Find a basis for the null space of the transpose operator

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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.
 
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Answers and Replies

  • #2
maajdl
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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
146
4
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?
 
  • #4
maajdl
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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
maajdl
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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
146
4
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.
 
  • #7
pasmith
Homework Helper
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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 [itex]V[/itex] is a vector space over [itex]\mathbb{R}[/itex], then the dual space [itex]V^{*}[/itex] is the set of linear functions from [itex]V[/itex] to [itex]\mathbb{R}[/itex]. It is not necessary for [itex]V[/itex] to be an inner product space or normed space for the dual space to be defined.

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

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

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 [itex]v_j = x^j[/itex].

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 [itex]f_n[/itex] is, other than "the dual vector such that [itex]f_n(x^k) = \delta_{kn}[/itex]".

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

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