Let $V=P_n(F)$ (the vector space of polynomials with coefficients in R of degree at most n) and let $c_0,c_1,…,c_n$ be distinct scalars in F. For $0 \leq i \leq n$, define $f_i \in V^{*}$ by $f_i(p(x))=p(c_i)$. Prove ${f_0,f_1,…,f_n}$ is a basis for $V^{*}$.
I feel like for this one I am not so familiar with the notations here, so $f_i$ are elements of $V^{*}$, and I am quite confused with $f_i(p(x))=p(c_i)$. Can someone first clarify that?
Best Answer
Degree-$n$ polynomials can have at most $n$ roots. If $f_0(p), \dots, f_n(p)$ are all equal to $0$, then $p$ has $n+1$ roots. Thus, $p$ is the zero polynomial. This proves that $f_0, \dots, f_n$ spans $V^\ast$.
As for linear independence, assume that $\alpha_0, \dots, \alpha_n$ are coefficients such that $\sum \alpha_i f_i(p)=0$ for all $p\in V$. This means $$ \sum \alpha_{i} p(c_i) =0 $$ no matter the choice of $p$. Try to find some contradiction, or choose a particular $p$ to work with.