Linear Space Question

1. Dec 9, 2005

jwsiii

If you define a linear space as the set X of all functions from a nonempty set T into a field F with addition and scalar multiplication defined, would a subspace be the set X restricted to a certain subset of T?

2. Dec 9, 2005

jwsiii

I have a better way to ask what I want to know. Supposed I use that same definition to define the temperature space of the surface for the Earth using the real numbers as my field F and the set of all points on the surface of the Earth as my T. What would be a subspace of this temperature space? Would it be the temperature space of a hemisphere? Would it be the temperature space of the surface of the Earth restricted to temperature configurations existing from 1990 to 2000? Thanks in advance for any help.

3. Dec 10, 2005

JasonRox

I have a hard time understanding what you are describing in the first post. The second doesn't do any good.

Are you describing X as a linear (vector) space of linear transformations from T to F?

If that is the question, then I would have to say no. A subspace of X does not have to be restricted to a subset of T. The space has nothing to do with the linear transformations (sort of does), you can create a subspace X such that all the linear transformations are one-to-one and onto from T to F, which implies the entire non-empty set is being used. We know this is a subspace because it is closed under composition. The composition of two one-to-one and onto functions remain one-to-one and onto, which satisfies the condition to be a subspace.

Note: I never knew that the set of linear transformations from one space to another space created a vector space itself. This is something new to me. I could be wrong, and I could be talking a lot of crap. I have no idea, but it seems to make sense to me.

Last edited: Dec 10, 2005
4. Dec 10, 2005

rachmaninoff

jwsiii:
You've inadvertently wandered into the realm of functional analysis. The zero vector is the zero function, which maps all points to 0 K (or 0 C, however you define it). A set of vectors (functions) is linearly independent if you can express one of them as a linear combination of the other functions.

The dimension of the space is countably infinite. On a closed interval in R, example basis sets are the Fourier series and Legendre polynomials. In your example, a basis set is the Spherical harmonics without the radial dependence; products of associated Legendre polynomials in the cosine of the polar coordinate and trigonometric functions in the azimuthal coordinate.

5. Dec 10, 2005

matt grime

The dimension, as a vector space of C(R) is not countably infinite. I can easily write down uncountably many linearly independent elements.

The reason for your confusion is that you're conflating denseness in an analytic sense with spanning in the algebraic sense.

Those examples of sets are not bases in the linear algebra sense of the word. They are a dense (usually orthonormal with respect to some inner product) set of vectors. Ie the closure of the subspace they span is the whole set. Moreover, if you look the (initial) question is about all functions from T to F, not continuous, or L^1 and F is not even necessarily real.

To get to the orginal question.

No I don't think a subspace is Fun(T,F) restricted to a subset. An example of a subspace would be: let t be in T, then set K(T,F,t) to be the space of functions that send t to zero. That is a subspace, but I don't see it as being the set of functions restricted to a subset.

I don't even see how taking an element in Fun(T,F) and restricting it to a function of a subset even defines an element of Fun(T,F) ie your notional definition of a 'subspace' doesn't even define a subset of Fun(T,F), never mind a subspace. And the planetary temperature thing makes even less sense: where did the time dependence come from? I don't think you want or need to consider linear algebra for this.

You can certianly define a quintuple for this case of yours: (x,y,h,t,T) of lattitude, longitude, altitude, time, temperature, you could even define this to lie in a vector space, though the assignment is fairly meaningless. You can then describe the sets of data you want in terms of entries in these tuples.

Last edited: Dec 10, 2005
6. Dec 10, 2005

jwsiii

Here's a better way of asking my question: Does a subspace of a space have the same dimensions as the original space? For example, would V3(R) (the space of all vectors of three real numbers) be a subspace of V4(R) (the space of all vectors of four real numbers)? Or is a subspace a restriction of a space to certain elements of that subspace? For example, would V4(R) where the first element is always zero be a subspace of V4(R)? Forget the whole temperature thing. Thanks.

Edit:

Which of these would be a subspace of {a1, a2, a3, a4}?:

{a1, a2, a3}
{0, a2, a3, a4}

Last edited: Dec 10, 2005
7. Dec 10, 2005

matt grime

The first option is certainly not a subset never mind anything else.

Look up the definitions.

You seem to be coming up with the idea that there is only one real vector space of 3 dimensions. There isn't: there are infinitely many of them, not equal but all isomoprhic.

Only the second set you describe is a subspace, since a subspace is necessarily a subset, but it can be identified with a copy of R^3 via the map

{a_1,a_3,a_4} ---> {0,a_2,a_3,a_4}

Or if you prefer, although the first set obvisouly describes something isomorphic to a subspace the description given is not of that subspace.

8. Dec 10, 2005

rachmaninoff

True, I just said continuous functions on a closed interval in R were countably infinite.

9. Dec 10, 2005

shmoe

C[0,1] doesn't have an countable basis is the usual linear algebra sense, that is a countable (independant) set where every element in C[0,1] can be a written as a linear combination of these elements. Linear combination implies finite sum. Allowing infinite sums is beyond the algebraic sense, and moving into normed spaces, your countable set generates a dense subspace of C[0,1].

10. Dec 10, 2005

matt grime

And your clarification (original point) is still obviously false. Neither the functions on [0,1], nor its basis are countably finite.

11. Dec 10, 2005

rachmaninoff

Sorry! I got confused. I'd forgotten that usual vector-space linear combinations were by definition finite.

Right, C[0,1] is a Banach space with a countable dense subspace. I got mixed up thinking about the $L^2\left[a,b\right]$ space, which is a Hilbert space and does have a countable orthonormal basis - but does not include all continuous functions on [a,b] (only the subset which are L^2 integrable). Analysis is never intuitive.

12. Dec 11, 2005

matt grime

I'm afraid you're still confused. These are the function spaces that do not have countable algebraic bases, and L^2[a,b] certainly does include all continuous functions since L^2 is the space of square integrable functions on, and on a compact set [a,b] that includes all continuous functions.

13. Dec 11, 2005

matt grime

That's obviosuly false. Exercise construct uncountably many continuous functions on the interval [0,1]

14. Dec 11, 2005

rachmaninoff

Easy enough, $$\{ f(x)=c \, | \, c\in\mathbb{R}\}$$. I'd meant to say their basis was countable, but that's false too.

So, let's see if I got it now (skimming back through my Analysis text); for an L^2 Hilbert space we can create a countable subset which is dense . The norm is induced from the inner product $$\left< f, g\right> = \int_X f^* g d\mu$$; so for a subset to be (topologically) dense, for element of the space f and epsilon > 0 there is an element g in the subset and $\|f-g\| < \epsilon$; equivalently there is a sequence g_n such that $$\lim \int_X (f-g_n)^*(f-g_n) d\mu = 0$$ or (f-g)*(f-g)=0 a.e. (g=lim g_n) (since it's positive). Aha! This implies f=g a.e.

So the measure of the domain where f != g is 0, but it is not an empty set - so the subset is not an algebriac basis.

Am I getting this?

Last edited by a moderator: Dec 11, 2005
15. Dec 11, 2005

matt grime

Erm, the reason that e^{inx} are not an algebraic basis is because not very element can be written as a finite linear combination of them. I don't know what your argument was an attempt to prove, I suppose it might be some part of the proof of that fact at some point.