full measure. As a special case, considering $\alpha _{1}=\alpha _{2}=1$ we get the phase retrieval frame $\phi =\{\delta _{1},\delta _{2},\delta _{1}+\delta _{2}\}$ and we can see that dual frames of $\Phi$ do phase retrieval for all $(x,y)\in {\mathbb {R}}^{2}$ except on the lines $x=\dfrac{-1}{3}$, $y=\dfrac{-1}{3}$ and $y=\dfrac{1}{3}-x$. Put $x=0$, $y=2/3$ we get a phase retrieval dual $\Psi$ and $x=1, y=\dfrac{-2}{3}$, satisfying in $y=\dfrac{1}{3}-x$, gives a non-phase retrieval dual frame G. See Fig. 1.

Fig. 1

(Non) phase retrieval dual frames

Example 2.6

Let $\phi =\{\phi _{i}\}_{i=1}^{4}$ be a full spark frame for ${\mathbb {R}}^{3}$. In this case $\phi$ is equivalent to the frame

$$\begin{aligned} \phi := \{\delta _{1},\delta _{2}, \delta _{3}, \alpha _{1}\delta _{1}+\alpha _{2}\delta _{2}+\alpha _{3}\delta _{3}\quad \alpha _{1},\alpha _{2},\alpha _{3}\ne 0\}. \end{aligned}$$

The set of all dual frames of $\phi$ is in the form of $D_{\phi }=\{g_{1},g_{2},g_{3},g_{4}\}$ where

$$\begin{aligned} g_{1}= \left[ \begin{array}{ccc} \alpha (1+\alpha _{2}^{2}+\alpha _{3}^{2})-\alpha _{1}x\\ \\ -\alpha \alpha _{1}\alpha _{2}-\alpha _{1}y \\ \\ -\alpha \alpha _{1}\alpha _{3}-\alpha _{1}z\\ \end{array} \right] , g_{2}=\left[ \begin{array}{ccc} -\alpha \alpha _{1}\alpha _{2}-\alpha _{2}x\\ \ \\ \alpha (1+\alpha _{1}^{2}+\alpha _{3}^{2})-\alpha _{2}y \\ \\ -\alpha \alpha _{2}\alpha _{3}-\alpha _{2}z\\ \end{array} \right] , \end{aligned}$$

$$\begin{aligned} g_{3}=\left[ \begin{array}{ccc} -\alpha \alpha _{1}\alpha _{3}-\alpha _{3}x\\ \\ -\alpha \alpha _{2}\alpha _{3}-\alpha _{3}y\\ \\ \alpha (1+\alpha _{1}^{2}+\alpha _{3}^{2})-\alpha _{3}z\\ \end{array} \right] , g_{4}=\left[ \begin{array}{ccc} \alpha \alpha _{1}+x\\ \\ \alpha \alpha _{2}+y\\ \\ \alpha \alpha _{3}+z \end{array} \right] \end{aligned}$$

where $\alpha =\dfrac{1}{1+\alpha _{1}^{2}+\alpha _{2}^{2}+\alpha _{3}^{2}}$ and $x,y,z\in {\mathbb {R}}$ are arbitrary. All dual frames of $\Phi$ are full spark except the cases $det [g_{i}\vert g_{j} \vert g_{k}]=0$, for $i,j,k\in I_{4}$, in which (x, y, z) is belong to the four distinct planes $x=-\alpha \alpha _{1}$, $y=-\alpha \alpha _{2}$, $z=-\alpha \alpha _{3}$ and $\alpha _{1}x+\alpha _{2}y+\alpha _{3}z=\alpha$. In a similar way, by choosing

$$\begin{aligned} (x,y,z)\in {\mathbb {R}}^{3}\setminus \{(x,y,z): (x+\alpha \alpha _{1})(y+\alpha \alpha _{2})(z+\alpha \alpha _{3}) (\alpha _{1}x+\alpha _{2}y+\alpha _{3}z-\alpha )=0\} \end{aligned}$$

we can say that full spark dual frames are translations of the canonical dual by a generic choice of the family $\{u_{i}\}_{i\in I_{4}}$.

Example 2.7

Consider $\phi =\{\delta _{1},\delta _{2},\delta _{3},\sum _{i=1}^{3}\delta _{i}, \delta _{1}-\delta _{2}+\delta _{3}\}$ as a full spark frame for ${\mathbb {R}}^{3}$. In this case, all dual frames are constructed by a generic choice of $\left[ \begin{array}{ccc} u_{4}\\ u_{5} \\ \end{array} \right] \in {\mathbb {R}}^{6}$. In fact, by putting $u_{4}=[x_{1}\quad y_{1} \quad z_{1}]^{T}$ and $u_{5}=[x_{2}\quad y_{2} \quad z_{2}]^{T}$, the set of all dual frames of $\phi$, are given by

$$\begin{aligned} g_{1}= \left[ \begin{array}{ccc} \dfrac{3}{5}-x_{1}-x_{2}\\ \\ -y_{1}-y_{2} \\ \\ \dfrac{-2}{5}-z_{1}-z_{2}\\ \end{array} \right] , g_{2}=\left[ \begin{array}{ccc} x_{2}-x_{1}\\ \\ \dfrac{1}{3}+y_{2}-y_{1} \\ \\ -z_{2}-z_{1}\\ \end{array} \right] , \end{aligned}$$

$$\begin{aligned} g_{3}=\left[ \begin{array}{ccc} \dfrac{-2}{5}-x_{1}-x_{2}\\ \\ -y_{1}-y_{2} \\ \\ \dfrac{3}{5}-z_{1}-z_{2}\\ \end{array} \right] , g_{4}=\left[ \begin{array}{ccc} \dfrac{1}{5}+x_{1}\\ \\ \dfrac{1}{3}+y_{1}\\ \\ \dfrac{1}{5}+z_{1}\\ \end{array} \right] , g_{5}=\left[ \begin{array}{ccc} \dfrac{1}{5}+x_{1}\\ \\ \dfrac{-1}{3}+y_{2}\\ \\ \dfrac{1}{5}+z_{2}\\ \end{array} \right] \end{aligned}$$

where $x_{1},x_{2},y_{1},y_{2},z_{1},z_{2}$ are obtained arbitrarily from ${\mathbb {R}}$. And all cases in which a dual frame fails to do phase retrieval is associated to the roots of a 6-variable polynomial in ${\mathbb {R}}^{6}$ given by multiplying of the polynomials as $det [g_{i_{1}}\vert g_{i_{2}}\vert g_{i_{3}}] =0,$ for all index set $\{i_{j}\}_{j=1}^{3}\subset I_{5}$. As one case, $det [g_{2}\vert g_{4} \vert g_{5}] =0$ deduces that

$$\begin{aligned} z_{1}+x_{2}-\dfrac{3x_{1}}{5}-\dfrac{3z_{2}}{5}+3x_{2}z_{1}-3x_{1}z_{2}=0, \end{aligned}$$

and the roots of this equation, which are associated to a family of non-phase retrieval dual frames, constitute a surface in ${\mathbb {R}}^{3}$.

3 Weak Phase Retrieval

The main result of this section is to obtain some equivalent conditions on a family of vectors to do phase retrieval in terms of weak phase retrieval. This also derives a relationship between, phase, weak phase and norm retrieval. First we present some properties of a family of vectors to do weak phase retrieval.

Proposition 3.1

Assume that $\Phi =\{\mathcal {\phi }_{i}\}_{i\in I_{m}}$ is a frame in ${\mathcal {H}}_{n}$. Then $\Phi$ does weak phase retrieval if and only if $P\Phi$ does weak phase retrieval, for every 2-dimensional orthogonal projection P on ${\mathcal {H}}_{n}$.

Proof

In case $\Phi$ does weak phase retrieval, it is simple to see that $P\Phi$ does weak phase retrieval, for every orthogonal projection P on ${\mathcal {H}}_{n}$, as well, see also [1]. Conversely, let $x,y\in {\mathcal {H}}_{n}$ and $\vert \langle x,\phi _{i}\rangle \vert =\vert \langle y, \phi _{i}\rangle \vert$. Then by the assumption that $P\Phi$ does weak retrieval for the projection P of ${\mathcal {H}}_{n}$ onto the closed subspace $span\{x,y\}$, implies that x and y weakly have the same phase. $\square$

It is shown that [1] a weak phase retrieval family in ${\mathbb {R}}^{n}$ spans the space that means such a family constitutes a frame for ${\mathbb {R}}^{n}$. In the complex case ${\mathbb {C}}^{n}$ there is no result available. In the following, we present sufficient condition in this regard which is also useful for the main result of this section.

Proposition 3.2

Let $\{\phi _{i}\}_{i\in I_{m}}$ be a family of vectors in ${\mathbb {C}}^{n}$ so that $U\Phi$ does weak phase retrieval for every unitary operator U on ${\mathbb {C}}^{n}$. Then $\{\phi _{i}\}_{i\in I_{m}}$ is a frame for ${\mathbb {C}}^{n}$.

Proof

Assume that there exists an element $x\in \{\phi _{i}\}_{i\in I_{m}}^{\perp }$, and get an ONB for ${\mathbb {C}}^{n}$ as $\{e_{i}\}_{i\in I_{n}}$ with $e_{1}=\dfrac{x}{\Vert x \Vert }$. Also, take $0\ne y\in span\{\phi _{i}\}_{i\in I_{m}}$ so there exists $\{\beta _{i}\}_{i=2}^{n}\subset {\mathbb {C}}$, with some non-zero elements such that $y=\sum _{i=2}^{n}\beta _{i}e_{i}$. Define

$$\begin{aligned} \mu : {\mathbb {C}}^{n}\rightarrow {\mathbb {C}}^{n}; \quad \mu (e_{i})=\delta _{i} \end{aligned}$$

where $\{\delta _{i}\}_{i\in I_{n}}$ is the standard ONB of ${\mathbb {C}}^{n}$. Clearly $\mu$ is a unitary operator and

$$\begin{aligned} \mu (x+y)=\mu (x)+\mu (y)=\Vert x\Vert \delta _{1}+\sum _{i=2}^{n}\beta _{i}\delta _{i}, \quad \mu (-x+y)=-\Vert x\Vert \delta _{1}+\sum _{i=2}^{n}\beta _{i}\delta _{i}. \end{aligned}$$

On the other hand,

$$\begin{aligned} \vert \langle \mu (x+y),\mu \phi _{i}\rangle \vert= & {} \vert \langle x+y,\phi _{i}\rangle \vert \\= & {} \vert \langle -x+y, \phi _{i}\rangle \vert \\= & {} \vert \langle \mu (-x+y),\mu \phi _{i}\rangle \vert , \end{aligned}$$

for all $i \in I_{m}$. The assumption that $\mu \Phi$ does weak phase retrieval deduces that $\mu (x+y)$ and $\mu (-x+y)$ weakly have the same phase that is impossible, unless $x=0$. Therefore, $\Phi$ is a frame for ${\mathbb {C}}^{n}$. $\square$

Theorem 3.3

Let $\Phi =\{\mathcal {\phi }_{i}\}_{i\in I_{m}}$ be a family of vectors in ${\mathcal {H}}_{n}$. Then the followings are equivalent;

• (i) $\Phi$ does phase retrieval.

• (ii) $U\Phi$ does phase retrieval for every invertible operator U on ${\mathcal {H}}_{n}$.

• (iii) $U\Phi$ does norm retrieval for every invertible operator U on ${\mathcal {H}}_{n}$.

• (iv) $U\Phi$ does weak phase retrieval for every invertible operator U on ${\mathcal {H}}_{n}$.

Proof

The equivalency of (i), (ii), and (iii) was proven in [9]. Also, we clearly have $(i) \Rightarrow (ii) \Rightarrow (iv)$. So, it is sufficient to show that $(iv) \Rightarrow (i)$. Assuming that $U\Phi$ does weak phase retrieval for all invertible operators on ${\mathcal {H}}_{n}$, specially we imply that $\mathcal {\phi }$ does weak phase retrieval. Moreover, $\Phi$ is a frame for ${\mathcal {H}}_{n}$, i.e., $\Phi$ spans ${\mathcal {H}}_{n}$ by Proposition 3.2. In the sequel, we are going to show that $\Phi$ yields phase retrieval. On the contrary, let there exist non-zero elements x and y in ${\mathcal {H}}_{n}$ so that

$$\begin{aligned} \vert \langle x,\phi _{i}\rangle \vert =\vert \langle y,\phi _{i}\rangle \vert , \quad (i\in I_{m}) \end{aligned}$$

(3.1)

but $y \ne c x$ for any $\vert c\vert = 1$. Since $\Phi$ does weak phase retrieval, there exists some $\vert \theta \vert =1$ so that $\theta phase(x_{j})=phase (y_{j})$ for all $j\in I_{n}$. The assumption that $y\ne \theta x$ implies $\vert x_{j}\vert \ne \vert y_{j}\vert$ for some $j\in I_{n}$. Since $\Phi$ is a frame, the equality in (3.1) is non-zero for some i then $y= cx$ immediately implies that x and y are linearly independent and we can consider a basis for ${\mathcal {H}}_{n}$ containing x and y as $\{x,y, e_{3},..., e_{n}\}$. We face the following cases

Case 1. If x and y are disjointly supported, then there exist no non-zero common coordinates. Without loss of generality we let $x_{1}, y_{2} \ne 0$, and so $x_{2}= y_{1} = 0$. Define $U:{\mathcal {H}}_{n}\rightarrow {\mathcal {H}}_{n}$ so that $Ux= x- y$, $Uy= x+ y$ and $Ue_{k}=e_{k}$, $3\le k\le n$. Then, U is a bounded invertible operator on ${\mathcal {H}}_{n}$ and

$$\begin{aligned} \vert \langle Ux,(U^{-1})^{*}\phi _{i}\rangle \vert =\vert \langle Uy,(U^{-1})^{*}\phi _{i}\rangle \vert , \quad (i \in I_{m}), \end{aligned}$$

by (3.1) and the invertibility of U. However, Ux and Uy have not weakly the same phase due to $phase(U x)_{1}=phase (Uy)_{1}$ and $phase(U x)_{2}= e^{i\pi }phase (Uy)_{2}$. This contradicts the assumption that $(U^{-1})^{*}\Phi$ does weak phase retrieval.

Case 2. Let x and y have one non-zero coordinate in common in ith index. If $y_{l}= 0=x_{l}$ for all $l\ne i$ then by (3.1) and using the assumption $\vert x_{i}\vert =\vert y_{i}\vert$, i.e., $y=\theta x$ that is a contradiction. Thus, there exists $l\ne i$ so that $y_{l}\ne 0$ or $x_{l}\ne 0$. Without loss of generality we let

$y_{l}\ne 0$ and $\dfrac{\vert x_{l}\vert }{\vert y_{l}\vert }< \dfrac{\vert x_{i}\vert }{\vert y_{i}\vert }$. In fact, if in all common non-zero elements $\dfrac{\vert x_{i}\vert }{\vert y_{i}\vert }=c$ then $y=\dfrac{\theta }{c}x$, which contradicts by the assumption.

So, we get $\epsilon >0$ so that $\dfrac{\vert x_{l}\vert }{\vert y_{l}\vert }<\epsilon < \dfrac{\vert x_{i}\vert }{\vert y_{i}\vert }$. Define $U:{\mathcal {H}}_{n}\rightarrow {\mathcal {H}}_{n}$ so that $Ux=\theta x-\epsilon y$, $Uy=\theta x+\epsilon y$ and $Ue_{k}=e_{k}$, $3\le k\le n$. Moreover,

$$\begin{aligned} (Ux)_{j}=(\vert x_{j}\vert - \epsilon \vert y_{j}\vert )\alpha _{j}\theta , \quad (Uy)_{j}=(\vert x_{j}\vert +\epsilon \vert y_{j}\vert )\alpha _{j}\theta . \quad (j\in I_{n}) \end{aligned}$$

where $\alpha _{j}=phase(x_{j})$. Hence, we obtain $phase (Ux)_{i}=phase (Uy)_{i}$, however $phase (Ux)_{l} =e^{i\pi }phase (Uy)_{l}$. That leads to a contradiction similar to the previous case, as required. $\square$

The following result gives a sufficient condition on a family of frames so that their canonical dual also yields weak phase retrieval.

Proposition 3.4

Let $\Phi =\{\mathcal {\phi }_{i}\}_{i\in I_{m}}$ be a frame in ${\mathcal {H}}_{n}$ with diagonal frame operator. Then $\Phi$ does weak phase retrieval if and only if its canonical dual does so.

Proof

Suppose that $\alpha _{1},..., \alpha _{n}$ be the diagonal elements of $S_{\Phi }$, respectively. Obviously, $\alpha _{i}> 0$, for all $i\in I_{n}$. Let $\Phi$ does weak phase retrieval, take $x,y\in {\mathcal {H}}_{n}$ such that $\vert \langle x, S_{\Phi }^{-1}\phi _{i}\rangle \vert =\vert \langle y, S_{\Phi }^{-1}\phi _{i}\rangle \vert$. Then, we get $S_{\Phi }^{-1}x=(x_{1}/ \alpha _{1},...,x_{n}/ \alpha _{n})$ and $S_{\Phi }^{-1}y=(y_{1}/\alpha _{1},..., y_{n}/\alpha _{n})$ weakly have the same phase. Consequently, x and y weakly have the same phase, as well. The converse is implied by a similar explanation. $\square$