Linearly independence of a family of linear functionals I'm solving the following exercise of linear algebra

Recall that $\Bbb{R}_n[X]$ denotes the space of polynomials of degree inferior or equal to $n$. Let $a_0,...,a_n\in\Bbb{R}$ be distinct real numbers. Show that there exists a unique $(n+1)$-uple $(\lambda_0,...,\lambda_n)\in\Bbb{R}^{n+1}$ such that $$P'(0)=\sum_{k=0}^{n}\lambda_kP(a_k),\ \text{for all}\ P\in\Bbb{R}_n[X].$$

My teacher sends to us the solution, but i get stacked in a part that we need to show that a family of linear functionals will be LI. I will try to explain whats going on:
We consider the evaluation maps, for all $i=0,...,n$, $$ev_{a_i}\colon\Bbb{R}_n[X]\to\Bbb{R},\ \text{such that}\ ev_{a_i}(P):=P(a_i).$$ Well, here we have $n+1$ functionals in $\Bbb{R}_n[X]^*$ (dual of $\Bbb{R}_n[X]$). Since $\dim\Bbb{R}_n[X]^*=n+1$, we wanted to show that the family $(ev_{a_i})_{i=0}^n$ is a basis for $\Bbb{R}_n[X]^*$ by proving that this family is LI. And here is my problem, i'm taking the linear combination $$\sum_{k=0}^{n}\alpha_kev_{a_k}=0$$ and trying to conclude that $\alpha_k=0,\forall k=0,...,n$, but i can't see how to obtain it. Maybe applying the equality above to some particular polynomial that gives to us what we want.
Anyway, thanks for any help!
 A: If someday anyone faces this question, the solution is here.
Solution: for each $i=0,1,...,n$ we consider the evaluation map $$\text{ev}_{a_i}\colon\Bbb{R}_n[X]\to\Bbb{R},\ \text{given by}\ \text{ev}_{a_i}(P):=P(a_i).$$ Notes that each $\text{ev}_{a_i}$ is a linear functional that belongs to $\Bbb{R}_n[X]^*$. Since $\dim{\Bbb{R}_n[X]^*}=n+1$, if we show that the family $(\text{ev}_{a_i})_{i=0}^{n}$ is linearly independent then we have that it is a basis for $\Bbb{R}_n[X]^*$. Consider the equation \begin{equation}\sum_{k=0}^{n}\alpha_k\text{ev}_{a_k}=0,\end{equation} we need to show that each $\alpha_k$ equals to zero. Usying the hypothesis that $a_0,...,a_n\in\Bbb{R}$ are all distinct and considering the polynomials $$P_0(X)=(X-a_1)\cdots(X-a_n)\quad\text{and}\quad P_k(X)=\frac{P_0(X)}{X-a_k},k=1,...,n$$ by applying the equation above for each of polynomial we get $\alpha_k=0$, for all $k=0,1,...,n$, as we wanted. Finally, take the "derivative operator in zero" $$D\colon\Bbb{R}_n[X]to\Bbb{R},\ \text{such that}\ D(P):=P'(0).$$ Is easy to check that $D$ is linear, so we get $D\in\Bbb{R}_n[X]^*$. Since $(\text{ev}_{a_i})_{i=0}^{n}$ is a basis for $\Bbb{R}_n[X]^*$, there is uniques $\lambda_0,...,\lambda_n\in\Bbb{R}$ such that $$D=\sum_{k=0}^{n}\lambda_k\text{ev}_{a_k},$$ so for all $P\in\Bbb{R}_n[X]$ $$D(P)=\sum_{k=0}^{n}\lambda_k\text{ev}_{a_k}(P)\Longleftrightarrow P'(0)=\sum_{k=0}^{n}\lambda_kP(a_k).$$
