show that there exist unique polynomials $p_0(x),...,p_n(x)$ such that $p_i(c_j)=\delta_{ij}$ for $0 \leq i,j \leq n$.

Question

Let $V=P_n(F)$ (the vector space of polynomials with coeffficients in R of degree at most n), and let $c_0,c_1,...,c_n$ be distinct scalars in F. show that there exist unique polynomials $p_0(x),...,p_n(x)$ such that $p_i(c_j)=\delta_{ij}$ for $0 \leq i,j \leq n$.

Suppose not unique and we can find another polynomial $q_i(c_j)=\delta_{ij}$. I am told to fix i to be any real number between ${0,...,n}$ in defining the new polynomial q. Why do I need to fix it?

Answer 1

(I assume you already showed that such polynomials $p_i$ exist and that you now want to prove unicity.)

Let $V=\mathbf{K}[X]_{\leqslant n}$ be space of polynomials of degree $\leqslant n$ with coefficients in $\mathbf{K}$.

I claim that $\{p_i\}_{i=0}^n$ form a basis for $V$. We have $n+1=\dim V$ distinct polynomials and they are linearly independent. Indeed, let $\sum \lambda_i p_i=0$. Now evaluating at $c_j$ gives $\lambda_j=0$ for all $j$.

Now let $\{q_i\}_{i=0}^n$ be polynomials that also satisfy $q_i( c_j)=\delta_{ij}$. For all $0\leqslant k\leqslant n$, we can write uniquely (since $\{p_i\}_{i=0}^n$ is a basis),

$$q_k=\sum_{i=0}^n \mu_i p_i$$

Now evaluating in $c_k$ gives $1=q_k(c_k)=\mu_k$. Evaluating in $c_\ell$ for $\ell\neq k$ gives $0=\mu_\ell$. Hence $q_k=p_k$.

Answer 2

No linear algebra is requried.
Consider the polynomial $p_i(x) - q_i(x)$. This is a degree $n$ polynomial with $c_i$ giving the $n+1$ roots. But a polynomial of degree $n$ only has $n+1$ roots if the polynomial is identically $0$. So $p_i - q_i = 0$.