# Subspaces of ℝ^3 and their bases

1. Apr 6, 2013

### Smazmbazm

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

Determine whether each of the following is a subspace of ℝ3, and if so find a basis for it:

a) The set of all vectors (x, y)
b) The set of all vectors of the form (sin2t, sintcost, 3sin2t)

Can someone please explain to me how you determine whether these are subspaces and what their bases are? I find this topic quite confusing and don't really know if my ideas about it are correct.

Can (x,y) be a subspace of ℝ3? Don't there have to be 3 variables (e.g. x, y and z) for it to be a subspace?

Thanks

Last edited by a moderator: Apr 7, 2013
2. Apr 6, 2013

### Vorde

You are correct that (x,y) cannot be a subspace of R^3, just for the reason you named.

Now as for b), remember that for sets of vectors to be a subspace there are generally three things that need to be true about the vectors.

Do you remember what they are?

3. Apr 6, 2013

### Smazmbazm

Yea the set has to contain the 0 vector and must be closed under scalar multiplication and matrix addition. So for part b, it's a valid subspace of ℝ3 but doesn't form a basis because the 3rd vector, 3sin2t is a linear combination of vector 1, sin2t?

4. Apr 6, 2013

### Vorde

True, so then what is a basis for the subspace? (hint: there is one more thing to do to find the true basis; it's not obvious, but remember the formula from trig that sin(2A)=2sinAcosA)

5. Apr 6, 2013

### Smazmbazm

I'm really not sure what the next step is. I suppose you could rewrite it as (2sint*cost, sint*cost, 3sin2t) but then vector 1 is a linear combination of vector 2, isn't it?

EDIT: Can be written as (2sint*cost, sint*cost, 6sint*cost). vector 1 and 3 are combinations of vector 2?

EDIT2: So wikipedia define a basis as "a set of linearly independent vectors that, in a linear combination, can represent every vector in a given vector space" so then is the basis simply sint*cost?

Last edited: Apr 6, 2013
6. Apr 7, 2013

### HallsofIvy

Staff Emeritus
There are a number of errors here-
"sin(2t)", "sin(t)cos(t)", and "3sin(2t)" are NOT three vectors, they are the three components of a single vector. You are given a formula for finding "vectors": if t= 0, this is (0, 0, 0), if $t= \pi/4$, it is (1, 1/2, 3).

But your basic error is in thinking it is a vector space! 6(1, 1/2, 3)= (6, 3, 18). Is there a "t" that gives that? That is, is there a t such that $sin(2t)= 6$, $sin(t)cos(t)= 3$, and $3sin(2t)= 18$?

7. Apr 7, 2013

### Vorde

Completely correct, sorry to the OP; I skimmed that line and assumed you meant a subspace of real-valued functions spanned by those three functions.

8. Apr 7, 2013

### Smazmbazm

So it's not a subspace? To disprove that a given formula cannot represent a subspace of a vector space, does one just have to find values that are impossible to get?

9. Apr 8, 2013

### Vorde

One must show that there is a vector v in the spanning set and a constant c such that cv is not in the spanning set.

Or, alternatively, that one of the other two tenets are not fulfilled (the zero vector is not in the set or there is a sum of vectors not in the set).

10. Apr 8, 2013

### Smazmbazm

Great. Got it now. Thanks for your help Vorde and HallsofIvy, much appreciated.