# Prove: If $y_0,…,y_n$ are pairwise different real numbers, then the vectors $f_{y_0},…,f_{y_n}$ form a basis of the dual space $V^*$

could you help me with this task in linear-algebra? I do not know what to do to prove the two following statements in (i) and (ii). I would appreciate it, if you would explain to me the solution in detail, because I want to try to understand the task. I tried my best to translate the task from German into English.

Let $n \in \mathbb N$ and let $V$ be the real vector field of the polynomials of degree less or equal to $n$. Let $y \in \mathbb R$ and $f_y:V \to \mathbb R$ the function, which maps a polynomial $p(x)$ on $p(y)$ and $g_y:V \to V$ a function, which maps a polynomial $p(x)$ on $p(x-y)$. Prove the following two statements.

(i) If $y_0,...,y_n$ are pairwise different real numbers, then the vectors $f_{y_0},...,f_{y_n}$ form a basis of the dual space $V^*$.

(ii) We have $(g_{y_1})^*(f_{y_2}) = f_{y_2-y_1}$ for all $y_1,y_2 \in \mathbb R$, where $(g_{y_1})^*:V^* \to V^*$ describes the dual mapping to $g_{y_1}$.

The dimension of $V$ is $n+1$ since the polynomials $\pi_i$, $0\leq i\leq n$ wit $\pi_i(x):=x^i$ form a basis of $V$ (I hope that this has been proved prior to the excersise.) So the dimension of $V^*$ is also $n+1$. Therefore it is enough to show that the functions $f_{y_j}$, $0\leq j\leq n$ are linearly independent. So let $\sum_{j=0}^n \lambda_j f_{y_j}=0$. Then in particular $\sum_{j=0}^n \lambda_j f_{y_j}(\pi_i)=0$ for all $0\leq i\leq n$. Thus $\sum_{j=0}^n \lambda_j y_j^i=0$ for $0\leq i\leq n$, which (Vandermonde) implies that all $\lambda_j$ vanish. This proves (i).
As for (ii) note that $g_{y_1}^*(f_{y_2})=f_{y_2}\circ g_{y_1}$ which implies $g_{y_1}^*(f_{y_2})(p)=(f_{y_2}\circ g_{y_1})(p)=f_{y_2}(p(x-y_1))=p(y_2-y_1)=f_{y_2-y_1}(p)$ for all $p\in V$.