Lemma. Two-triangle interpolation [cn-0029]

Status:

Statement. For any distinct real nodes \(x_1,\ldots ,x_{2s+1}\), the \(s^2\) evaluation vectors [cn-0028] \(h_{pq}\), \((p,q)\in Q_s\), form a basis of \(\mathbb R^{s^2}\). Equivalently, evaluation on the two triangles determines a bivariate polynomial of degree at most \(s-1\) in each variable uniquely.

Proof. There are \(s^2\) positions and coefficients, so it is enough to show that vanishing on the pattern forces the polynomial \(P\) to be zero. Induct on \(s\). For \(s=1\), \(P\) is constant and vanishes at \((x_2,x_1)\).

For \(s>1\), the row \(p=s+1\) supplies \(s\) roots of \(P(x_{s+1},v)\), whose degree is at most \(s-1\). Hence it vanishes identically, and \(P=(u-x_{s+1})P_1\) with degree at most \(s-2\) in \(u\). The remaining positions \((p,1)\), \(2\leq p\leq s\), give \(s-1\) roots of \(P_1(u,x_1)\), since their first factors are nonzero. Therefore

\[P(u,v)=(u-x_{s+1})(v-x_1)R(u,v),\qquad \deg _u R,\deg _v R\leq s-2.\]

On the remaining triangles the two factors are nonzero, so \(R(x_p,x_q)=0\) for \(2\leq q<p\leq s\) and for \(s+2\leq p<q\leq 2s+1\). Relabel the surviving nodes as

\[(y_1,\ldots ,y_{2s-1})=(x_{s+2},\ldots ,x_{2s+1},x_2,\ldots ,x_s),\qquad S(u,v)=R(v,u).\]

Swapping the variables interchanges the remaining triangles, and this relabeling makes them exactly \(Q_{s-1}\). The induction hypothesis gives \(S=0\), hence \(R=0\) and \(P=0\). Thus the square evaluation matrix is invertible.

Proof dependencies
Bivariate space and pattern [cn-0028]
Provenance
OpenAI, 23 September 2026 [cn-0042], Lemma 4.1, equation (4.2); reconstructed as a reading account on 9 October 2026. Lean source: TwoTriangleUnisolvence.