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Math Help - is it possible for a basis to have 1 vector or am I doing something wrong?

  1. #1
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    is it possible for a basis to have 1 vector or am I doing something wrong?

    question: let W be a subspace spanned by the vectors: { v_1=(2,0,-1,3) v_2=(1,2,2,-5) v_3=(3,2,1,-2) v_4=(7,2,-1,4)} Find the basis W^{\perp}

    work:let u=(a,b,c,d) be in W^{\perp} then u \cdot v_1=0, u \cdot v_2 = 0, u \cdot v_3 = 0, u \cdot v_4 = 0
    So we've got the following matrix augmented with a column of 0s
    \left [ \begin{array}{cccc} 2 & 0 & -1 & 3 \\ 1 & 2 & 2 & -5 \\ 3 & 2 & 1 & -2 \\ 7 & 2 & -1 & 1 \end{array} \right] in rref form is  \left [ \begin{array}{cccc} 1 & 0 & \frac{-1}{2} & 0 \\ 0 & 1 & \frac{5}{4} & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array} \right] so that gives me a=\frac{r}{2} b=\frac{-5r}{4} c=r d=0 therefore (\frac{1}{2},\frac{-5}{4},1,0) spans W^{\perp}

    question: did I do that right how d=0 or does d=s?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by superdude View Post
    question: let W be a subspace spanned by the vectors: { v_1=(2,0,-1,3) v_2=(1,2,2,-5) v_3=(3,2,1,-2) v_4=(7,2,-1,4)} Find the basis W^{\perp}

    work:let u=(a,b,c,d) be in W^{\perp} then u \cdot v_1=0, u \cdot v_2 = 0, u \cdot v_3 = 0, u \cdot v_4 = 0
    So we've got the following matrix augmented with a column of 0s
    \left [ \begin{array}{cccc} 2 & 0 & -1 & 3 \\ 1 & 2 & 2 & -5 \\ 3 & 2 & 1 & -2 \\ 7 & 2 & -1 & 1 \end{array} \right] in rref form is  \left [ \begin{array}{cccc} 1 & 0 & \frac{-1}{2} & 0 \\ 0 & 1 & \frac{5}{4} & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array} \right] so that gives me a=\frac{r}{2} b=\frac{-5r}{4} c=r d=0 therefore (\frac{1}{2},\frac{-5}{4},1,0) spans W^{\perp}

    question: did I do that right how d=0 or does d=s?
    Since the rank of the row reduced form is 3, the dimension of the solution space of the equations is 1.

    CB
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