# Linear transformations

1. Feb 3, 2004

### tinfoilhat

Let g(x) belonging to Pn-1(R) be an arbiitrary polynomial of degree n-1 or less. Show that there exists a polynomial f(x) belonging to Pn(R) such that xf''(x)-f'(x)=g(x)"

I interpreted this question as having to prove the linear transformation T: Pn(R) --> Pn-1(R) where f(x) |--> xf''(x)-f'(x) is onto.

If I let f(x)=a+bx+cx^2+...+zx^n
T(f(x))=-b

Therefore the range of T is {x} which has dimension 1. By the dimension theorem (and just looking at the results), the nullspace has dimension n. This of course is not onto.