Showing a vector space is infinite dimensionalby tylerc1991 Tags: dimensional, infinite, showing, space, vector 

#1
Oct3011, 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!!! 



#2
Oct3011, 07:32 PM

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P: 21,063





#3
Oct3011, 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).




#4
Oct3011, 07:35 PM

P: 166

Showing a vector space is infinite dimensional 



#5
Oct3011, 07:41 PM

P: 166

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
Oct3011, 07:43 PM

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P: 21,063

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
Oct3011, 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 ...




#8
Oct3011, 08:40 PM

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#9
Oct3011, 08:45 PM

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P: 21,063

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, x^{2}, ...}. 


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