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Math Help - Basis and dimension

  1. #1
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    Basis and dimension

    Find the basis for and the dimension of W=span(\mathbf{v_1, v_2, v_3}), where

    \mathbf{v_1}=\begin{pmatrix}1\\ 2\\ 3\end{pmatrix}, \mathbf{v_2}=\begin{pmatrix}1\\ 1\\ -1\end{pmatrix}, \mathbf{v_3}=\begin{pmatrix}-1\\ 0\\ 5\end{pmatrix}

    I'm not too sure how to start this question off
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  2. #2
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    If the three given vectors are independent, then they are a basis for their own span and the dimension is 3. If they are not, then one of the vectors can be written as a linear combination of the other two and so those two form a basis and the dimension is 2. (It clear that these vectors are not all multiples of one another so the dimension is not 1.)

    So start by checking if they are independent or not: do there exist numbers, a, b, and c such that
    av_1+ bv_1+ cv_2= a\begin{pmatrix}1 \\ 2 \\ 3\end{pmatrix}+ b\begin{pmatrix}1 \\ 1 \\ -1\end{pmatrix}+ c\begin{pmatrix}-1 \\ 0 \\ 5\end{pmatrix}= \begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix}

    That is the same as
    \begin{pmatrix}a \\ 2a \\ 3a\end{pmatrix}+ \begin{pmatrix}b \\ b \\ -b\end{pmatrix}+ \begin{pmatrix}-c \\ 0 \\ 5c\end{pmatrix}= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}

    \begin{pmatrix}a+ b- c \\ 2a+ b \\ 3a- b+ 5c\end{pmatrix}= \begin{pmatrix}0 \\ 0 \\ 0 \end{pmatrix}

    and that is the same as the three equations a+ b- c= 0, 2a+ b= 0, and 3a- b+ 5c= 0. Obviously a= b= c= 0 is a solution. If it is the only solution, then the three vectors form a basis and the space is three dimensional.

    If there exist another solution, we can solve for one of the vectors in terms of the other two. If we had av_1+ bv_2+ cv_3= 0 with, at least, c non-zero, then we can write cv_3= -av_1- bv_2 so v_3= -(a/c)v_1- (b/c)v_2. Since v_1, v_2, and v_3 span the space every vector in it can be written as a linear combination, v= pv_1+ qv_2+ rv_3. With v_3= (-a/c)v_1- (b/c)v_3 we can replace v_3 and get v= pv_1+ qv_2+ r(-(a/c)v_1- (b/c)v_2)= (p- ar/c)v_1+ (q- br/c)v_2. Then \{v_1, v_2\} is a basis and the space has dimension 2.

    That's the idea. I'll leave the actual calculation to you.
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