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Math Help - Finding a basis for an orthogonal

  1. #1
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    Finding a basis for an orthogonal

    Let S be the subspace of R^{4} spanned by x_1=(1, 0, -2, 1)^{T} and x_2=(0, 1, 3, -2)^{T}. Find a basis for S^{\bot}.

    I was thinking about multiplying the vectors by a set of vectors [x_1,x_2,x_3,x_4]^T \epsilon S^{\bot} but then since there are two vectors I am lost on where to begin. Help would be appreciated!
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    Quote Originally Posted by pakman View Post
    Let S be the subspace of R^{4} spanned by x_1=(1, 0, -2, 1)^{T} and x_2=(0, 1, 3, -2)^{T}. Find a basis for S^{\bot}.

    I was thinking about multiplying the vectors by a set of vectors [x_1,x_2,x_3,x_4]^T \epsilon S^{\bot} but then since there are two vectors I am lost on where to begin. Help would be appreciated!
    Can you explain what the notation S^{\bot} means?
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    Quote Originally Posted by mr fantastic View Post
    Can you explain what the notation S^{\bot} means?
    S^{\bot} is the orthogonal component of S

    From my book...

    Y^{\bot}=x \; \epsilon \; R^{n}\mid\;x^{\top}y=0 for every y\;\epsilon\;Y
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    Quote Originally Posted by pakman View Post
    Let S be the subspace of R^{4} spanned by x_1=(1, 0, -2, 1)^{T} and x_2=(0, 1, 3, -2)^{T}. Find a basis for S^{\bot}.

    I was thinking about multiplying the vectors by a set of vectors [x_1,x_2,x_3,x_4]^T \epsilon S^{\bot} but then since there are two vectors I am lost on where to begin. Help would be appreciated!
    the general method is to put your vectors in a matrix: A=\begin{pmatrix}1 & 0 & -2 & 1 \\ 0 & 1 & 3 & -2 \end{pmatrix}\ . then a basis for the set of the solutions

    A \bold{x}=0 is a basis for S^{\bot}. let \bold{x}=(x_1 \ x_2 \ x_3 \ x_4)^T, and A \bold{x}=0. then x_1-2x_3+x_4=x_2+3x_3-2x_4=0. therefore:

    x_3=2x_1+x_2, \ x_4=3x_1+2x_2. hence \bold{x}=x_1(1 \ \ 0 \ \ 2 \ \ 3)^T + x_2(0 \ \ 1 \ \ 1 \ \ 2)^T. so a basis for S^{\bot} is:

    \{(1 \ \ 0 \ \ 2 \ \ 3)^T, (0 \ \ 1 \ \ 1 \ \ 2)^T \}. \ \ \ \ \square
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    Quote Originally Posted by NonCommAlg View Post
    the general method is to put your vectors in a matrix: A=\begin{pmatrix}1 & 0 & -2 & 1 \\ 0 & 1 & 3 & -2 \end{pmatrix}\ . then a basis for the set of the solutions

    A \bold{x}=0 is a basis for S^{\bot}. let \bold{x}=(x_1 \ x_2 \ x_3 \ x_4)^T, and A \bold{x}=0. then x_1-2x_3+x_4=x_2+3x_3-2x_4=0. therefore:

    x_3=2x_1+x_2, \ x_4=3x_1+2x_2. hence \bold{x}=x_1(1 \ \ 0 \ \ 2 \ \ 3)^T + x_2(0 \ \ 1 \ \ 1 \ \ 2)^T. so a basis for S^{\bot} is:

    \{(1 \ \ 0 \ \ 2 \ \ 3)^T, (0 \ \ 1 \ \ 1 \ \ 2)^T \}. \ \ \ \ \square
    Did you take the transpose of the x_1 and x_2 to get that A?
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    Quote Originally Posted by pakman View Post
    Did you take the transpose of the x_1 and x_2 to get that A?
    yes! your vectors make the rows of A.
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