Proving that a linear space is infinite-dimensional

In summary, the conversation discusses proving that the subset S of the linear space V of all real functions continuous on the interval [-\pi,\pi] is infinite-dimensional. It is mentioned that S contains infinitely many functions such as sine and cosine, but it is not clear if these functions are independent. The discussion also touches on the definition of independence and the fact that mutually orthogonal functions are also independent. It is suggested to show that sin(px) and sin(qx) are independent for distinct positive integers p and q, which would prove that S is infinite-dimensional.
  • #1
scotto3394
6
0

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.
 
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  • #2
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
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
CompuChip said:
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.
 

FAQ: Proving that a linear space is infinite-dimensional

What does it mean for a linear space to be infinite-dimensional?

An infinite-dimensional linear space is one that has an infinite number of basis vectors, meaning that any vector in the space can be written as a linear combination of an infinite number of basis vectors.

How do you prove that a linear space is infinite-dimensional?

To prove that a linear space is infinite-dimensional, you can show that it contains an infinite linearly independent set of vectors. This can be done by constructing an infinite sequence of linearly independent vectors, or by showing that the space contains a subspace that is isomorphic to an infinite-dimensional space.

Can a finite-dimensional linear space also be infinite-dimensional?

No, a finite-dimensional linear space cannot also be infinite-dimensional. By definition, a finite-dimensional space has a finite number of basis vectors, while an infinite-dimensional space has an infinite number of basis vectors.

Are there any visual representations or examples of infinite-dimensional linear spaces?

One example of an infinite-dimensional linear space is the space of all polynomials with real coefficients. Another example is the space of all continuous functions on a closed interval. These spaces are infinite-dimensional because they have an infinite number of basis functions.

How does the concept of dimensionality apply to infinite-dimensional linear spaces?

Infinite-dimensional linear spaces still have a notion of dimensionality, but it is not a finite value like in finite-dimensional spaces. Instead, the dimension of an infinite-dimensional space is defined as the cardinality of its basis, which is the number of basis vectors. This can be either a countably infinite or uncountably infinite value.

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