# Show this is a basis

1. Nov 17, 2013

### SMA_01

Let β be a basis for ℝ over Q (the set of all rational numbers) and let a$\in$ℝ, a≠1.

Show that aβ={ay|y$\in$β} is a basis for ℝ over Q for all a≠0.

So I need to show (1) Linear independence, and (2) spanning. I am a little confused, especially because the dimension for the vector space is uncountably infinite.
Here is what I have for linear independence:

I took a finite subset of aβ, {az|z$\in$S$\subset$β, |S|<|ℝ|} (not sure if I wrote that correctly), so if

$\sum_{z\in S}$az=0

then z must be zero since a≠0 (initial assumption). Therefore, it follows that aβ is linearly independent.

Is this correct?
I don't know how to approach spanning.

2. Nov 18, 2013

### Staff: Mentor

This seems like armwaving to me.
Do you have a sense of what this problem is about? In particular, what a basis over Q that spans R might look like? For example, can you list a few of the basis elements of such a basis?

Given that β is some basis, you need to show that aβ is also a basis, where a ≠ 0.

To show that any set is a basis for some space, you need to do two things:
1. Show that the set of basis elements is linearly independent.
2. Show that the set of basis elements spans the space.

The latter requirement means that if x $\in$ R, then x can be written as a linear combination of the basis elements.

3. Nov 18, 2013

### SMA_01

Honestly, I can't fully grasp it intuitively. I know how a basis is for ℝ with n-dimension, but not ℝ.
I just can't see how something can span ℝ, is the standard basis applicable here?

If I can't show linear independence/spanning the usual way, how should I do this?

Thanks.

4. Nov 18, 2013

### vela

Staff Emeritus
You do show independence the usual way. You need to show that for any finite subset $\{v_1,v_2,\dots,v_n\} \subset a\beta$ that
$$q_1 v_1 + q_2 v_2 + \cdots + q_n v_n = 0$$ implies $q_1=q_2=\cdots=q_n=0$, where $q_i \in \mathbb{Q}$. What you wrote above doesn't show this. For one thing, there's no mention of the $q_i$'s.

5. Nov 18, 2013

### SMA_01

But a basis for ℝ over Q is uncountable, so how can I take a finite, countable subset?

6. Nov 18, 2013

### Dick

Because that's the definition of linear independence. To show it's linearly independent you need to show every finite subset is linearly independent.

7. Nov 18, 2013

### SMA_01

Okay, correct me if I'm wrong, but if I take a finite subset, I thought it won't be countable? Like I can't order the terms...

8. Nov 18, 2013

### vela

Staff Emeritus
Finite implies countable.

9. Nov 18, 2013

### SMA_01

So if I take a finite subset of ℝ, then I can order the terms? Sorry, I just want to clarify.

10. Nov 18, 2013

### vela

Staff Emeritus
What's the precise definition of finite? If you know this basic definition, you should be able to answer your own question.