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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?

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    Okay, now I see that it doesn't go to$0$for all the vectors in basis B.2012-10-11

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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!2012-10-11