# Dimension of set S, subspace of R3?

1. Apr 14, 2014

### concon

1. The problem statement, all variables and given/known data
Determine whether set S = {2a,-4a+5b,4b| aε R ^ bε R} is a subspace of R3?

If it is a subspace of R3, find the dimension?

2. Relevant equations
dimension= n if it forms the basis of Rn, meaning that its linear independent and span(S) = V

3. The attempt at a solution
I am confused on really where to start on this problem.
What do I do with the a's and b's? I have only done problems like this with real values for the vectors. What is the first step in determining if this is a subspace?

2. Apr 14, 2014

### jbunniii

First step is to recall the definition of a subspace: it must be closed under addition and scalar multiplication.

So start by taking two arbitrary elements of $S$. An arbitrary element looks like $v_1 = (2a_1, -4a_1 + 5b_1, 4b_1)$. A second one looks like $v_2 = (2a_2, -4a_2 + 5b_2, 4b_2)$. What do you get when you add $v_1$ and $v_2$?

3. Apr 14, 2014

### concon

well if you add them wouldn't you get a new vector
that is
(2a1 + 2a2, -4a1-4a2 + 5b1 +5b2, 4b1 + 4b2)
Which is not in the same form as v1 and v2 thus it is not a subspace right?
So you cannot give it a dimension? Is this correct?

Thanks for the reply! Can you look at my other posts and help with those?

4. Apr 14, 2014

### jbunniii

What if I rearrange the expressions a bit: $(2(a_1 + a_2), -4(a_1 + a_2) + 5(b_1 + b_2), 4(b_1 + b_2))$. Is this an element of $S$?

5. Apr 14, 2014

### concon

Okay gotcha so when you re-arrange like that them yes it is an element of S and thus is closed under addition. So it is a subspace?

6. Apr 14, 2014

### jbunniii

You also need to check that it is closed under scalar multiplication.

7. Apr 14, 2014

### HallsofIvy

Staff Emeritus
As far as the dimension is concerned, notice that (2a,-4a+5b,4b)= (2a, -4a, 0)+ (0, 5b, 4b)= a(2, -4, 0)+ b(0, 5, 4).

8. Apr 14, 2014

### concon

Multiplication

So if I take v1 = (2a, -4a+5b , 4b)
and multiply by the scalar k=3
you would get
v1 = (6a, -12a +15b, 12b)
So it isn't closed under scalar multiplication or is it?

9. Apr 14, 2014

### jbunniii

We can rewrite that expression as $(2(3a), -4(3a) + 5(3b), 4(3b))$. Is this in $S$? What happens if you replace $k=3$ by something more general, i.e. just $k$?

10. Apr 14, 2014

### concon

Well I guess as long as k isn't zero then it is also closed under scalar multiplication right?

11. Apr 14, 2014

### concon

Okay so it is subspace. Now how do I find the dimension?

12. Apr 14, 2014

### jbunniii

$k=0$ works too:
$$0(2a, -4a + 5b, 4b) = (2(0), -4(0) + 5(0), 4(0)) = (2a', -4a' + 5b', 4b')$$
where $a' = b' = 0$.

13. Apr 14, 2014

### jbunniii

See HallsofIvy's note above.

14. Apr 15, 2014

### concon

Okay so dimension is 2?

15. Apr 15, 2014

### jbunniii

That's right. As Halls noted,
$$(2a,-4a+5b,4b)= (2a, -4a, 0)+ (0, 5b, 4b)= a(2, -4, 0)+ b(0, 5, 4)$$
and this observation allows you to immediately see a basis for the subspace. Do you see why?