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Thread: Orthogonal Projections

  1. #1
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    Orthogonal Projections

    Find the 2x2 matrix A of the orthogonal projection onto the line L in R^2 that consists of all scalar multiples of the vector

    (5, 1)
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  2. #2
    Super Member PaulRS's Avatar
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    Remember that in general, given a linear space $\displaystyle V$ and a subspace $\displaystyle S$ with $\displaystyle
    \dim \left( S \right) =n< + \infty
    $ the orthogonal projection of a vector $\displaystyle \bold{v}$ over $\displaystyle S$ is given by $\displaystyle
    P_S \left( \bold{v} \right) = \sum\limits_{k = 1}^n {\left\langle {\bold{v},\bold{w}_k } \right\rangle \cdot \bold{w}_k }
    $ where $\displaystyle
    \left\{ {\bold{w}_1 ,...,\bold{w}_n } \right\}
    $ is an orthonormal base of $\displaystyle S$ ( Such a base exists by Gram-Schmidt)

    So in our case let $\displaystyle
    \bold{w} = \tfrac{1}
    {{\sqrt {26} }} \cdot \left( {\begin{array}{*{20}c}
    5 & 1 \\

    \end{array} } \right)

    $ this vector belongs to the line L and its norm is 1, thus ( since the dimension of the line is 1) it forms an orthonormal base of L.

    Thus: $\displaystyle
    P_L \left( \bold{u} \right) = \left\langle {\bold{u},\bold{w}} \right\rangle \cdot \bold{w}
    $ from there: $\displaystyle
    P_L \left( {\begin{array}{*{20}c}
    x & y \\

    \end{array} } \right) = \tfrac{1}
    {{26}} \cdot \left( {\begin{array}{*{20}c}
    {25x + 5y} & {5x + y} \\

    \end{array} } \right)
    $ and note that this can be re-written as (with column vectors): $\displaystyle
    \left( {\begin{array}{*{20}c}
    {\tfrac{{25}}
    {{26}}} & {\tfrac{5}
    {{26}}} \\
    {\tfrac{5}
    {{26}}} & {\tfrac{1}
    {{26}}} \\

    \end{array} } \right) \cdot \left( {\begin{array}{*{20}c}
    x \\
    y \\

    \end{array} } \right) = P_L \left( {\begin{array}{*{20}c}
    x \\
    y \\

    \end{array} } \right)
    $
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