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Finding an orthogonal complement without an explicitly defined inner product

  1. Apr 12, 2012 #1
    1. The problem statement, all variables and given/known data

    P5 is an inner product space with an inner product. We applied the Gram Schmidt process to the basis {1,x,x^2,x^3,x^4} and obtained the following result. {f1,f2,f3,f4,x^4+2}

    What is the orthogonal complement of P3 in P5 with respect to this inner product?


    2. Relevant equations

    http://tutorial.math.lamar.edu/Classes/LinAlg/OrthonormalBasis.aspx

    has everything on the gram schmidt process

    Definition of orthogonal complement:
    Suppose that W is a subspace of an inner product space V. We say that a vector u from V is orthogonal to W if it is orthogonal to every vector in W. The set of all vectors that are orthogonal to W is called the orthogonal complement of W.

    3. The attempt at a solution

    I'm not really sure. I feel like I need a defined inner product to actually find which vectors are orthogonal. I feel like I need to do something with the x^4+2, but honestly I am entirely lost.
     
  2. jcsd
  3. Apr 12, 2012 #2

    Dick

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    Of course you need to know what the inner product is to say what is orthogonal. They must have told you at some point.
     
  4. Apr 13, 2012 #3
    Nope, I copied this question straight from my homework page. We are given the result of the Gram Schmidt process, which is a set of orthonormalized vectors. Considering the orthogonal complement is the set of all vectors orthogonal to some subspace W, does that mean my answer is just a linear combination of the given result of the gram schmidt process? They are orthogonal vectors, and the only way I could think to get every orthogonal vector from that information would be to linearly combine them.
     
  5. Apr 13, 2012 #4

    Dick

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    Ok, maybe I'm beginning to see what the question is. So f1 is a constant term, f2 has the form a+bx, f3 is a quadratic, d+e*x+f*x^2 etc. Does that give you a hint?
     
  6. Apr 13, 2012 #5
    I think so. Thank you
     
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