# Infinite dim vector space

1. Dec 14, 2005

### Ratzinger

We have x=x1(1,0,0) + x2(0,1,0) + x3(0,0,1) to represent R^3. That's a finite dimensional vector space. So what do we need infinite dimensional vector space for? Why do we need (1,0,0,...), (0,1,0,0,...), etc. bases vectors to represent R^1 ?

2. Dec 14, 2005

### TD

To represent R^1, you only need 1 vector. A vector space of dimension n is spanned by a basis of n vectors, just as in your example of R^3. This is because a basis needs to span the vector space (which means you need *at least* n vectors) and has to be linearly independant (which means you can only have *at most* n vectors) which makes the number of vectors in the basis exactly n.

Then, what do we need vector spaces with infinite dimension? Consider the vectorspace $\mathbb{R}\left[ X \right]$ which is the vector space of all polynomials in x over R. This is trivially an infinite dimensional vector space since a finite number of vectors in a basis contains a vector with a maximum degree r, meaning that x^(r+1) and higher cannot be formed.

3. Dec 14, 2005

### HallsofIvy

"Functional Analysis" makes intensive use of "function spaces"- infinite dimensional vector spaces of functions satisfying certain conditions. TD gave a simple example- the space of all polynomials. Perhaps the most important is L2(X), the vector space of all functions whose squares are Lebesque integrable on set X.

4. Dec 14, 2005

### NateTG

Another familiar example of an infinite dimensional vector space is functions from an infinite domain to a ring
Consider that the space of functions
$$f:A \rightarrow R$$
from some set $A$ to a ring $R$
is a vector space with dimensions indexed on $A$
since we have a vector
$$f(a)=r$$
or
$$f_a=r$$
Scalar multiplication, and vector addition are performed using the ring.

Last edited: Dec 14, 2005
5. Dec 14, 2005

### matt grime

I think you should reconsider that example, NateTG. How is *a* function a vector space? Over what field? And what are its elements