Better proof of $f_{i}(P)=P(i)\left(x_{i}\right)$ for arbitrary scalars $x_{0}, \ldots, x_{n}$ is a basis for $\left(\mathbb{K}_{n}[X]\right)^{*}$

Question

I proved the above fact by using $P=\sum_{k=0}^{n} \frac{\left(D^{k} P\right)}{k !}(0)(x)^{k}$ and showing that each ${\left(D^{k} P\right)}(0)$ is a linear combination of $f_{i}(P)$ for i=k to n, starting from k=n to 0. Since I know that ${\left(D^{k} P\right)}(0)$ is a basis for $\left(\mathbb{K}_{n}[X]\right)^{*}$, I know that $f_{i}(P)$ is also a basis. However, I think this proof is messy and ugly. Can anyone give me a better proof?

Answer 1

I assume that you mean $P(x_i)$ (and not $P(i)(x_i)$, which does not make sense), and that $x_0,\ldots,x_n$ are pairwise distinct (otherwise the result is obviously false).

Since $(f_0,\ldots,f_n)$ has cardinality $n+1$, which is the dimension of $(\mathbb{K}_n[X])^*$, it is enough to prove that $f_0,\ldots,f_n$ are linearly independent.

Let $a_1,\ldots,a_n\in\mathbb{K}$ such that $a_0f_0+\cdots+a_nf_n=0$.

Fix $0\leq i\leq n$, and let $L_i=\displaystyle\prod_{k\neq i}\dfrac{(X-x_k)}{x_i-x_k}$.

Then $f_j(L_i)=\delta_{ij}$ since $x_j$ is a root of $L_i$ if $j\neq i$, and $L_i(x_i)=1$ by definition.

Evaluation the dependence relation at $L_i$ yields $a_i=0$, and we are done.