If K-independent x₁, x₂ and y₁, y₂ have products spanning a 3-dimensional K-space, then span(x) = K·a + K·ah and span(y) = K·b + K·bh for one h ∉ K

Let $K$ be a subfield of $\mathbb{C}$, and let $x_1, x_2$ and $y_1, y_2$ be complex numbers, each pair linearly independent over $K$, such that the four products $x_iy_j$ span a $K$-space of dimension $3$. Then there are $h \notin K$ and non-zero $a, b$ with $\operatorname{span}_K(x_1, x_2) = Ka + Kah$ and $\operatorname{span}_K(y_1, y_2) = Kb + Kbh$; the products span the three-term progression $ab\,(K + Kh + Kh^2)$.

The certificate $(1, u) \otimes (1, \bar u)$ of a point of a circle is of this kind, with $h = u$.

Proof. Four vectors in a space of dimension $3$ are dependent, so $x_1u' + x_2v = 0$ with $u' = g_{11}y_1 + g_{12}y_2$, $v = g_{21}y_1 + g_{22}y_2$ and $g \neq 0$; then $v \neq 0$. Put $h = x_2/x_1 \notin K$, $a = x_1$, $b = v$; then $bh = -u'$, the pair $v, -u'$ is independent (a relation $sv - tu' = 0$ gives $v(sx_1 + tx_2) = 0$), and it spans the same $2$-dimensional space as $y_1, y_2$.

Novelty. Known in general: the dimension-2 case of Bachoc, Serra and Zémor (2017), Lemmas 4–5 (there for $K$ algebraically closed in the ambient field), and of their linear Vosper theorem (Th. 33) over algebraically closed fields. Here over any subfield of $\mathbb{C}$. Not claimed.