Linear independence in Q[x]

1. Dec 3, 2011

autre

1. The problem statement, all variables and given/known data

Show that the set {1,x,x^2,...,x^n,...} is linearly independent in Q[x].

3. The attempt at a solution

Since an infinite set of vectors is linearly independent if each finite subset is also linearly independent, I think I need to show that every subset of {1,x,x^2,...,x^n,...} is linearly independent. It's easy to show any given subset is linearly indep. but I'm not sure how to show ALL subsets are linearly independent.

Any ideas?

2. Dec 3, 2011

micromass

Staff Emeritus
You only need to show that all subsets of the form

$$\{1,x,...,x^m\}$$

are linearly independent. Do you see why?? Do you see why it is independent?

3. Dec 3, 2011

autre

For any given subset of Q[x] whose degree is less than or equal to m, we can write an element a = a_0+a_1x+a_2x^2+...+a_mq^m. a = 0 iff for all i, a_i = 0, thus the set {1,x,...,x^m} is linearly independent in Q[x]?

4. Dec 3, 2011

micromass

Staff Emeritus
That is OK. I have a few follow up questions though!

Why is this true?? Did you prove it?

OK, but why does this imply that each finite subset of $\{1,x,...,x^n,...\}$ is linearly independent?

5. Dec 3, 2011

autre

Well if for some i, a_i=/=0, then a=a_ix^i by def of a in Q[x], right?

Since our choice of a finite subset of {1,x,...,x^n,...} was arbitrary?

6. Dec 3, 2011

micromass

Staff Emeritus
Hmm, it's not that easy...

You have to prove for example that

$$2x^3-4x^2+6x+9$$

is never constantly 0. Furthermore, you must prove it for all ai.

It was not arbitrary. We chose the subset $\{1,x,...,x^m\}$. This is indeed sufficient, but why?

7. Dec 3, 2011

autre

But our choice of m+1 as the degree of the subset was arbitrary. Anyway, since all we need to show is that finite subsets are linearly independent, we have shown that.

8. Dec 3, 2011

micromass

Staff Emeritus
Yes, our choice of m was arbitrary. But we still have not shown it for arbitrary polynomials then!

For example, we also need to show it for the set $\{1,x^2,x^4,x^{14}\}$. You have not yet considered such a set.

9. Dec 3, 2011

Deveno

there are two ways of looking at this:

algebraic way:

$$p(x) = a_0 + a_1x + a_2x^2 + \dots + a_nx^n \equiv 0 \text{ iff } a_i = 0\ \forall \ i$$

by the definition of equality of polynomials in Q[x].

analytic way:

$$p(x) = a_0 + a_1x + a_2x^2 + \dots + a_nx^n \equiv 0 \text{ iff } \forall x, p(x) = 0$$

in one way, "x" is just a formal symbol, it doesn't really stand for anything. in another way, x stands for some value which might be a number of some sort.