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Math Help - Orthonormal basis

  1. #1
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    Orthonormal basis

    Find an orthonormal basis for the vector subspace of \mathbb R^3,\,\mathbb W=\{(x,y,z)\in\mathbb R^3|x+y+z=0\}.

    Sketch of work:

    Spoiler:

    i'll take x=-y-z, so we have that

    \mathbb W=\{(-y-z,y,z)\in\mathbb R^3,\,\forall\,y,z\in\mathbb R\}, thus \mathbb W=\langle(-1,1,0),(-1,0,1)\rangle.

    these are linearly independent, so i have a basis for \mathbb W which is B_{\mathbb W}=\{(-1,1,0),(-1,0,1)\}.

    well, here's where i dunno what to do, don't know how to get the orthonormal basis for the subspace.
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  2. #2
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    Quote Originally Posted by Morgan View Post
    Find an orthonormal basis for the vector subspace of \mathbb R^3,\,\mathbb W=\{(x,y,z)\in\mathbb R^3|x+y+z=0\}.

    Sketch of work:

    i'll take x=-y-z, so we have that

    \mathbb W=\{(-y-z,y,z)\in\mathbb R^3,\,\forall\,y,z\in\mathbb R\}, thus \mathbb W=\langle(-1,1,0),(-1,0,1)\rangle.

    these are linearly independent, so i have a basis for \mathbb W which is B_{\mathbb W}=\{(-1,1,0),(-1,0,1)\}.

    well, here's where i dunno what to do, don't know how to get the orthonormal basis for the subspace.
    first, you don't need to hide your own work! second, you're right so far. now you just need to apply Gram-Schmidt process. let's give a name to the elements of the basis you found:

    u_1=(-1,1,0), \ u_2=(-1,0,1). now find v_1=u_2-\frac{<u_1,u_2>}{<u_1,u_1>}u_1. the set \{u_1,v_1 \} is an orthogonal basis for W. so you just need to normalize them, i.e. find e_1=\frac{u_1}{||u_1||}, \ e_2=\frac{v_1}{||v_1||}.

    then \{e_1,e_2 \} would be an orthonormal basis for W.
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