= Solution
The <dual space> is $V^*=\operatorname{Hom}_F(V,F)$, the <vector space> of all <linear functionals> with pointwise operations. For finite $d=\dim V$, choose a basis $v_1,\ldots,v_d$ and define its <dual basis> by $\varepsilon_j(v_i)=\delta_{ij}$. Every <linear functional> is $\ell=\sum_j\ell(v_j)\varepsilon_j$, and evaluation on each $v_i$ proves that the $\varepsilon_j$ are <linearly independent>. Thus \b[$\dim V^*=\dim V=d$]. This dimension formula is a finite-dimensional assertion.
On the space of <polynomials> of degree at most $n$, define the evaluation map $E(p)=(p(a_0),\ldots,p(a_n))$. Its kernel is zero: a nonzero degree-at-most-$n$ <polynomial> cannot have $n+1$ distinct roots. Since domain and codomain both have dimension $n+1$, $E$ is an <isomorphism>. The evaluations therefore form a basis of the <dual space>. The <linear functional> $p\mapsto p'(0)$ has a unique expansion in that basis, proving the existence and uniqueness of the requested weights.
Explicitly, use the <Lagrange interpolation polynomial> basis
$$
\ell_j(x)=\prod_{i\ne j}\frac{x-a_i}{a_j-a_i},\qquad p(x)=\sum_{j=0}^np(a_j)\ell_j(x).
$$
The <differentiation weights from nodal evaluations> are \b[$\lambda_j=\ell_j'(0)$]. If a node is zero, no singular reciprocal formula is needed: differentiate the finite product directly. For $n=0$ the sole weight is zero.
Back to article page
