Linear functionals: Φ(u)=0 implies Φ(v)=0, then u=kv.

In summary: Phi(u)=\mathrm{const}##.Thus, in summary, the problem can be solved by defining linear functionals ##\tilde u## and ##\tilde v## as described above and using the fact that ##\mathrm{ker}\,\tilde u\subseteq\mathrm{ker}\,\tilde v## and Theorem 2 to show that ##v=ku## for some scalar ##k##. This approach does not require the use of a basis and works for both finite and infinite dimensional vector spaces.
  • #1
Adgorn
130
18

Homework Statement


Suppose u,v ∈ V and that Φ(u)=0 implies Φ(v)=0 for all Φ ∈ V* (the duel space). Show that v=ku for some scalar k.

Homework Equations


N/A

The Attempt at a Solution


I've managed to solve the problem when V is of finite dimension by assuming u,v are linearly independent, expanding them to a basis of V and defining a function such that Φ(v)=0, Φ(u)=1 and Φ(wi)=0 where wi are all the other basis vectors. This contradicted the original statement in the problem (since Φ(v)=0 but Φ(u)=1) which meant u,v must be linearly dependent, which implies u is a scalar multiple of v.

However I don't know how to prove it in the general case where V could also be of infinite dimension, the intuitive solution I had was an argument of linear dependence since we have a scalar multiple but I cannot seem to use it in the general case, so this is where I got stuck. Any help would be greatly appreciated.
 
Physics news on Phys.org
  • #3
I assume you mean in my proof for a vector space of finite dimension. I needed to expand u,v to a basis to show that Φ is indeed an element of the duel space by defining its action on a basis.
 
  • #4
Adgorn said:
I assume you mean in my proof for a vector space of finite dimension. I needed to expand u,v to a basis to show that Φ is indeed an element of the duel space by defining its action on a basis.
But you didn't use finiteness, nor did you use the entire basis. All you used, is a complementary space of ##\mathbb{F}\cdot u## on which you defined a functional (that isn't identically zero) to vanish.
 
  • Like
Likes FactChecker
  • #5
fresh_42 said:
But you didn't use finiteness, nor did you use the entire basis. All you used, is a complementary space of ##\mathbb{F}\cdot u## on which you defined a functional (that isn't identically zero) to vanish.
I don't recall defining any complimentary space where Φ vanishes. If V is of dimension n, I merely expanded {v,u} to a basis S={v,u,w1,w2,...,wn-2} and defined a function Φ from V to its field K defined by Φ(v)=0, Φ(u)=1, Φ(w1)=0, Φ(w2)=0,..., Φ(wn-2)=0.
This function is obviously a linear functional since it's a part of the basis of V* duel to the basis S, and this conflicted with the original statement.

All I did here was show that if u,v are linearly independent, there exists a function which contradicts the conditions of the question, and so u,v must be linearly dependent.
 
  • #6
There is very important and simple theorem. Let ##X,Y,Z## be vector spaces say over field ##\mathbb{R}## or ##\mathbb{C}##. Perhaps some of these spaces are infinite dimensional. And let ##A:X\to Y,\quad B:X\to Z## be linear operators such that ##A(X)=Y## and ##\mathrm{ker}\,A\subset\mathrm{ker}\,B##.

Theorem 1. There is a linear operator ##\Lambda:Y\to Z## such that ##B=\Lambda A##.
This theorem particularly implies thee followin proposition.

Theorem 2. If ##f,f_1,\ldots, f_n\in X^*## and ##\bigcap_{i=1}^n\mathrm{ker}\, f_i\subseteq \mathrm{ker}\,f## then there exist numbers ##\lambda_1,\ldots,\lambda_n## such that ##f=\lambda_1f_1+\ldots+\lambda_nf_n##.

By the way this idea is a source of all Lagrange multipliers theorems.

Now turn to the problem of OP. Define linear functionals ##\tilde u,\tilde v:V^*\to\mathbb{R}## as follows ##\tilde u(\Phi)=\Phi(u)## and the same for ##\tilde v##. From conditions of the OP's problem it follows that ##\mathrm{ker}\,\tilde u\subseteq\mathrm{ker}\,\tilde v##. By Theorem 2 it follows that ##\tilde v=k\tilde u##. That is ##\Phi(v)=k\Phi(u)## for all ##\Phi\in V^*##. Thus ##v=ku##.

This prompts that there is no need to employ the whole space ##V^*## it is sufficient to use only its total subspace
 
Last edited:

1. What is a linear functional?

A linear functional is a mathematical function that takes in a vector as its input and returns a scalar value as its output. It is a type of linear transformation that maps a vector space to its underlying field.

2. What does the equation Φ(u)=0 implies Φ(v)=0 mean?

This equation means that if the linear functional Φ of vector u equals 0, then the linear functional Φ of vector v must also equal 0. In other words, if two vectors have the same linear functional value, then they are equivalent.

3. What is the significance of Φ(u)=0 implies Φ(v)=0 in relation to linear functionals?

This equation is significant because it shows that linear functionals are injective, meaning that different inputs will always result in different outputs. This is important in many mathematical applications, such as optimization problems.

4. How does the equation u=kv relate to linear functionals?

This equation shows that if the linear functional of u is equal to 0, then the linear functional of v must also be equal to 0. In other words, if a scalar multiple of u is equal to 0, then the linear functional of that scalar multiple must also be equal to 0.

5. Can you give an example of a linear functional where Φ(u)=0 implies Φ(v)=0, then u=kv?

One example of this is the zero functional, where the output is always 0 regardless of the input vector. This satisfies the condition Φ(u)=0 implies Φ(v)=0 for any vectors u and v, and also satisfies the condition u=kv for any scalar k.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
0
Views
419
  • Calculus and Beyond Homework Help
Replies
7
Views
339
  • Calculus and Beyond Homework Help
Replies
14
Views
532
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
3K
  • Calculus and Beyond Homework Help
Replies
1
Views
422
  • Calculus and Beyond Homework Help
Replies
24
Views
673
  • Calculus and Beyond Homework Help
Replies
8
Views
558
  • Calculus and Beyond Homework Help
Replies
28
Views
3K
Back
Top