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The sharp real phase-retrieval count for full-spark measurements

Statement

Let d≥1, m≥0, and let Φ=(ai)i∈[m] be a finite family in Rd, with [m]={1,…,m} and [0]=∅. Write ⟨u,v⟩=∑k<dukvk. If m≤2d−2, then there are x,z∈Rd with z≠x and z≠−x but ⟨ai,x⟩2=⟨ai,z⟩2 for every i∈[m]. If m≥2d−1 and every list of d distinct measurement labels indexes linearly independent vectors, then equality of all these squared measurements implies z=x or z=−x. For every d≥1, this count is attained at m=2d−1: for any distinct real numbers t1,…,tm, the vectors ai=(1,ti,ti2,…,tid−1)T satisfy the stated independence condition and hence the squared measurements determine the signal up to global sign.

Facts & Assumptions

Given: A finite labeled family Φ=(ai)i∈[m] in Rd, with d≥1 and m≥0.

[F1]

Real phase retrieval is equivalent to the complement property: real intensity measurements are injective modulo global sign exactly when every partition has one side spanning Rd.

[F2]

Complement property for a real finite measurement family: the complement property is the stated condition over every subset of labels and its labeled complement.

[F3]

Full-spark real measurement families: full spark means every list of d distinct measurement labels gives a linearly independent list of vectors.

[F4]
[F8]

The Vandermonde determinant for real moment-curve vectors: for distinct real nodes t1,…,td, the d moment-curve columns (1,tj,…,tjd−1)T are linearly independent.

[F9]

The reals form a field: the real coordinate and coefficient operations satisfy the field axioms used in the finite linear maps and the Vandermonde construction.

Proof

1.1givenF4F5F9algebra

Let I⊆[m] have q labels and enumerate them in increasing order as i0,…,iq−1. Define TI:Rq→Rd by TI(c)=∑j<qcjaij. This map is linear by distributivity of finite sums. Its image is a subspace: it contains zero, and if u=TI(c) and v=TI(c′), then αu+βv=TI(αc+βc′). For each j<q, the jth standard unit vector ej satisfies TI(ej)=aij, so the smallest-subspace property gives span⁡(ΦI)⊆im⁡TI. Conversely every TI(c) is a finite linear combination of the listed measurement vectors, so [F4] gives im⁡TI⊆span⁡(ΦI). Thus the image equals the span, also when q=0 because both are {0}.

2.1step 1.1F4F5

If q<d, rank-nullity and [F5] give dim⁡(im⁡TI)≤q<d, so by step 1.1 this span cannot equal Rd.

3.1step 2.1F1F2constructalgebra

Suppose m≤2d−2. Choose S⊆[m] with ∣S∣=min⁡(m,d−1). Then ∣S∣≤d−1; if m>d−1, the bound on m gives ∣[m]∖S∣=m−d+1≤d−1, and if m≤d−1 the complement is empty. Applying step 2.1 to both sides shows that neither side spans, so the complement property fails by [F2]. Hence phase retrieval fails by [F1].

4.1step 1.1step 3.1givenF1F2F3F4F5F6F7F9choosealgebra

The lower count is established in step 3.1. For the sufficient count, suppose m≥2d−1 and Φ is full spark. For any S⊆[m], the two sides cannot both have fewer than d labels, since that would imply m≤2d−2. Thus one side contains d distinct labels. By [F3], their vectors are independent. The coefficient map T:Rd→Rd for these d columns has trivial kernel by [F6]; by step 1.1 its image is their span, and rank-nullity [F4] and [F5] give that image dimension d. Hence the image, and therefore the span of the chosen vectors, is Rd by [F7]. The containing side also spans Rd. Since S was arbitrary, the complement property holds, and [F1] gives phase retrieval.

5.1step 3.1step 4.1F3F8constructalgebra∎

Set m=2d−1 and choose distinct real nodes t1,…,tm. Every d-label subfamily is independent by [F8], so the family is full spark by [F3]. Since m≥2d−1, step 4.1 proves it performs phase retrieval. For d=1, the same argument uses one node and the one-entry vector (1); for all d, step 3.1 rules out any smaller count.

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