Lagrange interpolation

Let $𝑆 = {𝑥 𝑖} 𝑛 𝑖 = 1 \subseteq 𝕂$ be a finite indexed set. The Lagrange polynomials

𝐿 𝑖 (𝑥) = \prod 1 \leq 𝑗 \leq 𝑛 𝑗 \neq 𝑖 𝑥 - 𝑥 𝑗 𝑥 𝑖 - 𝑥 𝑗

and have the the property that

𝐿 𝑖 (𝑥 𝑗) = 𝛿 𝑖 𝑗

for $1 \leq 𝑖, 𝑗 \leq 𝑛$ .

Proof

On the one hand

𝐿 𝑖 (𝑥 𝑖) = \prod 1 \leq 𝑗 \leq 𝑛 𝑗 \neq 𝑖 𝑥 𝑖 - 𝑥 𝑗 𝑥 𝑖 - 𝑥 𝑗 = 1 .

On the other hand, if $𝑘 \neq 𝑖$ then

𝑥 𝑘 - 𝑥 𝑗 𝑥 𝑖 - 𝑥 𝑗 = 0

for $𝑘 = 𝑗 \neq 𝑖$ , so $𝐿 𝑖 (𝑥 𝑘) = 0$ .

More generally, we can define a polynomial evaluating $𝑥 𝑖 \mapsto 𝑦 𝑖$ for $𝑦 𝑖 \in 𝕂$ by

𝑓 (𝑥) = 𝑛 \sum 𝑖 = 1 𝑦 𝑖 𝐿 𝑖 (𝑥) .

Properties

$\sum 𝑛 𝑖 = 1 𝐿 𝑖 (𝑥) = 1$ .

Proof of 1

First note that each $𝐿 𝑖 (𝑥)$ has order $𝑛 - 1$ , and therefore so does $𝑓 (𝑥) : = \sum 𝑛 𝑖 = 1 𝐿 𝑖 (𝑥)$ . Since $𝑓 (𝑥 𝑖) - 1 = 0$ for $𝑛$ distinct $𝑖$ , it follows from ^P1 that $𝑓 (𝑥) - 1 = 0$ , wherefore $𝑓 (𝑥) = 1 .$

Corollaries

If $𝕂$ is a Galois field then evaluation gives a surjection $𝕂 [𝑥] ↠ 𝕂 𝕂$ .
If $𝑉$ is a finite-dimensional vector space and $𝐴 \in E n d 𝕂 (𝑉)$ is diagonalizable with eigenvalues $𝜆 𝑖$ , then $𝐿 𝑖 (𝐴)$ gives a projection operator onto the eigenspace $E 𝜆 𝑖$ .

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