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Let $\mathcal{P}_n$ be a vector space of polynomials of degree less than or equal to $n$.

I have shown that the evaluation map $Eval_x : f \in \mathcal{P}_n \mapsto f(x) \in \mathbb{R}$ is a linear functional.

Next I have to show that for distinct $x_0 \dots x_n$ the functionals $Eval_{x_0} \dots Eval_{x_n}$ are linearly independent and form a basis for the dual space of $\mathcal{P}_n$

I would like to ask whether the following argument is sound

Suppose

$ \displaystyle 0 = \sum_{i=0}^n \lambda_i Eval_{x_i} $

Then for any $ f \in \mathcal{P}_n $ ,

$ \displaystyle 0 = \sum_{i=0}^n \lambda_i Eval_{x_i}(f) = \sum_{i=0}^n \lambda_i f(x_i) $

Choosing $ \displaystyle f(x) = \prod_{i \neq j} x_j (x-x_i)$ for successive $j$ we have that

$\displaystyle 0 = \lambda_j x_j \cdot k $

for non-zero $k$ and so $\lambda_j = 0$ for all $j$ provided $x_j \neq 0$ and the linear functionals are independent.

Also how do I deal with the case where $x_j = 0$?

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  • $\begingroup$ Change $x_j$ to $x_j+1.$ $\endgroup$ – Ehsan M. Kermani Jan 23 '15 at 1:18
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    $\begingroup$ Why do you want $x_j$ in there? Why not take $f(x) = \prod_{i \neq j}(x - x_i)$? $\endgroup$ – Jim Jan 23 '15 at 1:20
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Change $f$ to $f(x) = \prod_{i \neq j}(x - x_i)$. The rest of the proof is the same but now $x_j = 0$ will not be a special case.

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  • $\begingroup$ Maybe I'm dense but there is now no longer any mention of $j$ so what should $\prod_{i\neq j}(x-x_i)$ mean? $\endgroup$ – Cameron Williams Jan 23 '15 at 21:08
  • $\begingroup$ I'm just using the OP's notation. Really $f$ should be $f_j$. For a given $j$ it's a polynomial that vanishes on all $x_i$ for $i \neq j$ and does not vanish on $j$. $\endgroup$ – Jim Jan 23 '15 at 21:10
  • $\begingroup$ That was what I was about to ask.. if $f_j$ is what was meant. $\endgroup$ – Cameron Williams Jan 23 '15 at 21:10

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