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Linear Algebra- Spanning Sets definition

  1. Mar 17, 2010 #1
    Linear Algebra- Basis

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

    Is {e1,e2} a basis for R3 ?

    2. Relevant equations



    3. The attempt at a solution

    I know that {e1,e2,e3} is a basis for R3

    same here,

    Is this one a basis for R3
    {(1,1,2)T,(2,2,5)T}

    I know that
    {(1,1,2)T,(2,2,5)T,(3,4,1)T}
    is a basis for R3.

    Perhaps, I am a little unclear with the definition.

    Would appreciate any help.

    Roni
     
    Last edited: Mar 17, 2010
  2. jcsd
  3. Mar 17, 2010 #2
    If you know that {e1,e2,e3} is a basis, then, by the definition of a basis...what can you say about {e1,e2}?
     
  4. Mar 17, 2010 #3

    Mark44

    Staff: Mentor

    Following Tinyboss's lead, what is the definition of a basis?
     
  5. Mar 17, 2010 #4
    This is the part I don't understand...
     
  6. Mar 17, 2010 #5
    the vectors must be independent and they must span R^3

    but say I have {e1,e2} only. they are independent and they span R^3 (they do span R^3, right?)
     
  7. Mar 17, 2010 #6

    Mark44

    Staff: Mentor

    They are linearly independent but there are only two of them. How many vectors must there be in a basis for R^3?
     
  8. Mar 17, 2010 #7
    Well, the logic says 3, but I don't understand why 3 ?

    and I guess if it's there {e1,e2} is missing 1 vector.
    but why 3 ?
    here is another example {(1,1,2)^T,(2,2,5)^T}

    Thanks for your help.
     
  9. Mar 17, 2010 #8

    Mark44

    Staff: Mentor

    The number of vectors in a basis for a vector space is equal to the dimension of the vector space. For R^1, any basis must have 1 vector, for R^2, any basis must have two vectors, and so on. If a set contains fewer vectors than the dimension of the vector space it needs to span, it can't possibly span the space, hence can't be a basis.

    With your first example, where e1 = (1, 0, 0)^T and e2 = (0, 1, 0)^T, is there some linear combination of these vectors that generates (0, 2, 5)^T?
     
  10. Mar 17, 2010 #9
    I think I understand it...
    I guess I'll need to sit on it a little more.

    Thank you very much for your help:)
     
  11. Mar 17, 2010 #10

    Now this part is confusing me.

    So,
    {e1,e2} can't span R^3 ?

    Oh they must span R^2 with two rows only, right?

    what about {(1,1,2)^T,(2,2,5)^T}, what do they span ? R^3?! ( they have two vectors) , R^2 ?! (they have 3 rows so they can't really span R^2, right?!)

    Thanks.
     
  12. Mar 17, 2010 #11

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Not exactly. In this context, e1 and e2 are elements of R^3, so they are NOT elements of R^2 and therefore cannot span R^2. They do span something that is very similar to R^2, though. Can you see what it is?

    Same deal here, the vectors are in R^3, so they aren't in R^2 and can't span R^2. Have you learned what a "subspace" is yet?
     
  13. Mar 17, 2010 #12
    We studied all this, but the professor wasn't really helpful, so everything is on me...
    Let me see if I undertand.
    {e1,e2} does span R^3 but it doesn't include ALL vectors in the plane that's created.
    However,
    {e1,e2,e3} also spans R^3 but it forms a solid which is spread throughout all R^3, therefore, it includes ALL vectors in R^3

    and {e1} also spans R^3 but it forms a line and any combination of it must be on the line. However, there lots of vectors that are not on the line, therefore, this one cat be a basis for R^3 either.

    Right?

    Thanks a lot for your help.
     
  14. Mar 17, 2010 #13

    jbunniii

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    Science Advisor
    Homework Helper
    Gold Member

    No, it does not span R^3. You need at least three vectors to do that.

    {e1,e2} spans a 2-dimensional plane (subspace) contained within R^3.

    Think of the x-y plane in Euclidean 3-space. That plane is spanned by two vectors, for example, the unit vectors in the x- and y-directions. The coordinates of these vectors are (1,0,0) and (0,1,0). This plane is not the same as R^2, where the vectors have only two coordinates. But it's equivalent in a certain mathematical sense (we say that the x-y plane within R^3 is ISOMORPHIC to R^2).

    This is what is meant by "spanning" R^3. ANY vector in R^3 can be formed by an appropriate linear combination of e1, e2, and e2. This is NOT true if you only use e1 and e2, and thus {e1,e2} does not span R^3.

    No. It spans a one-dimensional line (subspace) contained in R^3. That line is isomorphic to R^1.

    Right, it's not a basis for R^3, BECAUSE it does not span R^3.

    There are two things that must be true in order for a set of vectors to be a basis for a vector space:

    (1) the vectors must span the whole space
    (2) the vectors must be linearly independent.

    {e1,e2,e2} are linearly independent, and so are any subset of them, so (2) is not the problem here. But in order to achieve (1) you need to have enough vectors. R^3 has dimension 3, and that means you need at least 3 vectors to span it. Furthermore, you can't have MORE than 3 linearly independent vectors in a 3-dimensional space, so if the vectors are to form a basis (a stronger condition than simply spanning the space) you need to have exactly three (linearly independent) vectors.
     
  15. Mar 17, 2010 #14
    Oh, I think I got it...
    {e1,e2} span some subspace which is part of R^3, but it doesn't span R^3 because I can't create any vector in R^3 with the linear combination of e1 and e2.

    Thank you so much for your help.
     
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