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For some complex numbers $\lambda_1, \lambda_2, ..., \lambda_k$, define $$p_i(\lambda) = \prod\limits_{j=1, j\neq i}^k \frac{\lambda - \lambda_j}{\lambda_i - \lambda_j}$$ We now observe that, for any polynomial $P(\lambda)$ of degree less than or equal to $(k-1)$, we have $$P(\lambda) = \sum\limits_{i=1}^k p_i(\lambda)P(\lambda_i).$$

Text is from 'Functional Analysis', by Bachman.

I can't see why we should have this representation. I get that $p_i(\lambda_j) = \delta_{ij}$ - does this have any relevance?

man_in_green_shirt
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1 Answers1

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Let $Q(\lambda) = \sum_{i=1}^k p_i(\lambda) P(\lambda_i)$. If $p_i(\lambda_j) = \delta_{ij}$ then $Q(\lambda_j) = \sum_{i=1}^k p_i(\lambda_j) P(\lambda_i) = P(\lambda_j)$ for any $j$. (Do you see why?) Two polynomials of degree $k-1$ must be equal if they are equal if they agree on $k$ points, so $P(\lambda) = Q(\lambda)$ for all $\lambda$.

Jair Taylor
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  • When you write $Q(\lambda_i) = \sum_{i=1}^k p_i(\lambda_i) P(\lambda_i) = P(\lambda_i)$, is the $i$ of the first $\lambda_i$ fixed? Because if we're summing over that index, wouldn't we get $\sum_{i=1}^k P(\lambda_i)$? – man_in_green_shirt Jun 21 '16 at 23:32
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    @man_in_green_shirt Ah, you're right, I was being sloppy. I should have used a different variable, $Q(\lambda_j)$. It's edited now. – Jair Taylor Jun 22 '16 at 00:54