Let $a_0,a_1,...,a_n\in\mathbb{R}$ distinct numbers and the polynomials:


$k$ from $0$ to $n$.

find coordinates of an arbitrary polynomial $Q\in\mathbb{R}_{\leq n}[X]$ with respect to that basis formed by those polynomials.

Let $b_0,b_1,...,b_n\in\mathbb{R}$ not necessarily distinct. show that there exists a unique polynomial of degree $n$ such that $P(a_k)=b_k\forall k=0$ to $k=n$

Which is $$P(X)=b_0P_0(X)+...+b_nP_n(X).$$

My attempt for the first one:

We can observe that $P_k(a_j)=1$ iff $k=j$ and $P_k(a_j)=0$ otherwise. If we try to associate a vector in $\mathbb{R}^n$ for every polynomial $P_k$ and form a matrix with all polynomial from $k=1$ to $k=n$, we can see that we get a matrix with $a_{i,i}=1\implies$ it's determinant its $\neq0.$ so the vectors are linear independent $\implies$ the polynomials form a basis.

Now to determine the coordinates I thought to take $Q(x)= q_0+q_1x+...+q_nx^n=\lambda_0P_0+...+\lambda_nP_n$ but that's pretty hard to compute isn't there an easier way?


Take a polynomial $Q$ of degree at most $n$, and suppose there exist scalars $\lambda_0, \ldots, \lambda_n$ such that $Q = \lambda_0 P_0 + \cdots + \lambda_n P_n$. Then it is easy to tell what these scalars are, since $$ Q(a_k) = \lambda_0 P_0(a_k) + \cdots + \lambda_n P_n(a_k) = \lambda_k P_k(a_k) = \lambda_k$$ and hence we must have that $\lambda_0 = Q(a_0), \ldots, \lambda_n = Q(a_n)$.

The above argument applied to $Q = 0$ shows that the $P_0, \ldots, P_n$ are linearly independent, and hence they are spanning since we know that $\dim \mathbb{R}_{\leq n}[x] = n + 1$.


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