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Thread: Gram-Schmidt procedure

  1. #1
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    Gram-Schmidt procedure

    I need some help in applying the Gram-Schmidt procedure to the basis {v1; v2; v3} of R^3 ,where where
    $\displaystyle v1 = (1; 1; 1) ; v2 = (0; 1; 1) ; v3 = (0; 0; 1);,$in order to obtain an orthogonal basis of R^3 containing (1; 1; 1) and an orthonormal
    basis of R^3 containing $\displaystyle (\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} ,\frac{1}{\sqrt{3}}) $
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by StefanM View Post
    I need some help in applying the Gram-Schmidt procedure to the basis {v1; v2; v3} of R^3 ,where where
    $\displaystyle v1 = (1; 1; 1) ; v2 = (0; 1; 1) ; v3 = (0; 0; 1);,$in order to obtain an orthogonal basis of R^3 containing (1; 1; 1) and an orthonormal
    basis of R^3 containing $\displaystyle (\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} ,\frac{1}{\sqrt{3}}) $
    1) Let A, B, C be the orthogonal basis of R^3 containing (1, 1, 1).

    $\displaystyle A=v_1$

    $\displaystyle B=v_2-\frac{A^{T}v_2}{A^TA}A$

    $\displaystyle C=v_3-\frac{A^Tv_3}{A^TA}A-\frac{B^Tv_3}{B^TB}B$

    2) Divide A, B, C by their lengths to convert to unit vectors.
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  3. #3
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    -can you explain to me why you calculated like this?
    Divide A, B, C by their lengths to convert to unit vectors.
    you are refering to the current A,B,C vectors right?
    -$\displaystyle A^{T},B^{T}$ represent the transpose of a vector?
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  4. #4
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    We can rewrite in slightly another form.

    Let a be a unit vector of A.

    The projection of $\displaystyle \ v_2 \ $ to a is $\displaystyle \ (a \cdot v_2) \ $ and

    the vector parallel to a with length $\displaystyle \ (a \cdot v_2) \ $ is

    $\displaystyle \ (a \cdot v_2) \ a$.

    So the vector

    $\displaystyle
    B \ = \ v_2 \ - \ (a \cdot v_2) \ a
    $

    is perpendicular to a.
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