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Determining Basis of R^5

  1. Dec 6, 2014 #1
    I've attached the question to this post. The answer is true, but I'm trying to figure out why.

    Using Gram-Schmidt, I can only necessarily find 3 orthogonal vectors given 3 linearly independent vectors from ## R^5 ##. How then is it possible to extend this set of 3 vectors that are linearly independent to form a basis of ## R^5##? Unless I'm missing something here, the question is asking if this set can be extended to span ALL of ## R^5 ## (i.e. a basis of this space) and not just a subspace, correct? How exactly can 3 vectors do this? I believe recalling an analog to the cross product in ##R^5## but we surely did not cover this in our class. Any help would be greatly appreciated!
     

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  3. Dec 6, 2014 #2

    mathman

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    Gold Member

    The point is that it is possible to find two vectors which added to the given three form a set of five linearly independent vectors. Example: (1,0,0,0,0) and (0,1,0,0,0) should work.
     
  4. Dec 6, 2014 #3

    statdad

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    Homework Helper

    Have you checked to see whether these three vectors are independent? If they are, what would that tell you? (Hint: you should know a very simple basis for [itex] \mathbb{R}^5[/itex]. If these are independent throw these vectors in with that basis and use your knowledge about how a basis works). Note that nothing in the question says you need to end up with an orthonormal basis at the end.

    If you haven't checked to see if these are independent, do so first: Try starting
    [tex]
    \begin{align*}
    x = \begin{bmatrix} 1 & 1 & 1 & 1 & 5 \end{bmatrix} & \\
    y = \begin{bmatrix} 1 & 2 & 3 & 4 & 5 \end{bmatrix} & \\
    z = \begin{bmatrix} 1 & 4 & 9 & 16 & 25 \end{bmatrix}
    \end{align*}
    [/tex]

    Then see whether the equation [itex] ay + bz = x [/itex] has any non-trivial solutions.
     
  5. Dec 6, 2014 #4
    Ahh, ok. Thank you. I misinterpreted what they were asking.
     
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