Subspaces of ℝ^3 and their bases

  • Thread starter Thread starter Smazmbazm
  • Start date Start date
  • Tags Tags
    Bases Subspaces
AI Thread Summary
The discussion centers on determining whether specific sets of vectors are subspaces of ℝ^3 and identifying their bases. It is established that the set of vectors (x, y) cannot be a subspace of ℝ^3 because it lacks a third component. For the second set, (sin2t, sintcost, 3sin2t), it is confirmed to be a valid subspace but does not form a basis due to linear dependence among its vectors. The conversation highlights the importance of verifying the zero vector's inclusion and the closure properties under scalar multiplication and vector addition to confirm subspaces. Ultimately, the participants clarify that to disprove a subspace, one must find values that cannot be represented by the given vectors.
Smazmbazm
Messages
45
Reaction score
0

Homework Statement



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:
Physics news on Phys.org
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?
 
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?
 
Smazmbazm said:
but doesn't form a basis because the 3rd vector, 3sin2t is a linear combination of vector 1, sin2t?

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)
 
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:
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?
 
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.

My bad.
 
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?
 
Smazmbazm said:
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?

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

Similar threads

Replies
11
Views
2K
Replies
4
Views
1K
Replies
1
Views
2K
Replies
10
Views
2K
Replies
17
Views
10K
Back
Top