Real phase retrieval is equivalent to the complement property

To prove that the complement property implies injectivity, assume the complement property [F3] and suppose ${\mathcal{I}}_{\mathrm{\Phi }}([x])={\mathcal{I}}_{\mathrm{\Phi }}([z])$. At every label $i$, equality of the intensity coordinates gives $⟨a_i,x{⟩}^2=⟨a_i,z{⟩}^2$.

Factoring the difference of squares over $\mathbb{R}$ yields $(⟨a_i,x-z⟩)(⟨a_i,x+z⟩)=0$ for every $i\in [m]$. This factorization also holds when $x=0$, $z=0$, or $a_i=0$; no nonzero-signal or nonzero-measurement hypothesis is used.

Put $S=\{i\in [m]:⟨a_i,x-z⟩=0\}$. For $i\in S$, $a_i\perp (x-z)$; for $i\mathrm{\notin }S$, the first factor in step 2.1 is nonzero, so $a_i\perp (x+z)$. By [F5], each vector in either span is a finite linear combination of its generators, and bilinearity extends each orthogonality statement to the whole span.

By the complement property [F3], either $\mathrm{span}({\mathrm{\Phi }}_S)={\mathbb{R}}^d$ or $\mathrm{span}({\mathrm{\Phi }}_{[m]\setminus S})={\mathbb{R}}^d$. In the first case step 3.1 makes $x-z$ orthogonal to every vector of ${\mathbb{R}}^d$, hence to itself; positive definiteness [F4] gives $x-z=0$. In the second case it gives $x+z=0$. Thus $[x]=[z]$ by [F2], proving injectivity.

Step 4.1 proves that the complement property implies injectivity. For injectivity implying the complement property, argue contrapositively and suppose the complement property fails; then some $S\subseteq [m]$ has the proper subspaces $U=\mathrm{span}({\mathrm{\Phi }}_S)$ and $V=\mathrm{span}({\mathrm{\Phi }}_{[m]\setminus S})$. We prove that every proper subspace $W\le {\mathbb{R}}^d$ has a nonzero orthogonal vector. Let $r=\dim W<d$ by [F6], choose a basis $w_0,\dots ,w_{r-1}$ of $W$, and define $L_W:{\mathbb{R}}^d\to {\mathbb{R}}^r$ by $L_W(x)=(⟨x,w_j⟩{)}_{j<r}$. Bilinearity makes $L_W$ linear. Its kernel is $W^{\perp }$: vanishing against the basis implies vanishing against every vector of $W$ because the basis spans $W$ and [F5] expresses each span vector as a finite linear combination; the converse is immediate. By [F7], the image is a subspace of ${\mathbb{R}}^r$, so [F6] gives $\mathrm{rank}{L}_W=\dim (\mathrm{im}{L}_W)\le r$. Rank-nullity [F7] and $\dim {\mathbb{R}}^d=d$ then give $\dim W^{\perp }=\dim \ker L_W=d-\mathrm{rank}{L}_W\ge d-r>0$. By [F6] a zero subspace has dimension zero, so $W^{\perp }$ contains a nonzero vector. Apply this argument to $U$ and $V$ and choose nonzero $u\in U^{\perp }$ and $v\in V^{\perp }$.

Set $x=u+v$ and $z=u-v$. Since $v\mathrm{\ne }0$, $x=z$ would imply $2v=0$, impossible by [F8]; since $u\mathrm{\ne }0$, $x=-z$ would imply $2u=0$, also impossible. Hence $[x]\mathrm{\ne }[z]$ by [F2].

If $i\in S$, then $a_i\in U$ and $u\perp a_i$, so $⟨a_i,x⟩=⟨a_i,v⟩$ and $⟨a_i,z⟩=-⟨a_i,v⟩$. If $i\mathrm{\notin }S$, then $a_i\in V$ and $v\perp a_i$, so both measurements equal $⟨a_i,u⟩$. If $a_i=0$, it lies in $U$ and both measurements are zero. Squaring shows equality of the intensity coordinates at every label, including when either component of the partition is empty.

When $m=0$, the only partition has both sides empty, and both spans equal $\{0\}\mathrm{\ne }{\mathbb{R}}^d$ by [F5] and $d\ge 1$. The complement property fails, while steps 5.1–6.2 give distinct global-sign classes with the same unique empty data. Thus the zero-measurement case is included.

Steps 6.1 and 6.2 give distinct quotient classes with the same intensity data, so ${\mathcal{I}}_{\mathrm{\Phi }}$ is not injective. Together with step 4.1 and the definition of phase retrieval [F1], this proves the stated equivalence.