February 14, 2020
From Axler’s Linear Algebra Done Right, chapter 4, exercise 5:
Suppose is a nonnegative integer, are distinct elements of , and . Prove that there exists a unique polynomial such that for
[This result can be proved without using linear algebra. However, try to find the clearer, shorter proof that uses some linear algebra.]
If then as is at most degree , there are at most distinct ’s such that . Equivalently, if had at least distinct ’s such that , then . Letting
then the previous statement implies that , so is injective. As is square, [Axler 3.69] implies that is invertible. Hence, if
then the polynomial is the unique polynomial such that for .