# What is the dim(N(T))?

1. Mar 30, 2013

### Mr.Tibbs

Define ne a linear operator T : Pn $\rightarrow$ Pn by
T(p(x)) = $\frac{d^{k}x}{dx^{n}}$p(x)
where 1 $\leq$k $\leq$n. Show that dim(N(T)) = k.

i. The null space of a linear operator is all vectors v $\in$ 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).

2. Mar 30, 2013

### LCKurtz

I assume you mean $\frac {d^k x}{dx^k}p(x)$. Or for easy display, $D^kp(x)$.

No. The null space isn't "any k derivative". The null space is a subspace of your nth degree polynomials.

No again, the null space is not a "summation of derivatives". For example if $n=5$ so you are considering polynomials of degree $5$, and you operate on them with $D^2$, which ones would map to the zero polynomial?

3. Mar 30, 2013

### Mr.Tibbs

Awesome! So I've been sitting here and had an epiphany. . . I think. . . so here is what I came up with.

Let B = {1,x,x$^{2}$,...,x$^{n}$} be the standard basis for P$_{n}$. Since p(x) is in the N(T) iff it's degree is < k, then the null space is all the polynomials with degree k-1 or less, thus [1,x,x$^{2}$,...,x$^{k-1}$] is a basis for N(T),

and the dim(N(T)) = k

:) I think I have it what do you think?

4. Mar 30, 2013

### LCKurtz

I think that's much better.