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Proving that a linear space is infinite-dimensional

  • Thread starter scotto3394
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Homework Statement


Let [tex]V[/tex] denote the linear space of all real functions continuous on the interval [tex] [-\pi,\pi][/tex]. Let [tex]S[/tex] be that subset of [tex]V[/tex] consisting of all [tex]f[/tex] satisfying the three equations [tex]\[\int_{-\pi}^{\pi} f(t) dt = 0, ~~~\int_{-\pi}^{\pi} f(t) cos(t) dt = 0, ~~~ \int_{-\pi}^{\pi} f(t)sin(t) dt = 0\] [/tex]

(c) Prove that [tex]S[/tex] is infinite-dimensional.


Homework Equations


From parts (a) and (b) of the question we know that [tex]S[/tex] is a subspace of [tex]V[/tex] and that [tex]S[/tex] contains the functions [tex]f(x) = cos (nx), ~~ f(x) = sin(nx)[/tex] for [tex]n=2,3,...[/tex]. Also I'm not sure whether it's relevant, but the sections before this group of exercises covered null spaces, rank, and linear transformations. So if it is relevant I would imagine that dim N + dim T = dim V would be useful where N is the null space, T is the range, V is the domain (which is a linear space), and dim is dimension.


The Attempt at a Solution


My idea is that you would try a proof by contradiction where you would assume that [tex]S[/tex] is finite-dimensional. From there I had a few random ideas, such as trying to show that the null space is infinite-dimensional and then use the formula above to show that [tex]S[/tex] would somehow be infinite-dimensional. Also I was thinking of assuming that there are [tex]n[/tex] independent elements and somehow showing that they are not actually independent using part b or by applying the fact that they are in [tex]S[/tex], though I'm not really sure how that would work.


Really I'm not looking for an answer, so much as helpful hints. I'm actually working through this book on my own so I don't really have a teacher to ask questions to. Also, I hope that I provided enough relevant data. Thank you for your time.
 

Answers and Replies

  • #2
CompuChip
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You already mention infinitely many functions that you know are in S.
So, correct me if I'm wrong, but if you show directly that they are all independent then you are done, right?
 
  • #3
If I could show that it would work, but the functions that I know are there (sine and cosine) aren't actually independent I think. Let's say within the elements I chose [tex] sin(2x) [/tex] and [tex] sin(4x)[/tex]. [tex] sin(4x) = (2cos(2x))sin(2x)[/tex] which is a multiple of [tex] sin(2x)[/tex] making it dependent. Then again, I might be wrong...
I'll definitely give it a try though.
 
  • #4
CompuChip
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My definition of independent is that
[tex]\langle f, g \rangle = 0[/tex]
where the inner product for function spaces is usually
[tex]\langle f, g \rangle := \int_{-\pi}^\pi f(x) g(x) \, dx[/tex]

But whichever definition you use, I am pretty sure that, sin(px) and sin(qx) are independent in either of them, for p and q (distinct) primes.
 
  • #5
Dick
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My definition of independent is that
[tex]\langle f, g \rangle = 0[/tex]
where the inner product for function spaces is usually
[tex]\langle f, g \rangle := \int_{-\pi}^\pi f(x) g(x) \, dx[/tex]

But whichever definition you use, I am pretty sure that, sin(px) and sin(qx) are independent in either of them, for p and q (distinct) primes.
That's not a definition of independent. It's a definition of orthogonal. Independent means c1*v1+...+cn*vn=0 implies c1=...=cn=0. You shouldn't have any trouble showing that a mutually orthogonal set of vectors is also independent. sin(px) and sin(qx) being orthogonal doesn't have anything to do with p and q being prime. p and q just need to be distinct positive integers. Show that.
 

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