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Change of basis in R^n and dimension is <n

  1. Aug 24, 2010 #1
    Suppose I have a basis for a subspace V in [tex]\mathbb{R}^{4}[/tex]:

    [tex]\mathbf{v_{1}}=[1, 3, 5, 7]^{T}[/tex]
    [tex]\mathbf{v_{2}}=[2, 4, 6, 8]^{T}[/tex]
    [tex]\mathbf{v_{3}}=[3, 3, 4, 4]^{T}[/tex]

    V has dimension 3, but is in [tex]\mathbb{R}^{4}[/tex]. How would one switch basis for this subspace, when you can't use an invertible transition matrix?

    Thanks!
     
  2. jcsd
  3. Aug 24, 2010 #2
    I don't really understand the question very well. "Changing basis" is a process of converting the representation of vectors in a space from one basis to another. But here you have two different spaces. It's kind of like asking "how do you change basis between [itex] \mathbb{R}^{3} [/itex] and [itex] \mathbb{R}^{2} [/itex]?"...well, you can't.
     
  4. Aug 25, 2010 #3
    Hmm, I guess I'm asking how one would switch to another another basis that spans V. Given co-ordinates in terms of v1, v2, v3, how would one find co-ordinates with respect to a new spanning set of vectors?

    If V spanned R4, then I could just use a transition matrix. It's the fact that V is dimension 3, but each basis vector has 4 co-ordinates, that is confusing me.
     
  5. Aug 25, 2010 #4
    Well, if they gave you coordinates in terms of v1, v2, and v3, then you'd have a vector with 3 components, right? And any other basis for the same space would also have 3 components, because the dimension of the subspace is 3. So then you have your usual 3x3 thing for changing basis between those.
     
  6. Aug 25, 2010 #5
    But how can you use a square matrix when the co-ordinate vectors have 4 components? This is what's confusing me...
     
  7. Aug 25, 2010 #6

    Office_Shredder

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    You don't have four components, because instead of writing out vectors of v in terms of the standard basis in R4, you write them out in coordinates in terms of v1, v2 and v3. Then you have three coordinates as required. For example, the vector (1,3,1) in this coordinate system would be 1*v1+3v2+1v3=(10,18,27,35)
     
  8. Aug 25, 2010 #7
    FFFFFFFFFFFUUUUUUUUUUUUUUUUU-

    Great explanation!!

    Thanks for the responses!
     
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