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Math Help - Orthonormal basis and linear combinations

  1. #1
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    Orthonormal basis and linear combinations

    Let \theta be a fixed real number and let \textbf{x}_1=\left[\begin {array}{cc}\cos\theta\\\noalign{\medskip}\sin\thet  a\end{array}\right] and \textbf{x}_2=\left[\begin {array}{cc}-\sin\theta\\\noalign{\medskip}\cos\theta\end{array  }\right].

    Given a vector \textbf{y} in R^2, write it as a linear combination c_1\textbf{x}_1+c_2\textbf{x}_2.

    It also asked me to prove that (\textbf{x}_1,\textbf{x}_2) was an orthonormal basis for R^2 so I did that.

    However,
    \textbf{y}=c_1\textbf{x}_1+c_2\textbf{x}_2. <\textbf{y},\textbf{x}_1>=\textbf{y}\left[\begin {array}{cc}\cos\theta\\\noalign{\medskip}\sin\thet  a\end{array}\right] and <\textbf{y},\textbf{x}_2>=\textbf{y}\left[\begin {array}{cc}-\sin\theta\\\noalign{\medskip}\cos\theta\end{array  }\right]

    Can someone assist me on where to go from here? Thank you!
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by pakman View Post
    Let \theta be a fixed real number and let \textbf{x}_1=\left[\begin {array}{cc}\cos\theta\\\noalign{\medskip}\sin\thet  a\end{array}\right] and \textbf{x}_2=\left[\begin {array}{cc}-\sin\theta\\\noalign{\medskip}\cos\theta\end{array  }\right].

    Given a vector \textbf{y} in R^2, write it as a linear combination c_1\textbf{x}_1+c_2\textbf{x}_2.

    It also asked me to prove that (\textbf{x}_1,\textbf{x}_2) was an orthonormal basis for R^2 so I did that.

    However,
    \textbf{y}=c_1\textbf{x}_1+c_2\textbf{x}_2. <\textbf{y},\textbf{x}_1>=\textbf{y}\left[\begin {array}{cc}\cos\theta\\\noalign{\medskip}\sin\thet  a\end{array}\right] and <\textbf{y},\textbf{x}_2>=\textbf{y}\left[\begin {array}{cc}-\sin\theta\\\noalign{\medskip}\cos\theta\end{array  }\right]

    Can someone assist me on where to go from here? Thank you!
    I'm not sure what you're trying to get ?
    Did you prove that it was an orthonormal basis ?

    If you're looking for <y,x_1> and <y,x_2>, then... :

    <y,x_1>=<c_1x_1+c_2x_2,x_1>

    Because the scalar product is bilinear, this is equal to :

    <c_1x_1,x_1>+<c_2x_2,x_1>=c_1<x_1,x_1>+c_2<x_2,x_1  >

    But <x,x>=||x||^2. Here, ||x_1||=1 because it's an orthonormal basis.

    And <x_2,x_1>=0 because it's an orthonormal basis.


    Therefore, <y,x_1>=c_1 \cdot 1^2+c_2 \cdot 0=c_1


    If you prefer, the scalar product of y with the n-th vector of an orthonormal basis represents the n-th coordinate of y in this basis. (this is extra info and I'm not sure I explained it well )
    Last edited by Moo; May 30th 2008 at 12:41 AM.
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