My question is the following:
Let $\{a_0,a_1,...,a_n\}$ be (pairwise) distinct, real numbers. Let $V$ be the vector spaces of all polynomials of degree at most $n$, ie $V = \Bbb P_n$. Let $\phi_j : V^* \rightarrow \Bbb R$ ($V^*$ is the dual basis of $V$) be given by $\phi_j(p) = p(a_j)$.
My question is to show that $\{\phi_1, \phi_2, ..., \phi_n\}$ forms a basis for $V^*$. So far, it boils down to showing that, given $\{\nu_1,...,\nu_n\}$, $$\sum_{j=0}^n \nu_j a_j^r = 0 \ \ \ \text{for} \ r = 0, 1, ..., n \\ \Rightarrow \nu_j = 0 \ \ \ \text{for all $j$}. $$ The issue is that we have, for example, $1^2 = (-1)^2$, so when trying to show the matrix $M = (M_{ij})$ given by $M_{ij} = a_j^i$ has non-zero determinant, I am having issues.
Any help would be appreciated! Thank you! (=
[This is part of a past paper exam question - it doesn't explicitly ask me to do this, but this is part of the method that I am using. The above must be a basis because the required result implies that it is a basis - shame I can't use the answer to prove itself =P!]