What makes the set {1,x,x^2,...,x^n,...} linearly independent in Q[x]?

[autre]

Homework Statement

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

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?

 

[micromass]

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?

 

[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]?

 

[micromass]

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

a = a_0+a_1x+a_2x^2+...+a_mq^m. a = 0 iff for all i, a_i = 0,

Why is this true?? Did you prove it?

thus the set {1,x,...,x^m} is linearly independent in Q[x]?

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

 

[autre]

Why is this true?? Did you prove it?

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

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

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

 

[micromass]

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

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 a_i.

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

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

 

[autre]

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

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.

 

[micromass]

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.

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.

 

[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.