CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)

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A two-vector real family with a failed partition and an ambiguous pair

Statement refuted

In $R^{2}$ with $u \cdot v = u_{0} v_{0} + u_{1} v_{1}$ , take the real measurement family $Φ = (a_{1}, a_{2})$ with $a_{1} = (1, 0)$ and $a_{2} = (0, 1)$ , and define its intensity data by $I_{Φ} (x) = ((a_{1} \cdot x)^{2}, (a_{2} \cdot x)^{2})$ . The partition $S = {1}$ has neither side spanning $R^{2}$ . The signals $x = (1, 1)$ and $z = (- 1, 1)$ satisfy neither $z = x$ nor $z = - x$ , but $I_{Φ} (x) = I_{Φ} (z)$ .

Statement references

Real intensity data and global-sign classes

Complement property

Sign classes and well-defined intensity data

Radu Balan, Pete Casazza, and Dan Edidin, converse partition construction, Theorem 2.8 proof

Radu Balan, partition construction, Theorem 3.2 proof

Facts & Assumptions

Given: $R^{2}$ with the standard dot product, the family $a_{1} = (1, 0)$ and $a_{2} = (0, 1)$ , and the global-sign quotient used for real phase retrieval.

[F1]

Real phase retrieval via finite real measurements: intensity coordinates are $⟨ a_{i}, x ⟩^{2}$ , and the target identifies $x$ with $- x$ .

[F2]

Complement property for a real finite measurement family: the complement property requires one side of every label partition to span the full ambient space.

[F3]

The real sign quotient and its induced measurement maps: two real signal classes agree exactly when their representatives are equal or negatives.

[F4]

Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ , Linear subspace of a vector space: each span is the smallest linear subspace containing its measurement vectors; in particular, coordinate axes are subspaces and spans are closed under scalar multiplication.

Counterexample

1.1givenF2F4algebra

Let $E_{1} = {(t, 0) : t \in R}$ and $E_{2} = {(0, t) : t \in R}$ ; coordinatewise addition and scalar multiplication show these are subspaces. Since $a_{1} \in E_{1}$ , minimality of the span gives $span {a_{1}} \subseteq E_{1}$ , while every $(t, 0) = t a_{1}$ belongs to the span because it is a subspace containing $a_{1}$ . Thus $span (Φ_{{1}}) = E_{1}$ . The same two inclusions for $a_{2}$ give $span (Φ_{{2}}) = E_{2}$ . Both are proper subspaces of $R^{2}$ , so [F2] says the complement property fails.

1.2givenF3algebra

The vectors $(1, 1)$ and $(- 1, 1)$ are neither equal nor negatives: equality fails in the first coordinate, while equality with the negative fails in the second coordinate. Thus they define distinct signal classes by [F3].

2.1step 1.1step 1.2F1algebra∎

Their measurements are $((1, 1) \cdot a_{1}, (1, 1) \cdot a_{2}) = (1, 1)$ and $((- 1, 1) \cdot a_{1}, (- 1, 1) \cdot a_{2}) = (- 1, 1)$ . Squaring coordinatewise gives $(1, 1)$ in both cases, so [F1] gives equal intensity data on the distinct classes from step 1.2. Together with the failed partition in step 1.1, this exhibits the required ambiguity when the complement property fails.

Depends on

Used by

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Sources

Radu Balan, Pete Casazza, and Dan Edidin, On signal reconstruction without phase, Applied and Computational Harmonic Analysis 20(3) (2006), the partition construction in the converse proof of Theorem 2.8 (standard reference, not scraped)
Radu Balan, Frames and Phaseless Reconstruction, full lecture notes (2015), Theorem 3.2 proof (standard reference, not scraped)