Find a basis for the null space of the transpose operator

[nateHI]

Homework Statement

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 $D^t: V^*\to V^*$.

Homework 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 $f_i(v_j)=\delta_{ij}$

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.

 

[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* ?

 

[nateHI]

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* ?

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

 

[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.

 

[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?

 

[nateHI]

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?

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.

 

[pasmith]

Homework Statement

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 $D^t: V^*\to V^*$.

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* ?

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
<br /> f^t : \phi \mapsto (\phi \circ f)<br />

Here we are asked for a basis of \ker D^t = \{\phi \in P_n^{*}: D^t(\phi) = 0\}, where D: v \mapsto v&#039; is the differentation operator on V.

Homework 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 $f_i(v_j)=\delta_{ij}$

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 $f_i(v_j)=\delta_{ij}$

You need to define v_j = x^j.

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.

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.