- #1

- 24

- 0

T(p(x)) = [itex]\frac{d^{k}x}{dx^{n}}[/itex]p(x)

where 1 [itex]\leq[/itex]k [itex]\leq[/itex]n. Show that dim(N(T)) = k.

i. The null space of a linear operator is all vectors v [itex]\in[/itex] V such that T(v) = 0

ii. The dimension of a vector space V, is the number of vectors in any basis of V.

My logical understanding is that the null space of T is any k derivative that gives 0 i.e. k is bigger than n. Thus the dimension has k number of possible vectors. My problem is I don't know how to write the proof. . . I'm assuming that I just have to write the null space as a summation of derivatives from 1 to k thus the dimension of the null is k (the largest possible number).