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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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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
up vote 1 down vote accepted

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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