The peripheral eigenvalue locus is defined over Q

Fix an automorphism $\sigma$ of $\mathbb{C}$ over $\mathbb{Q}$. If generator matrices $A_1,\dots ,A_n$ define a representation of the given presentation, then their entrywise images $\sigma A_1,\dots ,\sigma A_n$ do too: determinants and all relator equations are integer-polynomial equations by [F1]. Word evaluation commutes with $\sigma$, including inverse letters because their adjugate formulas have integer coefficients. Thus $\sigma \rho$ is again a representation, with peripheral matrices $\sigma (\rho (\mu ))$ and $\sigma (\rho (\lambda ))$.

Suppose $(M,L)\in {\mathcal{E}}_K$. A common eigenline can be represented by one nonzero vector $v\in {\mathbb{C}}^2$, with $\rho (\mu )v=Mv$ and $\rho (\lambda )v=Lv$. Applying $\sigma$ gives the same equations for $\sigma \rho$, the nonzero vector $\sigma v$, and nonzero eigenvalues $\sigma M,\sigma L$. Hence $(\sigma M,\sigma L)\in {\mathcal{E}}_K$. Applying the inverse automorphism proves equality $\sigma ({\mathcal{E}}_K)={\mathcal{E}}_K$. This treats both eigenline charts at once; it does not assume a preferred eigenvector, an algebraic section of eigenlines or a closed projection image.

By [F3], the radical vanishing ideal of the closure of ${\mathcal{E}}_K$ is the extension of its rational contraction, and that contraction is radical. This is exactly the stated meaning of being defined over $\mathbb{Q}$. In particular reduction of the image introduces no unproved field-of-definition step: it is the point-set vanishing ideal to which the finite-dimensional row argument was applied. The statement retains its full-AC branch through [F3]; all matrix and eigenvector operations in steps 1.1–2.1 are finite.