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Math Help - Inner Product - Functions.

  1. #1
    Junior Member pearlyc's Avatar
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    Inner Product - Functions.

    Show that the following does not define an inner product on R^3:
    <br />
<(x,y,z),(a,b,c)> = xa + 2xb + 2ay + yb + 4zc

    Define an inner product on P_2 by

    <p,q> = \int^1_{-1} p(x)q(x) dx

    Let U \subseteq P_2 be the subspace spanned by { x,x^2}.

    (i) Find an orthornormal basis for U.
    (ii) Find the polynomial in U that is as close as possible to 1. (for the norm corresponding to the above inner product)



    -----------------------------------------

    And a little vector space question,

    Determine whether the polynomial

    p(x) = x^2 + x + 2

    belongs to span { {p_1(x), p_2(x),p_3(x)}} where

    p_1(x) = 2x^2 + x  + 2, p_2(x) = x^2 - 2x, p_3(x) = 5x^2 - 5x + 2

    Thanks for your help guys. (:
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  2. #2
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    Quote Originally Posted by pearlyc View Post
    Determine whether the polynomial

    p(x) = x^2 + x + 2

    belongs to span { {p_1(x), p_2(x),p_3(x)}} where

    p_1(x) = 2x^2 + x  + 2, p_2(x) = x^2 - 2x, p_3(x) = 5x^2 - 5x + 2
    Your set of polynomials (vectors) span p(x) if p(x) can be written as a linear combination of the vectors  p_1(x), p_2(x), p_3(x), i.e.
    p(x) = ap_1(x) + bp_2(x) + cp_3(x)
    x^{2} + x + 2 = (2a + b + 5c)x^{2} + (a - 2b - 5c)x + 2(a+c)

    So you have to determine whether or not the system of equations (constructed by equating the coefficients of each term) is consistent or not:
    \begin{array}{ccccccc} 2a & + & b & + & 5c & = & 1 \\ a & - & 2b & - & 5c & = & 1 \\ a & & & + & c & = & 1 \end{array}

    Note that you don't actually have to find the solutions. The determinant will determine whether or not your system is consistent (implying your set of vectors span p(x)).
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    Quote Originally Posted by pearlyc View Post
    Show that the following does not define an inner product on R^3:
    <br />
<(x,y,z),(a,b,c)> = xa + 2xb + 2ay + yb + 4zc
    All of the axioms should hold except this one: \langle\textbf{v},\;\textbf{v}\rangle\geq 0. Think about it. It should be fairly easy to find a counterexample.
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