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Inner product change of basis

  1. Dec 12, 2013 #1
    1. The problem statement, all variables and given/known data
    Given ##S = \{1, x, x^2\}##, find the coordinates of ##x^2 + x + 1## with respect to the orthogonal set of S.


    2. Relevant equations
    Inner product on polynomial space:
    ##<f,g> = \int_{0}^{1} fg \textrm{ } dx##

    3. The attempt at a solution
    I used Gram-Schmidt to make ##S## orthogonal and got ##S' = \{1, x - \frac{1}{2}, x^2 - x + \frac{1}{6}\}##.

    So the change of basis matrix I got was $$ \left( \begin{array}{ccc}
    1 & \frac{1}{2} & \frac{1}{3} \\
    0 & 1 & 1 \\
    0 & 0 & 1 \end{array} \right)$$

    But ##x^2 + x + 1## looks exactly like ##S##, so it would seem like it's the identity matrix so then it wouldn't change anything, which is where I'm stuck.
     
  2. jcsd
  3. Dec 12, 2013 #2

    ShayanJ

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    Gold Member

    The problem says "with respect to the orthogonal set of S",and because S is not an orthogonal set,so you should find the components in S' basis.So forget about S and calculate the components w.r.t. S' directly!
    Also,[itex] x^2+x+1 [/itex] doesn't look exactly like S,only its coordinates w.r.t. S is (1,1,1).
     
  4. Dec 12, 2013 #3

    jbunniii

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    Science Advisor
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    Gold Member

    Think about what the change of basis matrix does for you. Consider, for example, the polynomial ##x##. Expressed in terms of the original basis ##S##, we can write ##x = 0(1) + 1(x) + 0(x^2)##, so its coefficient vector in terms of the original basis is ##(\begin{array}{ccc}0& 1& 0\end{array})^T##. To express it in terms of the new basis, we simply multiply the matrix by this vector:
    $$\text{new coefficient vector} = \left( \begin{array}{ccc}
    1 & \frac{1}{2} & \frac{1}{3} \\
    0 & 1 & 1 \\
    0 & 0 & 1 \end{array} \right)
    \left( \begin{array}{c} 0 \\ 1 \\ 0 \end{array}\right) = \left( \begin{array}{c} \frac{1}{2}\\ 1 \\0\end{array} \right)$$
    This means that our polynomial ##x## can be expressed in terms of the orthogonal basis using the new coefficients:
    $$x = \frac{1}{2}(1) + 1\left(x-\frac{1}{2}\right) + 0\left(x^2 - x + \frac{1}{6}\right)$$
    and we can easily see that the left hand side does indeed equal the right hand side.

    You can use exactly the same technique for any other polynomial. Start by finding the coefficients of ##1 + x + x^2## in terms of the original basis ##S##.
     
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