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

## 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.

Last edited:

maajdl
Gold Member
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* ?

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
Gold Member
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
Gold Member
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?

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 Helper

## 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
$$f^t : \phi \mapsto (\phi \circ f)$$

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 $V$.

## 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 $c$.

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