Infinite Dimensional Vector Space

jimmypoopins
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Homework Statement


Prove that V is infinite dimensional if and only if there is a sequence v_1, v_2,... of vectors in V such that (v_1,...,v_n) is linearly independent for every positive integer n.


Homework Equations


A vector space is finite dimensional if some list of vectors in it spans the space.


The Attempt at a Solution


We need to prove two directions. For the forward direction, we assume V is infinitely dimensional (and therefore is not finite dimensional). It's really giving me a headache. All of the theorems in my book involve finite-dimensional vector spaces, and none of the proofs seem to give me any information pertaining to the forward direction of this problem. Do I want to use induction somehow? For example, the infinite dimensional vector space F^(infinity) has elements that can be written in the form a_1e_1+a_2e_2+...+a_ne_n+... with the a_i's scalars in the field and the e_i's the vector with all zeroes in every spot except for a 1 in the ith place. That is a linearly independent list, that works for all n, but i don't know how to explain it for every infinite dimensional vector space.

For the backward direction, first I wanted to assume V is finite dimensional and use contradiction. Then V has a basis that is linearly independent spans V, and has a dimension, let's say m=n-1. Then consider the linearly independent list (v_1,...,v_n). It has dimension n, but every linearly independent list of vectors in a finite-dimensional vector space can be extended to a basis of the vector space, so since dim(V) < dim(v_1,...,v_n) we have a contradiction and V must be infinite dimensional.

I think the backward direction is almost complete, but I'm having trouble with the forward. Any hints, please?
 
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V is infinite dimensional. Pick a vector v1 in V. If v1 spanned V, V would be finite dimensional. It's not. So there's another independent vector v2. If {v1,v2} spanned then V would be finite dimensional. It's not. So there's an independent v3. If {v1,v2,v3} spanned then etc. It's sort of induction. The point is that any set of linearly independent vectors can always be extended.
 
Dick said:
V is infinite dimensional. Pick a vector v1 in V. If v1 spanned V, V would be finite dimensional. It's not. So there's another independent vector v2. If {v1,v2} spanned then V would be finite dimensional. It's not. So there's an independent v3. If {v1,v2,v3} spanned then etc. It's sort of induction. The point is that any set of linearly independent vectors can always be extended.

Thank you very much, I get it.

How about the backward direction? Is it correct? Does it need tweaking? Or is it completely wrong?
 
Yeah, I think it's basically ok. The point is that if V has dimension n, then it can't have n+1 linearly independent vectors in it.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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