Find a basis for the null space of the transpose operator

That's why the dual basis is \{f_n\}.In summary, we are asked to find a basis for the null space of the transpose operator D^t: V^*→V^*. This is equivalent to finding a basis for the set of linear functions from V to R that result in zero when composed with the differentiation operator D. By using the fact that the dual basis satisfies f_i(v_j)=δ_ij, where v_j=x^j, we can see that the only linear function that will result in zero when composed with D is the dual vector f_n. Therefore, our basis for the null space of D^t is {f_n}.
  • #1
nateHI
146
4

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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  • #2
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
maajdl said:
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
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
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
maajdl said:
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
nateHI said:

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^*##.

maajdl said:
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].

nateHI said:

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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1. What is a null space?

The null space of a matrix or operator is the set of all vectors that when multiplied by the matrix or operated on by the operator give a result of zero. In other words, it is the solution to the equation Ax=0, where A is the matrix or operator and x is a vector.

2. Why is finding a basis for the null space important?

Finding a basis for the null space is important because it allows us to understand the solutions to the equation Ax=0 and the properties of the matrix or operator A. It also helps us to solve other equations and systems of equations involving A more easily.

3. What is the transpose operator?

The transpose operator is an operation that switches the rows and columns of a matrix. This means that the rows of the original matrix become the columns of the transposed matrix and vice versa. The transpose operator is denoted by adding a superscript "T" to the original matrix.

4. How do you find a basis for the null space of the transpose operator?

To find a basis for the null space of the transpose operator, we can follow these steps:

  • Write the transpose of the original matrix.
  • Set up the equation Ax=0, where A is the transposed matrix and x is a vector of variables.
  • Use row reduction techniques to solve for the variables in terms of free parameters.
  • Write the solution in vector form, with the free parameters as coefficients.
  • The resulting vectors will form a basis for the null space of the transpose operator.

5. Can the null space of the transpose operator be empty?

Yes, it is possible for the null space of the transpose operator to be empty. This would mean that there are no vectors that when multiplied by the transposed matrix give a result of zero. In other words, the transpose operator is invertible and has no non-trivial solutions to the equation Ax=0.

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