# Infinite-dimensional vector space question

1. Feb 15, 2012

### Master J

I'm learning about rings, fields, vector spaces and so forth.

The book I have states:

"Real-valued functions on R^n, denoted F(R^n), form a vector space over R. The vectors can be thought of as functions of n arguments,

f(x) = f(x_1, x_2, ... x_n) "

It then says later that these vectors cannot be specified by a finite number of scalars and so the space is infinite-dimensional. Am I missing something here? From what I see, there are only a finite number of elements x_n in the vector, so why is the space infinite-dimensional?

Last edited: Feb 15, 2012
2. Feb 15, 2012

Doesn't it say that the vectors are functions? If yes, then could you write any real-valued multivariable function as the linear combination of a finite number of specific functions in F(R^n)? In other words, could you find a set that is linearly independent and spans F(R^n) with only n elements in it?
That's why it is infinite dimensional.

3. Feb 15, 2012

### HallsofIvy

You seem to be confusing the n variables with basis vectors. The real valued functions on a single variable, f(x), include functions of the form $x^n$ for any integer n, none of which can be written in terms of lower powers so you will have to have an infinite basis right there.

4. Feb 16, 2012

### Master J

Ah, right, a minor confusion I believe.

So the jist is, there is no finite linearly independent set of functions on R^n that allows one to write any other function in R^n as a combination of them?

Also, let's say I have some random fucntion on R^n:

f(x) = 3x^3 + sin(x^2)

How is that representable as a vector? Is it that one would have to break it up into linearly independent parts (in this, expand out the sin function, which would be infinite)?

Last edited: Feb 16, 2012
5. Feb 16, 2012

### Fredrik

Staff Emeritus
Correct. "V is infinite-dimensional" means that for all positive integers n, there's a linearly independent subset of V with n members.

No. (You seem to have chosen n=1, but that doesn't matter for what I'm about to say). The f defined this way is said to be a vector because it's a member of a vector space. The vector space is the set of all functions from $\mathbb R^n$ (in this case with n=1) into $\mathbb R$ with the standard definitions of addition and scalar multiplication. If we call that set V, the standard definitions look like this:
Let $a,b\in\mathbb R$ and $f,g\in V$ be arbitrary. f+g and af are respectively defined by
\begin{align} (f+g)(x) &=f(x)+g(x) &\text{for all }x\in \mathbb R^n.\\ (af)(x) &=a(f(x)) &\text{for all }x\in \mathbb R^n. \end{align}

6. Feb 16, 2012

Correct.

Well, no. Any set of mathematical objects with two operations defined on them (namely 'vector addition' and 'scalar multiplication') that satisfies the axioms of vector space is called a vector space. The members of the set are called 'vectors', and they can be any mathematical object. The term vector space is used because these mathematical properties were first abstracted to study real vectors in physics and analytic geometry, but there are many other sets of different mathematical objects that also satisfy the axioms of vector space. We call those sets as vector spaces and their members as vectors. You could use the term 'linear space' if you think the term 'vector space' sounds confusing.
In particular, the set of all functions from Rn to R forms a vector space because the way we define vector addition and scalar multiplication on them (as Fredrik has defined them for you) make them vector spaces.

7. Feb 16, 2012

### Master J

When I think of a vector, I tend to think of a column or row matrix (having been first introduced to them that way). For instance, the set of objects

x = [x_1, x_2, x_3]

where x is an element of R is a vector right. But the idea of functions of one or many variables being vectors has nothing to do with this reprensentation? It is simply the case that both sets of objects satistfy the axioms of a vector space.

8. Feb 16, 2012

### Fredrik

Staff Emeritus
That's right. A vector is a member of a vector space. It doesn't matter what it looks like. Even when the vectors are finite collections of numbers, the dimension of the vector space may have nothing to do with how many numbers there are in each collection.

9. Feb 16, 2012

### lavinia

You can think of an n dimensional vector like R^n as the set of functions on a finite set. That is why it is finite dimensional. But the set of functions on R^n is defined on an infinite set and so is infinite dimensional. So for instance the set of functions on the positive integers is an infinite dimensional vector space. In fact a good way to see why is to think of R^n as the set of functions on the integers 1 to n. Then let n go to infinity and you see that the dimension increases without bound.

10. Feb 16, 2012

### Deveno

a silly example:

consider the set of all constant sequences of real numbers:

S = {(x,x,x,x,........), x in R}

these sequences lie in what one might call R, which is infinite-dimensional.

now R can be given a vector-space structure in the expected way, by defining addition and scalar multiplication "coordinate-wise" (or in this case, perhaps "term-wise" is a better word).

it turns out that S, even though it lives in an infinite-dimensional vector space, is finite-dimensional: a basis for S is:

{(1,1,1,1,.........)}, the constant sequence of 1's.

a not-so-silly example:

R[x], the space of all real polynomials, is infinite-dimensional (no finite set of polynomials generates every polynomial, just pick one of greater degree that the polynomial of maximal degree in the finite set).

but the space of polynomials of degree < n, is an n-dimensional subspace (we just typically don't write down the infinite string of 0 coefficients at the "tail").

(this is a common trick in mathematics, btw...to embed a finite thing in an infinite thing, we often add "fake infinity" by appending infinite bits of nothing. sometimes we do the opposite thing, too....which is to treat infinite things that are "eventually 0" as if they were finite).

11. Feb 16, 2012