Proofs from Data fitting and polynomial interpolation

Suppose a linear combination of the columns is zero. Its coefficients define

and the zero combination says $p(t_i)=0$ for every $i$. By the factor theorem, each distinct root contributes a factor $(t-t_i)$, so a non-zero polynomial of degree at most $n-1$ cannot have all $n$ roots. Hence every $\lambda_i$ is zero. The $n$ columns are independent in $\mathbb R^n$, so they also span $\mathbb R^n$, giving a unique solution to $Ax=y$.

If

evaluate at $t=t_k$. Every term vanishes except $\lambda_k\ell_k(t_k)=\lambda_k$, so $\lambda_k=0$. This holds for every $k$, so the Lagrange polynomials are linearly independent; with $n$ polynomials in $P_{n-1}$, they form a basis.