Beginner's doubt regarding vector spaces.

[ask_LXXXVI]

I have just started a study of linear algebra and I have a doubt regarding vector spaces.

Consider the vector space spanned by the 3 dimensional vectors [1,0,0] and [0,1,0] , this would be a 2-dimensional vector space no doubt.But it also is a subspace of $$\mathbb R$$ ^3.I have no problem in this.

But consider the vector space spanned by the functions cos x and sin x. This is also a 2-dimensional vector space. But what bigger vector space is it a subspace of? And what is the dimension of that bigger vector space? (Is it always important to know ,of what bigger vector space is a vector space we are considering, a subspace of? )

As I gather,the vector space spanned by cos x and sin x is solution space of a differential equation such as y'' + y = 0.

P.S. I hope this doesn't come under homework question and that I have posted in apt forum.If not please sorry for the impertinence on my part owing to my ignorance :uhh:

 

[Ben Niehoff]

Well, there are lots of bigger vector spaces that it could be a subspace of. Every vector space doesn't have to be a subspace of something else; they can just be defined as is. But it's important to understand the notion of a subspace, because it is useful for other things.

If you look at the differential equation

y'''' + 5y'' + 4y = 0

you will find its solution space is spanned by

cos x, sin x, cos 2x, sin 2x

So, that's an example of a 4-dimensional vector space, a 2-dimensional subspace of which is spanned by {cos x, sin x}.

Another example would be the space of Fourier modes on the interval [0, 2pi]. This is an infinite-dimensional vector space spanned by

cos x, sin x, cos 2x, sin 2x, cos 3x, sin 3x, etc.

 

[ask_LXXXVI]

Well, there are lots of bigger vector spaces that it could be a subspace of. Every vector space doesn't have to be a subspace of something else; they can just be defined as is. But it's important to understand the notion of a subspace, because it is useful for other things.

Thanks , that really helped.

If you look at the differential equation

y'''' + 5y'' + 4y = 0

you will find its solution space is spanned by

cos x, sin x, cos 2x, sin 2x

So, that's an example of a 4-dimensional vector space, a 2-dimensional subspace of which is spanned by {cos x, sin x}.

Yes, but consider [1,0,0,0] and [0,1,0,0] as an example. The subspace it spans will only be part of R^4 and it will be a 2-dimensional subspace in it. We can't say it is subspace of any vector space other than 4. Isn't it? We can't say it is part of R^3 nor R^5 or anything because it has 4 components. In that way isn't a finite dimension vector different from a function such as cos x.(can we think of cos x as a vector with uncountably infinite components ?)

I guess it is all in what vector space we are defining.You rightly pointed out the 4th order diff eqn , it showed me a new angle of thought.


Anyways what's the biggest dimensional vector space that cos x can be a subspace of?
Does that vector space dimension have cardinality of natural numbers(N₀,i.e aleph-nought) or that of the continuum (C)? I hope this last question isn't mumbo-jumbo.

 

[Spartan Math]

Perhaps a little more intriguing is that upon expanding sin(x) and cos(x) into their respective Taylor series, we can regard sin(x) and cos(x) as elements of the formal power series ring over the rationals, which is also a vector space (over the rationals)!

 

[Fredrik]

Anyways what's the biggest dimensional vector space that cos x can be a subspace of?

The biggest one I can think of without any effort is the set of functions from $\mathbb R$ into $\mathbb R$, with the vector space structure defined by

$$(f+g)(x)=f(x)+g(x)$$

$$(af)(x)=a(f(x))$$

for all functions f,g in that set, and all real numbers a.

Does that vector space dimension have cardinality of natural numbers(N₀,i.e aleph-nought) or that of the continuum (C)? I hope this last question isn't mumbo-jumbo.

The dimension is either infinite or not. It's infinite if there for every integer n exists a linearly independent set with n members. You could however ask for the cardinality of a basis for that vector space. Hm, now things are getting complicated. I could be wrong, but I think bases in the usual sense (a linearly independent set that isn't a proper subset of another linearly independent set) are often uncountable, even on vector spaces where orthonormal bases are countable. So I'm pretty sure that you can find an uncountable basis for this one, and also that there's no countable one.

 

[ask_LXXXVI]

The biggest one I can think of without any effort is the set of functions from $\mathbb R$ into $\mathbb R$, with the vector space structure defined by

$$(f+g)(x)=f(x)+g(x)$$

$$(af)(x)=a(f(x))$$

for all functions f,g in that set, and all real numbers a.

Thanks for the reply.

The dimension is either infinite or not. It's infinite if there for every integer n exists a linearly independent set with n members. You could however ask for the cardinality of a basis for that vector space.

Yeah, you understood what I wanted to say here. I intended to say about cardinality of the basis only.

Hm, now things are getting complicated. I could be wrong, but I think bases in the usual sense (a linearly independent set that isn't a proper subset of another linearly independent set) are often uncountable, even on vector spaces where orthonormal bases are countable. So I'm pretty sure that you can find an uncountable basis for this one, and also that there's no countable one.

I on my limited knowledge also am inclined to think that the basis for the set of functions from $\mathbb R$ into $\mathbb R$ has to be uncountable.

My logic (intuitive) is that any function can be expressed as its Fourier Transform.

When we consider only the set of periodic functions of period T , all the functions can be expressed in terms of harmonics of exp(2*pi/T) (Fourier Series).So these harmonics are basis for this set. And the set of harmonics has cardinality which is countably infinite.

When we consider the basis for the set of functions the set of functions from $\mathbb R$ into $\mathbb R$ , the basis can be said to be set of exponentials which can have any arbitrary frequency(Fourier Transform). Such a set consisting of the exponentials with any arbitrary frequency will have cardinality which is uncountably infinite.Is my process of thinking correct ? I am yet to reach these topics in linear algebra. Have only done some basic concepts of finite dimensional vectors and somewhat worked with vector spaces of functions in my digital communication course , which I consider was only skimming the surface .

Anyways thanks everybody for helping.

 

[jbunniii]

When we consider the basis for the set of functions the set of functions from $\mathbb R$ into $\mathbb R$ , the basis can be said to be set of exponentials which can have any arbitrary frequency(Fourier Transform). Such a set consisting of the exponentials with any arbitrary frequency will have cardinality which is uncountably infinite.

Loosely speaking, you have the right idea. But there are some technicalities.

Consider the definition of the Fourier transform. If you use the integral definition, then only functions in $L^1(\mathbb{R})$ (i.e., absolutely integrable functions) have a Fourier transform. And the complex exponentials are not even in $L^1(\mathbb{R})$, so they can't be a basis for $L^1(\mathbb{R})$.

Also, the Fourier transform of an $L^1$ function isn't necessarily in $L^1$, so the inversion formula need not apply. Thus the question arises: is it always possible to recover the original function after projecting it onto the complex exponential "basis"?

A lot of these shortcomings can be overcome by defining a suitable generalization of the Fourier transform that applies to a much broader class of functions (and other objects that are not even functions). The tempered distributions are one such class.

 

[VKint]

For an infinite-dimensional space $$V$$, there are some subtleties that have to be kept in mind whenever you talk about a "basis" for $$V$$. The most natural extension of the finite-dimensional concept of a basis to infinite-dimensional spaces is the Hamel Basis, which is defined as a maximal set independent under finite linear combinations, i.e., a set maximal under the property that any finite subset is linearly independent. Using Zorn's Lemma, you can then show that every vector space has a basis (in fact, this proposition is equivalent to the Axiom of Choice and Zorn's Lemma), and that every Hamel basis has the same cardinality (I think). The Hamel basis thus has the virtue of not requiring any "extra structure" on the vector space to be defined (there are no infinite sums that have to converge). Also, even in the context of Hilbert and Banach spaces, such as function spaces, "bases" of the type you've described do not necessarily all have the same cardinality, as noted below, whereas the Hamel bases of a vector space all have the same cardinality, and thus give a well-defined notion of dimension (again, if memory serves). In particular, if you want to talk about the "dimension" of an infinite-dimensional vector space (barring topological notions of dimensionality, of which I'm pretty ignorant), regardless of whether or not you can make sense of infinite sums of vectors, you need to use Hamel bases.

However, Hamel bases tend to be very large, and more or less impossible to work with explicitly in any cases that we actually care about (e.g., functions over the real numbers)--in fact, a Hamel basis of any infinite-dimensional, complete, normed vector space must be uncountable. In particular, the "dimension" of the space of function on $$\mathbb{R}$$ must be uncountable--in fact, I think a Hamel basis of the space of all functions (or even continuous functions) on $$\mathbb{R}$$ might have cardinality strictly greater than that of the continuum. However, this doesn't necessarily correspond intuitively to any useful fact about functions on the reals.

Also, a more minor point: The two-dimensional subspace of $$\mathbb{R}^4$$ that you've described is, in fact, a subspace of $$\mathbb{R}^5$$ (or is at least isomorphic to one), as well as $$\mathbb{R}^6$$, $$\mathbb{R}^7$$, and so on. This is because $$\mathbb{R}^4$$ is isomorphic to a subspace of $$\mathbb{R}^n$$ for any $$n \geq 4$$. The way you've chosen to describe your subspace of $$\mathbb{R}^4$$ (i.e., in terms of vectors with 4 components) doesn't change the geometric fact that $$\mathbb{R}^4$$ "lives inside" $$\mathbb{R}^5$$. In fact, you can always extend a vector space $$V$$ by simply adding elements to its basis, i.e., constructing the free vector space on a strict superset of a basis for $$V$$.

As an aside, Hamel bases can be quite useful theoretically, even if somewhat uninteresting in the context of function spaces--for example, you can use a Hamel basis of $$\mathbb{R}$$ over $$\mathbb{Q}$$ to prove that $$\mathbb{R}$$ can be partitioned into arbitrarily (but countably) many disjoint, nonempty subsets $$S_i$$, each of which is closed under addition and none of which is equal to the set of positives or negatives. Strange...but interesting. At least to me.

 

[ask_LXXXVI]

Also, a more minor point: The two-dimensional subspace of $$\mathbb{R}^4$$ that you've described is, in fact, a subspace of $$\mathbb{R}^5$$ (or is at least isomorphic to one), as well as $$\mathbb{R}^6$$, $$\mathbb{R}^7$$, and so on. This is because $$\mathbb{R}^4$$ is isomorphic to a subspace of $$\mathbb{R}^n$$ for any $$n \geq 4$$. The way you've chosen to describe your subspace of $$\mathbb{R}^4$$ (i.e., in terms of vectors with 4 components) doesn't change the geometric fact that $$\mathbb{R}^4$$ "lives inside" $$\mathbb{R}^5$$. In fact, you can always extend a vector space $$V$$ by simply adding elements to its basis, i.e., constructing the free vector space on a strict superset of a basis for $$V$$.

Even if a 2-D subspace in $$\mathbb{R}^4$$ is isomorphic to a 2-D subspace in $$\mathbb{R}^5$$ , aren't the two different? I mean ,yeah there is one-to-one correspondence between the two sets.I may be way off the mark,but I think the fact that there are 4 components in one case and 5 in another make the elements of the two different.