Show set is a basis for dual space of a Vector Space

Let $V$ be a Vector Space of polynomials of degree less than or equal to n. Define $\alpha_{k}: V \rightarrow \mathbb{R}$ by $\alpha_{k}(p)=\int_{-1}^1$ ${t^{k}}p(t)dt, p\in V$

Show that $\{{\alpha_{0},\alpha_{1}...\alpha_{n}}\}$ is a basis for the dual space of V.

I have a hint that $dimD(V)=n+1$ and so I only need to prove linear independence.

Thought process: Let $B$ be the basis for $V$ s.t $B=\{1,x,x^{2},...x^{n}\}$ This is a linearly independent set, $\sum_{i=0}^{n}$$b_{i}x_{i} \forall b \in F, and has dimension of n+1. I need to show that \sum_{i=0}^{n}$$a_{i}\alpha_{0}=0$ $\forall a \in F$ Applying $\alpha$ to each vector gives us 0 each time by the way the integral is constructed (assuming I evaluated p=1 correctly and generalizing), hence $\sum_{i=0}^{n}$$a_{i}\alpha_{0}=\sum_{i=0}^{n}$$b_{i}x_{i}$$=0 Is this decent? Also somewhat unrelated, but should this integrand remind me of the Laplace Transformation? - Getting 0 as every linear combination of the \alpha_k is more or less the exact opposite of what you want here. Let's recall the definitions: a set of vectors \{\alpha_i\} over F is linearly independent if for all a=(a_0,...,a_n)\in F, \sum_{i=0}^n a_i\alpha_i=0 only if a_0=a_1=...=a_n=0! Luckily for the sake of getting the problem worked out, you've jumped to conclusions about the values of \alpha_k(p). Indeed \alpha_k(1)=0 for k odd, but not even; now for k odd consider \alpha_k(x). – Kevin Carlson Oct 11 '12 at 19:33 @KevinCarlson, I'm having trouble evaluating the integral at \alpha_{k}(x) – Edgar Aroutiounian Oct 11 '12 at 19:59 I should perhaps have called it t rather than x! Then it's just \int_{-1}^1 t^{k+1}, no? – Kevin Carlson Oct 11 '12 at 20:00 My confusion is what would p(t) be if we choose say p=x^{2}? – Edgar Aroutiounian Oct 11 '12 at 20:04 Nothing but t^2. – Kevin Carlson Oct 11 '12 at 20:06 1 Answer OK, now suppose we had$$0=\sum_{i=0}^n c_i\alpha_i=\sum_{i=0}^n c_i \int_{-1}^1 t^ip(t)dt$$for every$p$. Let$p$have degree$k$. Then$\int_{-1}^1 t^np(t)dt$begins with a term of degree$t^{n+k+1}$, applying the ordinary power rule for integration. But no other term in the sum can reach such a high power, so if the sum is to be$0$, we must have$c_i=0$. Now assuming we had the inductive hypothesis that if$\sum_{i=0}^{n-1}c_i\alpha_i=0,$all the$c_i$are$0$, we get that$c_i=0$all the way from$0$to$n$. All that's left is the base case,$n=0\$, which is nothing since a single vector is always linearly independent.

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Thank you! I was just doing something similar when you posted that. Much appreciated! – Edgar Aroutiounian Oct 11 '12 at 20:33
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