Let \(\mathcal{F}\) be the family of sets \(I \subseteq E(K_V)\) that are row-independent at some placement. It is finite, since \(E(K_V)\) is, and a placement at which every member of \(\mathcal{F}\) is row-independent is generic. We find one as in the proof of lemma 10.18. There each \((2,3)\)-sparse edge set is row-independent at some framework by Pollaczek-Geiringer’s and Laman’s characterization (theorem 10.10); here each member of \(\mathcal{F}\) is row-independent at some placement by definition.

We show, by induction on \(|\mathcal{F}'|\), that for every \(\mathcal{F}' \subseteq \mathcal{F}\) there is a placement at which every member of \(\mathcal{F}'\) is row-independent. For \(\mathcal{F}' = \emptyset \) any placement will do. Otherwise let \(I_0 \in \mathcal{F}'\), let \(p_0\) be a placement at which every member of \(\mathcal{F}' \setminus \{ I_0\} \) is row-independent, and let \(q\) be a placement at which \(I_0\) is row-independent. For \(t \in {\mathbb {R}}\), let \(p^{t} = p_0 + t\, (q - p_0)\). The rigidity row of an edge \(uv\) at a placement \(p'\) is \(x \mapsto \langle p'_u - p'_v,\, x_u - x_v \rangle \) (lemma 6.4), which depends linearly on \(p'\).

Let \(I\) be a member of \(\mathcal{F}' \setminus \{ I_0\} \), or \(I_0\), and let \(t_0\) be \(0\) or \(1\) accordingly. A set of edges is row-independent at a placement exactly when its rigidity rows there are linearly independent (lemma 9.6). Fix a basis of \((({\mathbb {R}}^{d})^{V})^{*}\). The coordinate vectors of the rows at \(p^{t}\) of the edges in \(I\) are the rows of a matrix \(A + t\, B\), where \(A\) and \(B\) do not depend on \(t\). These rows are linearly independent at \(t_0\), so some \(|I| \times |I|\) minor of \(A + t_0 B\) is nonzero. The same minor of \(A + t\, B\) is a polynomial in \(t\) that is nonzero at \(t_0\), so it has finitely many roots. At every other \(t\), the rows at \(p^{t}\) of the edges in \(I\) are linearly independent. Choose \(t\) outside the roots of these finitely many polynomials, one for each member of \(\mathcal{F}'\). Then every member of \(\mathcal{F}'\) is row-independent at \(p^{t}\).

Let \(p\) be the placement in the definition of \(\mathcal{R}_d(V)\). A set is independent in \(\mathcal{R}_d(V) = \mathcal{L}_p(V)\) if and only if it is row-independent at \(p\) (lemma 10.17). If \(I\) is row-independent at \(p\), then it is row-independent at some placement, namely \(p\). Conversely, if \(I\) is row-independent at some placement, then it is row-independent at \(p\), since \(p\) is generic (definition 24.1).

Both matroids have ground set \(E(K_V)\), so it suffices to show that they have the same independent sets. A set \(I \subseteq E(K_V)\) is independent in \(\mathcal{L}_p(V)\) if and only if it is row-independent at \(p\) (lemma 10.17). Since \(p\) is generic, this holds if and only if \(I\) is row-independent at some placement, that is, if and only if \(I\) is independent in \(\mathcal{R}_d(V)\) (lemma 24.4).

Both matroids have ground set \(E(K_V)\). A set \(I \subseteq E(K_V)\) is independent in \(\mathcal{R}_2(V)\) if and only if it is row-independent at some placement in \({\mathbb {R}}^{2}\) (lemma 24.4). The same holds in \(\mathcal{R}(V)\) (theorem 10.13), so the two matroids are equal.

The matroid \(\mathcal{R}_d(V)\) is \(\mathcal{L}_p(V)\) (lemma 24.5). It is represented by the extended rigidity rows \(\tilde r_e\) at \(p\) (definition 10.16), and \(\tilde r_e = r_e\) for every edge \(e\) of \(K_V\) (definition 10.14). In a represented matroid, the vectors of a set span a space whose dimension is the rank of the set (lemma 16.7). For the set \(E(H)\), this is the claim.

The map \(R(H, p)\) has the same rank as its transpose, whose image is spanned by the rigidity rows of \(H\) at \(p\) (lemma 9.5). For an edge \(uv\) of \(H\), the rigidity row of \(H\) at \(uv\) is \(x \mapsto \langle p_u - p_v,\, x_u - x_v \rangle \) (lemma 6.4), which is also the row of \(K_V\) at \(uv\). So the rank of \(R(H, p)\) is the dimension of the span in lemma 24.8, which is \(r_d(H)\).