Showing a vector space is infinite dimensional

In summary, a vector space is a mathematical concept that allows for the addition and multiplication of vectors to form new vectors. It is used to represent and manipulate various mathematical objects. A vector space can be infinite dimensional if it has an infinite number of basis vectors, which are linearly independent vectors used to form all other vectors in the space. It is important to show that a vector space is infinite dimensional because it allows for the use of infinite series and functions in mathematical applications. Common methods for showing that a vector space is infinite dimensional include using linear independence, dimension arguments, spanning sets, and proofs by contradiction.
  • #1
tylerc1991
166
0

Homework Statement



This is only part of a problem I am working on, but the only part that I have questions about is the following:

Show that [itex]\mathcal{F}(\mathbb{R})[/itex] is infinite dimensional.

Homework Equations



[itex]\mathcal{F}(\mathbb{R})[/itex] is the set of all functions that map real numbers to real numbers.
[itex]f_n[/itex] is the function defined by the rule [itex]f_n(x) = e^{nx}[/itex] for [itex]n \in \mathbb{N}[/itex]

The Attempt at a Solution



Suppose [itex] \mathcal{F}(\mathbb{R}) [/itex] is finite dimensional.
This means that there exists a finite basis for [itex] \mathcal{F}(\mathbb{R}) [/itex].
Consider the set of vectors [itex] \mathcal{E} = \{ f_1, f_2, \dotsc, f_n \} [/itex] for some [itex] n \in \mathbb{N} [/itex].
Suppose [itex] \mathcal{E} [/itex] is a basis for [itex] \mathcal{F}(\mathbb{R}) [/itex], and consider the vector [itex] f_{n+1} [/itex] in [itex] \mathcal{F}(\mathbb{R}) [/itex].
Since [itex] \mathcal{E} [/itex] spans [itex] \mathcal{F}(\mathbb{R}) [/itex], we see that [itex] f_{n+1} \in \text{span}\{\mathcal{E}\} [/itex].
This means that
[itex] f_{n+1} = \sum_{k=1}^n a_k f_k = a_1 f_1 + a_2 f_2 + \dotsb + a_n f_n [/itex]
for some [itex] a_1, a_2, \dotsc, a_n \in \mathbb{R} [/itex].
Equivalently, this means that
[itex] a_1 f_1 + a_2 f_2 + \dotsb + a_n f_n - f_{n+1} = 0. \quad \quad (1) [/itex]
It has been shown that [itex] \{ f_1, f_2, \dotsc, f_{n+1} \} [/itex] is linearly independent.
This means that each coefficient of [itex] f_i [/itex] equals 0 for [itex] i = 1, 2, \dotsc, n+1 [/itex].
But this is impossible, as [itex] -1 \neq 0 [/itex].
Hence, [itex] \mathcal{E} [/itex] does not span [itex] \mathcal{F}(\mathbb{R}) [/itex].

I feel like I have't shown that [itex]\mathcal{F}(\mathbb{R})[/itex] is infinite dimensional.
It seems like I have shown that *some* finite bases do not work for [itex]\mathcal{F}(\mathbb{R})[/itex].
I think that I am close, but I think there is something missing.
Could someone point me in the right direction to finish this proof?

Thank you!
 
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  • #2
tylerc1991 said:

Homework Statement



This is only part of a problem I am working on, but the only part that I have questions about is the following:

Show that [itex]\mathcal{F}(\mathbb{R})[/itex] is infinite dimensional.

Homework Equations



[itex]\mathcal{F}(\mathbb{R})[/itex] is the set of all functions that map real numbers to real numbers.
[itex]f_n[/itex] is the function defined by the rule [itex]f_n(x) = e^{nx}[/itex] for [itex]n \in \mathbb{N}[/itex]

The Attempt at a Solution



Suppose [itex] \mathcal{F}(\mathbb{R}) [/itex] is finite dimensional.
This means that there exists a finite basis for [itex] \mathcal{F}(\mathbb{R}) [/itex].
Consider the set of vectors [itex] \mathcal{E} = \{ f_1, f_2, \dotsc, f_n \} [/itex] for some [itex] n \in \mathbb{N} [/itex].
Suppose [itex] \mathcal{E} [/itex] is a basis for [itex] \mathcal{F}(\mathbb{R}) [/itex], and consider the vector [itex] f_{n+1} [/itex] in [itex] \mathcal{F}(\mathbb{R}) [/itex].
Since [itex] \mathcal{E} [/itex] spans [itex] \mathcal{F}(\mathbb{R}) [/itex], we see that [itex] f_{n+1} \in \text{span}\{\mathcal{E}\} [/itex].
This means that
[itex] f_{n+1} = \sum_{k=1}^n a_k f_k = a_1 f_1 + a_2 f_2 + \dotsb + a_n f_n [/itex]
for some [itex] a_1, a_2, \dotsc, a_n \in \mathbb{R} [/itex].
Equivalently, this means that
[itex] a_1 f_1 + a_2 f_2 + \dotsb + a_n f_n - f_{n+1} = 0. \quad \quad (1) [/itex]
It has been shown that [itex] \{ f_1, f_2, \dotsc, f_{n+1} \} [/itex] is linearly independent.
Where has this been shown? Certainly not in your work leading up to this statement. What you have is that {f1, f2, ..., fn} is a basis, so those vectors are a linearly independent set, but you haven't shown that {f1, f2, ..., fn, fn+1} is a linearly independent set.
tylerc1991 said:
This means that each coefficient of [itex] f_i [/itex] equals 0 for [itex] i = 1, 2, \dotsc, n+1 [/itex].
But this is impossible, as [itex] -1 \neq 0 [/itex].
Hence, [itex] \mathcal{E} [/itex] does not span [itex] \mathcal{F}(\mathbb{R}) [/itex].

I feel like I have't shown that [itex]\mathcal{F}(\mathbb{R})[/itex] is infinite dimensional.
It seems like I have shown that *some* finite bases do not work for [itex]\mathcal{F}(\mathbb{R})[/itex].
I think that I am close, but I think there is something missing.
Could someone point me in the right direction to finish this proof?

Thank you!
 
  • #3
You can also do this problem by showing there exist a linearly independent subset that has infinite many members (countably infinite is fine).
 
  • #4
Mark44 said:
Where has this been shown? Certainly not in your work leading up to this statement. What you have is that {f1, f2, ..., fn} is a basis, so those vectors are a linearly independent set, but you haven't shown that {f1, f2, ..., fn, fn+1} is a linearly independent set.

I mentioned at the beginning of the post that this question was part of a greater question that I am working on. I have shown that they are linearly independent, but for the sake of practicality I decided to leave it out.
 
  • #5
deluks917 said:
You can also do this problem by showing there exist a linearly independent subset that has infinite many members (countably infinite is fine).

True, but I have to appeal to the definition of a basis that we are using. By definition, a basis for a vector space V is a set that
i) is linearly independent
ii) spans V

The definition of span for an infinite set is the set of all finite linear combinations of elements of the set, so it would be awkward (I am thinking) so show that this infinite set spans the space.
 
  • #6
OK then. Since you have already shown that the n+1 functions are lin. independent, which contradicts the (tacit) assumption that the dimension of your space is n.
 
  • #7
Sheesh, taking mathematics to a whole new level of coldness :). Thank you for your (terse) help on the matter. The question is moot now anyway, as I can't assume what the functions are that span the space. Back to square 1 ...
 
  • #8
Mark44 said:
OK then. Since you have already shown that the n+1 functions are lin. independent, which contradicts the (tacit) assumption that the dimension of your space is n.

tylerc1991 said:
Sheesh, taking mathematics to a whole new level of coldness :). Thank you for your (terse) help on the matter. The question is moot now anyway, as I can't assume what the functions are that span the space. Back to square 1 ...
Sorry, I didn't mean to seem brusque. All I meant was that with that new information (that you already showed that the n + 1 functions were linearly independent), your proof looked OK.
 
  • #9
Do you have any more information than you've shown about the functions in your space? IOW, is there anything more given than they map a real to a real?

For example, if the functions are continuous and infinitely differentiable, they can be represented by Taylor series, with one possible basis being {1, x, x2, ...}.
 

What is a vector space?

A vector space is a mathematical concept that describes a set of objects (vectors) that can be added together and multiplied by scalars (numbers) to form new vectors. It is a fundamental concept in linear algebra and is used to represent and manipulate a wide range of mathematical objects, such as matrices, functions, and polynomials.

How can a vector space be infinite dimensional?

A vector space is said to be infinite dimensional if it has an infinite number of basis vectors. This means that there are an infinite number of linearly independent vectors that can be combined to form any other vector in the space. In contrast, a finite dimensional vector space has a finite number of basis vectors.

What does it mean to show that a vector space is infinite dimensional?

To show that a vector space is infinite dimensional, we must prove that it has an infinite number of linearly independent vectors. This can be done by finding a set of vectors that cannot be expressed as a linear combination of a finite number of other vectors in the space.

Why is it important to show that a vector space is infinite dimensional?

Knowing that a vector space is infinite dimensional can be useful in many mathematical applications. For example, it allows us to work with infinite series and functions, which are often used in mathematical models and theories. It also has applications in areas such as physics, computer science, and engineering.

What are some common methods for showing that a vector space is infinite dimensional?

There are several methods for showing that a vector space is infinite dimensional. One common method is to use the concept of linear independence to find a set of vectors that cannot be expressed as a linear combination of a finite number of other vectors. Another method is to use dimension arguments, where we compare the number of basis vectors in the space to the dimension of the space itself. Other techniques include using the concept of spanning sets and proofs by contradiction.

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