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Inner product space and orthonormal basis.

  1. Nov 22, 2011 #1
    1. The problem statement, all variables and given/known data
    Assume the inner product is the standard inner product over the complexes.
    Let W=
    Spanhttp://img151.imageshack.us/img151/6804/screenshot20111122at332.png [Broken]

    Find an orthonormal basis for each of W and Wperp..

    3. The attempt at a solution
    Obviously I need to use Gram-Schmidt orthogonalization here, to find the orthonormal basis.
    So I've applied the process to W, and normalized each vector to get an orthonormal basis. But I'm confused here as to what I would do for Wperp.. Don't I use G-S on W to find Wperp.? Am I finding the basis of W or am I finding Wperp. by applying G-S, and how am I supposed to find the other one?
    If my guess is correct, do I find Wperp. by applying G-S to W, and then the orthonormal basis is each of these vectors normalized? Then I would apply G-S to Wperp. and normalized each of those vectors.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 22, 2011 #2

    lanedance

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    Gram schimdt uses a set of linearly independent vectors to generate a set of orthornormal vectors with teh same span.

    You have 3 vectors that span W. Are they linearly independent? If so use gram-schmidt to generate a set of 3 orthonormal vectors that span W.

    Now consider W perp. As W is spanned by 3 linearly independent vectors (check?), Wperp will be spanned by 2 linearly independent vectors. Note every vector in W perp is orthogonal to the all vectors in W. You could use this fact with dot products to find basis vectors for Wperp.
     
  4. Nov 22, 2011 #3
    How am I supposed to check they are linearly independent? Besides that, if they aren't, how would it be possible to use gram-schmidt? Also, how come Wperp will only have 2 vectors? I think I'm missing what you mean when you say use the dot product. There are 6 possible combinations if Wperp has 2 vectors, and each one would equal zero. But wouldn't it take ridiculously long to write them all down and work out each one?
     
  5. Nov 22, 2011 #4

    lanedance

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    you should probably understand how to check for linear independence prior to using gram-schmidt
    Actually you can still attempt gram schimdt and it will come out in the wash that they were and you will end up with a zero vector (when?)
    how many vectors span [itex]\mathbb{C}^5[/itex]
    its not that bad - you could write the vectors as rows in a matrix, say {w_1,w_2,w_3} span W, and u is a vector in W perp
    [tex]
    \begin{pmatrix}
    \textbf{w}_1^{T} \\ \textbf{w}_2^{T} \\ \textbf{w}_3^{T}
    \end{pmatrix} \textbf{u} = \textbf{0}
    [/tex]

    Hope this helps, but at the end of the day you have to try some work and ideas out
     
  6. Nov 22, 2011 #5
    Ok I just didn't understand since I had already applied GS (and not gotten zero)
    So Wperp should be the span of the "remaining" vectors? But why could it not be another set of 3 linearly independent vectors?

    Should those w's be the conjugates? (since we are using the standard inner product). I'm not too sure what you're getting at. I know the product should be zero, but how am I supposed to do that algebraically? Do I need to multiply that w matrix by an arbitrary vector an+bni where n is 1 to 5, and then solve? How would I end up with two different sets?
     
  7. Nov 22, 2011 #6

    lanedance

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    wouldn't you then have 6 linearly independent vectors?
    yes, it should be conjugates

    write u=(a,b,c,d,e), then you have 5 unknowns whilst the matrix gives you 3 constraints, leading to a subspace spanned by 2 vectors - if you try some of these things it should become obvious
     
  8. Nov 22, 2011 #7
    Oh yea.
    I was going to do that for u, but shouldn't each a, b, c, d, e be in a complex form? In that case I would have 10 unknowns, and that just wouldn't work.
    Also, I'm checking and re-checking my work, but I've ended up with
    v1 = [1, 1+i, 1, 1-i, 1]
    v2 = [2+i, -2+2i, -2+i, 2+2i, 2i]
    v3 = (1/15)[-28+36i, 61-27i, -80i, -27-61i, -28-44i]
    After apply G-S, which should give me an orthogonal basis, except these vectors are not orthogonal. I've checked it so many times, why are they not orthogonal?
     
  9. Nov 23, 2011 #8

    lanedance

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    are you normalizing your vectors first? the norm of |v1|>1
     
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