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Basis vector

  1. Mar 12, 2008 #1
    1. The problem statement, all variables and given/known data

    Suppose we have,

    Let V=span(v1, v2, v3, v4). Find a basis for V.
    (there are actual vectors given, however I can't exactly write them in an easy to read form)

    2. Relevant equations
    N/A


    3. The attempt at a solution

    My initial thought is, if I require a basis in which all linear combinations of v1, v2, v3, v4 that can be written, couldn't the v1, v2, v3, v4 be the basis so long as they are linearly independent? As well, wouldn't the standard basis vector (this is for matrices) be valid? However if it was that easy...well I'm probably wrong. I don't really need a direct solution; really just a clarification.
     
  2. jcsd
  3. Mar 12, 2008 #2

    morphism

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    Well, what's a basis? It's a linearly independent spanning set. So how about you try to discard the linearly dependent vectors from {v1, v2, v3, v4} (i.e. the ones in this set that can be written as a linear combination of the others)? This will certainly make this set linearly independent, but will it stay spanning?
     
  4. Mar 12, 2008 #3

    Dick

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    If V is the span of {v1,v2,v3,v4} and they are linearly independent, then they are a basis. If there are actual vectors given and this is a problem then it's likely that they aren't linearly independent and you are supposed to eliminate the ones that are linear combinations of the others.
     
  5. Mar 12, 2008 #4
    So I should expect my answer to be 3 out of the 4 vectors, or perhaps 2 then? That would make sense.

    I also need help with comparing vectors as matrices. With vectors in R space I know what to do, but for matrices I'm not quite sure.

    Suppose I have matrices A, B, C, D such that

    aA+bB+cC+dD=0, thus I can solve the system and find values of a, b, c, d. Would this be the right approach to find if its linearly independent? If it is, I'm not quite sure how to interpret my results since I get a=0, b=0, c=0, d=t, where t is a free variable. Meaning that there isn't just the trivial solution and its linearly dependent...however how would I go about finding which one I should remove?

    Also, technically, is it wrong to say that the span is the standard basis for matrice vectors? Because A, B, C, and D are just subspaces of all matrices right?
     
  6. Mar 12, 2008 #5

    Dick

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    How can you get a solution like that unless D=0?
     
  7. Mar 12, 2008 #6
    A, B, C, and D are matrices, so I get multiple equations in which I'm solving by row reducing the matrix.
     
  8. Mar 12, 2008 #7
    So I'm getting

    1 0 0 0
    0 1 0 0
    0 0 1 0
    0 0 0 0

    Sorry for this double post, I clicked edit and didn't realize where I was typing.
     
  9. Mar 13, 2008 #8
    your basis is
    1 0 0 0
    0 1 0 0
    0 0 1 0

    the zero lines says that the 4th vector is composed out of these 3 vectors
     
  10. Mar 13, 2008 #9

    HallsofIvy

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    Saying "aA+bB+ cC+ dD= 0" and then "I get a= 0, b= 0, c= 0, d= t, where t is a free variable" means that tD= 0 so either t= 0 (and your matrices are independent) or D= 0. I don't believe either of those is true.

    I can't even make sense out of that last paragraph! No, the "span" is NOT "the standard basis". A "span" is a subspace, a "basis" is a collection of independent vectors.
    And, since you had already stated that A, B, C, D are matrices they are certainly not "subspaces of matrices".
     
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