# Find a basis for the null space of the transpose operator

1. Mar 19, 2014

### nateHI

1. The problem statement, all variables and given/known data

Let $n$ be a positive integer and let $V = P_n$ be the space of polynomials over $R$. Let D be the differentiation operator on $V$ . Find a basis for the null space of the transpose operator $D^t: V^*\to V^*$.

2. Relevant equations

Let $T:V\to W$ be a linear transformation. Then $(im T)^0=kerT^t$
Let let $\{f_0,f_1,...f_n\}$ be a dual basis for $\{v_0,...,v_n\}$ then $f_i(v_j)=\delta_{ij}$

3. The attempt at a solution
Let $\{1,x,...x^n\}$ be a basis for $V$ and let $\{f_0,f_1,...f_n\}$ be the dual basis. The image of $D$ is the set of polynomials with degree less than n. We know from $f_i(v_j)=\delta_{ij}$ that $f_n$ will annihilate any polynomial without an $x^n$ term. But that is the entire image of $D$ so $\{f_n\}$ is the basis we seek, where I used $(im T)^0=kerT^t$.

I think this is correct, I'm just looking for tips to make it cleaner.

Last edited: Mar 19, 2014
2. Mar 19, 2014

### maajdl

What does that mean:

Find a basis for the null space of the transpose operator Dt:V∗→V∗.

What is the relation between D and DT?
What is the definition of V* ?

3. Mar 19, 2014

### nateHI

Sorry for my lack of understanding, are you asking or prompting me to think harder because I missed something?

4. Mar 19, 2014

### maajdl

No, I just ask because I think I might answer if I understood the question a bit more.
By the way, informal discussion is often helpful.
That's my own experience.

5. Mar 19, 2014

### maajdl

I just understood that your question involved the dual space and Dt is probably the equivalent of D in this space.
On wikipedia, the dual space is defined on the basis of "coordinates" , which is not very encouraging for me.
How is the norm defined, how can I construct the dual space?

6. Mar 19, 2014

### nateHI

Hmm, very good questions. I'll take a crack at answering tomorrow night. I just finished my homework and need some sleep. Unless of course someone smarter than me (of which there are many to choose from) gets to it first.

7. Mar 19, 2014

### pasmith

If $V$ is a vector space over $\mathbb{R}$, then the dual space $V^{*}$ is the set of linear functions from $V$ to $\mathbb{R}$. It is not necessary for $V$ to be an inner product space or normed space for the dual space to be defined.

A linear map $f: V \to W$ has a corresponding dual map $f^{t} : W^{*} \to V^{*}$ given by
$$f^t : \phi \mapsto (\phi \circ f)$$

Here we are asked for a basis of $\ker D^t = \{\phi \in P_n^{*}: D^t(\phi) = 0\}$, where $D: v \mapsto v'$ is the differentation operator on $V$.

You need to define $v_j = x^j$.

I think you have actually to say what $f_n$ is, other than "the dual vector such that $f_n(x^k) = \delta_{kn}$".

The short proof is to write down
$$D^t(\phi)\left(\sum_{k=0}^n a_kx^k\right) = \sum_{k=1}^n ka_k\phi(x^{k-1})$$
and see that $D^t(\phi) = 0$ requires $\phi(1) = \phi(x) = \dots = \phi(x^{n-1})= 0$, so that $\phi(v) = cv^{(n)}(0)$ for some scalar $c$.