 Solution

The dual space is $V^{\ast }={\mathrm{Hom}}_F(V,F)$, the vector space of all linear functionals with pointwise operations. For finite $d=\dim V$, choose a basis $v_1,\dots ,v_d$ and define its dual basis by ${\epsilon }_j(v_i)={\delta }_{ij}$. Every linear functional is $\ell ={\sum }_j\ell (v_j){\epsilon }_j$, and evaluation on each $v_i$ proves that the ${\epsilon }_j$ are linearly independent. Thus $\dim V^{\ast }=\dim V=d$. This dimension formula is a finite-dimensional assertion.

On the space of polynomials of degree at most $n$, define the evaluation map $E(p)=(p(a_0),\dots ,p(a_n))$. 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 $p\mapsto p^{\prime }(0)$ has a unique expansion in that basis, proving the existence and uniqueness of the requested weights.

Explicitly, use the Lagrange interpolation polynomial basisThe differentiation weights from nodal evaluations are ${\lambda }_j={\ell }_j^{\prime }(0)$. If a node is zero, no singular reciprocal formula is needed: differentiate the finite product directly. For $n=0$ the sole weight is zero.