how to prove span of Lagrange basis functions is equivalent to${\rm span}\{1,x,\dots,x^n\} $

Question

Suppose we have n+1 points $x_0, x_1 , \dots , x_n$ and the following is Lagrange basis function. $$ L_i(x):=\prod_{j=0, j \neq i}^n \frac{x-x_j}{x_i-x_j} $$

How to prove that $$ \operatorname{span}\{1,x,\dots,x^n\} = \operatorname{span}\{ L_0(x),L_1(x),\dots,L_n(x) \} $$

I've only ever taken a linear algebra course, so I can not understand the span of a set of scalars (since they are just some real numbers).

Answer 1

Define:V=span$\left\{ 1,x,x^2,\cdots ,x^n \right\} $=$\left\{ f\,\,\mathrm{is} ~\mathrm{a} ~\mathrm{polynomial}~|~deg\left( f \right) \leqslant n\,\,or\,\,f=0 \right\} $

Firstly,

$\forall f\in V$,we can get $f\left( x \right) =\sum_{i=0}^n{f\left( x_i \right) L_i\left( x \right)}$.

Because $f\left( x \right) -\sum_{i=0}^n{f\left( x_i \right) L_i\left( x \right)}$ has $n+1$ roots $x_0,x_1,\cdots,x_n$, so $f\left( x \right) -\sum_{i=0}^n{f\left( x_i \right) L_i\left( x \right)}=0$.

Secondly:

$L_0(x),L_1(x),\cdots,L_n(x)$ are linearly indenpendent.

If $~k_0L_0(x)+k_1L_1(x)+\cdots k_nL_n(x)=0$, then $$~k_0L_0(x_i)+k_1L_1(x_i)+\cdots k_nL_n(x_i)=0~,~\forall 0\le i\le n.$$

Because $L_i\left( x_j) =\delta _{ij} \right) $ ,we can get $k_i=0$,$\forall 0\le i \le n$.

So we complete the proof.

Answer 2

You’re missing context. Let $F$ be a field and $x_0,x_1,…,x_n\in F$. Let $V=\{f\in F[x]|\text{deg}(f)\leq n\}$. Define $P_i:V\to F$ such that $P_i(f)=f(x_i)\in F$, $\forall 0\leq i\leq n$. It’s easy check, each $P_i$ is a linear map. So $P_i\in L(V,F)=V^*$, $\forall i\in J_n \cup \{0\}$. Our goal is to show $\{P_0,P_1,…,P_n\}$ is basis of $V^*$, dual space of $V$.

Since $\text{deg}(L_i)=n$, we have $L_i=\prod_{j\neq i}\frac{x-x_j}{x_i-x_j}\in V$. We claim $\{L_0,…,L_n\}$ is linearly independent. If $\sum_{i=0}^nc_i\cdot L_i=0$, then $$(\sum_{i=0}^nc_i\cdot L_i)(x_j)= \sum_{i=0}^nc_i\cdot L_i(x_j)= \sum_{i=0}^nc_i\cdot \delta_{ij}=c_j=0,\ \forall j\in J_n.$$ Thus $\{L_0,…,L_n\}$ is independent. Since $\{1,x,x^2,..,x^n\}$ is basis of $V$, we have $\text{dim}(V)=n+1$. Since $|\{L_0,…,L_n\}|=n+1$ and $\{L_0,…,L_n\}$ is independent, we have $\{L_0,…,L_n\}$ is basis of $V$. So $\text{span}(\{1,x,..,x^n\})$ $=V$ $=\text{span}(\{L_0,…,L_n\})$. Your desired result.

It’s easy to check, $P_j(L_i)$ $=L_i(x_j)$ $=\delta_{ij}$, $\forall (i,j)\in J_n\times J_n$. Hence $\{P_0,…,P_n\}$ is dual basis of $\{L_0,…,L_n\}$.