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Showing a vector space is infinite dimensional

by tylerc1991
Tags: dimensional, infinite, showing, space, vector
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tylerc1991
#1
Oct30-11, 06:54 PM
P: 166
1. The problem statement, all variables and given/known data

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.

2. Relevant 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]

3. 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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Mark44
#2
Oct30-11, 07:32 PM
Mentor
P: 21,281
Quote Quote by tylerc1991 View Post
1. The problem statement, all variables and given/known data

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.

2. Relevant 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]

3. 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.
Quote Quote by tylerc1991 View Post
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!!!
deluks917
#3
Oct30-11, 07:35 PM
P: 367
You can also do this problem by showing there exist a linearly independent subset that has infinite many members (countably infinite is fine).

tylerc1991
#4
Oct30-11, 07:35 PM
P: 166
Showing a vector space is infinite dimensional

Quote Quote by Mark44 View Post
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.
tylerc1991
#5
Oct30-11, 07:41 PM
P: 166
Quote Quote by deluks917 View Post
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.
Mark44
#6
Oct30-11, 07:43 PM
Mentor
P: 21,281
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
#7
Oct30-11, 07:48 PM
P: 166
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 ...
Mark44
#8
Oct30-11, 08:40 PM
Mentor
P: 21,281
Quote Quote by Mark44 View Post
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.
Quote Quote by tylerc1991 View Post
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.
Mark44
#9
Oct30-11, 08:45 PM
Mentor
P: 21,281
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, ...}.


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